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Reference Reading from Solomons, Organic Chemistry 6th edition
Acid-Base Theory (Unit 1) Acid-Base Theory (Unit 2) Isomerism (Unit 1) Isomerism (Unit 2) Nomenclature
90-93, 96-101 102-118, 320, 433-434, 795-796, 903-905, 970-972 59-61 178-185, 188, 193-198, 200 41-47, 65-75, 128-137, 284-286, 288-289, 415-417, 615-617, 705-706, 792-793, 797-800, 899-900 87-90, 94-96 224-233, 238-252, 256-259 260 913-914 438-443 923-924, 926-927, 966 260-262, 265-268, 300, 302-305 308, 312, 318-319, 443-444 719 470-472, 716-719, 729-734, 759-761 708-711 805-817, 820-824 919-921 711, 825-827, 918 327-333 336-339, 346-351, 422-425 617-618, 655-658 658-664, 930-931, 974 930-931, 974 334-335, 366-379, 393-397 1146-1148 165-166, 473-475, 690-691, 915-916
Reaction Mechanism (Unit 1) - Introduction to Mechanism Reaction Mechanism (Unit 2) - Nucleophilic substitution Reaction Mechanism (Unit 3) - Nucleophilic substitution Reaction Mechanism (Unit 4) - Nucleophilic substitution Reaction Mechanism (Unit 5) - Nucleophilic substitution Reaction Mechanism (Unit 6) - Nucleophilic substitution Reaction Mechanism (Unit 7) - Elimination Reaction Mechanism (Unit 8) - Elimination Reaction Mechanism (Unit 9) - Nucleophilic addition Reaction Mechanism (Unit 10) - Nucleophilic addition Reaction Mechanism (Unit 11) - Nucleophilic addition Reaction Mechanism (Unit 12) - Nucleophilic addition Reaction Mechanism (Unit 13) - Nucleophilic addition Reaction Mechanism (Unit 14) - Nucleophilic addition Reaction Mechanism (Unit 15) - Electrophilic addition Reaction Mechanism (Unit 16) - Electrophilic addition Reaction Mechanism (Unit 17) - Electrophilic substitution Reaction Mechanism (Unit 18) - Electrophilic substitution Reaction Mechanism (Unit 19) - Electrophilic substitution Reaction Mechanism (Unit 20) - Radical reactions Amino acids Oxidation and Reduction Uses of compounds with different functional groups Structure determination (Unit 1) Structure determination (Unit 2) Structure determination (Unit 3) Structure determination (Unit 4) Organic synthesis Organic Laboratory Technique (Unit 1) Organic Laboratory Technique (Unit 2) Organic Laboratory Technique (Unit 3) Organic Laboratory Technique (Unit 4)
295-297 359-360, 742 541-547 169-171
Summary of Organic Reactions (I. Alkanes)
I.
Alkanes
A.
Cracking
Page 1
heat , catayst
very long alkane → long alkane + short alkene
B.
Combustion y y CxHy + (x + 4 ) O2(g) → xCO2(g) + 2 H2O(l)
C.
Chlorination Chain initiation (chain initiating step) Cl
hυ
Cl
Cl
+
Cl
Cl· radical is generated.
Chain propagation (chain propagating step) H Cl
H
H C
H
Cl
H +
H
C
H Cl
Cl
H
Cl· radical is consumed.
H
H
C
H
Cl
+ Cl
H
C
H
Cl· radical is regenerated.
H
Chain termination (chain terminating step) Cl
+
Cl
Cl
Cl H
H Cl
C
Cl
H
C
H
H
H H
H
H C
C
H
H
H H H
H C C H H H
A mixture of products would be obtained.
+
Energy
radical is destroyed.
Summary of Organic Reactions (II. Alkenes and alkynes)
II.
Alkenes and alkynes
A.
Addition reactions
1.
Reaction with bromine in CCl4 Br2 + R C C R R in CCl4
Note :
2.
Br Br
in dark
R
R C
C
R
R
R
1,2-dibromoalkane
Test for C=C or C≡C bond
Reaction with bromine water Br2(aq) + R C C R R
in dark
R
Br OH
R C
C
R
R
R
halohydrin
Note :
3.
Test for C=C or C≡C bond
Reaction with hydrogen bromide
H Br H
H C H H
H C
C C H
+
C C
H
major product
H H H 2-bromopropane
H Br
Br H H
H H propene
H C
C C
H
minor product
H H H 1-bromopropane
Note :
4.
Follows Markownikov's orientation
Reaction with conc. sulphuric(VI) acid OH
conc. H2SO4(l) +
R C C R
O
R
R
S
H
O
R C
C
R
R
O R
alkyl hydrogensulphate(VI)
Note :
A reversible reaction OH O
S
H
O
R C
C
R
R
H
O + H2O(l)
C
R
R
R
alcohol
alkyl hydrogensulphate(VI)
Note :
5.
Laboratory preparation of alcohol
Reaction with water CH2 CH2 + H2O ethane
Note :
6.
H3PO4 300 ºC
CH3 CH2 OH ethanol
Industrial preparation of alcohol
Catalytic hydrogenation of oil Note :
OH
R C
hardening of oil
R + H2SO4(aq)
Page 2
Summary of Organic Reactions (II. Alkenes and alkynes) B.
Ozonolysis of alkene CH3 CH3CHCH CH2 3-methylbut-1-ene
Note :
C.
CH3 O
(1) O3, CH2Cl2, -78 ºC
O +
CH3CH C H
(2) Zn / H2O
2-methylpropanal
H C H methanal
Can be used to determine the structure of the alkene with characterization of the derivatives of carbonyl compounds formed.
Oxidation cleavage of double bond CH3CH2
CH3
C C
CH3CH2
(1) KMnO4, OH-, hot (2) H+
CH3
CH3 CH3CHCH CH2 3-methylbut-1-ene
Note :
O
O H3C C
CH3CH2 C CH2CH3 + pentan-3-one
3-ethyl-2-methylpent-2-ene
a)
Page 3
propanone
CH3 O
(1) KMnO4, OH-, hot (2) H+
CH3CH C OH + CO2
+ H2O
2-methylpropanoic acid
The outcomes are different from the ones from ozonolysis.
Reaction with cold acidic or alkaline permanganate solution H H
H H H C C H ethane
H C C H
cold KMnO4(aq) H3O+ or OH-
O O H H ethane-1,2-diol
D.
CH3
Oxymercuration of alkyne H
R C C terminal alkyne
H
HgSO4(aq) H2SO4(aq)
R C
C H
O H Markovinkov orientation
keto-enol tautomerization
H R C C
H
O H methyl ketone
Summary of Organic Reactions (II. Alkenes and alkynes) E.
Page 4
Polymerization of alkenes Chain initiating step O
O
R C O O C
O
heat
R
2R
2 R C O
+ 2 CO2
alkyl radical
Diacylperoxide
Chain propagating step H R
C H
H H
H
R C
C H
H H
H
H H H H
H C
R C C
C
H
H H
C
H H
R C C C H
C
etc.
long chain polymer
H H H H
Chain terminating step H H R
R
(CH2 CH2)n C
H H
C
C C
H H
H H
H H
H H
(CH2 CH2)n C C H H
C C
(CH2 CH2)n R
(CH2 CH2)n R
combination
disproportionation
H H H H R
(CH2 CH2)n C C C C H H H H
H H R
(CH2 CH2)n C
H H oxidation product
(CH2 CH2)n R
H H
C
H C C
H
H H
(CH2 CH2)n R
reduction product
Summary of Organic Reactions (III. Aromatic hydrocarbons)
III.
Aromatic hydrocarbons
A.
Substitution reactions of benzene
1.
Nitration
NO2
conc. HNO3(aq) conc. H2SO4(l) 55°C benzene
Note :
2.
nitrobenzene
No reaction if conc. H2SO4(l) is not added.
Halogenation
X
X2 FeX3 or AlX3 halobenzene
benzene
3.
Sulphonation
O
conc. H2SO4 heat
O
S
O H
benzene benzenesulphonic acid
O O
S
O-Na+
O H NaOH(s) 350°C
benzenesulphonic acid
Note :
4.
H3O+
sodium phenoxide / sodium phenolate
Industrial preparation of phenol
Alkylation
R
RX and AlX3
benzene
B.
alkylbenzene
Oxidation of alkylbenzene
O
R
C OH 1) KMnO4, OH-, heat 2) H3O+
alkylbenzene
benzenecarboxylic acid
OH
phenol
Page 5
Summary of Organic Reactions (IV. Halogeno-compounds)
IV.
Halogeno-compounds
A.
Nucleophilic substitution
1.
Reaction with sodium hydroxide R X haloalkane
NaOH(aq)
R OH alcohol
X
OH NaOH(aq) heat
halobenzene
Note :
2.
Reaction with potassium cyanide R X haloalkane
3.
phenol
Prolonged heating is required for the alkaline hydrolysis of halobenzene.
KCN(aq)
R CN nitrile
Reaction with ammonia H N H + R X H ammonia haloalkane
H N
H N
R N
R
H 1° amine R N
R
H 2° amine
R
+ HX
H 1° amine
+ R X
R
+ HX
H 2° amine
haloalkane
R N
+ R X
R
+ HX
R 3° amine
haloalkane
R R N
R
R 3° amine
Note :
4.
5.
R X-
R N
+ R X
R quaternary ammonium halide
haloalkane
A mixture of products would be obtained.
Reaction with alkoxide R' O-Na+ R X + haloalkane sodium alkoxide
R O R' ether
Reaction with carboxylate ion
O
O -
R' C O Na sodium haloalkane carboxylate R X +
+
R' C O ester
R
Page 6
Summary of Organic Reactions (IV. Halogeno-compounds) B.
Elimination reaction
1.
Reaction with alcoholic sodium hydroxide to form alkene X H
R C
2.
C
R
NaOH(alc.)
R C
C
R R
R R
haloalkane
alkene
R
Reaction with alcoholic sodium hydroxide to form alkyne X X
R C
C
R
boiling NaOH(alc.)
R C C
H H 1,2-dihaloalkane
alkyne
R
Page 7
Summary of Organic Reactions (V. Hydroxy compounds)
V.
Hydroxy compounds
A.
Reactions of alcohols
1.
Formation of halide
a)
by halogenating agent PCl5 / PCl3 / SOCl2
R OH alcohol
b)
R Cl chloroalkane
by heating with halide salt in acidic medium + R OH Na X , conc. H2SO4(l) heat alcohol
R X haloalkane
2.
Formation of alkoxide
3.
Oxidation of alcohols
2 C2H5OH(l) + 2 Na(s) → 2 C2H5O-Na+(alc.) + H2(g C2H5OH(l) + NaOH(s) d C2H5O-Na+(alc.) + H2O(l)
H R C O H
O
[O]
R C
H 1° alcohol
H
R C O H
aldehyde
R R C O H
carboxylic acid
O
[O]
R C
H 2?alcohol
O
[O]
R
ketone
R R C O H
[O]
Resistant to oxidation
R 3?alcohol
4.
Dehydration of alcohols H H
H C C H conc. H SO 2 4 H O heat H
H H H C
C H
ethene
ethanol
5.
Esterification O
R OH + R' C OH carboxylic alcohol acid
O conc. H2SO4(l) R' C O R heat ester
+
H2O
Page 8
Summary of Organic Reactions (V. Hydroxy compounds) 6.
Page 9
Triiodomethane formation H H
O
NaOH(aq) + I2(aq)
R C C H
R C
O H
O
H
1. 2.
+ H C -
I
I
carboxylate iodoform ion (yellow ppt.)
alcohol with 1-hydroxyethyl group
Note :
I
Test for the presence of 1-hydroxyethyl group. Shorten the carbon chain by 1 C.
Details : H H
R C C H + 2 I- + 2 H2O O H
+ I2 + 2 OH-
O H
H alcohol with 1-hydroxyethyl group O H R C C
H
oxidation
R C C H
methyl ketone
NaOH(aq) + I2(aq)
H
O
I
R C
+ H C
O-
I
H methyl ketone
I
carboxylate iodoform ion (yellow ppt.)
7.
Distinction between primary, secondary and tertiary alcohols
a)
With potassium dichromate(VI) H R C O H
O
[O]
[O]
R C
H 1° alcohol
H
aldehyde
R
R C
H 2?alcohol
carboxylic acid
O
[O]
R C O H
O R C O H
R
ketone
R [O]
R C O H R 3?alcohol
b)
Resistant to oxidation
Lucas' test H
R C OH H 1° alcohol H
R C OH R 2° alcohol R R C OH R 3° alcohol
conc. HCl(aq) ZnCl2(aq)
no reaction
H conc. HCl(aq) ZnCl2(aq)
R C
Cl slow reaction
R R
conc. HCl(aq) ZnCl2(aq)
R C
Cl fast reaction
R
Note : 3º alcohol will turn cloudy quickly, 2º alcohol will turn cloudy slowly but 1º alcohol will not turn cloudy at all.
Summary of Organic Reactions (V. Hydroxy compounds) B.
Reactions of phenol
1.
Reaction with sodium
O-Na+
OH Na
sodium phenoxide / sodium phenolate
phenol
2.
+ H2
Reaction with sodium hydroxide O-Na+
OH NaOH(aq)
sodium phenoxide / sodium phenolate
phenol
Note :
3.
+ H2O
Phenol is soluble in sodium hydroxide solution.
Esterification OH
O
O R C Cl O
C O R
or O
R C O C
R ester
phenol
4.
Reaction with bromine
OH
OH Br2(aq)
phenol
Note :
Br
Br
Br 2,4,6-tribromophenol (a white ppt.)
A test for phenol.
Page 10
Summary of Organic Reactions (VI. Carbonyl compounds)
VI.
Carbonyl compounds
A.
Nucleophilic addition
1.
Reaction with hydrogen cyanide O H
O R C R
2.
+ HCN
R C R C N
Reaction with sodium hydrogensulphate(IV) R
O C
R
+
+ Na HSO3
-
CH3/H
Note:
SO3-Na+
H O C
CH3/H
1. 2.
white precipitate
Sensitive to steric hinderance and reacts with aldehyde and methyl ketone only. Can be used to purify aldehyde or methyl ketone.
B.
Addition-elimination (condensation)
1.
Reaction with hydroxylamine O R C R
+
R
H N OH
C N OH + H2O
H
R
hydroxylamine
Note :
2.
1. 2.
oxime (white ppt.)
A test for aldehyde or ketone. Identify the carbonyl compound by characterization.
Reaction with 2,4-dinitrophenylhydrazine H H
R
O C + H N N R
NO2
O2N 2,4-dinitrophenylhydrazine
Note :
3.
1. 2.
R C C
H
H methyl ketone
Note :
1. 2.
H
NO2 + H O 2
C N N
R
O2N
2,4-dinitrophenylhydrazone (orange ppt.)
Test for aldehyde or ketone but no reaction with carboxylic acid and its derivatives Identify the carbonyl compound by characterization.
Triiodomethane formation O H
R
NaOH(aq) + I2(aq)
O
I
R C
+ H C
O-
I
I
carboxylate iodoform ion (yellow ppt.)
Test for methyl ketone. Shortens the carbon chain by 1 C
Page 11
Summary of Organic Reactions (VI. Carbonyl compounds) C.
Oxidation and reduction
1.
Oxidation of aldehyde with potassium dichromate(VI) O
[O]
R C H
carboxylic acid
aldehyde
Note :
2.
Orange acidified potassium dichromate will turn to green chromium(III) ion.
Oxidation of aldehyde with Tollens' reagent O
Test for aldehyde
Oxidaton of aldehyde with Fehling's reagent O
O
Fehling's reagent
R C O
R C H
Test for aldehyde
Resistance of ketones to oxidation O
R C
[O]
R
Note :
5.
no reaction
Ketone is stable to general oxidation.
Reduction with sodium tetrahydridoborate / lithium tetrahydrioaluminate O-
O
+ NaBH4(aq)
R C R/H
R C R/H
+ LiAlH4(ether)
R C R/H
H
R C R/H
O H
H3O+
R C R/H H
H alkoxide
aldehyde or ketone
alcohol
LiAlH4 cannot be used in aqueous medium.
Haloform reaction O H
R C C
H
H methyl ketone
Note :
R C R/H alcohol
O-
O
Note :
O H
H3O+
H alkoxide
aldehyde or ketone
6.
Cu2O brick red ppt.
+
carboxylate ion
aldehyde
Note :
4.
+
-
R C O NH4 + Ag(s) [Ag(NH3)2]OH(aq) ammonium silver mirror carboxylate
aldehyde
Note :
O
Tollen's reagent
R C H
3.
O
R C OH
1. 2.
NaOH(aq) + I2(aq)
O
I
R C
+ H C
O-
I
I
carboxylate iodoform ion (yellow ppt.)
Test for methyl ketone. Shorten the carbon chain by 1 C.
Page 12
Summary of Organic Reactions (VII. Carboxylic acids and their derivatives)
VII.
Carboxylic acids and their derivatives
A.
Formation of carboxylic acid
1.
Hydrolysis of nitrile
R C OH
nitrile
Note :
2.
O
dil. H2SO4(aq) reflux
R C N
carboxylic acid
Complete hydrolysis of nitrile
Oxidation of alcohol H
O
[O]
R C O H
R C
H 1° alcohol
Note :
3.
R C O H
H
aldehyde
carboxylic acid
Aldehyde intermediate can be separated from the reacting mixture by distillation.
Oxidation of aldehyde O
[O]
R C H
O
R C OH carboxylic acid
aldehyde
4.
O
[O]
Oxidation of alkylbenzene R
O C OH
1) KMnO4, OH-, heat 2) H3O+ alkylbenzene
benzenecarboxylic acid
B.
Reactions of carboxylic acid
1.
Formation of salt O
NaOH(aq)
R C OH carboxylic acid
2.
O R C O-Na+ sodium carboxylate
Formation of acyl chloride O
R C OH alcohol
PCl5 / PCl3 / SOCl2
O
R C
Cl
chloroalkane
Page 13
Summary of Organic Reactions (VII. Carboxylic acids and their derivatives) 3.
Formation of anhydride O
H C
140°C
C
O + H2O
C OH
H
H
O butenedioic anhydride
O cis-butenedioic acid
Note :
Intramolecular dehydration
O CH3
O
C OH + HO C
ethanoic acid
Note :
CH3
ethanoic acid
O -
+
C O Na
+ Cl
O
C CH3
CH3
CH3
O R C Cl acyl chloride
R' C O
R C O C acid anhydride O R C OH carboxylic acid
R
R
O R C O
alcohol O
C O C
ester
+ R OH
O
O
O
O R' C O-Na+ sodium haloalkane carboxylate
R
ester
+ R OH
O R C O
alcohol
+ R OH
O
heat
R C O
alcohol
Formation of amide
R
ester
H3O+
cold dil. H2SO4(aq) R C N or hot conc. H2SO4(l) nitrile
6.
O
C O C
ethanoic anhydride
Formation of ester
Note :
CH3
Intermolecular dehydration
R X +
5.
O
P2O5 (P4O10)
CH3
O
4.
O
H
C OH
R
ester
O R C NH2 amide
Partial hydrolysis of nitrile.
Reduction with lithium tetrahydridoaluminate O R C OH carboxylic acid
(1) LiAlH4(ether) (2) H3O+
H
R C OH H 1° alcohol
CH3 + NaCl
Page 14
Summary of Organic Reactions (VII. Carboxylic acids and their derivatives) C.
Reactions of acyl chlorides and anhydrides
1.
Reaction with water O
H2O(l)
R C Cl acyl chloride
O
O
R C O C acid anhydride
2.
R'
HCl
R C OH carboxylic acid
+
O R' C OH
Reaction with alcohol O
O
ROH
R C O C acid anhydride
R C Cl acyl chloride O O
O
ROH
R C OR
R'
+
O R' C OH
ester
conc. NH3(aq)
O
R C NH2 amide O
conc. NH3(aq)
R C NH2 amide
Reaction with amine O
HCl
ester
R C O C R' acid anhydride
R C Cl acyl chloride O O
+
R C OR
Reaction with ammonia O
4.
+
O
H2O(l)
R C Cl acyl chloride O O
3.
O
R C OH carboxylic acid
RNH2
O R R C NH
N-substituted amide O R
RNH2
R C O C R' acid anhydride
R C NH N-substituted amide
D.
Reactions of ester
1.
Acid and base-catalysed hydrolyses O
R C O
R
ester
H3O+ heat
O R C O ester
Note :
2.
R
NaOH(aq) heat
O
R C OH + R OH carboxylic acid O
alcohol
R C O- + R OH carboxylate alcohol ion
Rate of hydrolysis : alkaline medium > acidic medium > neutral medium
Reduction with lithium tetrahydridoaluminate O R C O ester
R
(1) LiAlH4(ether) (2) H3O+
H
R C OH H 1° alcohol
Page 15
Summary of Organic Reactions (VII. Carboxylic acids and their derivatives) E.
Reactions of amide
1.
Hydrolysis of amide O
R C NH2 amide O R C NH2
-
OH reflux
amide
2.
R C OH carboxylic acid O R C Ocarboxylate ion
Dehydration of amide O R C NH2
3.
O
H3O+ reflux
P2O5 or (CH3CO)2O or SOCl2
R C N
Hofmann degradation of non-substituted amide O R C NH2
Br2 / NaOH(aq) R N H H 1?amine with 1 C less
amide
Note : Only applicable to non-substituted amide.
4.
Reduction with lithium tetrahydridoaluminate O R C NH2 amide
1) LiAlH4 / ether 2) H3O+
H R C N H H H 1° amine
Note : Amide is the only acid derivative which will not be reduced to alcohol.
Page 16
Summary of Organic Reactions (VIII. Nitrogen compounds)
VIII. Nitrogen compounds A.
Formation of amine
1.
From nitrile
H H
(1) LiAlH4(ether) (2) H3O+
R C N
R C N H H 1° amine
nitrile
Note :
2.
LiAlH4 could not be used in aqueous medium.
From amide O
H
1) LiAlH4 / ether
R C NH2
R C N H
2) H3O+
H H
amide
1° amine
Note : Amide is the only acid derivative which will not be reduced to alcohol.
3.
1º, 2º, 3º amines and quaternary ammonium compounds by alkylation H N H + R X H ammonia haloalkane
H N
H N
R N
R
H 1° amine R N
R
H 2° amine
R
+ HX
H 1° amine
+ R X
R
+ HX
H 2° amine
haloalkane
R N
+ R X
R
+ HX
R 3° amine
haloalkane
R R N
R
R 3° amine
R N
+ R X haloalkane
R X-
R quaternary ammonium halide
Note : A mixture of products would be obtained.
4.
Phenylamine from nitrobenzene NH2
NO2 1) Sn, conc. HCl(aq) 2) OH-(aq)
B.
Base properties of amine
1.
Salt formation R N H H
+ HCl
H R N H ClH
Page 17
Summary of Organic Reactions (VIII. Nitrogen compounds) C.
Reactions of amine
1.
Reactions with ethanoyl chloride and benzoyl chloride R N H H
O
H
+
R
N-substituted ethanamide O
O H
C Cl
C N R
benzoyl chloride
amine
2.
CH3 C N
ethanoyl chloride
amine R N H
O H
+ CH3 C Cl
N-subsituted benzenecarboxamide
Reaction with nitric(III) acid H
HNO2 0-5°C
R N H
R+
carbocation
aliphatic primary amine
N2
+
nitrogen bubbles
+ N N
NH2 HNO2 0-5°C
aromatic primary amine
Note :
3.
benzenediazonium ion
A test used to distinguish aliphatic primary amine from aromatic primary amine.
Coupling reaction of benzenediazonium ion with naphthalen-2-ol and phenol HO
+ N N
OH
N N
naphthalen-2-ol benzenediazonium ion + N N
red orange ppt.
OH
N N
OH
phenol benzenediazonium ion
Note :
orange ppt.
Tests for benzenediazonium ion. Indirect tests for aromatic primary amine.
Page 18
Summary of Organic Reactions (IX. Other reactions)
IX.
Page 19
Other reactions Radical addtion of X2 to benzene X H
H
X
X2 uv X
H
H
X X
H X
H
Reactions of benzenediazonium ion + N N
CN CuCN
+ N N H3PO2
Preparation of vinyl chloride H
H C
Cl2 Electrophilic addition
C
H
H
H H
H
alcoholic NaOH
H C C H Cl Cl
Anti-Markownikov's addition
Cl
H
C C H
H Cl + R O O
+
H H propene
H
C C H
minor product
H H H 2-chloropropane
H C
C
H Cl H H C
H
H C
R
Cl H H
alkyl peroxide
H C
C C H
major product
H H H 1-chloropropane
Test for terminal alkyne +
( NH ) R–C≡CH Cu → R–C≡C-Cu+(s) (red ppt.) 3 2
+
( NH ) R–C≡CH Ag → R–C≡C-Ag+(s) (white ppt.) 3 2
Test for phenol [Fe(H2O)6]3+(aq) + from neutral iron(III) chloride
2C6H5O-(aq) → [Fe(H2O)4(C6H5O)2]+(aq) + 2H2O(l) from phenol purple complex
Nomenclature of Organic compounds I.
Nomenclature of organic compounds A.
B.
II.
1.
Methyl, primary, secondary and tertiary carbon
2.
α carbon, β carbon and γ carbon
Skeletal formula / Stick formula
Nomenclature of hydrocarbon A.
III.
Labelling of carbon
Alkyl and aryl group
Other functional group A.
Halogeno-compounds
B.
Alkanols and phenol
C.
Aldehydes and ketones
D.
Carboxylic acids 1.
Carboxylic acid derivatives – esters, acid chlorides, anhydrides and amides
E.
Nitriles
F.
Amines
G.
Amino-acids
H.
Ethers
IV.
Priority of functional groups
V.
Examples
VI.
Other common abbreviations
Acid-Base Theory
I.
Definition of acid-base A. B. C.
II.
Strength of acid and base A.
III.
Arrhenius definition Bronsted-Lowry definition Lewis definition Leveling effect
Factors determining the strength of a Bronsted-Lowry acid A.
Relative stability of the conjugate base comparing with the acid 1. Inductive effect (1)
2.
Resonance effect a) Resonance effect on a benzene ring (1) (2)
3. 4. 5. 6.
Amine vs Amide
Phenol Aromatic amine
Intramolecular hydrogen bond Solvation / hinderance to solvation a) Steric hinderance to solvation b) Effect of solvent Strength of the bond between the proton and the conjugate base Electronegativity of the atom a) Effect of hybridization
Isomerism I.
Structural isomerism A. B. C. D.
II.
Chain isomerism Position isomerism Functional group isomerism Tautomerism / Tautomerization
Stereoisomerism A.
B.
Diastereomerism / Geometrical isomerism 1. Physical properties of cis-/trans-geometrical isomers of 1,2-dichloroethene 2. Physical and chemical properties of cis-/trans-geometric isomers of butenedioic acid a) Physical properties b) Chemical properties Optical isomerism / Enantiomerism 1. Physical properties 2. Optical activity a) Plane polarized light b) Measurement of optical activity by polarimeter c) Specific rotation 3. Chiral centre 4. Racemic mixture / Racemic modification / Racemic form / Racemate
Reaction Mechanism
Reaction Mechanism I.
Bond breaking A.
II.
Types of reactive species A. B. C.
III.
Bond breaking of a bond to carbon Free radical (electron deficiency) Electrophile (electron deficiency) Nucleophile (electron rich)
Classification of reaction A.
Types of reaction 1. Substitution 2. Addition 3. Elimination
Page 1
Reaction Mechanism
IV.
Page 2
Nucleophilic substitution A. B. C. D. E. F. G.
Nucleophilic substitution 1. Nucleophile 2. Leaving group SN2 reaction (1 step reaction) SN1 reaction (2 steps reaction) Competition between SN1 and SN2 reactions Alkanol from haloalkane (RX → ROH) 1. Alkaline hydrolysis of haloalkane 2. Hydrolysis of haloalkane Rate of hydrolysis of haloalkane, haloalkene and halobenzene Other relevant reactions 1. Reactions of haloalkane a) Nitrile from haloalkane (RX → RCN) b) Alkylation of ammonia and amine (NH3 → RNH2) c) Use of SN2 reaction in organic synthesis 2. Reactions of alkanol a) Haloalkane from alkanol (ROH → RX) (1)
3.
(1) (2)
b)
V.
Use of chlorinating and brominating reagent
b) Luca's test to distinguish 1º, 2º and 3º alkanol Reactions of amine a) Action of nitric(III) acid on 1º amine 1º aliphatic amine 1º aromatic amine
Laboratory preparation of phenol from benzenamine
Elimination reaction A.
B.
C. D.
Elimination reaction 1. Stability of elimination product 2. E2 reaction (not required in A-Level) 3. E1 reaction (not required in A-Level) Competition between substitution and elimination reaction 1. Effect of temperature 2. Effect of bulkiness of the substrate and base 3. Effect of bascity of the nucleophile Conditions favouring substitution and elimination reaction Other relevant reactions 1. Reaction of haloalkanes with alcoholic sodium hydroxide to alkene, diene and alkyne 2. Preparation of vinyl chloride 3. Dehydration of alkanol
Reaction Mechanism
VI.
Page 3
Nucleophilic addition (Nucleophilic addition-elimination) A.
AdN reaction 1. Addition of HCN to carbonyl compound 2. Rate of nucleophilic addition a) Electronic effect (1) (2)
3.
Inductive effect (a) Effect of protonation Resonance effect
b) Steric effect Other relevant reactions a) Reactions of carbonyl group (1) (2) (3) (4) (5)
b)
Reduction of carbonyl compound by LiAlH4 and NaBH4 Addition of NaHSO3 to carbonyl compound Condensation reaction with hydroxylamine Condensation reaction with 2,4-dinitrophenylhydrazine Haloform reaction / Iodoform reaction
Reactions of carboxylic acid and its derivatives (1)
(2)
Carboxylic acid and its derivatives (a) Difference between carbonyl compound and carboxylic acid and its derivatives (b) Reactivities of carboxylic acid and its derivatives Formation of different acid derivatives (a) Use of chlorinating agent to prepare acyl chloride (b) Formation of ester (Esterification with alkanol and phenol) (c) Formation of acid anhydride (i) (ii) (iii)
(3) (4)
c)
(d) Formation of amide (Acylation and benzoylation of amine) Reaction of ester (a) Hydrolysis of ester Reaction of amide (a) Reduction of amide and other acid derivatives (b) Hofmann degradation of amide
Reactions of nitrile (1) (2)
VII.
Through intramolecular dehydration by heating Through intermolecular dehydration by a very strong dehydrating agent From acyl chloride
Hydrolysis of nitrile and amide (a) Dehydration of amide Reduction of nitrile
Electrophilic addition A. B.
Addition of HBr to alkene 1. Markownikoff's rule. 2. Reactivity of alkene towards electrophilic addition Other relevant reaction 1. Addition of Br2 to alkene 2. Addition of H2SO4 to alkene a) Preparation of alkanol from alkene 3. Hydration of alkene 4. Ozonolysis of alkene 5. Preparation of ethane-1,2-diol a) Oxidative Cleavage of double bond b) Comparision of ozonolysis and oxidative cleavage 6. Oxymercuration of alkyne
Reaction Mechanism
VIII. Electrophilic substitution A. B.
IX.
Representation of arenium ion Other relevant reaction 1. Sulphonation of benzene a) Preparation of phenol 2. Nitration of benzene a) Reduction of nitrobenzene 3. Halogenation of benzene 4. Alkylation of benzene (Friedel-Crafts alkylation) 5. Diazocoupling of diazonium ion a) Colour of a substance 6. Bromination of phenol
Free radical Reaction A. B.
C.
Formation of free radical Free radical Substitution 1. Chain reaction e.g. chlorination of methane a) Chain initiation (chain initiating step) b) Chain propagation (chain propagating step) c) Chain termination (chain terminating step) 2. Reaction between H2(g) and Cl2(g) Free radical Addition 1. Chain reaction e.g. Polymerization of alkene 2. Anti-Markownikoff orientation
Page 4
Amino acids I.
Amino acids A. B.
Zwitterion Polypeptides
Oxidation and Reduction I.
II.
Oxidation A.
Combustion of alkane
B.
Oxidation of alkanol and aldehyde
C.
Oxidation of aromatic side chain
Reduction A.
Reduction of nitrobenzene
B.
Catalytic hydrogenation (Hydrogenation of alkene)
Uses of Different Compounds I.
Uses of halogeno-compounds A. B.
II.
Use as solvent Manufacture of polymer 1. Preparation of vinyl chloride 2. Physical properties of PVC and Teflon
Uses of alcohols A. B. C. D.
III.
Use as solvent Alcoholic drink Blending agent Ethan-1,2-diol
Uses of carbonyl compounds A. B.
IV.
Preparation of urea-methanal Use of propanone
Uses of carboxylic acids and their derivatives A. B. C.
V.
Food preservatives Manufacture of nylon and terylene Use of ester
Uses of amines and their derivatives A. B.
Azo compounds as dyes in dyeing industries Amine derivatives as drugs
Determination of Structure I.
II.
Determination of empirical formula and molecular formula A.
Different kinds of formula
B.
Determination of empirical formula
C.
Determination of molecular formula
Degree of unsaturation A.
Determination of degree of unsaturation
B.
Meaning of degree of unsaturation
III.
Sodium fusion test
IV.
Test for different functional groups by wet chemistry
V.
Introduction to IR and NMR spectroscopy A.
Use of infra-red (IR) spectrum in the identification of functional groups 1.
More examples
Organic synthesis I.
Retrosynthetic analysis
II.
Structural analysis A.
III. IV.
Chain length 1.
Carbon chain
2.
Nitrogen and Oxygen containing chain
B.
Degree of unsaturation
C.
Oxidation or Reduction
D.
Position of functional group
Systematic approach to organic synthesis Examples
Organic Laboratory Technique I.
II.
III.
Purification of organic compound A.
Solvent extraction
B.
Steam distillation
C.
Chromatography
D.
Recrystallization
E.
Filtration and Suction filtration
Use of quickfit apparatus A.
Handling of quickfit apparatus
B.
Different setup of quickfit apparatus 1.
Reflux setup
2.
Distillation setup
Testing for purity A.
Determination of melting point
B.
Determination of boiling point
Nomenclature of Organic compounds I.
Nomenclature of organic compounds A.
B.
II.
1.
Methyl, primary, secondary and tertiary carbon
2.
α carbon, β carbon and γ carbon
Skeletal formula / Stick formula
Nomenclature of hydrocarbon A.
III.
Labelling of carbon
Alkyl and aryl group
Other functional group A.
Halogeno-compounds
B.
Alkanols and phenol
C.
Aldehydes and ketones
D.
Carboxylic acids 1.
Carboxylic acid derivatives – esters, acid chlorides, anhydrides and amides
E.
Nitriles
F.
Amines
G.
Amino-acids
H.
Ethers
IV.
Priority of functional groups
V.
Examples
VI.
Other common abbreviations
Nomenclature of organic compounds
Page 1
Topic
Nomenclature of organic compounds
Reference Reading
13 Organic Chemistry, Fillans, 3rd Edition pg. 50–65 Guidelines for systematic chemical nomenclature, HK Exam. Authority pg. 7–17 Chemical nomenclature, symbols and terminology for use in school science. pg. 58–65
Unit 1
Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 380–381, 400–401, 414, 428, 432, 438, 439, 446, 453, 467, 484, 498, 516–517 Reading Organic Chemistry, 6th Edition, Solomons, pg. 41-47, 65-75, 128-137, 284-286, 288-289, 415-417, 615-617, 705-706, 792-793, 797-800, 899-900
Syllabus
Nomenclature of organic compounds Classification of hydrocarbons Priority of functional group in IUPAC naming
Notes
I.
Nomenclature of organic compounds
The basic idea of the IUPAC (International Union of Pure and Applied Chemistry) system of naming is that by following some rules, a name can be assigned to a molecule. And from the assigned name, the original molecule can be reconstructed unambiguously. A given structure may have several different names but a name will only give 1 structure. H H H H
For example, the structure is called H C C C C O H is called butan-1-ol or 1-butanol but no matter H H H H
butan-1-ol or 1-butanol refer to the same structure. IUPAC system of naming is a kind of substitutive nomenclature. In substitutive nomenclature, hydrogen atoms in a named 'parent' hydrocarbons are considered to be substituted by other groups. Actually, the IUPAC system of naming is not very strict. It is possible to have more than 1 name for a given molecule. But no matter which name is assigned, only 1 structure can be rebuilt. A. Labelling of carbon To facilitate discussion, carbon atoms on a carbon chain can be labelled by 2 different methods. 1.
Methyl, primary, secondary and tertiary carbon
The first way is based on the no. of carbon atoms attached to the carbon to be concerned. If no other carbon is attached to the carbon, the carbon atom is called a methyl carbon. If 1 carbon atom is attached to the carbon, the carbon atom is called a primary carbon, 1º carbon If 2 carbon atoms are attached to the carbon, the carbon atom is called a secondary carbon, 2º carbon If 3 carbon atoms are attached to the carbon, the carbon atom is called a tertiary carbon, 3º carbon CH3
H H C
X
H C
X
H
H
methyl carbon
1° carbon
CH3
CH3 CH3
C
X
H 2° carbon
CH3
C
X
CH3 3° carbon
e.g. CH3CH2OH is called a primary alkanol since the hydroxyl group is attached to a 1º carbon.
Nomenclature of organic compounds 2.
Page 2
α carbon, β carbon and γ carbon
Another way is according to the position of the carbon relative to a substituent (functional group). If the carbon is immediately next to the substituent, it is called a α carbon. The carbon attached to the α carbon is called a β carbon. The carbon attached to the β carbon is called a γ carbon. O C
C
γC
C
βC
X
αC
C γC
C βC
C
C αC
carbonyl carbon
B. Skeletal formula / Stick formula / Bond line formula In order to make the presentation simple, sometimes skeletal formula is used instead of the full structural formula or condensed structural formula. In the skeletal formula, all atomic symbols of carbon and hydrogen are omitted.
Full structural formula
H H H C
C H
H H
Condensed structural formula
H3C
CH3
CH3
CH3
H
CH3
HH C HH H C
C
C H
H
H
H
C
H C
C
H
H
CH3CHCH3 CH3CH(CH3)CH3
H
CH2 CH2
H
CH2 CH2 CH3CHCH2
H H C C
C
H
H H H C H C
H
H
CH2 CH
H C
C
CH2 CH
H
C C
CH3C(CH2)CH3
Skeletal formula / Stick formula / Bond line formula
H
H H
C
C C
C
H H
H H H H H H C C C H H C C C H H H H H H H H H H C C C
C H
H H O H
CH3CH2CH(OH)CH3
OH
H
H O H
O
H H
H C C O C C
CH3CH CH2
H H H
H C C
C
H
H
H
Condensed structural formula
H C
CH3CH2CH3
H H H
H C
H
H
H C C C
H
Full structural formula
CH3CH3
H H H
H
Skeletal formula / Stick formula / Bond line formula
H H
H
CH3COOCH2CH3
O
Nomenclature of organic compounds
Page 3
II. Nomenclature of Hydrocarbon Some examples H H H H H C
C C
C O H is called butan-1-ol (or 1-butanol). (a primary alcohol)
H H H H
H H3C
C
CH2 CH3 is called 2-methylbutane (or isopentane). iso- means 1 C is attached to the next-to-end C.
CH3
CH3
CH3 H3C
C
H3C
CH2 CH3 is called 2,2-dimethylbutane (or tert-butylethane).
C CH3
is called the tert-butyl group.
CH3
CH3
CH2 CH2 CH CH2 CH2 CH3 H3C
CH CH3
is called 4-(1-methylethyl)heptane (or 4-isopropylheptane).
CH3 NH2 is called methanamine (or methylamine).
O C
H benzenecarbaldehyde (or benzaldehyde)
O C
OH benzenecarboxylic acid (or benzoic acid)
Nomenclature of organic compounds
Page 4
The basis of IUPAC substitutive nomenclature is that every name consists of a root, one suffix and as many prefixes as necessary. prefix - root - suffix Root represents the longest carbon chain containing the principal functional group as indicated by the suffix and also it should contain the largest no. of substituent. Suffix represents the functional group with the highest priority among all the functional groups present. Prefix represents other functional group and arranged in alphabetical order. 3-amino is a prefix 2-chloro is a prefix butan- is a root -ol is a suffix
Cl CH3 CH CH2 CH2 OH 3-amino-2-chlorobutan-1-ol NH2
e.g.
The numbers are called locants used to indicate the position of the functional group. The numbers are arranged to give (a) the lowest possible number to the group cited by a suffix (principal functional group), then (b) the lowest possible individual numbers (not sum) to those cited as prefixes. CH3 CH2 CH CH CH2 CH2 CH2 CH2 CH CH3 CH3 CH3
CH3
2,7,8-trimethyldecane (not 3,4,9-trimethyldecane) N.B.
hyphen (-) is used to separate number from letter. comma (,) is used to separate number from number.
Name of the first 10 straight chain hydrocarbons 1. methane (meth-) 2. ethane (eth-) 3. propane (prop-)
4. butane (but-) 5. pentane (penta-) 6. hexane (hex-/hexa-)
7. heptane (hept-/hepta-) 8. octane (oct-/octa-) 9. nonane (nona-)
10. decane (dec-/deca-)
For the parent chain containing double bond or triple bond, suffixes -ene and -yne are used respectively. C
C ethene
C
C C
C
C C C
C ethyne
C
C C propyne
C
C C
C but-1-ene
C
C C
C but-2-yne
C hepta-1,3-diene
If the chain contains both double bond and triple bond, the name will be ended with -yne. C
C C
C
C pent-3-en-1-yne
C
C C
C
C C hex-3-ene-1,5-diyne
N.B.
C
C C
C
C pent-1-en-3-yne
-e- is omitted if it is followed by a vowel e.g. y, a, o.
If both double bond and triple bond has the same lowest possible number, double bond will be given a higher priority. C
C C
C
C pent-1-en-4-yne
C
C C
C
C C hexa-1,3-dien-5-yne
Nomenclature of organic compounds
Page 5
Classification of hydrocarbons
Hydrocarbons
Aliphatic (Acyclic)
Alkane
Alkene
Alkyne
CH3 CH3
CH2 CH2
CH CH
ethane
ethene
ethyne
Alicyclic (Cyclic)
Aromatic
cyclohexane
benzene (Arene)
Aliphatic (Acyclic) : Hydrocarbon with no ring structure. It can be further classified into straight chain and branched hydrocarbon. Alicyclic (Cyclic) : Aromatic :
Hydrocarbon with ring structure.
A special kind of ring compound with extra stability due to delocalization of electrons.
and naphthalene are aromatic. Originally, 'aromatic' means having a sweet smell since Benzene the firstly discovered aromatic compounds possess special smell. A. Alkyl, vinyl, allyl and aryl group When a hydrogen is removed from an alkane, the structure is called an alkyl group R–. It is cited by adding -yl. H H C
methyl group
H
When a hydrogen is removed from an alkene, the structure is called a vinyl group. It is cited by adding -enyl. H H H H C
C C
prop-2-enyl group
H
When a hydrogen is removed from an alkyne, the structure is called an allyl group. It is cited by adding -ynyl. H H C
C C
prop-1-ynyl group
H
N.B.
The carbons are counted from the point joining to the main chain.
When a hydrogen is removed from a benzene, the structure is called a phenyl group Ph– or
.
phenyl group However if a phenyl group is attached to a –CH2–, it is called the benzyl group. CH2
benzyl group
Phenyl group and other substituted phenyl group (e.g. chlorophenyl group) are collectively called aryl group Ar–. NH2 phenylamine (or more accurate benzenamine).
Nomenclature of organic compounds
Page 6
III. Other functional group A. Halogeno-compounds The presence of halogen (–X) is always cited by the prefixes (fluoro-), (chloro-), (bromo-) and (iodo-). Cl
Br Br H chloromethane
H C
H C
H
C H 1,2-dibromoethane
H H
B. Alkanols and phenol The presence of hydroxyl group (–OH) is cited by suffix (-ol) or prefix (hydroxy-) H H
OH OH
C OH ethanol
H C
H C
H H
N.B.
H
H O
C H ethane-1,2-diol
HO C
C OH hydroxyethanoic acid
H
H
(-e-) of the root is omitted if it is followed by an vowel e.g. -o- or -y-.
When a hydrogen on the benzene is substituted by a hydroxyl group, the compound is called phenol. OH phenol
O2N
OH 4-nitrophenol
C. Aldehydes (Alkanals) and ketones (alkanones) Aldehyde (Alkanal) contains a terminal carbonyl group It is cited by suffix (-al) or prefix (oxo-). O
O C
. It has the general formula
O R C
H
.
O O
CH3 C H ethanal
H C
C OH oxoethanoic acid O
O
Ketone contains a non-terminal carbonyl group C . It has the general formula R C R It is cited by suffix (-one) or prefix (oxo-) if the C is counted in the root or prefix (carbonyl-) if the C is not counted in the root. O
O
CH3 C CH2 CH2 CH3 pentan-2-one
O
CH3 C CH2 C OH 3-oxobutanoic acid
D. Carboxylic acids (Alkanoic acid) Carboxylic acid contains a carboxyl group prefix (carboxy-). O CH3
O C OH
O
C OH ethanoic acid
HO C CH2
. It is cited by suffix (-oic acid) or (-carboxylic acid) or by
O C OH propanedioic acid.
O C
OH benzenecarboxylic acid (common name : benzoic acid)
O H H HO C
C N+ H Cl- (carboxymethyl)ammonium chloride H H
N.B.
1. 2.
locant is not necessary for dioic acid since the carboxyl groups must occupy the ends of the C chain. When suffix (-carboxylic acid) or prefix (carboxy-) is used, the C in the carboxyl group is not counted as a part of the parent chain.
Nomenclature of organic compounds 1.
Page 7
Carboxylic acid derivatives – esters, acid chlorides, anhydrides and amides
Esters, acid chlorides, anhydrides and amides are all derivatives of carboxylic acid. The derivative is formed by replacing –OH hydroxyl group with other functional groups. –OH replaced by
General formula
Carboxylic acid
Example O
O R C OH
Ester
–OR –Cl
Anhydride (formed by dehydration of alkanoic acid)
–OOCR
R'
CH3
O
C O CH2CH2CH3 propyl ethanoate O
O R C
C OH ethanoic acid O
O R C O
Acid chloride (acyl chloride)
CH3
Cl
CH3
O
O
R C O C
C
R'
CH3
O
C O C O
CH3
Cl ethanoyl chloride
CH3 ethanoic anhydride
O
C O C
CH2CH2CH3
butanoic ethanoic anhydride Amide
–NH2
O
O
R C NH2
O
O
R C
is called the acyl group. e.g. CH3 C
CH3
C NH2 ethanamide
ethanoyl group.
E. Nitriles Nitrile contains cyano group –C≡N. It is cited by suffix (-nitrile) or (-carbonitrile) for cyclic nitrile or by prefix (cyano-) CH3
C N ethanenitrile
C N benzenecarbonitrile H O
C N cyclohexanecarbonitrile
N
C C
C OH cyanoethanoic acid
H
N.B.
When suffix (-carbonitrile) or prefix (cyano-) is used, the C in the cyano group is not counted as a part of the parent chain.
Nomenclature of organic compounds
Page 8
F. Amines Amine contains amino group –NH2. Amine is alkyl or aryl derivative of ammonia, with the hydrogen replaced by alkyl group or aryl group. It is cited by suffix (-amine) or prefix (amino-). CH3 CH CH2 NH2 CH3
CH3 NH2 methanamine (methylamine)
2-methylpropan-1-amine
CH3 N CH2CH3 H
N-methylethanamine (ethylmethylamine)
CH3 N CH2CH3 CH2CH2CH3
N-ethyl-N-methylpropan-1-amine (ethylmethylpropylamine)
NH2 benzenamine (phenylamine). N
N-phenylbenzenamine (diphenylamine).
H
H2N CH2 CH2 OH 2-aminoethanol
N.B.
The locant N- is used to indicate the alkyl or aryl group is attached to the N, not the parent C chain.
G. Amino-acids Amino-acid contains both amino group –NH2 and carboxyl group –COOH. O H2N CH2 CH2 C OH 3-aminopropanoic acid
H. Ethers Ether contains –O–R group. It is cited by prefix (R-oxy-) e.g. methoxy-. CH3 O CH2 CH3 methoxyethane (ethyl methyl ether) CH3CH2
O CH2CH3 ethoxyethane (diethyl ether)
H3C O CH2CH2 O CH3 1,2-dimethoxyethane
N.B.
In some circumstances, -oxy- is used alone to indicate the presence of oxygen.
IV. Priority of functional groups If a compound has more than 1 functional group. The functional group with the highest priority would be named as the suffix and all others would be named as prefixes. e.g. O HO
C CH3 O O
O O C OH
2-ethanoyloxybenzenecarboxylic acid
HO C
C O CH2 CH3
ethoxycarbonylmethanoic acid
N C
O CH2CH3
4-ethoxy-2-hydroxybenzenecarbonitrile
Nomenclature of organic compounds
Page 9
V. Examples
Steps in writing IUPAC name 1. 2. 3. 4. 5.
3. 4. 5.
3. 4. 5.
CH3 OH
CH2 CH CH3 OH
CH3 CH3
C
CH3
CHCH2CH2CHCH2CHO
hydroxyl group (–OH)
Number the parent chain so that the principal functional group is on the lowest numbered carbon. Label other substituents as prefixes. Group the prefixes, arrange them in alphabetic order and join with the root.
pentane-2,4-diol
oct-6-enal
2-methyl 2-methylpentane-2,4-diol
3-methyl, 7-methyl 3,7-dimethyloct-6-enal
C
C C
C
OH
OH
alkanal group (–CHO) O
C C
C C
C2H5C(CH3)=CHCH2CH2CO2H C
C C
C
C C
C H
C
C OH
C C
C
carboxyl group (–COOH)
Number the parent chain so that the principal functional group is on the lowest numbered carbon. Label other substituents as prefixes. Group the prefixes, arrange them in alphabetic order and join with the root.
hept-4-enoic acid
propan-1-ol
5-methyl 5-methylhept-4-enoic acid
1-phenyl 1-phenylpropan-1-ol
Identify the principal functional group Identify the parent chain containing the principal functional group Number the parent chain so that the principal functional group is on the lowest numbered carbon. Label other substituents as prefix. Group the prefixes, arrange them in alphabetic order and join with the root.
C
OH
O C C
C
Identify the principal functional group Identify the parent chain containing the principal functional group
Steps in writing IUPAC name 1. 2.
C
Identify the principal functional group Identify the parent chain containing the principal functional group
Steps in writing IUPAC name 1. 2.
CH3
C C
C
C C
C
hydroxyl group (–OH)
O
OH
C OH
C
C C
C C
C C
C
C
C C
C C
C O C
C
C
double bond (C=C) C
C C
C
Nil C
C C
C
but-1-ene
butane
3-methyl, 3-methyl 3,3-dimethylbut-1-ene
1-ethoxy, 3-methyl, 3-methyl 1-ethoxy-3,3-dimethylbutane
Nomenclature of organic compounds
Page 10
VI. Other common abbreviations Abbreviation
Functional group
R–
Alkyl group
Me–
Et–
Methyl group
Ethyl group
Structure
Example
H
H
H C
H C
H
H
H H
H H
H C
C
H H Ph–
H C
H
Me–H
C H
Et–H
H H
Phenyl group
Ph–H
H H
Ar–
Alkyl group
X–
Halogeno group
Ar–H
H
Cl–, Br–, I–
Glossary
nomenclature IUPAC parent chain prefix root functional group aliphatic (acyclic) alicyclic (cyclic) alkyl group vinyl group allyl group aryl group hydroxyl group aldehyde (alkanal) ketone (alkanone) carboxylic acid (alkanoic acid) carboxyl group ester amide acyl group nitrile amine amino group
Past Paper Question
94 2C 8 c i ii iii 96 2C 9 c i ii 97 2B 5 d i ii iii 98 2B 7 c i ii
suffix locant alkane alkene alkyne phenyl group alkanol phenol carbonyl group acid chloride (acyl chloride) anhydride ether
94 2C 8 c i ii iii 8c Give a systematic name to each of the following compounds. CH3 CH3 i
1
CH3CH CH CH2CH CH2
4,5-dimethylhex-1-ene ii
CH3CH2CH2 O C
1 CH2CH2CH3
propyl butanoate iii
1 mark
O
1 mark 1
O CH3CH2 C CHCH2CH3 OH
C
4-hydroxyhexan-3-one Badly-answered. Many candidates did not know the general rules for systematic naming.
1 mark
Nomenclature of organic compounds 96 2C 9 c i ii 9c Give a systematic name to each of the following compounds : (Deduct ½ marks for each minor mistake. Max. Deduction = 2 marks) i CH3 OH
Page 11
1
CH3 CH CH2 CH CH3
4-methylpentan-2-ol / 4-methyl-2-pentanol ii
1 mark
CH3
1
CH3 CH3 O
2,5,5-trimethylcyclohex-2-enone 1 mark Disappointing answers for (i). Some candidates did not know the basic rules of nomenclature. Very few gave the correct name for (ii).
C
97 2B 5 d i ii iii 5d Give a systematic name to each of the following compounds: i CH3 CH3
3
N
ii
O CH3CH2
iii
H HO2C
C C
C
CH(CH3)CH2CH3
CH3 H
98 2B 7 c i ii 7c Give a systematic name to each of the following compounds : i CH3 CH2CH3
ii
CH3CH=CHCH2CHO
2
Acid-Base Theory
I.
Definition of acid-base A. B. C.
II.
Strength of acid and base A.
III.
Arrhenius definition Bronsted-Lowry definition Lewis definition Leveling effect
Factors determining the strength of a Bronsted-Lowry acid A.
Relative stability of the conjugate base comparing with the acid 1. Inductive effect (1)
2.
Resonance effect a) Resonance effect on a benzene ring (1) (2)
3. 4. 5. 6.
Amine vs Amide
Phenol Aromatic amine
Intramolecular hydrogen bond Solvation / hinderance to solvation a) Steric hinderance to solvation b) Effect of solvent Strength of the bond between the proton and the conjugate base Electronegativity of the atom a) Effect of hybridization
Acid-base Theory
Unit 1
Page 1
Topic
Acid-base Theory
Reference Reading
14.1 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 41–43, 138–140, 185–186, 252–254 Organic Chemistry, Solomons, 5th Edition pg. 86–109, 830–834, 942–943 Organic Chemistry, Fillans, 3rd Edition pg. 218–220, 301–313 Organic Chemistry, Morrison Boyd, 6th Edition pg. 33–35, 729–736, 849–852, 903–905 Organic Chemistry, 6th Edition, Solomons, pg. 90-93, 96-101
Syllabus
Acid-base Theory Expression of Ka and pKa Relative strength of acids and bases
Notes
A. Acid-base Theory
Nevertheless, scientists find that acids and bases bear some other properties. According to these properties, they developed a series of definitions to describe acid and base. They are called Arrhenius definition, Brønsted Lowry definition and Lewis definition. The coverage of the latter definition is boarder than the former one. I.
B rφ n A rr h s
ef wis d initio Le Lowry d n e d te nius defi f e
n it io in tion ni
At the early stage of development of acid-base theory, acid was only described as a sour substance. And base is only a substance which neutralizes the sour taste of an acid.
Unit 1
Common Sense
Definition of acid-base
A. Arrhenius definition In 1884, Swedish chemist Svante Arrhenius proposed that Acid – a hydrogen-containing compound that, when dissolved in water, produces hydroxonium ions, H3O+(aq). Base – a substance that, when dissolved in water, produces hydroxide ions, OH-. Examples of Arrhenius acid e.g. H2SO4(l) + H2O(l) → H3O+(aq) + HSO4-(aq) HSO4-(aq) + H2O(l) → H3O+(aq) + SO42-(aq) Examples of Arrhenius base e.g. NaOH(s) + aq → Na+(aq) + OH-(aq) KOH(s) + aq → K+(aq) + OH-(aq) This is the definition learned in certificate level.
Acid-base Theory
Unit 1
Page 2
B. Brønsted-Lowry definition Arrhenius theory is criticized that 1. Acid is restricted to hydrogen-containing species and base is restricted to hydroxide-containing species. 2. The theory is only applicable to aqueous medium where a lot of acid-base reaction takes place in the absence of water. In 1923, Danish chemist Johannes Brønsted and British chemist Thomas Lowry proposed, a boarder definition, that Acid – a proton donor Base – a proton acceptor e.g. HCl(aq) + proton donor
CuO(s) → Cu2+Cl-2(aq) proton acceptor
e.g. NH3(aq) + proton acceptor (base)
H2O(l) d proton donor (acid)
NH4+(aq) + proton donor (conjugate acid of NH3(aq))
OH-(l) proton acceptor (conjugate base of H2O(l))
e.g. HCl(aq) + proton donor (acid)
H2O(l) d proton acceptor (base)
H3O+(aq) + proton donor (conjugate acid of H2O(l))
Cl-(aq) proton acceptor (conjugate base of HCl(aq))
e.g. H2O(l) + proton acceptor (base)
H2O(l) d proton donor (acid)
H3O+(aq) + proton donor (conjugate acid of H2O(l))
OH-(aq) proton acceptor (conjugate base of H2O(l))
+
H2O(l)
When a Brønsted-Lowry acid, a proton donor, donates a proton, it becomes a potential proton acceptor and is called the conjugate base of the acid. Stronger a proton donor, weaker will be the conjugate base. Conversely, when a Brønsted-Lowry base, a proton acceptor, accepts a proton, it becomes a potential proton donor and is called the conjugate acid of the base. Stronger a proton acceptor, weaker will be the conjugate acid. All Arrhenius acids and bases can be classified into Brønsted-Lowry acid and base accordingly.
Acid-base Theory
Unit 1
Page 3
C. Lewis definition The American chemist Gilbert N. Lewis (1923) proposed an even broader definition for acid and base. He proposed that Acid – an electron acceptor Base – an electron donor This definition offers many advantages, including i. The acids are not limited to compounds containing hydrogen. ii. It works with solvents other than water. iii. It does not require formation of a salt or of acid-conjugate pairs. Since all chemical species are potential electron acceptor or electron donor, virtually, all chemical species are either Lewis acid or Lewis base. e.g. H3N:(g) + BF3(g) → H3N→BF3(s) electron electron donor acceptor (Lewis base) (Lewis acid) Most of the chemical reaction is caused by redistribution of electrons which leads to rearrangement of atoms and formation of a new substance. Therefore, all kind of reactions may be considered as Lewis acid-base reaction. Concept of Lewis acid-base is very useful in describing reaction mechanisms. II. Strength of acid and base In measuring the strength of acid, Brønsted-Lowry definition is usually used. Strength of an acid can be measured by an equilibrium constant called acidity constant or acid dissociation constant, Ka. For an acid, HA in water HA(aq) + H2O(l) d H3O+(aq) + A-(aq)
[H3O+(aq)][A-(aq)] Equilibrium constant, Keq = [HA ][H O ] (aq) 2 (l)
Acidity constant, Ka is defined as Ka =
[H3O+(aq)][A-(aq)] = Keq[H2O(l)] [HA(aq)]
where [H2O(l)] is a constant since water is the solvent in large excess.
For a stronger acid, more HA(aq) molecules will dissociate into H3O+(aq) ion and A-(aq) ion and give a larger value for K a. Similar to concentration of H+(aq) ion, Ka can also be expressed in a negative log scale. pH = - log [H+]
A low pH means a high concentration of H+(aq) ion.
pKa = - log Ka
A low pKa means a larger value for Ka and a stronger acid.
Acid-base Theory
Unit 1
Page 4
Relative strength of selected acids and their conjugate bases Stronger an acid, weaker will be its conjugate base. Stronger a base, weaker will be its conjugate acid. N.B. Redox reaction can be considered as a competition for electron. oxidizing agent + e- d reducing agent In redox reaction, only strong oxidizing agent reacts with reducing agent. Similarly, acid base reaction can be considered as a compeition for proton. cojugate base + H+ d conjugate acid In acid base reaction, only strong acid react with strong base. The strength of an acid or a base is also an indicator of their stability. A strong acid or base is less stable than a weak acid or base. A strong acid tends to react with a strong base to form a weak conjugate base and weak conjugate acid. A. Leveling effect An acid stronger than hydroxonium ion H3O+(aq), does not show difference in acidity in water. The water will convert all those acid molecules into H3O+(aq) ions. HCl(aq) + strong acid
H2O(aq) → H3O+(aq) + strong weak base acid
Cl-(aq) weak base
HBr(aq) + strong acid
H2O(aq) → H3O+(aq) + strong weak base acid
Br-(aq) weak base
A base stronger than hydroxide ion, e.g. NH2- amide ion, also does not exist in aqueous medium. Water will convert all those base molecules into OH-(aq) ions. NH2- + H2O(l) → NH3(aq) + OH-(aq) This is known as leveling effect of solvent.
Glossary Past Paper Question
Arrhenius acid-base
Brønsted-Lowry acid-base
Lewis acid-base
Acid-base Theory
Unit 2
Page 1
Topic
Acid-base Theory
Unit 2
Reference Reading
14.0 14.2–14.4 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 41–43, 138–140, 185–186, 252–254 Organic Chemistry, Solomons, 5th Edition pg. 86–109, 830–834, 942–943 Organic Chemistry, Fillans, 3rd Edition pg. 218–220, 301–313 Organic Chemistry, Morrison Boyd, 6th Edition pg. 33–35, 729–736, 849–852, 903–905
Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 471, 499–501, 519–520 Organic Chemistry, 6th Edition, Solomons, pg. 102-118, 320, 433-434, 795-796, 903-905, 970-972 Reading Syllabus
Factors determining the strength of an acid
Notes
III. Factors determining the strength of a Bronsted-Lowry acid A. Relative stability of the conjugate base comparing with the acid 1. Inductive effect 2. Resonance effect 3. Intramolecular hydrogen bond 4. Solvation / hinderance to solvation 5. Size of the conjugate base 6. Electronegativity of the atom A. Relative stability of the conjugate base comparing with the acid The strength of an acid is mainly determined by the relative stability of the conjugate base comparing with the acid. The equilibrium constant is related to the standard free energy change (∆Go) equation Ka ∆Go = -2.303 RT log Keq = -RT ln Keq = -RT ln [H O ] 2 (l)
where ∆Go = ∆Ho - T∆So
∆Go is an indicator of the relative stability of the product comparing with the reactant. A very negative value means the product is more stable than the reactant. For a reaction with small change in entropy, i.e. reaction without change in physical state, the ∆Go is roughly the same as ∆Ho. Since Ka = Keq [H2O(l)], a very negative value of ∆Go means a very strong acid. The acidity of the acid increase as the relatively stability of the conjugate base increase comparing with the acid molecule.
Acid-base Theory I.
Unit 2
Page 2
Inductive effect H O H C
C O H
H O +
H2O
H C
H
C O-
+
H3O+
H
Upon ionization in water, an ethanoic acid molecule forms a carboxylate ion and a hydroxonium ion. An electronwithdrawing (-I) groups stabilize the negative charge of the carboxylate anion by dispersing the negative charge. This increases the acidity of the acid.
Inductive effects of halogen atoms on acid strength Acid Structure Ethanoic CH3CO2H Iodoethanoic ICH2CO2H Chloroethanoic ClCH2CO2H Trichloroethanoic Cl3CCO2H Trifluoroethanoic F3CCO2H
pKa (H2O) at 25ºC 4.76 3.12 2.85 0.7 0.23
Butanoic 2-Chlorobutanoic 3-Chlorobutanoic 4-Chlorobutanoic
4.81 2.81 4.05 4.52
Note :
CH3CH2CH2CO2H CH3CH2CHClCO2H CH3CHClCH2CO2H ClCH2CH2CH2CO2H
Relatively weak acid
Relatively strong acid Relatively strong acid Relatively weak acid
pKa is a log scale, increase by 1 in pKa means 10 folds decrease in acid strength.
As inductive effect is transmitted through σbond, it decreases quickly with increasing distance from the negative centre. It is only a short range effect. Electron-donating : positive inductive effect (+I) Electron-withdrawing : negative inductive effect (-I)
Inductive effects of various Groups
Acid-base Theory
Unit 2
2.
Page 3
Resonance effect ethanoic acid CH3COOH 4.76
Acid pKa
phenol Ph-OH 9.9
ethanol CH3CH2OH 16
Experimentally, ethanoic acid is 1.7 × 1011 times more acidic than ethanol. This cannot be explained satisfactorily by the presence of extra oxygen atom. Comparing choroethanoic acid and ethanoic acid, the presence of the chlorine atom only contribute a 81 times difference in acidity. The exceptionally high acidity of ethanoic acid can be explained by the resonance stabilized ethanoate ion formed from ethanoic acid. Comparing the structure of ethanoic acid and ethanoate ion, The ethanoate ion has identical resonance structures. In contrast, the two resonance structures of the carboxylic acid molecules are not identical. One of the resonance structure even involves charge separation and makes it less stable. Species with identical resonance structure implies a more evenly distributed electron cloud which is relatively more stable. This makes ethanoate exceptionally stable and ethanoic acid exceptionally acidic.
Free energy
O CH3
C
O OH
CH3
C O-
f O CH3
CH3
C OH
+
H+
OC O
pKa = 4.76
As the product is stabilized more than the reactant, the Ka of the acid will increase. Inductive effect and Resonance effect are collectively called Electronic effect.
∆G' ∆G' is more negative (less positive) than ∆G Reaction coordinate
A carboxylic acid yields a resonance-stabilized anion; it is a stronger acid than an alcohol. (The plots are aligned with each other for easy comparison.)
Acid-base Theory
Unit 2
Page 4
(1) Amine vs Amide For a similar reason, amine is more basic than amide. H
H CH3 CH2
+ H+
N H
CH3 CH2 > N H + H
Alkyl group is an electron-donating group, it will stabilize the conjugate acid formed and make the amine more basic. O CH3 C
O
H N H
+ H
+
H
CH3 C < N H + H
Just the opposite, acyl group is an electron-withdrawing group, it will destabilize the conjugate base formed and make the amide less basic. a) Resonance effect through a benzene ring (1) Phenol H
H
H
O
O+
O+
Resonance effects of various groups - H+
-
-
O
O
O
O
-
-
Since the phenoxide ion is more stabilized by resonance than the phenol molecule, phenol is much more acidic than other alcohols. e.g. pKa of phenol = 9.9 pKa of CH3CH2OH = 16 (This is a wrong explanation used in HK A-level, for more upto-date explanation, please read Solomon pg.942–943) Though phenol is acidic, phenoxide ion does not have identical resonance structures. Phenol is less acidic than carboxylic acid. Thus, phenol reacts with strong alkali, e.g. NaOH, but not with weak alkali, e.g. Na2CO3 or NaHCO3. This can be used as a test to distinguish phenol and ethanoic acid. Vorlander's rule of resonance effect on benzene ring 1) Saturated groups (containing only single bonds between all atoms) are electron donating (+R). O
F ,
O C
R (has a lone pair on the atom) e.g. 2) Unsaturated groups (containing multiple bonds between any atoms) are electron withdrawing (-R). Cl ,
O
O
e.g.
C N
,
C
N
,
O (has a multiple bond between the first and the second atom)
Acid-base Theory
Unit 2
Page 5
The acidity of phenol increases as the benzene ring is substituted by nitro group, –NO2. pKa of 4-nitrophenol = 7.15 pKa of 2,4,6-trinitrophenol = 0.42 Since the nitro group is very far from the hydroxyl group, the inductive effect is very negligible. Comparing with inductive effect, reasonance effect has a much longer range. O
-
O
O -
N+ O O-
O
N+ O-
-O
N+ O-
By resonance, the negative charge on the 4-nitrophenoxide ion is more dispersed. This gives extra stability to the 4-nitrophenoxide ion and make nitro substituted phenol more acidic. (2) Aromatic Amine The basicity of amine can also be explained in the same way. H N H
H N CH2CH2CH2CH3 H
pKb
3.2
H N H H
4.8
9.4
basicity decreases
The high basicity of butan-1-amine can be explained by the presence of the electron-donating alkyl group. This stabilizes the conjugate acid CH3CH2CH2CH2NH3. Consider the equilibrium H
H N H
H N H +
H+
H H N H
is an electron withdrawing group by inductive effective -I. It destabilizes the conjugate acid H
H N H
and make
H N H
the least basic one. Furthermore,
H N H
is also stabilized by resonance but
Therefore, the equilibrium position will be lying on the left hand side.
H N H
H N H
H N H
is not.
Acid-base Theory
Unit 2
3.
Page 6
Intramolecular hydrogen bond
Existence of intramolecular hydrogen-bond may have positive or negative effect on the acidity of an acid. Example 1 – butenedioic acid H C
CO2H
C H
CO2H H C
(Z)-butenedioic acid / cis-butenedioic acid (maleic acid) pKa1 = 1.83
CO2H
(E)-butenedioic acid / trans-butenedioic acid (fumaric acid) pKa1 = 3.03
C HO2C
pKa2 = 6.07
H
pKa2 = 4.44
For a dibasic acid, there are 2 pKa values. The first dissociation of maleic acid is more complete but the second dissociation is just the reverse. Comparatively, only the acid molecule that undergoes 1st dissociation may undergo 2nd dissociation. Therefore, the value of pKa2 is not as important as pKa1 in determining the acidity of an acid. i.e. maleic acid is more acidic that fumaric acid. This can be explained by a H-bond stabilized conjugate base of maleic acid. O H
O
C O H
-
C
C
C
C
H
C O H
H
O
O
C Oδ
H
C O-
H C
H + H+
+ 2H+
C
C OδO
H
C OO
H-bond stabilized The first conjugate base is stabilized by formation of intramolecular H-bond which causes a stronger 1st dissociation and weaker 2nd dissociation.
O
O H
H O C
H
C O H C
C
C
C H O C
H
O
C O
H + H+
H
O C O
Not H-bond stabilized
2H+ + A2-
2H+ + A2-
H+ + HA-
H+ + HA-
H2A Dissociation of maleic acid
+ 2H+
C
O
O
C O C
H2A Dissociation of fumaric acid
H
Acid-base Theory
Unit 2
Page 7
Example 2 – hydroxybenzenecarboxylic acid H O
C
O
H O
O C HO
2-hydroxybenzenecarboxylic acid
OH
4-hydroxybenzenecarboxylic acid
2-hydroxybenzenecarboxylic acid is about 40 times more acidic than that of 4-hydroxybenzenecarboxylic acid since the conjugate base is once again stabilized by intramolecular H-bond. H O
C
O
δO
H O
4.
C
δO
-O
H O
C
O O-
Solvation / hinderance to solvation
a) Effect of solvent Acid shows higher acidity in polar solvent. e.g. Hydrogen chloride is a strong acid in water but not acidic in methylbenzene. Ions can be stabilized in polar solvent but not in non-polar solvent. b) Steric hinderance to solvation
Bulky acid ⇒ hindered negative centre ⇒ less solvation stabilization ⇒ unstable conjugate base ⇒ lower acidity The theory regarding the strength of acid is also applicable to the strength of base. The strength of a base is also depending on the relative stability of the conjugate acid comparing with the parent base. Amines exhibit a similar steric hinderance to solvation. In gas phase where there is no solvation, the order of base strength is (CH3)3N > (CH3)2NH > CH3NH2 [3º > 2º > 1º] which is governed by the positive inductive effect (+I) of methyl group only. But in water, the order of base strength of methylamines is (CH3)2NH > CH3NH2 > (CH3)3N [2º > 1º > 3º]. The result is a combination of inductive effect and steric hinderance to solvation. The bulky conjugate acid formed, aminium ion, is less stabilized by solvation.
Acid-base Theory 5.
Unit 2
Page 8
Size of the conjugate base - (Strength of bond between the proton and the conjugate base)
On moving down a group (from F–H to I–H), the bond length increases and bond strength decreases tremendously (from 568 kJmol-1 to 298 kJmol-1). Acidity pKa bond dissociation energy / kJmol-1 Ionic size of the conjugate base
H–F< 3.2 568
H–Cl -7 432
(X-)
<
H–Br -9 366
<
H–I -10 298
sin g increa →
As the size of the conjugate base gets larger, the charge density on the X- gets lower and becomes more stable. As a result HI is most acidic one. Indeed, acidity is depending on relative stability instead of bond strength. However, an acid molecule possessing a weak bond is usually unstable, it tends to lose the proton to the base (water) to form a more stable system. H–I(aq) + H2O(aq) less stable combination
d
I-(aq) + H3O+(aq) more stable combination
The overall change can be considered as the breaking of a weak H–I bond and formation of a strong H–O bond, as a result the overall stability of the system is enhanced. N.B.
6.
Solely strength of bond would not be accepted as a valid explanation to the difference in acidity.
Electronegativity of the atom
Besides the dispersion of the negative charge on the conjugate base, an increase in attraction between the electrons and the nucleus also enhances the stability of the conjugate base. High electronegativity makes an atom attract the negative charge more strongly and form a more stable anion. Therefore, F- ion is more stable than OH- ion as F is more electronegativity than O. On moving across a period (from C–H to F–H), the bond strength only increases moderately from 425 kJmol-1 to 568 kJmol-1. The effect of difference in size is outweighed by the big difference in electronegativity. Acidity pKa bond dissociation energy / kJmol-1 Electronegativity of the atom (Pauling scale)
H3C–H < 50 425 2.5
H2N–H < 38 431 3.0
HO–H < 15.74 498 3.5
F–H 3.2 568 4.0
Acid-base Theory
Unit 2
Page 9
a) Effect of hybridization Ethyne
H C
Ethene
H C
hybridization
s character
C H
sp
50%
pKa 25
H
sp2
33.3%
44
sp3
25%
50
C
H
H H H
Ethane
H C
C H
H H The stronger the s character a carbon atom possesses, the closer will be the bonding electron to the nucleus. This is because s orbital is closer to the nuclei than the p orbital. The carbon atom possessing more s character behaves like a more electronegative atom and leads to a more stable carbanion. H
Stability
H C
C
>
H H
H C
H
C
>
H C
C
H H
Consequently, the acidity of hydrocarbon decreases with following order : alkyne > alkene > alkane.
Glossary
Arrhenius acid-base Brønsted-Lowry acid-base Lewis acid-base acidity constant leveling effect free energy change inductive effect resonance effect electronic effect Vorlander's rule intramolecular hydrogen bond steric hinderance
Past Paper Question
91 2C 9 a ii 92 1A 3 c 93 1A 3 f i ii 94 1B 4 a 95 1A 3 b 96 1A 2 e ii iii 99 2A 4 c i
92 2C 9 b
96 1A 3 b
91 2C 9 a ii 9a Outline chemical tests which would allow you to distinguish between the compounds in the following pairs. Describe what you would observe in each case. COOH OH ii
2
Add solution of sodium hydrogencarbonate to the compounds. PhCOOH will react to give CO2 bubble (effervescence observed). PhOH give no observable change. COO-Na+
COOH + NaHCO3
+ CO2
92 1A 3 c 3c In aqueous media, why does ammonia act as a base whereas hydrogen fluoride acts as an acid? Higher electronegativity of F than N on HF makes HF acidic because the proton experiences less electron sharing and is easier to release. 1 mark The prominent lone pair on N in NH3, comparing with the withheld electrons pair on F in HF, donates the lone pair more readily and favours basic properties. 1 mark C This question required reasons, rather than statements in the form of equations. Various explanations might have been provided; such as a discussion of the different shielding and ease of release or acceptance of protons when there is a change in the electronegativity of the atom to which they are bonded.
2
Acid-base Theory Unit 2 92 2C 9 b 9b Arrange the following hydroxy-compounds in order of increasing acidity. Explain your order. OH
OH
Page 10 5
OH
NO2 OH OH
OH <
< NO2
1 mark 1 mark ½ mark
Acidity depends on the equilibrium HA d H+ + APhenoxide ion is stabilized by resonance, O-
O
O
O
-
-
½ mark i.e. ∴phenol is more acidic than cyclohexanol. The nitro group in the nitrophenol is electron withdrawing which attracts electron and is capable of resonance, 1½ mark thus further stabilizes the phenoxide ion. O-
N O
O
O
O-
N O
½ mark Many candidates did not use the equilibrium between the acid and its conjugate base to interpret the acid strength. Candidates described –NO2 group incorrectly in this answer as a deactivating group rather than an electronwithdrawing group. A few candidates regarded the nitro-group as electron-donating.
C
93 1A 3 f i ii 3f Consider the following compounds: O H C
CH3
and OH
OH
Y
i
Z
Account for the fact that both Y and Z are acidic. Acidity depends on the equilibrium HB(aq) d H+(aq) + B-(aq) Both compounds have phenol functional group which can be involved in above equilibrium In cases of phenols, equilibrium lies to the right due to resonance stabilization of the phenoxide ion Explanation of resonance stabilization : O
O
O
-
O
½ mark ½ mark ½ mark
-
R
R
ii
R
R
Which compound, Y or Z, is the stronger acid? Explain. Y is the stronger acid.
½ mark 2 ½ mark
O
This is because the electron withdrawing O
group further stabilizes the phenoxide ion.
1 mark
O
O O H
C
C
H
½ mark Most candidates explained the acidity of phenols in terms of the weakening of the O–H bond by resonance, instead of a more formal consideration in terms of the stability of the phenoxide ion. Some could not write resonance structures correctly. Few candidates explained acidity in terms of equilibrium.
Acid-base Theory Unit 2 Page 11 94 1B 4 a 4a Describe a chemical test to show that ethanoic acid is a stronger acid than phenol. (Tests involving the use of a 2 pH meter or indicator paper are NOT accepted.) Addition of Na2CO3(aq) or NaHCO3(aq) to CH3COOH(aq) gives CO2(g) / effervescence. 1 mark 1 mark Addition of Na2CO3(aq) or NaHCO3(aq) to phenol does not give CO2(g) /effervescent Or Only CH3COOH(aq) reacts with Mg Insoluble base oxide e.g. CuO(s) is soluble only in CH3COOH(aq) C Some candidates did not read the question with enough care. Instead of giving a chemical test, they employed physical methods (e.g. conductivity measurement) and therefore were not credited with marks. Others erroneously stated that the stronger acid would require a larger volume of base for neutralization. 95 1A 3 b 3 3b Arrange the following carboxylic acids in the order of increasing acidity. Explain your arrangement. ClCH2CO2H, ClCH2CH2CO2H and FCH2CO2H ClCH2CH2CO2H < ClCH2CO2H < FCH2CO2H 1 mark Inductive effect by electron withdrawing group stabilizes the carboxylate anions more than the carboxylic acids. ½ mark F is more electronegative than Cl and closest to the carboxylate / carboxyl group ∴ FCH2CO2H is most acidic ½ mark the Cl substituted carboxylic acids, inductive stabilization operates most effectively with decreasing distance. ∴ the order of increasing acidity follows the decreasing distance. 1 mark C It was surprising to see that many candidates answered that the stronger the acid, the weaker the O–H bond. Candidates should know that bond strength does not have a direct bearing on acidity and only the relative stability of the acid and the conjugate base dictate the acidity. 96 1A 2 e ii iii 2e Which is the stronger acid in each of the following pairs of substances ? Briefly explain your choice. ii HClO3 (aq) , HClO4(aq) HClO4 ½ mark ½ mark the conjugate base ClO4½ mark possesses a large no. of oxo croup (O atoms) ½ mark is stabilized to a greater extent by resonance / mesomeric effect. Or, ½ mark In HClO4, the Cl has a higher oxidation state and ½ mark a larger number of oxo groups (O atoms). which can induce a higher positive charge density on H.♣ ½ mark iii HMnO4 (aq) , H2CrO4(aq) HMnO4 ½ mark ½ mark The conjugate base MnO4½ mark possesses a large no. of oxo group than HCrO4½ mark is stabilized to a greater extent by resonance / mesomeric effect. Or, ½ mark Mn has a higher O.S. than Cr, ½ mark MnO4- is singly charged while CrO42- is doubly charged, ½ mark thus the conjugate base MnO4- is more stable. Or, ½ mark Mn has a higher oxidation state ½ mark and a larger number of oxo groups, which can induce a higher positive charge density on H.♣ ½ mark (For answers marked with ♣, maximum 1 mark) C It was mentioned in the 1995 Subject Report that bond strength does not have a direct bearing on the acidity of a compound, and that the relative stability of an acid and its conjugate base dictates the acidity. However, many candidates still wrote that with the increase in polarization, the O−H bond in the oxoacid weakens and hence its acid strength increases. Most candidates correctly chose the stronger acid but gave an incorrect explanation. Many did not know that resonance stabilization in oxoacids is related to the number of oxogroups attached. Some candidates drew wrong structures for the oxoacids with H bonded to the central atom. As a result, they wrongly explained the acid strength in terms of the polarization of the Cl–H or Mn–H bond.
2
2
Acid-base Theory Unit 2 96 1A 3 b 3b Arrange the following compounds in the order of increasing basic strength. Explain your arrangement.
Page 12 3
NH2 NH2
,
and NH3 NH2 < NH3 <
Basicity :
NH2
1 mark
The basicity of an amino compound depends on the position of the following equilibrium RNH2 + H2O d RNH3+ + OH-
½ mark
The primary amine is a stronger base than NH3 Q Electron donating property / inductive effect of alkyl group stabilizes the RNH3+ to a greater extent. ∴ 1º ½ mark amine is more basic / the e- pair is more available. For the aromatic amine, overlapping of the orbital containing the lone pair in N with the e-electron cloud of the benzene ring makes the lone pair less available. ½ mark Thus, the equilibrium lies on the LHS. ½ mark NH2
∴
C
is the weakest base
Most candidates explained the relative strengths of the bases in terms of the availability of the lone pair of electrons on the N atom. Only a few candidates completed their answers with a discussion of the equilibrium position. Some wrongly chose NH2
as the strongest base, saying that the cation + NH3
is stabilized by resonance which leads to a dispersion of the positive charge. 99 2A 4 c i 4c i Arrange, with explanation, the compounds below in the order of increasing acidity. HCO2H, CH3CO2H and CF3CO2H
Isomerism I.
Structural isomerism A. B. C. D.
II.
Chain isomerism Position isomerism Functional group isomerism Tautomerism / Tautomerization
Stereoisomerism A.
B.
Diastereomerism / Geometrical isomerism 1. Physical properties of cis-/trans-geometrical isomers of 1,2-dichloroethene 2. Physical and chemical properties of cis-/trans-geometric isomers of butenedioic acid a) Physical properties b) Chemical properties Optical isomerism / Enantiomerism 1. Physical properties 2. Optical activity a) Plane polarized light b) Measurement of optical activity by polarimeter c) Specific rotation 3. Chiral centre 4. Racemic mixture / Racemic modification / Racemic form / Racemate
Isomerism
Unit 1
Page 1
Topic
Isomerism
Unit 1
Reference Reading
15.0–15.1 15.2.1 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 26–33 Organic Chemistry, Solomons, 5th Edition pg. 164–174, 179–186, 201–203, 175, 302, 836 Organic Chemistry, Fillans, 3rd Edition pg. 35–49 Organic Chemistry, Morrison Boyd, 6th Edition pg. 125–138, 144–146, 158–160 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 390–391 Organic Chemistry, 6th Edition, Solomons, pg. 59-61
Syllabus
Isomerism Structural isomerism Stereoisomerism Geometrical isomer (Diastereomer)
Notes
Isomerism Isomerism (same molecular formula)
Isomers are different compounds sharing the same molecular formula. Structural isomers are also called constitutional isomers. The atoms of the isomers are connected in a different order. They are said to have a different connectivity.
Structural / Constitutional Isomerism (different structural formula / connectivity)
Stereoisomers differs only in arrangement of their atoms in space but have the same connectivity.
Chain Position Functional Tautomerism Isomerism Isomerism Group Isomerism
Stereoisomerism (same structural formula / connectivity)
Diastereomerism / Geometrical Isomerism / cis-trans Isomerism (Each isomer is the mirror images of its own)
I.
Enantiomerism / Optical Isomerism (Each isomer is not the mirror image of its own)
Structural isomerism
A. Chain isomerism They have different carbon chains or skeletons. (Branched and unbranched chain) e.g. C4H10 CH3 CH CH3 CH3 CH2 CH2 CH3
CH3
butane
2-methylpropane
B. Position isomerism They are molecules which have a substituent in different positions on the same carbon skeleton. (Alkanol) e.g. C3H8O CH3 CH2 CH2 OH
CH3 CH CH3
propan-1-ol
OH
propan-2-ol (Disubstituted benzene) e.g. C8H10 CH3 CH3
1,2-dimethylbenzene
CH3 CH3
1,3-dimethylbenzene
CH3
CH3
1,4-dimethylbenzene
Isomerism
Unit 1
Page 2
C. Functional group isomerism They are isomers in different homologous series and have different functional groups (Ether and alcohol) e.g. C2H6O CH3 O CH3
CH3 CH2 OH
methoxymethane (dimethyl ether)
ethanol
(Acid and Ester) e.g. C4H8O2 O
O
CH3 CH2 CH2 C O H
CH3 CH2 C O
butanoic acid
CH3
methyl propanoate
D. Tautomerism / Tautomerization Tautomerism is a kind of dynamic isomerism. It is a rapid and reversible interconversion of isomers associated with the actual movement of electrons as well as of one or more hydrogen atoms. Tautomerism must not be confused with resonance. Each tautomer is capable of independent existence and potential isolation. Tautomers represent real compounds, whereas individual resonance structure does not. (Keto-enol tautomerism) e.g. C3H6O O
OH
CH3 C CH3
CH2 C CH3
keto form > 99 %
1.5 × 10-4 %
propanone
propen-2-ol
Interconvertible keto and enol forms are known as tautomers, and their interconversion is called tautomerizaton.
enol form
Usually the enol form contributes only a very small percentage in the equilibrium.
II. Stereoisomerism A. Diastereomerism / Geometrical isomerism By definition, a diastereomer is a stereomer which is superposable with its
own mirror image.
However, a pair of diastereomers are not mirror images of each other. cis-trans isomerism is a kind of diastereomerism, where its existence is due to hindered / restricted rotation of double bond. But-2-ene and hexa-2,4-diene are two of the examples. They have similar but not identical chemical properties and very different physical properties. (but-2-ene CH3CH=CHCH3) CH3 H
C
C
CH3 H
cis-but-2-ene / (Z)-but-2-ene
CH3 H
H C
C CH3
trans-but-2-ene / (E)-but-2-ene
For cis-isomer, the two identical groups (usually hydrogen atoms) are on the same side. (Latin: 'cis' = on this side) For trans-isomer, the two identical groups are on the opposite sides. (Latin: 'trans' = across)
Isomerism
Unit 1
Page 3
For trisubstituted or tetrasubstituted alkene, the terms cis and trans are either ambiguous or do not apply at all since there is no two identical groups. For this kind of structure, the configuration can be denoted by (E-Z) system which is based on the atomic number (bulkness). (Z) German: 'zusammen' = together (E) German: 'entgegen' = across Consider
Br C H
Based on the atomic number, Br has a higher priority than H Cl has a higher priority than F
Cl C F
Since Br and Cl are on the same side (together), the structure is called (Z)-1-chloro-1-fluoro-2-bromoethene.
(Z)-2-bromo-1-chloro-1-fluoroethene For further details, please read Solomons pg. 302 and 175 1.
Physical properties of cis-/trans-geometrical isomers of 1,2-dichloroethene Cl
Cl C C
H
H
cis-1,2-dichloroethene
melting point boiling point dipole moment
- 80 ºC 60 ºC 1.90 D
Cl
H C C
H
Cl
trans-1,2-dichloroethene
melting point boiling point dipole moment
-50 ºC 48 ºC 0D
The boiling point of the cis-isomer is higher than that of the trans-isomer because the intermolecular forces among the former one are stronger, i.e. it is polar. However, the melting point of the later one is higher because the transisomer is more symmetrical in shape and can be packed more regularly in the solid crystal. Because the melting point of a substance is affected by the packing of the molecule on top of the intermolecular forces, boiling point is a better indicator of the strength of the intermolecular forces.
Isomerism
Unit 1 2.
Page 4
Physical and chemical properties of cis-/trans-geometrical isomers of butenedioic acid
Butenedioic acid is another example exhibiting geometrical isomerism. H C C H
CO2H
H C
CO2H
C
CO2H (maleic acid)
HO2C
cis-butenedioic acid / (Z)-butenedioic acid
H
(fumaric acid)
trans-butenedioic acid / (E)-butenedioic acid
a) Physical properties Dipole moment Solubility Melting point
Maleic acid non-zero Higher 131 ºC
Fumaric acid zero Lower 287 ºC
Although the dipole moment of fumaric acid is zero, it is capable to form intermolecular hydrogen bond. Therefore, the melting point is higher. Dipole moment Maleic acid has a stronger dipole moment than fumaric acid.
H C
CO2H
C H
CO2H
Maleic acid has 2 dipole moments pointing to the same side and gives a nonzero resultant. Therefore, it is more soluble in polar solvent. i.e. water.
H C
CO2H
C HO2C
The 2 dipole moments of fumaric acid opposing each other and give a zero net dipole moment
H
Melting point Maleic acid has a lower melting point than fumaric acid. Although maleic acid has a stronger dipole moment, the formation of intramolecular hydrogen bond reduces the formation of intermolecular hydrogen bond. This reduces the extent of the hydrogen bond formed between adjacent molecules in the crystal and causes a lower melting point than that of fumaric acid.
Isomerism
Unit 1
Page 5
b) Chemical properties Acidic properties Both acids show acidic properties in water but have different acidity. H C C H
H
CO2H
CO2H
pKa1 = 1.83
C
maleic acid pKa2 = 6.07
CO2H
C HO2C
H
pKa1 = 3.03
fumaric acid pKa2 = 4.44
The difference is due to the formation of intramolecular H-bond stabilized conjugate base of maleic acid. Dehydration When maleic acid is heated to 140ºC, water is evolved and anhydride forms.
By contrast, no reaction takes place when fumaric acid is heated to 140ºC. At 290ºC, there is sufficient energy to overcome the barrier of rotation about the C=C double bond and the cis configuration required for anhydride formation is attained.
Interconversion between maleic acid and fumaric acid
Maleic acid can also be converted to fumaric acid by boiling with hydrochloric acid catalyst. Since fumaric acid is more stable than maleic acid, the equilibrium lies much towards the fumaric acid side. This is because in fumaric acid, the two bulky carboxyl groups are further apart and experience less steric repulsion.
Glossary
structural / constitutional isomer stereoisomerism geometrical isomerism diastereomerism
tautomerism / tautomerization
Isomerism
Unit 1 92 1A 1 a i 94 2C 7 a 96 2C 8 b i 97 1A 4 a i ii 98 1A 4 b i
Past Paper Question
Page 6
97 1A 4 b i 98 2B 5 a ii
92 1A 1 a i 1a There are several isomers of benzenedicarboxylic acid. i Draw the structures of all possible isomers of benzenedicarboxylic acid. CO2H CO2H CO2H CO2H
1½
CO2H
CO2H
½ mark each
94 2C 7 a 7a What do you understand by the terms “structural isomerism” and “stereoisomerism”? Structural isomerism – occurrence of more than one structure for a given molecular formula 1 mark Stereoisomerism – occurrence of more than one configuration (different arrangements of groups in space) for a give structural formula. 1 mark Examples to illustrate the two types of isomers. 1 mark C Badly-answered. Not many candidates gave precise definitions for 'structural isomerism' and 'stereoisomerism'. Instead, they just used examples (correctly or incorrectly drawn) to illustrate the various kinds of isomers. The following terms were often used incorrectly : molecular formula, structural formula, empirical formula. 96 2C 8 b i 8b The following compounds can exist in isomeric forms : In each case, state the type of isomerism and draw suitable representations for the isomers. i butenedioic acid, and Geometrical isomerism / cis-trans isomerism (Deduct ½ mark for spelling mistake) COOH HOOC
3
2 1 mark
COOH COOH
½ + ½ mark Many candidates did not assign trans- and cis-isomers. Some wrongly gave positional isomers as the answer.
C
97 1A 4 a i ii 4a The formula HO2CCH=CHCO2H can represent two compounds. i Draw a structure for each compound, clearly showing the difference between them. ii One of the compounds reacts with P2O5(s) to give compound A . Give the structure of A.
2
97 1A 4 b i 4b i Draw all possible isomeric structures of dimethylbenzene.
3
98 1A 4 b i 4b Alcohol E has the structure CH3CH(OH)C2H5. i Draw the structures of three structural isomers of E, all of which are alcohols.
1½
98 2B 5 a ii 5a Consider the following compound F.
5
CH3 a
b
CH CH C CH c
d
F
ii
Draw all possible three-dimensional structures for F, indicating the expected bond angles around the carbon atoms a, b, c, and d in one of the structures.
Isomerism
Unit 2
Page 1
Topic
Isomerism
Unit 2
Reference Reading
15.2.2 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 26–33 Organic Chemistry, Solomons, 5th Edition pg. 164–174, 179–186, 201–203, 175, 302, 836 Organic Chemistry, Fillans, 3rd Edition pg. 35–49 Organic Chemistry, Morrison Boyd, 6th Edition pg. 125–138, 144–146, 158–160 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 392–393, 420, 512 Organic Chemistry, 6th Edition, Solomons, pg. 178-185, 188, 193-198, 200
Syllabus
Optical isomerism / Enantiomerism Optical activities Chiral centre Racemic mixture
Notes
B. Optical isomerism / Enantiomerism Optical isomers are also known as enantiomers. Individual enantiomer is not the mirror images of its own. From another point of view, enantiomers exist in pairs. They are non-superposable mirror images of each other. In order to describe this properties, the molecules are said to be chiral. In Greek, 'chiral' means: 'cheir' = hand, since the two hands are non-superposable mirror images of each other.. e.g. butan-2-ol
CH2CH3 C H H3C OH
CH2CH3 H C HO CH3
(+)-2-butanol
≡
CH2CH3 C OH H3C H
(-)-2-butanol
Plane symmetry in diastereomer (cis-but-2-ene) Structurally, all chiral molecules do not have a plane of symmetry. Unlike enantiomer, diastereomer has at least one plane of symmetry. And because of this difference, only enantiomer shows optical activity.
H3C H
C
C
CH3
2nd plane of symmetry
H
1st plane of symmetry
Isomerism
Unit 2
Page 2
The 2 enantiomers may be denoted by several different methods i)
(+) or (-) based on the optical activity of the enantiomers.
ii) (R-S) system based on the absolute configuration which is similar to the (E–Z) system for geometrical isomers. For further details, please read Solomons, pg. 174–179 iii) By relative configuration comparing with a standard compound called Glyceraldehyde. For further details, please read Solomons, pg. 199–201 Methods ii) and iii) will not be discussed in the Hong Kong A-level syllabus. 1.
Physical properties
They have almost exactly the same chemical and physical properties except optical activity. The mixing of 2 enantiomers causes melting point depression proving that the 2 isomers are indeed different. (R)-butan-2-ol or (-)-butan-2-ol or (R)-(-)-butan-2-ol
(S)-butan-2-ol or (+)-butan-2-ol or (S)-(+)-butan-2-ol
99.5 ºC 0.808 1.397
99.5 ºC 0.808 1.397
Boiling point (1 atm) Density (gcm-3 at 20 ºC) Refractive index (20 ºC) 2.
Optical activity
Optical activity is the properties possessed by certain substances of rotating the plane of polarization of polarized light. a) Plane polarized light Light is an electromagnetic phenomenon. A beam of light consists of two mutually perpendicular oscillating fields: an oscillating electric field and an oscillating magnetic field.
An ordinary light beam consists of oscillations of the electric field occurring in all possible planes perpendicular to the direction of propagation. By contrast, a plane-polarized light has an electric field oscillating only in one plane. It can be produced by passing an ordinary light beam through a polarizer.
Because of the asymmetrical distribution of electron cloud in an enantiomer, an enantiomer is capable to interact with the electric field and alter the direction of oscillation of the plane-polarized light.
Isomerism
Unit 2
Page 3
b) Measurement of optical activity by polarimeter
If a substance rotates the plane-polarized light to the right hand side, it is said to be dextrorotatory. (Latin: 'dexter' = right). And the enantiomer is denoted by (+)- prefix. e.g. (+)-butan-2-ol Similarly, if the plane-polarized light is rotated to the left, the substance is said to be levorotatory, denoted by (-)prefix. (Latin: 'laevus' = left). e.g. (-)-butan-2-ol c)
Specific rotation (Details not required)
Since the degree of rotation is depending on a lot of factors including i) length of the polarimeter tube ii) concentration of the sample iii) wave length of the light source iv) temperature v) nature of solvent For the purpose of comparison, a specific rotation [α] is defined as [α] = where
α . c×l
[α] is the specific rotation α is the observed rotation c is the concentration of the solution in gcm-3 of solution or density in gcm-3 for pure liquid l is the length of the tube is dm 25
e.g. [α] D = +3.12º means the specific rotation is 3.12º in a clockwise direction at 25ºC using D line of a sodium lamp. (λ = 599.6 nm). In reporting a specific rotation, the solvent used should be quoted. N.B.
The direction of the optical rotation has no direct relationship with absolute configuration of the molecule. It can only be determined by the experiment. e.g. a molecule with (R) configuration may be dextrorotatory or levorotatory.
Isomerism
Unit 2 3.
Page 4
Chiral centre
The C-2 of butan-2-ol has four different groups attaching to it tetrahedrally. The C-2 is called the chiral centre or chiral carbon.
H CH3 C* CH2CH3 OH
By interchanging any two groups attached to the chiral centre, one enantiomer can be converted to another. A molecule contains one chiral centre only must be a chiral molecule. 4.
Racemic mixture / Racemic modification / Racemic form / Racemate
If a mixture contains equal amount (50%-50%) of (+)-isomer and (-)-isomer, the mixture will show no optical activity. Owing to the opposing effect of the two enantiomers, optically activity of one cancels that of another one. Such kind of mixture is called racemic mixture and denoted by (±)- prefix. e.g. (±)-butan-2-ol
Glossary
optical isomerism / enantiomerism plane polarized light polarizer chiral carbon racemic mixture
Past Paper Question
90 2C 8 a iv v 93 2C 9 c 96 2C 8 b ii 98 1A 4 a ii 99 2B 6 c i ii iii
chiral plane of symmetry polarimeter dextrorotatory
optical activity levorotatory
specific rotation
98 1B 8 a i
90 2C 8 a iv v 8a It is suggested that the structure of a compound having the molecular formula C12H11ClO4 is either A or B. COOH Cl
CH C
COOH CH3 CH C
CH3
CH2Cl
COOH A
iv
CH3
COOH B
Assuming the compound has the structure A, give the structural formula of the hydrogenation product from reaction with one mole of hydrogen. Would you expect the product to show optical activity? Explain your answer.
2
COOH Cl
CH2 CH
CH3 CH3
COOH
C v
not optically active, because there is no chiral center. Most candidates gave correct answers to this part. Repeat part (iv) assuming the compound has the structure B. COOH CH3 CH2 C H CH2Cl COOH
C
1 mark 1 mark
COOH CH2 C COOH
CH3 CH2Cl H
, 1 mark not optically active, because it is a racemic mixture. 2 marks Over 90 % of the candidates were not aware that the hydrogenation product was a racemic mixture and hence optically inactive.
3
Isomerism Unit 2 93 2C 9 c 9c 2-Bromobutane can exist in isomeric forms. Draw a suitable representation for each of these isomers. CH3 Br
H
Et
Et
C CH 3
C Br H
CH3
Br
Et H
2
CH3 Et
Br H
or C
Page 5
1 mark each
Candidates should pay more attention in providing a proper 3-dimensional structural formula.
96 2C 8 b ii 8b The following compounds can exist in isomeric forms : In each case, state the type of isomerism and draw suitable representations for the isomers. ii 2-aminopropanoic acid. enantiomerism / optical isomerism (Deduct ½ mark for spelling mistake) CO2H H
2 1 mark
CO2H
NH2
H2N
CH3
H CH3
½ + ½ mark (Accept any appropriate representation of enantiomerism) Some candidates showed weakness in three-dimensional representations. Enantiomerism instead of optical isomerism should have been given. Some wrongly gave stereoisomerism as the answer. Two incorrect representations given by candidates are shown below.
C
NH2 H C
NH2
CO2H
CH3
H
C
CO2H CH3
98 1A 4 a ii 4a Alcohol E has the structure CH3CH(OH)C2H5. ii What type of isomerism can be exhibited by E ? 98 1B 8 a i 8a Show how you would i determine whether a sample of C2H5CH(OH)CH3 is in the (+) form or (±) form. 99 2B 6 c i ii iii 6c State the relationship between each pair of structures shown below: i CH2CH3 CH2CH3 and Cl
Cl Cl
ii
CH3 H
iii
Cl H C C
and
CH3CH2
CH2CH3
Br CH2
C CH and 3 CH H CH2
C
H
C
CH3 H
H C CH 3 CH Br
Reaction Mechanism
Reaction Mechanism I.
Bond breaking A.
II.
Types of reactive species A. B. C.
III.
Bond breaking of a bond to carbon Free radical (electron deficiency) Electrophile (electron deficiency) Nucleophile (electron rich)
Classification of reaction A.
Types of reaction 1. Substitution 2. Addition 3. Elimination
Page 1
Reaction Mechanism
IV.
Page 2
Nucleophilic substitution A. B. C. D. E. F. G.
Nucleophilic substitution 1. Nucleophile 2. Leaving group SN2 reaction (1 step reaction) SN1 reaction (2 steps reaction) Competition between SN1 and SN2 reactions Alkanol from haloalkane (RX → ROH) 1. Alkaline hydrolysis of haloalkane 2. Hydrolysis of haloalkane Rate of hydrolysis of haloalkane, haloalkene and halobenzene Other relevant reactions 1. Reactions of haloalkane a) Nitrile from haloalkane (RX → RCN) b) Alkylation of ammonia and amine (NH3 → RNH2) c) Use of SN2 reaction in organic synthesis 2. Reactions of alkanol a) Haloalkane from alkanol (ROH → RX) (1)
3.
(1) (2)
b)
V.
Use of chlorinating and brominating reagent
b) Luca's test to distinguish 1º, 2º and 3º alkanol Reactions of amine a) Action of nitric(III) acid on 1º amine 1º aliphatic amine 1º aromatic amine
Laboratory preparation of phenol from benzenamine
Elimination reaction A.
B.
C. D.
Elimination reaction 1. Stability of elimination product 2. E2 reaction (not required in A-Level) 3. E1 reaction (not required in A-Level) Competition between substitution and elimination reaction 1. Effect of temperature 2. Effect of bulkiness of the substrate and base 3. Effect of bascity of the nucleophile Conditions favouring substitution and elimination reaction Other relevant reactions 1. Reaction of haloalkanes with alcoholic sodium hydroxide to alkene, diene and alkyne 2. Preparation of vinyl chloride 3. Dehydration of alkanol
Reaction Mechanism
VI.
Page 3
Nucleophilic addition (Nucleophilic addition-elimination) A.
AdN reaction 1. Addition of HCN to carbonyl compound 2. Rate of nucleophilic addition a) Electronic effect (1) (2)
3.
Inductive effect (a) Effect of protonation Resonance effect
b) Steric effect Other relevant reactions a) Reactions of carbonyl group (1) (2) (3) (4) (5)
b)
Reduction of carbonyl compound by LiAlH4 and NaBH4 Addition of NaHSO3 to carbonyl compound Condensation reaction with hydroxylamine Condensation reaction with 2,4-dinitrophenylhydrazine Haloform reaction / Iodoform reaction
Reactions of carboxylic acid and its derivatives (1)
(2)
Carboxylic acid and its derivatives (a) Difference between carbonyl compound and carboxylic acid and its derivatives (b) Reactivities of carboxylic acid and its derivatives Formation of different acid derivatives (a) Use of chlorinating agent to prepare acyl chloride (b) Formation of ester (Esterification with alkanol and phenol) (c) Formation of acid anhydride (i) (ii) (iii)
(3) (4)
c)
(d) Formation of amide (Acylation and benzoylation of amine) Reaction of ester (a) Hydrolysis of ester Reaction of amide (a) Reduction of amide and other acid derivatives (b) Hofmann degradation of amide
Reactions of nitrile (1) (2)
VII.
Through intramolecular dehydration by heating Through intermolecular dehydration by a very strong dehydrating agent From acyl chloride
Hydrolysis of nitrile and amide (a) Dehydration of amide Reduction of nitrile
Electrophilic addition A. B.
Addition of HBr to alkene 1. Markownikoff's rule. 2. Reactivity of alkene towards electrophilic addition Other relevant reaction 1. Addition of Br2 to alkene 2. Addition of H2SO4 to alkene a) Preparation of alkanol from alkene 3. Hydration of alkene 4. Ozonolysis of alkene 5. Preparation of ethane-1,2-diol a) Oxidative Cleavage of double bond b) Comparision of ozonolysis and oxidative cleavage 6. Oxymercuration of alkyne
Reaction Mechanism
VIII. Electrophilic substitution A. B.
IX.
Representation of arenium ion Other relevant reaction 1. Sulphonation of benzene a) Preparation of phenol 2. Nitration of benzene a) Reduction of nitrobenzene 3. Halogenation of benzene 4. Alkylation of benzene (Friedel-Crafts alkylation) 5. Diazocoupling of diazonium ion a) Colour of a substance 6. Bromination of phenol
Free radical Reaction A. B.
C.
Formation of free radical Free radical Substitution 1. Chain reaction e.g. chlorination of methane a) Chain initiation (chain initiating step) b) Chain propagation (chain propagating step) c) Chain termination (chain terminating step) 2. Reaction between H2(g) and Cl2(g) Free radical Addition 1. Chain reaction e.g. Polymerization of alkene 2. Anti-Markownikoff orientation
Page 4
Reaction Mechanism
Unit 1
Page 1
Topic
Reaction Mechanism
Unit 1
Reference Reading
16 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 40–41, 44–46 Organic Chemistry, Solomons, 5th Edition pg. 209–211 Organic Chemistry, Fillans, 3rd Edition pg. 82–94 Organic Chemistry, Morrison Boyd, 6th Edition pg. 165–167
Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 403–406, 440–441 Reading Assignment Organic Chemistry, 6th Edition, Solomons, pg. 87-90, 94-96
Syllabus
Introduction to reaction mechanism Types of bond breaking Types of reactive species Types of reactions
Notes
Reaction Mechanism A chemical reaction is a sequence of bond-breaking and bond-forming steps involving bonding and nonbonding electrons. A detailed description of a chemical reaction outlining each separate stage is called the Mechanism. Chemical reaction can be interpreted as redistribution of e- and rearrangement of atoms which leads to formation of new substance. I.
Bond breaking
All kinds of reactions are initiated by a bond breaking process. There are 2 kinds of bond breaking process : Homolytic fission (Homolysis) –
the electrons forming the bond broken are shared equally between the fragments formed. This leads to formation of 2 radicals. Heterolytic fission (Heterolysis) – the electrons forming the bond broken are shared unequally between the fragments formed. This leads to formation of 1 cation and 1 anion. Mode of bond breaking
Electronegativity
A
B
A + B
Homolytic fission (Symmetrical fission)
A≈B
A
B
A- + B+
Heterolytic fission (Unsymmetrical fission)
A>B
A
B
A+ + B-
Heterolytic fission (Unsymmetrical fission)
A secondary > primary. R R C O H R
ZnCl2
R R C O + ZnCl2 R H
r.d.s.
R R C+ R
Cl-
R R C
Cl
R
Since the chloroalkane formed is immiscible with water and alkanol, it will form an emulsion with water and alkanol. Depending on the rate of reaction, 3º alkanol will turn cloudy immediately, 2º alkanol will turn cloudy gradually while 1º alkanol will not turn cloudy at all. Because 3º alkanol is the most reactive one, it is believed that this is a SN1 reaction.
Glossary
Lewis acid
chlorinating / brominating agent
Luca's test
Reaction Mechanism
Past Paper Question
Unit 5
96 2C 7 b i ii iii 99 1B 8 b i ii
Page 3
96 2C 8 c i 99 2B 7 a i ii
96 2C 7 b i ii iii 7b In an experiment 25 g of (CH3)3COH react with 36 g of HCl to give 28 g of (CH3)3CCl. i Find the limiting reactant of the reaction showing clearly your calculation. Formula mass of (CH3)3COH = 74.12 Formula mass of HCl = 36.458 Formula mass of (CH3)3CCl = 92.562 25 No. of moles of (CH3)3COH = 74.12 = 0.337 ½ mark 36 No. of moles of HCl = 36.458 = 0.987 ½ mark HCl is in excess. ∴ (CH3)3COH is the limiting reactant. ½ mark C Well answered. Some candidates had no idea of a limiting reactant. Candidates calculated the relative molecular masses carelessly. Some used 35.0 or 36.0 for the relative atomic mass of chlorine without referring to the Periodic Table given in the examination paper. ii Calculate the percentage yield of (CH3)3CCl. 28 No. of moles (CH3)3CCl = 92.562 = 0.302 ½ mark 0.302 % yield = 0.337 × 100 = 89.6% (90%) 1 mark C Many did not know how to calculate the percentage yield and they used the mass ratio instead of the mole ratio in the calculation. The mechanism of reaction was poorly presented. iii Name the type of the reaction and outline the mechanism of the reaction. (Movement of electron pairs should be indicated by curly arrows.) Unimolecular nucleophilic substitution / SN1 ½ mark Mechanism 2½ marks + O
OH H Cl
H H
C
+
Cl
Cl-
Many candidates did not distinguish between SN1 and SN2 reactions. Many candidates did not know where the arrow should begin and where it should end. Many started the mechanism with protonation of the alcohol by free H+ instead of by HCl. The first step is shown below: (CH3)3COH
H Cl
Candidates should appreciate that H+ does not exist freely in solutions and it is always attached to a conjugate base which can be a solvent molecule.
1½
1½
3
Reaction Mechanism
Unit 5
Page 4
96 2C 8 c i 8c Identify K, L, M, N and P in the following reactions : (Deduct ½ mark for each minor mistake; max. Deduction for the whole question = 2 marks) i K
K: C
PCl5
1
(CH3)2CHCH2CH2Cl
OH or (CH3)2CHCH2CH2OH
1 mark
Common mistakes made by candidates are listed below: (CH3)2CHCH2CH3 for K;
99 1B 8 b i ii 8b In an experiment to prepare 1-bromobutane, a mixture of butan-1-ol, potassium bromide and concentrated sulphuric(VI) acid was heated under reflux for 30 minutes. i Draw a labelled diagram of the set-up used. ii Suggest how to isolate 1-bromobutane from the reaction mixture. 99 2B 7 a i ii 7a Lucas reagent, a mixture of ZnCl2 and concentrated HCl, can be used to distinguish the following alcohols from one another : (CH3)2CHCH2OH, CH3CH2CH(OH)CH3 and (CH3)3COH i State the expected observation when these alcohols are separately treated with Lucas reagent. ii Suggest why these alcohols behave differently towards Lucas reagent. (Hint: the zinc ion binds strongly with the oxygen atom of an alcohol, weakening the C–O bond and creating a better leaving group.)
Reaction Mechanism
Unit 6
Page 1
Topic
Reaction Mechanism
Unit 6
Reference Reading
17.3 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 49–53, 158–159 Organic Chemistry, Solomons, 5th Edition pg. 212–236, 240–243 Organic Chemistry, Fillans, 3rd Edition pg. 174–183 Organic Chemistry, Morrison Boyd, 6th Edition pg. 172–203, 208–210 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 521–523 Organic Chemistry, 6th Edition, Solomons, pg. 923-924, 926-927, 966
Syllabus
Nucleophilic substitution of amine
Notes
3.
Reactions of amine
NH2- is a very strong base, thus it is a very poor leaving group. Normally, 1º amine R–NH2 does not undergo nucleophilic substitution reaction. However, upon the action of nitric(III) acid, amino group –NH2 group will be converted to diazonium group –N+≡N. Diazonium group is a very good leaving group and leaves the substrate in form of very stable N2. a) Action of nitric(III) acid on 1º amine (1) 1º aliphatic amine Reagent : nitric(III) acid / nitrous acid (HNO2) or (H–O–N=O) or (HONO) H R N H aliphatic primary amine
+ HNO2
+ R N N
+ 2 H2O
alkyldiazonium ion
R+ carbocation
+
N2 nitrogen bubbles
Since HNO2(aq) is unstable, it is usually freshly prepared by dissolving NaNO2(s) in HCl(aq). Nevertheless, the HNO2(aq) prepared must be kept below 5ºC otherwise it will decompose into NO2(g) readily. NaNO2(s) + HCl(aq) → NaCl(aq) + HNO2(aq) The diazonium salts (diazonium chloride if HCl(aq) is used in the preparation of HNO2(aq)) intermediate formed is also highly unstable. Even at a temperature below 5ºC, the diazonium salt will still decompose gradually with evolution of nitrogen bubbles. This observation can be used to identify 1º aliphatic amine. The fate of the carbocation R+ depends on whether it will undergo elimination or substitution reaction. A lot of products may be obtained. e.g. alkene, haloalkane, alcohol and ether. e.g. R+ + Cl- → R–Cl R+ + H2O → ROH + H+ R+ + ROH → ROR + H+
Reaction Mechanism
Unit 6
Page 2
(2) 1º aromatic amine If a 1º aromatic amine is treated with NaNO2(s) in HCl(aq) below 5ºC, no bubble will evolve. Comparing with alkyldiazonium ion, benzenediazonium ion is more stable. This is because benzenediazonium ion is stabilized by resonance and the C–N bond possess certain double bond character. N
N-
N-
+N
N+
N+
H
H
H
+
N N H
H
+
Benzenediazonium ion can also be converted to a lot of other products. Since N2 is an extremely good leaving group, benzenediazonium ion is capable to form aryl cation by the departure of N2. Nu + N N
NH2 NaNO2
+
H + N2
HCl, 0-5 °C
H diazonium ion
benzenamine
H
Note : arenium ion will be studied in the chapter of electrophilic substitution. H E H
H
H
+
H
H
aryl cation (not arenium ion)
H H arenium ion
E : a positive electrophile attaching to the benzene ring
By using different nucleophile, different products can be obtained through a SN1 reaction.
b) Laboratory preparation of phenol from benzenamine The formation of aryl cation provide a very useful way to prepare phenol in laboratory. Upon heating, the solution of benzenediazonium ion will be converted to phenol. H O H + N N
NH2 NaNO2 HCl, 0-5 °C
H
+
+ O H
OH
H + N2
H benzenamine
H
diazonium ion
H H
aryl cation (not arenium ion)
In this reaction, water plays the roles of solvent and nucleophile. The water in the acid mixture behaves as an nucleophile and reacts with the aryl cation intermediate to form a phenol.
Glossary
nitric(III) acid / nitrous acid
alkyldiazonium ion
benzenediazonium
aryl cation
Reaction Mechanism
Unit 6
Page 3
92 1A 1 d 94 1A 3 a iii iv 95 1A 3 d i iii
Past Paper Question
92 1A 1 d 1d Give the reagents and reaction conditions required to convert phenylamine into benzene.
2
, HCl, 0° C NaNO → Stand in H 3 PO 2 → NaNO2 ½ mark HCl ½ mark or HONO 1 mark 0ºC ½ mark H3PO2 ½ mark or EtOH ½ mark Few (10-15 %) knew of H3PO2 reagent. 2
C
94 1A 3 a iii iv 3a Consider the two compounds: CH3CH2CHCH3
CH3CH2CH2NH2 and
OH C
D
iii Write equation(s) to show the reaction(s) (if any) of nitric(III) acid with C and D separately. ½ mark
D → CH3CH2CH2OH + N2
1½ mark
HNO2
or
iv
2
→ no reaction C HNO2
D
HNO2
H
+ CH3CH2CH2 N N
N2 + CH3CH2 C + H
unstable
CH3CH CH2 CH3CH2CH2Cl
etc.
CH3CH2CH2OH
If C is ignored, no mark for the first part. If “only D reacts” award ½ mark for C. If for D, “CH3CH2CH2N+≡N Cl-”, award ½ mark only The reaction(s) in (iii) can be used to distinguish between C and D. What observation(s) would you look for in this test? In the case of D, colourless gas bubbles wil be seen or N2 gas evolved with D 1 mark
95 1A 3 d i iii 3d i Give the reactants and conditions for the preparation of benzenediazonium chloride in the laboratory. Reactant : NaNO2 / HCl or HNO2; aniline 1 mark Condition : < 5 ºC or -10 to 5 ºC or ice bath ½ mark iii If an aqueous solution of benzenediazonium chloride is heated, a solid, which is soluble in dilute NaOH, can be obtained. Suggest a structure for the solid obtained and account for its solubility in dilute NaOH. OH
(no mark for name) 1 mark Phenol is weakly acidic ½ mark The following equilibrium lies to the right in the presence of NaOH and the ionic phenoxide is more soluble ½ mark OH OH- +
C
OH2O +
½ mark Most candidates correctly produced the structure of phenol. But many candidates did not know that the solubility of phenol in NaOH is due to the formation of a water soluble sodium phenoxide.
1
1½
2½
Reaction Mechanism
Unit 7
Page 1
Topic
Reaction Mechanism
Unit 7
Reference Reading
18.0 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 101–102, 108–109, 116, 159–160, 175–176 Organic Chemistry, Solomons, 5th Edition pg. 244–251, 317–319, 322–327, 332–333 Organic Chemistry, Fillans, 3rd Edition pg. 185–188, 206–210, 429–430 Organic Chemistry, Morrison Boyd, 6th Edition pg. 290–294, 300–304, 306–315 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 466–477, 486–491 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 462 Organic Chemistry, 6th Edition, Solomons, pg. 260-262, 265-268, 300, 302-305
Syllabus
Elimination Reaction Competition between Substitution and Elimination reaction
Note
V. Elimination reaction A. Elimination reaction In nucleophilic substitution, the nucleophile attacks the α carbon atom and leads to a substitution product. Br H CH3 C C
C2H5OH
H
55°C
H H
H CH3 C O CH2CH3 CH3 97%
But besides this substitution product, another elimination product is also obtained Br H CH3 C C
C2H5OH
H
55°C
H H
H CH3 C C
H
H 3%
This experimental evidence shows that the reaction takes place through 2 competing pathways : substitution vs elimination Substitution Br H α β CH3 C C H H H
H
H H + CH3 C O CH2CH3 + Br-
CH3 C O
CH3CH2 O
CH2CH3 + HBr
CH3
CH3
ether
H
In the substitution pathway, the ethanol molecule acts as a nucleophile. It displaces the Br atom from the haloalkane molecule. Elimination Br H CH3 C C
H
H H
H CH3 C C H
CH3CH2 O H
H H+
Br-
+ CH3CH2 O H +
H CH3 C C
H + HBr + CH3CH2OH
H alkene
In elimination pathway, the ethanol molecule has another role. It acts as a base and subtracts the β H atom from the haloalkane.
Reaction Mechanism
Unit 7
Page 2
The role of the base is also supported by another evidence. By using a stronger base, the proportions of the two products change dramatically. Br H CH3 C C H H
H
H
C2H5OH C2H5O-Na+, 55°C
H
CH3 C O CH2CH3 + CH3 C C CH3 21%
-
H
H 79%
+
Sodium ethoxide, C2H5O Na , is a very strong base which can be prepared by dissolved sodium metal in excess ethanol. The ethanol also serves as a solvent. 2 C2H5OH(l) + 2 Na(s) → 2 C2H5O-Na+(alc.) + H2(g) Alternatively, solid sodium hydroxide or potassium hydroxide dissolved in ethanol (alcoholic sodium hydroxide) may also be used. This involves an equilibrium of formation of ethoxide ion, the strong base. C2H5OH(l) + NaOH(s) d C2H5O-Na+(alc.) + H2O(l) 1.
Stability of elimination product
If there are two different kind of β hydrogen available for subtraction, the one which will lead to a more substituted alkene will be preferred. This is because double bond (sp2 hybridized C) and triple bond (sp hybridized C) are electron withdrawing by inductive effect, they will be stabilized by electron releasing alkyl group. H H Br H γ β α β H C C C C H
H C C
H H
H H H H
H
H C
H H
major product (di-substituted) CH3CH2O-
H H
C H + H C C
H C
C H
H H H minor product (mono-substituted)
Reaction Mechanism
Unit 7
Page 3
Similar to the SN2 and SN1 mechanism, there are also E2 and E1 reactions, depending on the number of molecule involved in the formation of the transition state. 2.
E2 reaction (not required in A-Level)
E2 stands for bimolecular elimination reaction. In E2 reaction, the rate of reaction depends on both the concentration of the substrate and the base. Comparing with SN2 reaction, E2 involves more bond breaking and bond formation, thus a higher activation energy. i.e. SN2 reaction only involves 2 bond while E2 reaction involves 4 bonds. Rate ∝ [substrate][base] -
B:
H r.d.s. C
H C
C
C
C
C
-
+ HB + L
Lδ transition state of E2 reaction -
L
3.
B δ
E1 reaction (not required in A-Level)
E1 stands for unimolecular elimination reaction. In E1 reaction, the rate of reaction depends on the concentration of the substrate only. Rate ∝ [substrate]
C
-
B:
H
H C L
r.d.s.
C
C
δ+
L δ− transition state of E1 reaction
H C
C +
+ L-
C
C
-
+ HB + L
intermediate
E1 and SN1 reaction share the same intermediate – carbocation. In SN1 reaction, the nucleophile is joined to the carbocation to yield a substitution product. In E1 reaction, the nucleophile acts as a base rather. It removes the β-H and leads to an elimination product. E1 reaction is favoured by a strong and bulky base. e.g. (CH3)3CO-Na+. Also if the α C is sterically hindered by bulky groups, e.g. 3º substrate, the E1 reaction will also be favoured over SN1 reaction.
Reaction Mechanism
Unit 7
Page 4
B. Competition between substitution and elimination reaction 1. Effect of temperature Elimination reaction has a higher activation energy than Substitution reaction. This is because more bond breaking and bond formation are involved in elimination than in substitution. As a result, an increase in temperature will cause a larger increase in rate of elimination than in the rate of substitution. 45蚓 Br CH3 CH CH3
53% NaOH / CH3CH2OH
CH2 CH CH3 + 100蚓
47% CH3 CH O CH2CH3
64%
CH3
36%
Elimination reaction is favoured by a higher temperature. 2. Effect of bulkiness of the substrate and base If the α carbon is sterically hindered, e.g. in 3º haloalkane and use of bulky base (CH3)3CO-Na+, elimination will be more favourable than the substitution reaction. 3. Effect of basicity of the nucleophile In general, all nucleopiles are also bases. If the basicity of the nucleophile is very high (e.g. CH3CH2O-Na+), elimination will compete with nucleophilic substitution. There are only very few strong nucleophile but with low basicity, e.g. I-, CN- and CH3COO-Na+. They only undergoes substitution reaction. C. Conditions favouring substitution and elimination reaction
Temperature Structure of the substrate Bulkiness of the nucleophile Basicity of nucleophile
Glossary Past Paper Question
elimination
Substitution Reaction Low (substitution has a lower activation energy) Non-hindered α carbon (e.g. 1º haloalkane) Low
Elimination Reaction High (elimination has a higher activation energy) Sterically hindered α carbon (e.g. 3º haloalkane) High
Low
High
steric hinderance
alcoholic sodium hydroxide
Reaction Mechanism
Unit 8
Page 1
Topic
Reaction Mechanism
Reference Reading
18.1–18.2 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 101–102, 108–109, 116, 159–160, 175–176 Organic Chemistry, Solomons, 5th Edition pg. 244–251, 317–319, 322–327, 332–333 Organic Chemistry, Fillans, 3rd Edition pg. 185–188, 206–210, 429–430 Organic Chemistry, Morrison Boyd, 6th Edition pg. 290–294, 300–304, 306–315 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 466–477, 486–491 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 473–478 Organic Chemistry, 6th Edition, Solomons, pg. 308, 312, 318-319, 443-444
Syllabus
Formation of alkene, diene and alkyne from haloalkane Preparation of vinyl chloride from dichloroethane Dehydration of alkanol
Note
D. Other relevant reactions 1.
Unit 8
Reaction of haloalkanes with alcoholic sodium hydroxide (sodium ethoxide) to alkene, diene and alkyne C2H5O
-
CH3
H H β β C C CH3 H Br
H CH3
C
C CH3
H more substituted alkene (major product)
By heating with a strong base (e.g. C2H5O-Na+), a haloalkane can be dehydrohalogenated to an alkene. H2N CH3
-
H C
C CH3
CH3
C
C CH3
Br C2H5O
-
alkene (very unreactive)
H H CH3
C C CH3 Br Br
alkyne
X H2N
-
H H C C
H
H C CH3
H Br alkene (very unreactive)
H C C C
CH3
H allene (highly unstable)
For a 1,2-disubstituted haloalkane, use of C2H5O-Na+ will only give a very unreactive haloalkene. C2H5O-Na+ is not strong enough to subtract the second hydrogen atom from the substrate. An even stronger base Na+NH2- (or boiling C2H5O-Na+) is required to cause further elimination of the molecule. Or, the 1,2-disubstituted haloalkane can be converted into alkyne by Na+NH2- (or boiling C2H5O-Na+) in one step. Further elimination of 1,2-disubstituted haloalkane will only give alkyne but not allene (1,2-diene). This is because allene is highly unstable where the two π bonds are perpendicular to each other without any overlapping. Orbital representation of allene
Reaction Mechanism
Unit 8
Page 2
Formation of alkyne versus formation of conjugated diene Normally, a 1,2-disubstituted alkane will undergo elimination to form an alkyne in the presence of a very strong base, e.g. Na+NH2- (or boiling C2H5O-Na+). But if the haloalkene does not possess β hydrogen, a conjugated diene would be produced instead. H H C H
H H C H
CH3 H
H
C
C
C
Br
Br
H
CH3 H
H
C
C
C
Br
Br
H
NaNH2
H
H H C H
H
CH3 C β
C
C
β
NaNH2
X
H
Br H haloalkene (no further elimination)
NaNH2
H
H C
CH3 H
H
C
C
C
Br
H
H
formation of alkyne impossible
NaNH2
H
H H C
CH3 C
C
H
C
H
H
formation of unstable allene is not favourable
CH3 H H C H
C
C
C
H
H
formation of more stable conjugated diene
Conjugated 1,3-diene is relatively stable comparing with allene (1,2-diene). It has an extensive delocalization of π electrons.
Orbital representation of conjugated 1,3-diene
2.
Preparation of vinyl chloride The major use of vinyl chloride (chloroethene) is to prepare the plastic poly(vinyl chloride), PVC, by polymerization. It can be prepared by chlorination of ethene followed by elimination in the laboratory. CH3CH2O H
H C C
H
H
Cl2 Electrophilic addition
-
H H H C C H Cl Cl
alcoholic NaOH
H
H C
Cl
C H
The mechanism of electrophilic addition of Cl2 to ethene will be discussed in the section of electrophilic addition.
Reaction Mechanism 3.
Unit 8
Page 3
Dehydration of alkanol H H
H C
H
C H
H C
+O H
H O
H C
C H
H
+ H2O
H +O H H
H
H
H H
H H
C H +
H H H C
C H +
H +O H H
H2O
Unlike other elimination reactions, dehydration of alkanol is done by heating with acid instead of base. In a basic medium, the leaving group of the alkanol would be OH- which is a very strong base and poor leaving group. This makes dehydration unfavorable. In acidic medium, the hydroxyl group –OH is protonated and transformed into a better leaving group water, H2O. Since there is no strong base present in the reaction medium, the reaction is mainly an E1 reaction. Therefore, the reactivity of alkanol toward dehydration has the following order : 3º alkanol > 2º alkanol > 1º alkanol, parallel to all other reactions with carbocation intermediate. Use of dehydrating concentrated sulphuric acid also helps to remove the water and shift the equilibrium to the product side. For the more reactive 3º alkanol and 2º alkanol, non-dehydrating and non-oxidizing concentrated phosphoric acid may be used to prevent charring of the substrate. Besides ethene (elimination product), ethoxyethane (substitution product) is also obtained. By varying the temperature, the two products can be selected. 180 °C (E1) CH3CH2 OH
conc. H2SO4(l) 140 °C (SN1)
CH2 CH2 CH3CH2 O CH2CH3
In the SN1 pathway, CH3CH2OH serves as the nucleophile and attacks the carbocation intermediate to form the ether (ethoxyethane). H H H C
H C
H H
H H
H H C O H
+C
H H O+
C
H
H H
H H H C
C O H
H H
Glossary
diene
allene
C H CH3CH2 O CH2CH3
C H
H H
H H
C
vinyl chloride
electrophilic addition
CH3CH2 O+ H H
Reaction Mechanism
Unit 8
Page 4
91 2C 7 b i 92 2C 9 a i 94 2C 9 b iii 98 1A 4 c ii 99 2B 5 b i ii
Past Paper Question
91 2C 7 b i 7b When compound P is treated with potassium hydroxide in ethanol, two products Q, C6H12, and R, C8H18O, are obtained. CH3 CH3
C CH2CH2Cl CH3 P
i
Give the structures and the systematic names for Q and R. CH3 CH3
CH3
C CH2 CH2 Cl
CH3
CH3
C CH CH2 CH3
P
4
CH3 +
CH3
C CH2 CH2 O CH2 CH3 CH3
Q
R
1 mark each Q 3,3-dimethylbut-1-ene 1 mark R 1-ethoxy-3,3-dimethylbutane 1 mark Most candidates could provide the correct structures for Q and R. However, they were generally weak in nomenclature, especially for R.
C
92 2C 9 a i 9a Give the structural formula(e) of the major organic product(s) from each of the following reactions : Br i
2
CH3CH2CHCHCH3 KOH (fused), 200°C Br
CH3CH2C CCH3 or CH3 CH CH CH CH2
2 marks
94 2C 9 b iii 9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC) respectively. PVC is more rigid and durable than PE, but incineration of PVC causes a more serious pollution problem. iii Show how ethene can be converted to chloroethene. CH2 CH2
Cl2
CH2Cl CH2Cl
CH2Cl CH2Cl
720 - 920 K
- HCl NaOH / EtOH
CH2 CHCl
CH2 CHCl
Or Some candidates wrongly gave the structure of chloroethene as CH3CH2Cl.
C
2 marks
98 1A 4 c ii Alcohol E has the structure CH3CH(OH)C2H5. 4c On treatment with dilute H2SO4(aq), E gives mainly two isomeric compounds, F and G, both of which have the formula C4H8. On treatment with bromine, both F and G give a product H with formula C4H8Br2. ii What is the isomeric relationship between F and G ? 99 2B 5 b i ii 5b The following reaction produces a mixture of K and L. N CH3 Br
i ii
NaOH(aq) heat
C12H17NO + C12H15N K
2
L
Give the structures of K and L. Name the types of reactions leading to their formation. Outline a mechanism for the formation of K.
1
Reaction Mechanism
Unit 9
Page 1
Topic
Reaction Mechanism
Unit 9
Reference Reading
19.0 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg.
Syllabus
Nucleophilic addition
Notes
VI. Nucleophilic addition (Nucleophilic addition-elimination)
220–221, 228–231, 234–237, 239–244, 251, 254–258, 259–263, 265–267 Organic Chemistry, Solomons, 5th Edition pg. 680–682, 691–693, 701–707, 724–726, 790–793, 845–846 Organic Chemistry, Fillans, 3rd Edition pg. 225–226, 229–238, 240–243, 247–248, 251–259, 262, 275 Organic Chemistry, Morrison Boyd, 6th Edition pg. 669–671, 677–680, 683–684, 697–698, 724, 737–740, 755–778, 797–799, 802–807,830–831 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 482, 484–485 Organic Chemistry, 6th Edition, Solomons, pg. 719
O
Besides substitution reaction with haloalkane, a nucleophile can also react with carbonyl group
C
and
C N through addition pathway. This type of reaction is called nucleophilic addition cyano group reaction, AdN.
In carbonyl group, O is more electronegative than C, therefore the C=O bond is polarized by inductive effect through the σ bond. The C=O bond is further polarized by mesomeric / resonance effect through the π bond. The carbonyl group can be represented as O
O
C
C +
or
δO C δ+
Similar to the carbonyl group, in a cyano group, the C≡N bond is polarized by inductive effect and mesomeric effect / resonance effect. A nitrile group can be represented as C N
+ C N
or
δ+ δC N
A. AdN reaction As a result, the C atom will carry a little positive charge and this makes the atom more vulnerable to the attack of a nucleophile.
Nucleophilic substitution
Nucleophilic addition δ-
O
C δ+
OC Nu+
Nu
δ+ δC N
Nu +
C δ+
L δ-
C Nu + L
C NNu+
Nu
Comparing with nucleophilic substitution, O and N are joined to C by a multiple bond in carbonyl and cyano group. Shifting of a bond pair will not cause departure of them.
Reaction Mechanism 1.
Unit 9
Page 2
Addition of HCN to carbonyl compound (Cyanohydrin formation) -
H C N
δ O
C + δ N C
-
CN consumed
O C
-
O H -
CN regenerated
-
C
+ N C-
C N cyanohydrin
C N
Since HCN molecule is not a very good nucleophile, the rate of direct addition of HCN will be rather slow. Alternatively, a mixture of cyanide salt and aqueous acid may be used e.g. NaCN(aq) and H2SO4(aq). The acid is added into the reaction mixture drop by drop to prevent formation of excess HCN(g) which is highly toxic and volatile. The experiment should be performed in the fume cupboard because of the high toxicity of the HCN(g). O H C H
+ NaCN
H2SO 4
O H H C H C N
-
As CN ion is regenerated in the reaction, it acts as a catalyst speeding up the reaction. This reaction has vital importance in organic synthesis. The product contains 1 more C atom than the substrate. It provides a useful way to increase the length of a carbon chain. The cyano group (–C≡N) can then be converted to other functional groups, e.g. carboxyl group, amide group, amino group, easily.
2.
Rate of nucleophilic addition The cyanohydrin formation is found to be first order with respect to the carbonyl compound and first order with respect to the cyanide. O
Rate = k [
C
][CN-]
It can be shown that the proton is not involved in the rate determining step of the reaction. Besides the concentration of the reactants, the susceptibility (reactivity) of the carbonyl carbon to the attack of nucleophile also has key effect on the reaction rate. The susceptibility depends on 2 factors 1. Electronic effect – inductive effect and resonance effect 2. Steric effect
Reaction Mechanism
Unit 9
Page 3
a) Electronic effect Attachment of different substituent affects the positiveness of the carbonyl carbon and easiness of the bond formation between the carbonyl carbon and the nucleophile.
(1) Inductive effect When an electron donating group is attached to the carbonyl carbon, the carbon is made less positive and will be less susceptible to the attack of the nucleophile. Therefore, propanone is less reactive than methanal. δO
δO
C H δ+ H
C H3C δ+ CH3
methanal
propanone
(a) Effect of protonation For most of the nucleophilic addition reactions, the reaction is catalyzed by acid. In acidic medium, the oxygen atom of carbonyl group is protonated. This makes the carbonyl carbon more positive and more reactive towards nucleophilic addition. Moreover, if the nucleophile is negatively charged, the product will be a neutral one instead negatively charged. O
H+
C
H
H
O+
O
C
C +
(2) Resonance effect 4-methoxybenzenaldehyde is found to be less reactive than benzenaldehyde. This is because methoxy group is an electron donating group which donates the electron to the carbonyl group. This makes the carbonyl carbon less positive and less susceptible to nucleophilic addition. O
O
C
CH3
H
benzenecarbaldehyde (more reactive)
O
C H
CH3
+ O
C
OH
4-methoxybenzenecarbaldehyde (less reactive)
(b) Steric effect The first step of the reaction is the close approach of nucleophile to the carbonyl carbon. If the carbon is sterically hindered by bulky groups surrounded, the reaction rate will be slowed down. Comparing the rate of nucleophilic addition of propanone with that of methanal : Methanal ⇒ less hindered carbonyl carbon ⇒ faster reaction rate
O C H
H
Propanone ⇒ more hindered carbonyl carbon ⇒ slower reaction rate
O C CH3
H3C
Nu
Nu
Besides the electronic effect, the steric effect also favours the reaction of an alkanal over that of a ketone.
Glossary
nucleophilic addition (AdN) carbonyl group nitrile group vulnerable cyanohydrin susceptibility electronic effect inductive effect resonance effect steric effect protonation
Reaction Mechanism Past Paper 92 2C 9 c i ii 93 2C 8 b ii Question 96 1A 3 f ii 98 1A 5 b 99 1A 5 b ii
Unit 9
Page 4
92 2C 9 c i ii 9c Hydrogen cyanide reacts with butanone to give an addition product. i Write a mechanism for the reaction. O-
O
H+
CN
OH or
CN
O
H+
CN
2 marks The addition product formed can exist in isomeric forms. State the type of isomerism, and draw suitable representations of the isomers. enantiomerism, (optical isomerism or stereoisomerism) 1 mark OH NC
C
2
OH
-CN
-CN
ii
+ OH
2
OH
CH2CH3 CH3
NC
CH3 CH2CH3
1 mark
Candidates were still weak in providing a proper 3-D structural formula, e.g. CN C CH3
OH C2H5
(no heavy and dotted lines)
93 2C 8 b ii 8b Outline a mechanism for each of the following reactions, using curly arrows to show electron pair displacement. In each case state what type of reaction takes place, and give the expected product. ii butanone with hydrogen cyanide. Mechanism: AdN Nucleophilic addition ½ mark Me C Et
C
Me
Me O
C
H+ NC-
Et
O +
H
NC
C Et
O
H
2 marks Wrong direction of curly arrows; ambiguity in indicating the starting and finishing location of the curly arrows. Wrong spelling for the terms, "nucleophile". Mixing up the meaning of terms, e.g. nucleophile and electrophile, addition and condensation.
2½
Reaction Mechanism Unit 9 96 1A 3 f i ii 3f i Give the structures of compounds D and E in the following organic synthesis :
Page 5 2
O C
HCN
H
+
D
H3O heat
E
OH D:
ii
OH E:
CN
COOH
1 + 1 marks (Deduct ½ mark for each minor mistake in the structures.) Outline a reaction mechanism for the formation of D. (Movement of electron pairs should be indicated by curly arrows.) ½
½
O
H CN
½
+ H O
OH CN
H
H ½
-
2
CN
4 × ½ marks
Or ½
½ O-
O H -
½
C
CN
CN
H+ ½ H
OH CN
4 × ½ marks The reaction mechanism was poorly presented. Many candidates wrongly used curly arrows to represent the direction of attack of reactants instead of the movement of electron pairs, e.g. O
-
+
H
C H CN
98 1A 5 b 5 Consider the five reactions of butanone (C4H8O) J shown in the reaction scheme below:
5b
State the type of reaction involved in the formation of L and outline the mechanism of the reaction.
3
Reaction Mechanism 99 1A 5 b ii 5b Consider the following reactions : CHO
ii
HCN
D
Unit 9
(1) LiAlH4 (2) H3O
+
E
Name the type of reaction involved in the formation of D and outline the mechanism of the reaction.
Page 6
Reaction Mechanism
Unit 10
Page 1
Topic
Reaction Mechanism
Unit 10
Reference Reading
19.1 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg.
Syllabus
Nucleophilic addition of carbonyl compound Reduction of carbonyl compound Addition of NaHSO3, NH2OH and 2,4-DNPH to carbonyl compound Iodoform reaction
Notes
3.
220–221, 228–231, 234–237, 239–244, 251, 254–258, 259–263, 265–267 Organic Chemistry, Solomons, 5th Edition pg. 680–682, 691–693, 701–707, 724–726, 790–793, 845–846 Organic Chemistry, Fillans, 3rd Edition pg. 225–226, 229–238, 240–243, 247–248, 251–259, 262, 275 Organic Chemistry, Morrison Boyd, 6th Edition pg. 669–671, 677–680, 683–684, 697–698, 724, 737–740, 755–778, 797–799, 802–807,830–831 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 485–487, 489 505 Organic Chemistry, 6th Edition, Solomons, pg. 470-472, 716-719, 729-734, 759-761
Other relevant reactions
a) Reactions of carbonyl group (1) Reduction of carbonyl compound by LiAlH4 and NaBH4 LiAlH4 is called lithium tetrahydridoaluminate or lithium aluminium hydride. LiAlH4 is a very powerful reducing agent. It provides hydride ion, H-, for the reaction to take place. Since H- ion is a stronger base than OH- ion, a solvent other than water must be used, otherwise H- ion will react with water and give H2. Ether (ethoxyethane or called diethyl either) is the most common solvent for this reaction. N.B.
H +
-
Li H Al H H
Since the reaction is very vigorous and potentially explosive, it is not advisable to conduct such an experiment in the laboratory of a secondary school.
H- is also a very powerful nucleophile. It reacts with and reduces carbonyl compound readily. O
Hfrom LiAlH4
OH
O-
C
C
carbonyl compound
H alkoxide ion
H
C
H O H + from aqueous acid
H
+ H2O
alkanol
An aqueous acid serves as a work up to remove all the unreacted LiAlH4 and converts the alkoxide ion to an alkanol. NaBH4 sodium tetrahydridoborate or sodium borohydride is similar to LiAlH4. NaBH4 is a weaker reducing agent which does not reacts with water. Therefore, it has the advantage that it can be used with aqueous solvent. O H H
H alkoxide ion
carbonyl + Na H B H compound
N.B.
-
O C
C
OH H
C
H O H + from aqueous acid
H
+ H2O
alkanol
Since hydride ion H- is a nucleophile, it does not react with electron rich C=C or C≡C. Furthermore H- is a polar reagent while C=C and C≡C are non-polar substrates.
Reaction Mechanism
Unit 10
Page 2
(2) Addition of NaHSO3 to carbonyl compound
O
O
O
O
C
C
S
O
HO
S
-
O O
O H
CH3
S
O
O hydrogensulphite addition product
O
O C
H
C
CH3 +
Na+HSO3-
H O C
CH3
CH3
SO3-Na+ white precipitate
In this reaction, sodium hydrogensulphite acts as a nucleophilic reagent. By the reason of the bulkiness of hydrogensulphite ion, addition of NaHSO3 is very sensitive to steric hinderance. A non-hindered aldehyde have about 90% yield and a slight more bulky methyl ketone have only 10 – 60 % yield. Other higher ketones do not form any addition product at all. aldehyde
2,2,4,4-tetramethylpropan-3-one
O H
O R C
methyl ketone
H
R C
C H H
CH3
CH3 O
CH3
C
C CH3
CH3
C
CH3
increase in steric hinderance about the carbonyl carbon
Since the addition product is an ionic crystalline compound, this reaction can be used to distinguish aldehyde and methyl ketone from other higher ketones. The latter do not give any precipitate at all. Furthermore, this reaction is reversible. The equilibrium position can be shifted to left by addition of either an acid or a base to remove the hydrogensulphite ion. Addition of acid HSO3- + H+ → SO2 + H2O Addition of base HSO3- + OH- → SO32- + H2O Therefore, this reaction can be used to separate an aldehyde or a methyl ketone from a mixture. The aldehyde or methyl ketone is first precipitated by sodium hydrogensulphite and filtered. It is regenerated upon addition of an acid or a base. N.B.
Although methanal and ethanal forms addition product with sodium hydrogensulphate(IV), the crystalline compounds formed are very soluble in water, therefore no ppt. will be obtained.
Reaction Mechanism
Unit 10
Page 3
(3) Condensation reaction with hydroxylamine Hydroxylamine is a derivative of ammonia. Other amines are alkyl derivatives of ammonia but hydroxylamine is a hydroxyl derivative of ammonia, with a hydrogen replaced by a hydroxyl group. H N H H ammonia
H N OH H hydroxylamine
H N
R
H 1° amine
Hydroxylamine is another nucleophile. It also reacts with a carbonyl compound through nucleophilic addition pathway. The difference is that the reaction intermediate formed has an acidic hydrogen atom attaching to the nitrogen atom. This makes the intermediate undergo elimination (dehydration) readily with the neighbouring hydroxyl group. The elimination product is called oxime. The reaction is known as a nucleophilic condensation (or addition-elimination) reaction because it consists of 2 steps, nucleophilic addition followed by elimination. Condensation reaction = Addition reaction + Elimination reaction elimination H H N OH H hydroxylamine
C O
HO N + C OH
carbonyl compound
H HO N C O H intermediate
addition reaction Condensation reaction
- H2O
HO N C oxime
Reaction Mechanism
Unit 10
Page 4
(4) Condensation reaction with 2,4-dinitrophenylhydrazine H H H H
NO2
H N N
H N N H hydrazine
O2N 2,4-dinitrophenylhydrazine
2,4-dinitrophenylhydrazine, or in short DNPH / DNP, is a derivative of a nitrogen containing compound, hydrazine. Pure DNPH is insoluble in water because it is not very polar. A little conc. H2SO4(l) is used to dissolve the orange DNPH solid and then diluted to prepare the DNPH solution. Reaction of 2,4-dinitrophenylhydrazine with carbonyl compound is similar to that of hydroxylamine. The condensation product formed is called hydrazone.
addition reaction H H H N N O C
H H NO2
O2N
-O
C N N + H O N 2
H NO2
HO C N N H
NO2
O2N
carbonyl 2,4-dinitrophenylhydrazine compound
- H2O
elimination reaction
H C N N
NO2
O2N 2,4-dinitrophenylhydrazone
Use of derivative of hydroxylamine and 2,4-dinitrophenylhydrazine for characterization Both oxime and hydrazone formed in the condensation reactions are high melting solid. By converting an unknown carbonyl compound into the corresponding oxime and hydrazone, the carbonyl compound can be identified. This is done by measuring the melting points of the oxime and hydrazone and comparing them with the literature values. Aldehyde or Ketone Ethanal Propanone Benzenecarbaldehyde 2-methylbenzenecarbaldehyde 3-methylbenzenecarbaldehyde 4-methylbenzenecarbaldehyde
m.p. of oxime 46.5 ºC 61 ºC 35 ºC 49 ºC 60 ºC 79 ºC
m.p. of 2,4-dinitrophenylhydrazone 168.5 ºC 128 ºC 237 ºC 195 ºC 211 ºC 233 ºC
Reaction Mechanism
Unit 10
Page 5
(5) Haloform reaction / Iodoform reaction Formation of enolate ion Both aldehyde and ketone show tautomerization. The keto form of the molecule coexists with the enol form of the molecule. Keto-enol tautomerization is an dynamic equilibrium involving the movement of a hydrogen atom. H O
- H+
C C α H
δ-
δO
+ H+
keto form
OH
+ H+
αC C H enolate ion
αC
- H+
C
enol form
An intermediate, enolate ion, is formed during the transition between the keto form and the enol form. Enolate ion can be considered as conjugate base of the keto or enol form.
C
O
O-
C
C
H
δ-
or
C
C
H
δO C
H
Since the enolate ion is stabilized by resonance, the acidity of the α-hydrogen is much higher than other hydrogen atoms attached to the carbon chain.
O
H H
C
Cα C β H H
α-hydrogen pKa ≈ 20
β-hydrogen pKa ≈ 50
It is observed that α-hydrogen is almost 1030 times more acidic than β-hydrogen. In the presence of a base e.g. NaOH, a carbonyl compound loses its α-hydrogen readily and forms an enolate ion. -O
H O
O
H H
C
Cα C β H H
Cα H
C
H C β H
+ H2O
Enolate ion possesses an electron rich carbon atom and behaves as a nucleophile. Haloform reaction Haloform has a general formula CHX3. Choroform CHCl3 and bromoform CHBr3 are liquid insoluble in water. Iodoform CHI3 is a bright yellow precipitate. Formation of yellow precipitate of iodoform serves as a O H
very useful test of methyl ketone
C
C H H
OH H
and 1-hydroxyethyl
C
C H
H H
group.
Reaction Mechanism
Unit 10
Page 6
Haloform reaction consists of two steps : 1. Formation of trihalo ketone by nucleophilic substitution 2. Formation of haloform by nucleophilic addition-elimination Formation of trihalo ketone from methyl ketone The presence of a base, e.g. NaOH(aq), promotes formation of enolate ion. The enolate ion formed acts as a nucleophile and displace the I- ion from the I2 molecule. Once the first α-hydrogen is substituted by an iodine atom, the acidity of the remaining α-hydrogens will be increased due to the negative inductive effect of the iodine atom. Consequently, the remaining two α-hydrogens will also be substituted by iodine atoms. enolate ion formation -O
nuleophilic substitution
H O
O H R C
R C
Cα H H
methyl ketone
I
I
O
C H α H
R C
further substitutions
I
OH-, I2
C H α H
OH-, I2
O
I
R C
C
I
α
I triiodo ketone
enolate ion -I effect of halogen make the H even more acidic
Formation of haloform from trihalo ketone The triiodo ketone formed is much more reactive than the original methyl ketone because –CI3 group is a better leaving group than the original –CH3 group, as -CI3 ion is a much weaker base than -CH3 ion does. O R C
I
O
I
C
R C
C
O
I
α
I
I
H O-
O I
H
triiodo ketone
R C
I C
O
I
I
H
O
I
R C
H C
Ocarboxylate ion
I
I iodoform
For 1-hydroxyethyl group, it is first oxidized to methyl ketone by the iodine in alkaline medium. Then, the methyl ketone formed is converted to carboxylate through the same mechanism mentioned above. H H C C H O H
H
oxidation + I2 + 2 OH-
H 1-hydroxyethyl group
C H + 2 I- + 2 H2O O H C
methyl ketone
Besides testing of methyl ketone and 1-hydroxylethyl group, iodoform reaction is also used in organic synthesis to shorten the carbon chain by one carbon atom. It also converts methyl ketone or 1hydroxyethyl group to carboxylate ion or carboxylic acid upon acidification. N.B.
1. 2. 3.
Haloform reaction can be used to test for methyl ketone and 1-hydroxyethyl group It can be used to cut the carbon chain short by 1 C. It oxidizes methyl ketone and 1-hydroxyethyl group to carboxylate.
Some chemistry text books say that NaOI(aq) is used to bring about iodoform reaction. In fact, according to the mechanism, the reaction is actually brought about by OH-(aq) ion and I2(aq) molecule. NaOI(aq) takes the advantages that it is prepared from I2(aq) and NaOH(aq). NaOH(aq) + I2(aq) d NaOI(aq) + NaI(aq) + H2O(l) As a personal preference, I prefer I2(aq) / NaOH(aq) to NaOI(aq) as the reagent used.
Reaction Mechanism
Unit 10
Page 7
Glossary
lithium tetrahydridoaluminate / lithium aluminium hydride ether acid work up sodium tetrahydridoborate / sodium borohydride sodium hydrogensulphite hydroxylamine oxime nucleophilic addition-elimination condensation reaction 2,4-dinitrophenylhydrazine hydrazone characterization keto-enol tautomerization enolate β-hydrogen haloform reaction / iodoform reaction methyl ketone 1-hydroxyethyl group trihalo ketone
Past Paper Question
90 2C 9 a ii iii 91 1A 1 b i 92 2C 8 b 93 2C 8 a i 94 2C 9 a i 95 2C 7 a ii 96 2C 8 c v 97 1A 5 a v 98 1A 5 a 5 c d i ii 99 2B 5 a i
91 2C 9 a iii 92 2C 9 a iii 95 2C 9 b i 97 2B 5 a i iii
97 2B 5 a i ii iii
97 2B 7 a ii
97 2B 7 b i
90 2C 9 a ii iii 9a Each of the following carbonyl compounds P, Q and R reacts with 2,4-dinitrophenylhydrazine in the presence of a strong acid to give a phenylhydrazone derivative. O
O
O
CH3 C CH2CH2CH3
CF3
P
C
CH2CH2CH3
CH3
CH3 C CH2 CH CH3
Q
R
ii
C
C
Under the same conditions, P, Q and R form phenylhydrazones at different rates. Predict with an explanation which will react the fastest, and which the slowest. Rate Q > P > R (slowest) Taking P as a reference. Q is faster than P because F substituent is electronegative. 1 mark ⇒ carbonyl carbon more electrophilic 1 mark 1 mark ⇒ more susceptible to nucleophilic attack by -NH2 R is slower than P because the -CH(CH3)2 group is more bulky and hindered the approach of the -NH2 to the 2 marks carbonyl carbon. (electron donating feature of -CH(CH3)2 is not significant due to long distance) Many candidates considered the electronic effect but not the steric effect. iii Explain why 2,4-dinitrophenylhydrazones are useful derivatives of carbonyl compounds. 2,4-dinitrophenylhydrazones of carbonyl compounds are high melting solids useful for characterization 2 marks OR are bright colour ppt useful for identifying carbonyl group Many candidates confused 2,4-DNPs with azo dyes, and wrongly stated that the former were important dyes in industry.
91 1A 1 b i 1b i CH3 H
C CH2 E
CH3
2
3 C O
H F
Explain why hydroxylamine reacts with F and not with E, and give the product expected from the reaction. C O
In F, the double bond is strongly polarized, polarization. OR
due to the electronegativity of oxygen. In E there is no such
C O + -
in which the carbon is electron deficient. 1 mark Hydroxylamine with N having lone electron pair is electron rich and takes part in nucleophilic addition at C.
The oxygen has a tendency to acquire electrons leading to
HONH2
5
C O
OR (This properties of NH2OH must be shown in some way) Product is Me-CH=NOH (no mark for Me-CHOH-NHOH) There are many ways to answer this question:
1 mark 1 mark
Reaction Mechanism Unit 10 For all the 3 marks, answer MUST explain WHY polarization difference is responsible for the difference in reactivity - answer MUST include the fact that NH2OH reacts as nucleophile or by nucleophilic addition. C A noticeable number of candidates did not know the structure of hydroxylamine.
Page 8
91 2C 9 a iii 9a Outline chemical tests which would allow you to distinguish between the compounds in the following pairs. Describe what you would observe in each case. iii O O CH3
C
CH3
O C
O CH3
CH3
will give yellow ppt of iodoform. No ppt observed for
Add iodine and NaOH, O C
2
-
.
+
COO Na + CHI3
CH3 + NaOH + I2
2 marks C
Many candidates wrote the formula of iodoform incorrectly as CH3I.
92 2C 8 b 8b Describe how 2,4-dinitrophenylhydrazine may be used to identify an unknown ketone such as butanone. Give an equation for the reaction involved. Take the unknown ketone to reacts with 2,4-DNPH; Get the solid derivative; Purified by recrystallization; Determine its m.p.; Compare with literature data. 2½ marks NHNH2 NO2
O CH3CH2 C CH3
C
CH3 NH N CCH2CH3 NO2
+ NO2
4
NO2
1½ mark
Candidates gave the wrong structure for 2,4-dinitrophenylhydrazine, e.g. NH2 NO2
NO2
Many candidates did not know how to make use of the m.p. of hydrazone as a means to identify the ketone. They just mentioned the use of 2,4-DNP as a test for carbonyl functional group. 92 2C 9 a iii 9a Give the structural formula(e) of the major organic product(s) from each of the following reactions : iii O C
CH3
COO-Na+
2
I2/NaOH(aq)
+ CHI3
2 marks
Reaction Mechanism Unit 10 93 2C 8 a i 8a Give a chemical test to distinguish one compound from the other in each of the following pairs. Your answer should include the reagents required, the observation expected, and the chemical equation(s) for each test. CH3CH2COCH2CH3 and CH3CH2OCH2CH3 i Reagent: 2,4-DNPH Observation: red/orange/brown/yellow precipitate Equation:
Page 9
3
1 mark 1 mark
NHNH2 O2N
NO2
Et
Et
C O+
C NNH
Et
NO2
Et NO2
C
1 mark
Incorrect structure for 2,4-dinitrophenylhydrazone.
94 2C 9 a i 9a Give the structural formula(e) of the major organic product(s) (P to V) formed in each of the following reactions. i O CH3 C CH2CH3
NH2OH
1
P
OH N
1 mark
P:
95 2C 7 a ii 7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer should include the reagents used, the observation expected and the chemical equation(s) for each test. ii O O CH3CH2CCH2CH3 and CH3CH2CH2CCH3 Treat compound with I2/OHO CH3CH2CH2CCH3
I2 KOH / NaOH
3
½ + ½ mark O CH3CH2CH2C O-K+ + CHI3
yellow precipitate observed.
½ + ½ mark ½ mark
O
No ppt. with CH3CH2CCH2CH3 C
½ mark
Many candidates failed to give the equations for the reactions. Some candidates erroneously gave the silver mirror test as the answer. One common mistake was the wrong formula for iodoform (CH3I instead of CHI3). Some candidates gave the wrong colour for CHI3. 'No observable change' instead of 'no reaction' should be given because pentan-2-one does undergo reaction (iodination) under the conditions.
95 2C 9 b i 9b Identify K, L, M, N, P, R and S in the following reactions: i O N
1
2
H2NHN
NO2
K
O2N NHN
NO2
O
K: C
/ cyclobutanone
candidates appeared to be unfamiliar with the cyclic structure and gave an acyclic ketone;
1 mark
Reaction Mechanism Unit 10 96 2C 8 c v 8c Identify K, L, M, N and P in the following reactions : (Deduct ½ mark for each minor mistake; max. Deduction for the whole question = 2 marks) v O NaHSO3
CH3CH2 C CH3
C
1
P
CH3 + SO3 Na or CH3CH2 C SO3-Na+ OH OH
P:
Page 10
1 mark
wrong structures for P, such as SO3H CH3CH2
SO3H
C CH3 CH3CH2
C CH3
ONa
OH
97 1A 5 a v 5a Give a structure for each of the compounds D, E, F. G and H: v I2 / NaOH(aq) C4H8O H
5
triiodomethane + sodium salt of a carboxylic acid
97 2B 5 a i ii iii 5a In an experiment, 10.0 g of butanone reacts with 5.0 g of hydrogen cyanide to give 11.0 g of 2-hydroxy-2methylbutanenitrile. i Find the limiting reactant of the reaction, showing clearly your calculation. ii Calculate the percentage yield of 2-hydroxy-2-methylbutanenitrile. iii Name the type of reaction and outline the mechanism involved. (Movement of electron pairs should be indicated by curly arrows.) 97 2B 7 a ii 7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Each test should include the reagent(s), the expected observation with each compound and the chemical equation(s). ii OH OH CH2CH3
and
H
J
12
CH CH3
97 2B 7 b i 7b Identify J, K, L, M and N in the following reactions. i O C
6
CH NOH
5
Reaction Mechanism Unit 10 98 1A 5 a c d i ii 5 Consider the five reactions of butanone (C4H8O) J shown in the reaction scheme below:
5a 5c 5d i ii
Suggest a reagent by which K may be formed and give the structure for K. Give structures for compounds M, N, P and R. S is a structural isomer of J. S also reacts with 2,4-C6H3(NO2)2NHNH2 to give a red precipitate. Draw the structure of S. How may J and S be identified by making use of their reactions with 2,4-C6H3(NO2)2NHNH2 ?
99 2B 5 a i 5a Identify D, E, F, G and J in the following reactions. i D O
NNH
Page 11
1 4 1 1
Reaction Mechanism
Unit 11
Page 1
Topic
Reaction Mechanism
Unit 11
Reference Reading
19.2.0 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg.
Syllabus
Carboxylic acid and its derivatives Preparation of acid derivatives Reaction of acid derivatives
Notes
b) Reactions of carboxylic acid and its derivatives
220–221, 228–231, 234–237, 239–244, 251, 254–258, 259–263, 265–267 Organic Chemistry, Solomons, 5th Edition pg. 680–682, 691–693, 701–707, 724–726, 790–793, 845–846 Organic Chemistry, Fillans, 3rd Edition pg. 225–226, 229–238, 240–243, 247–248, 251–259, 262, 275 Organic Chemistry, Morrison Boyd, 6th Edition pg. 669–671, 677–680, 683–684, 697–698, 724, 737–740, 755–778, 797–799, 802–807,830–831 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505 Organic Chemistry, 6th Edition, Solomons, pg. 708-711
(1) Carboxylic acid and its derivatives (a) Difference between carbonyl compound and carboxylic acid and its derivatives Carboxylic acid consists of a carbonyl group
O C
and a hydroxyl group –OH. Because of the presence of
carbonyl group, in certain extent, it shares the same reaction pattern of ordinary carbonyl compound. However, carboxylic acid does not resemble the properties of ketone entirely. There are 3 major differences. 1.
Carboxylic acid has reaction with sodium, phosphorus trichloride PCl3 and phosphorus pentachloride PCl5 while ketone or aldehyde haven't. Since carboxylic acid also possesses a hydroxyl group, it shows some typical properties of a hydroxy compound. e.g. 2 H–O–H(l) + 2 Na(s) → H2(g) + 2 NaOH(aq) O
2 CH3
C OH (l)
+ 2 Na(s) → H2(g) + 2 CH3
O C O-Na+ (acid)
The ethanoic acid used in this reaction must be dry, in which there is no free hydrogen ion. Otherwise, Na(s) will react with the hydrogen ion vigorously and may cause explosion. e.g. 3 H–O–H(l) + PCl3(l) → 3 HCl(g) + H3PO3(l) O
3 CH3
C OH (l)
+ PCl3(l) → 3
O CH3
C
Cl (l)
+ H3PO3(l)
e.g. 4 H–O–H(l) + PCl5(s) → 5 HCl(g) + H3PO4(l) O CH3
C OH (l)
+ PCl5(s) →
O CH3
C
Cl (l)
+ POCl3(l) + HCl(g)
In these reactions, the hydroxyl group –OH is substituted by –Cl atom. Another useful reagent for the same purpose is thionyl chloride SOCl2 which has an even higher yield than PCl3 or PCl5.
Reaction Mechanism
Unit 11 2.
Page 2
Carboxylic acid and its derivatives undergoes condensation (addition-elimination) reaction instead of addition reaction only. In carboxylic acid, a hydroxyl group is attached to carbonyl carbon. It can be converted to a group leaving group by protonation. Therefore, any nucleophilic addition is usually followed by elimination and leads to condensation reaction. O R C
L
Nu
O nucleophilic R C L addition Nu
elimination
O R C Nu + L
Condensation reaction = Addition reaction + Elimination 3.
Carboxylic acid and its derivatives have no reaction with 2,4-dinitrophenylhydrazine. –OH is a +R and -I group. Since resonance effect is usually stronger than inductive effect, –OH is an electron donating group in this case. It makes the carbonyl carbon less positive and less reactive than ketone and aldehyde. O
O CH3
C O H
CH3
δ−O
or
C O H +
CH3
C O H δ+
Similarly, all derivatives of carboxylic acid are with the resonance stabilized carbonyl group. They do not react with 2,4-dinitrophenylhydrazine. (b) Reactivities of carboxylic acid and its derivatives Carboxylic acid and its derivative shows the following order of reactivity. The trend is determined by the leaving ability of the substitutent attaching to the carbonyl group. Acid chloride (acyl chloride) O R C
Leaving group Basicity of the leaving group
Glossary
O Cl
Most reactive ClLow
Acid anhydride O
R C O C
RCOO
Carboxylic acid
-
O R'
R C OH
-
OH
Ester
Amide O
O R C O
RO
-
R'
R C NH2
Least reactive NH2-
––––––––––––––––––––––––––––––––––––––––––––––––––→
High
Reaction Mechanism
Unit 11
Page 3
95 2C 7 a iv 97 2B 7 a iii
Past Paper Question
95 2C 7 a iv 7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer should include the reagents used, the observation expected and the chemical equation(s) for each test. iv O O CH3CH2CNHCH2CH3 and CH3CH2CCH2NHCH3 Treat compound with a solution of 2,4-dinitrophenylhydrazine in methanol (or ethanol)
NHCH3
1 mark
O2N
O 2N O
3
H2NNH
NO2
NH
NO2
N NHCH3
Orange / red / brown / yellow precipitate is observed.
1 mark ½ mark
O
No ppt. with CH3CH2CNHCH2CH3 .
½ mark
OR -
O
OH (aq) heat
CH3CH2CNHCH2CH3 C2H5NH2
iv
HCl
O CH3CH2C O + C2H5NH2
1 mark
+ C2H5NH3Cl white fume
1½ mark Comparison ½ mark Many candidates failed to give the equations for the reactions. Very badly answered. Few candidates could identify the keto group probably because they were not familiar with compounds with more than one functional group. Some did not know that secondary amines do not react with HNO2 to give N2. Most could not recall the chemistry of amides that are susceptible to hydrolysis.
97 2B 7 a iii 7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Each test should include the reagent(s), the expected observation with each compound and the chemical equation(s). iii O O C2H5
C O(CH2)2CH3
and
C2H5
C
CH2OC2H5
12
Reaction Mechanism
Unit 12
Page 1
Topic
Reaction Mechanism
Unit 12
Reference Reading
19.2.1 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg.
Syllabus
Preparation of acid derivatives Reaction of acid derivatives
Notes
(2) Formation of different acid derivatives
220–221, 228–231, 234–237, 239–244, 251, 254–258, 259–263, 265–267 Organic Chemistry, Solomons, 5th Edition pg. 680–682, 691–693, 701–707, 724–726, 790–793, 845–846 Organic Chemistry, Fillans, 3rd Edition pg. 225–226, 229–238, 240–243, 247–248, 251–259, 262, 275 Organic Chemistry, Morrison Boyd, 6th Edition pg. 669–671, 677–680, 683–684, 697–698, 724, 737–740, 755–778, 797–799, 802–807,830–831 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 462–463, 472, 505–508, 510–511, 520 Organic Chemistry, 6th Edition, Solomons, pg. 805-817, 820-824
O R C
O nucleophilic R C L addition Nu
L
Nu
elimination
O
R C Nu + L
–L = –Cl, –COOR, –OH, –OR, –NH2 Nu: = RCOO-Na+, Na+OH-, ROH, NH3
In general, a more reactive acid derivative can be converted to a less reactive acid derivative by using a suitable nucleophilic reagent. The reaction is a simple nucleophilic addition-elimination reaction. A less reactive derivative can also be converted to a more reactive derivative but the yield of conversion would be rather low. Exception : 1. 2.
Amide cannot be prepared from carboxylic acid and ammonia. This is because carboxylic acid is an acid and ammonia is a base, they reacts to give a salt only. Amide is usually prepared from acid chloride with ammonia. Similarly, ester cannot be prepared from carboxylic acid and sodium alkoxide as they will neutralize each other.
(a) Use of chlorinating agent to prepare acyl chloride Acyl chloride is the most reactive acid derivative, it cannot be prepared from the other acid derivative. It is prepared by converting the –OH group of carboxylic acid to –Cl group by chlorinating agent. Chlorinating agent : SOCl2 (thionyl chloride), PCl5 (phosphorus pentachloride) or PCl3 (phosphorous trichloride) O R C OH
SOCl2 or PCl5 or PCl3
O R C Cl
Reaction Mechanism
Unit 12
Page 2
(b) Formation of acid anhydride (i) Through intramolecular dehydration by heating
Certain dioic acid can be dehydrated by heating through intramolecular dehydration to form cyclic acid anhydride. This is only possible if the two hydroxyl groups of the two carboxyl groups are facing each other spatially. (ii) Through intermolecular dehydration by a very strong dehydrating agent However, linear acid anhydride, e.g. ethanoic anhydride CH3COOCOCH3, cannot be prepared by heating alone. Since the formation of the anhydride involves intermolecular dehydration, if the acid is heated strongly, the thermal agitation will not allow the two acid molecules to get to each other for the dehydration. Therefore, a very strong dehydrating agent e.g. phosphorus pentaoxide P2O5(s) (or P4O10(s)) should be used instead of heating. O CH3
O C OH + HO C CH3
O
P2O5 CH3
O
C O C
CH3
(iii) From acyl chloride Similarly, it can also be prepared from reactive acyl chloride by mixing with a carboxylate salt. O CH3
O -
+
C O Na
+ Cl
C CH3
O CH3
O
C O C
CH3 + NaCl
Reaction Mechanism
Unit 12
Page 3
(c) Formation of ester (Esterification with alkanol and phenol) O carboxylic acid R' C O H
Or.d.s.
R' C O H R +O
alcohol
O R' C O
R + H2O water
ester
R O
H charge separation involved
R O H
O- H R' C O+ H
Esterification is also a typical condensation reaction. An alcohol molecule is added to a carboxylic acid molecule which elimination of water. The nucleophilic oxygen atom is added to the carbonyl carbon. It can be seen that an intermediate with charge separation in the structure is involved. This makes the reaction having a high activation energy, thus, a rather slow rate. Then, the proton possessed by the alcohol is shifted to the hydroxyl group to convert it to a better leaving group, water H2O. Contrastingly, esterification in the presence of any aqueous acid is found to be catalyzed. Acid Catalyzed Esterification
H
H O R C OH
H+
f
H
O
+O R C OH
r.d.s.
H O
R'
f
O
R C OH
f
+O H R'
H
R C O H + O
fR
O C OR'
+ H2O + H+
R'
no charge separation involved
In the acid catalyzed pathway, the carbonyl oxygen is first protonated. This activates the carbonyl group by increasing the polarity of the C=O bond and makes the carbonyl carbon more positive and more vulnerable to the attack of the nucleophile, alcohol. Furthermore, no reaction intermediate with charge separation is involved in the acid catalyzed mechanism and a much lower activation energy would be involved. Esterification of phenol However, phenol is a weaker base than aliphatic alcohol because of benzene ring is electron withdrawing.
H
H
O
O
A substrate more reactive than carboxylic acid is required to give a reasonable yield. Phenol reacts with acyl chloride and acid anhydride to form ester. O OH
O OH
O
R C Cl
O C O
R C O C
R
R + Cl
O O C
O R+R C O
Reaction Mechanism
Unit 12
Page 4
(d) Formatioin of amide (Acylation and benzoylation of amine) Acylation means substitution by an acyl group O
O
O C
R
. Two common reagents used in acylation are ethanoyl
O
chloride CH3 C Cl and ethanoic anhydride CH3 C O C CH3 . Both of them are reactive derivatives of carboxylic acid, which reacts with nucleophile, e.g. NH3 or RNH2, readily. When ethanoyl chloride or ethanoic anhydride O
are used, the reaction is also called ethanoylation since the acyl group involved is the ethanoyl group O CH3
C
-O
Cl CH3
C
CH3
H N H + H
H N H
NH3 excess
O H
Cl
C N H + H
C
CH3
.
O C NH2 + NH4+Clethanamide
-
CH3
+ Cl
H
In ethanoylation of ammonia, excess ammonia is used to neutralize the HCl formed. In the reaction, one H on O
ammonia molecule is substituted by an ethanoyl group C CH3 . Unlike esterification, catalysis by protonation is impossible in here. The presence of aqueous acid will neutralize the nucleophile NH3 to nonnucleophilic species NH4+ and stop the reaction. Furthermore, ethanoyl chloride or anhydride also reacts with water directly. O
. Benzoyl chloride is the Similar to acylation, benzoylation means substitution by an benzoyl group C commonly used reagent. It is prepared by treating benzenecarboxylic acid (benzoic acid) with phosphorus trichloride PCl3, phosphorus pentachloride PCl5 or thionyl chloride SOCl2. O
O C OH + PCl3
3
3
C
Cl + H3PO3
benzoyl chloride
The reactive benzoyl chloride reacts with the ammonia or amine through nucleophilic addition-elimination similar to the ethanoylation discussed above. O C
Cl
-O C
H N H H
Cl
H N H + H
O H C N H + H
NH3 excess
O + Cl-
C NH2 + NH4+Clbenzenecarboxamide or benzamide
Reaction Mechanism
Unit 12
Page 5
(3) Reaction of ester (a) Hydrolysis of ester Ester can be hydrolyzed in neutral, acidic or alkaline medium. Since OH- is a nucleophile stronger than H2O, the rate of hydrolysis will be fastest in alkaline medium. -
O R' C O
O R' C O -
R
-
O
carboxylic acid
H
H OH is regenerated
-
O
-
R' C O H + H O
O H
H O
H
OH
O + R' C O R
R
R
alkanol
-
OH is consumed
OH
O R' C O
-
carboxylate ion
OH- ion can be considered as a catalyst in the hydrolysis or it can be considered as a reagent which neutralizes the carboxylic acid to carboxylate ion. In acidic medium, hydrolysis is catalyzed by protonation of carbonyl group. H H
+O H+
O R'
R' C O
C O
O
R R'
R
C
O
R
H O + H
H O H
H
H
O R'
C
H O
O+ R H
O+ R'
+H O
C O H
R
O R'
C O H
+ H+ + H O
R
Therefore, the rate of hydrolysis of ester follows the following order : alkaline medium > acidic medium > neutral medium.
Glossary
Esterification
Past Paper Question
90 1B 4 c i ii 92 1A 1 a iii 93 2C 7 a iii 94 2C 9 a ii 95 2C 9 b ii 96 1A 3 d ii 97 1A 5 a iii 98 2B 7 a i 99 2B 6 a i
acylation
benzoylation
92 1B 5 a ii iv
90 1B 4 c i ii 4c Amides are generally prepared by reacting equimolar amounts of an amine and an acid halide. (Relative atomic masses : H, 1.0; C, 12.0; N, 14.0; O, 16.0; Cl, 35.5) Calculate: i the mass (in grams) of Y required to react with 4.0 g of benzoyl chloride (C6H5COCl), and Relative molecular mass of Y = 6 × 12.0 + 7 × 1.0 + 14.0 = 93.0 Relative molecular mass of C6H5COCl = 7 × 12.0 + 5 × 1.0 + 16.0 + 35.5 = 140.5 ½ mark 4.0 × 93.0 1 mark mass of Y = 140.5 = 2.65 g ii the percentage yield of the same reaction if 4.0 g of the amide product is obtained. Relative molecular mass of amide (C13H11NO) = 13 × 12.0 + 11 × 1.0 + 14.0 + 16.0 = 197.0 4.0 × 197 .0 = 5.6 g 100% yield = 140.5 4.0 percentage yield = 5.6 × 100% = 71% 1½ mark C The calculation of percentage yield was very, very poorly answered. It is clear that such work is rarely done by candidates. It is also obvious that this very simple concept is unknown to most candidates.
1½
1½
Reaction Mechanism Unit 12 92 1A 1 a iii 1a There are several isomers of benzenedicarboxylic acid. iii On heating, one of the isomers gives a compound W, with formula C8H4O3. Give the structure of W.
Page 6
1
O C O C O
1 mark
92 1B 5 a ii iv 5a Ethanolylation (acetylation) of phenylamine (C6H5NH2) may be carried out by refluxing for 1 hour with ethanoic anhydride (acetic anhydride) giving a product which is insoluble in cold water. (Ethanoic anhydride hydrolyses in water to give ethanoic acid. The b.p. of phenylamine and ethanoic anhydride are 184°C and 140°C respectively; the m.p. and b.p. of the product are 114°C and 304°C respectively.) ii Give the structure of the product from the ethanoylation reactions.
1
H O N C
iv
C
CH3
1 mark If 40g of phenylamine is to be ethanoylated, calculate the mass of ethanoic anhydride required. (Relative atomic masses: H 1.0; C 12.0; N 14.0; O 16.0) molecular mass of PhNH2 = 93 molecular mass of (CH3CO)2O = 102
→ PhNHCOMe + CH 3CO 2 H Since PhNH 2 + (CH 3CO)2 O 102 1 The mass of ethanoic anhydride required = 40 × × g = 43.9g 93 1 Calculation was very badly done, suggesting that few do experiments involving quantities.
2
2 marks
93 2C 7 a iii 7a Give the structural formula(e) of the major organic product(s) formed in each of the following reactions. NH2 iii CH3COCl
2
O H N C
CH3
1 mark C
This was generally well answered.
94 2C 9 a ii 9a Give the structural formula(e) of the major organic product(s) (P to V) formed in each of the following reactions. NH2 ii CH3CH2COCl
1
Q
N
Q:
1 mark
O
95 2C 9 b ii 9b Identify K, L, M, N, P, R and S in the following reactions: ii O O CH3CH2CH2COH
C
L
1
CH3CH2CH2CCl
L : phosphorus(V) chloride / PCl5 or phosphorus(III) chloride / PCl3 or thionyl chloride / SOCl2 Common mistakes included : HCl for L;
1 mark
Reaction Mechanism Unit 12 96 1A 3 d ii 3d Give the structure of the major organic product(s) in each of the following reactions : ii O O
Page 7
1
heat
HO C CH2 CH2 C OH O O
O
heat
HO C CH2 CH2 C OH
O O
1 mark
97 1A 5 a iii 5a Give a structure for each of the compounds D, E, F. G and H: iii SOCl2 F
CH3CH2CO2H
98 2B 7 a i 7a Briefly explain why each of the following proposed reactions cannot lead to the target molecule. In each case, using the same starting molecule, outline a feasible synthetic route to obtain the target molecule, with no more than three steps. In each step, give the reagent(s) used, the conditions required and structure of the product. i O O CH3OH
C NHCH3
C OCH3
99 2B 6 a i 6a i Suggest a reagent for the following conversion : O Cl
5
NH2
4-chlorophenylamine
Cl
NH CCH3
N-(4-chlorophenyl)ethanamide
12
Reaction Mechanism
Unit 13
Page 1
Topic
Reaction Mechanism
Unit 13
Reference Reading
19.2.2–19.2.3 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg.
Syllabus
Reduction of acid and its derivatives Hoffmann degradation of amide
Notes
(3) Reaction of amide
220–221, 228–231, 234–237, 239–244, 251, 254–258, 259–263, 265–267 Organic Chemistry, Solomons, 5th Edition pg. 680–682, 691–693, 701–707, 724–726, 790–793, 845–846 Organic Chemistry, Fillans, 3rd Edition pg. 225–226, 229–238, 240–243, 247–248, 251–259, 262, 275 Organic Chemistry, Morrison Boyd, 6th Edition pg. 669–671, 677–680, 683–684, 697–698, 724, 737–740, 755–778, 797–799, 802–807,830–831 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 505, 519 Organic Chemistry, 6th Edition, Solomons, pg. 919-921
(a) Reduction of amide and other acid derivatives Except amide, all other acid derivative are reduced by LiAlH4 to alcohol. The nucleophilic hydride ion, H-, serves as a strong reducing agent in this reaction. O R C
aldehyde O
O L
R C
H-
R C
L
H
O-
H + L
R C
H
H
H O H +
H
H-
OH acid R C H + H2O work up H
where L = –Cl, –OOCR, –OH, –OR The aldehyde intermediate cannot be isolated from the reaction mixture. It will be reduced directly to a primary alcohol. Exception : Reduction of amide Amide will be reduced by LiAlH4 to amine instead of alkanol. LiAlH4 can be considered to be comprised of three parts : Li+ : a spectator ion H- : a nucleophile, a very strong reducing agent AlH3 : a Lewis acid, a catalyst activates the reactivity of a carbonyl group The actual mechanism of the reaction is still not very clear, the following may be a possible one. Simplified Proposed Mechanism H
H H Al H
H Al H H-
O
O+
R C N H
R C N Li+
O
R C N Li+
H
H H-
H
acid-base reaction
activation
H H Al H O
R C N Li+ H H
reduction
H H O H +
HR C N H H imine (further reduction)
H R C N
-
H H R C N H H
H H acid work up
1° amine
Reaction Mechanism
Unit 13
Page 2
Depending on the original structure of the amide, it is reduced by LiAlH4 to 1º amine, 2º amine or 3º amine. Amide is the only acid derivative which will not be reduced to 1º alcohol by LiAlH4. O
1) LiAlH4 / ether 2) H3O+
R C NH2 amide
H H 1° amine
O R C N H R' N-substituted amide
O R C N
H R C N H
R''
R' N.N-disubstituted amide
1) LiAlH4 / ether 2) H3O+
1) LiAlH4 / ether 2) H3O+
H R C N H H R' 2° amine
H R C N
R''
H R' 3° amine
This reaction is commonly used to prepare 2º and 3º amine where direct alkylation of 1º and 2º amine will yield a mixture of 2º, 3º amine and quaternary ammonium ion. Reduction selectivity of LiAlH4 and NaBH4 In term of reducing power, LiAlH4 is much stronger than NaBH4. LiAlH4 can reduce all kinds of carbonyl compound to alcohol. This includes ketone, aldehyde and all kinds of acid derivatives except amide. In contrast, NaBH4 is only strong enough to reduce ketone and aldehyde to alcohol, but not any acid derivative.
Reaction Mechanism
Unit 13
Page 3
(b) Hofmann degradation of amide Reagent : a base – NaOH(aq) and an electrophile – Br2(l) O R C NH2 + Br2 + 4 NaOH
R N H + 2 NaBr + Na2CO3 + 2 H2O H
non-substituted amide
1° amine
Hofmann degradation is also known as Hofmann rearrangement. According to the mechanism, the 2 protons on the amino group are first subtracted by the base. Therefore, only a non-substituted amide RCONH2 can undergo Hofmann degradation.
Comparing with the reduction of amide by LiAlH4, Hofmann degradation provides a way to shorten the carbon chain by 1 carbon atom. O R C NH2
1) LiAlH4 / ether 2) H3O+
amide O R C NH2
R C N H H H 1° amine
Br2 / NaOH(aq) R N H
amide
Glossary
H
reduction selectivity
H 1?amine with 1 C less
Hofmann degradation / Hofmann rearrangement
Reaction Mechanism
Unit 13
92 2C 8 a i ii 93 2C 7 a iv 94 2C 9 a iii 95 2C 9 b vi vii 96 2C 8 a i ii 97 1A 5 a i ii 98 1B 8 b ii iii 99 2B 5 a iv
Past Paper Question
Page 4
96 2C 9 a iii 97 2B 7 b iii 98 2B 7 a iii
92 2C 8 a i ii 8a Amines can be prepared from an amide or an alkyl halide. i Illustrate this statement by the preparation of RCH2NHCH3. Both synthetic reaction sequences must start with RCO2H. In each sequence, give the reagent(s) for each step and the structures of intermediate compounds. (I)
NH RCOOH PCl → RCOCl CH → RCONHCH3 LiAlH → RCH 2 NHCH3 5
3
LiAlH 4
2
PBr3
4
CH 3 NH 2
(II) RCOOH → RCH 2 OH → RCH 2 Br → RCH 2 NHCH3 ii
7
3½ marks 3½ marks
Explain which of your two methods is more suitable for preparing RCH2NHCH3.
2
CH 3 NH 2
→ also gives (RCH2)2NCH3. 2 marks (II) is less appropriate than (I) in preparation, since RCH2 Br
C
The least well-answered part in IIC. Common mistakes in conversions include : 1. Preparing an amide by direct reaction between an acid with NH3 or amine, RCOOH + CH3NH2
RCONHCH3
RCOOH + NH3
2.
RCONH2
Converting a secondary amide to a secondary amine employing Br2/OH- (Hofmann Degradation applicable to primary (non-substituted) amide only) Br2 / OH RCONHCH3
3.
-
RCH2NHCH3
Alkylating a primary amide with alkyl halide to obtain a secondary amide CH3Br RCONH2
RCONHCH3
93 2C 7 a iv 7a Give the structural formula(e) of the major organic product(s) formed in each of the following reactions. O iv LiAlH4
2
CH3 C CH2CH2COOCH3 OH
CH3
C
CH2CH2CH2OH
H
C
reduction of ketone – 1 mark, reduction of ester – 1 mark Many candidates did not realize that esters could be reduced to alcohol by LiAIH4.
94 2C 9 a iii 9a Give the structural formula(e) of the major organic product(s) (P to V) formed in each of the following reactions. iii LiAlH4 CH3CH CHCH2CH2COOH
R: C
CH3CH CHCH2CH2CH2OH
R
1 mark
Candidates were casual / careless in writing structural formula, particularly in putting the number of H atoms to each C atom. Common mistakes included : for R (saturating C=C) CH3CH2CH2CH2CH2CH2OH
1
Reaction Mechanism Unit 13 95 2C 9 b vi vii 9b Identify K, L, M, N, P, R and S in the following reactions: vi O LiAlH4
C N(CH3)2
H C N
R: C
1
R
CH3
1 mark
CH3
H
Page 5
wrong structure for R. such as OH N
CH3 CH3
vii
1
CO2CH3 NaBH4
S
O CO2CH3
S: C
1 mark
OH
wrong structures for 5, such as CH2OH
OH
COOCH3
OH
COOCH3
O
Candidates did poorly in the cases of P. R and S. and were thus not familiar with the chemistry of exhaustive methylation of amine, the hydride reduction of amide, and the mild reducing reagent sodium borohydride.
Reaction Mechanism Unit 13 96 2C 8 a i ii 8a i An acyclic compound H of molecular formula C4H8O2 has a fruity smell. It does not produce a derivative with 2,4-dinitrophenylhydrazine nor with propanoyl chloride. Deduce the functional group(s) of H. Draw FOUR possible structures for H. H is not aldehyde / ketone ½ mark ½ mark because it does not form hydrazone derivative. or R
N NH
Page 6 4½
NO2
O2N
(½ mark)
H does not possess an –OH group / is not an alcohol because it does not form ester with propanoyl chloride. or
½ mark ½ mark
O
no R O C C2H5 formation
(½ mark)
H has a fruity smell, The functional group(s) in F is most likely an ester.
½ mark
Any FOUR of the following structures (½ marks for each structure)
2 marks
H
O
H
O
C ii
O
O
O
O
O O
O O
O
O
(Deduct ½ mark for each extra structure) Poorly answered. Many candidates did not appear to know the reaction of 2,4-dinitrophenylhydrazine and propanoyl chloride and hence the formation of the respective 2,4-dinitrophenylhydrazone and ester was omitted. Some did not know the reaction of ester with LiAIH4 or the structure of ethanoic anhydride. On reduction with an excess of LiAlH4, H gives only one product, J. which reacts with ethanoic anhydride. Deduce, giving equations, the structure for H and also the structure for J. O O O
O
ROH
R
O
Among the four carbon esters only ethyl ethanoate gives one reduction product or, other esters give a mixture of alkanols on reduction
1 mark ½ mark (½ mark)
O
∴ structure of H is
½ mark
O
and structure of J is OH Equation for the reduction of H O CH3 C OC2H5
LiAlH4
2 C2H5OH
½ mark
½ mark
3½
Reaction Mechanism Unit 13 96 2C 9 a iii 9a Each of the following conversions can be completed in not more than three steps. Use equations to show how you would carry out each of these conversions in the laboratory. For each conversion, give the reagent(s), conditions, and structure of the intermediate(s). (For answers with more than 3 steps, deduct 1 mark for each extra step.) iii COCl NH2
O
(1) NH3
NH2
Br2 / OH- (or OBr-) (½) heat (½)
(1)
3 marks Many candidates wrongly thought that acid amide can be prepared directly from the reaction of carboxylic acid with ammonia. Heating was omitted in the Hofmann degradation step.
C
97 1A 5 a i ii 5a Give a structure for each of the compounds D, E, F. G and H: (1) LiAlH4 i CH3CH2CONH2
ii CH3CH2CONH2
D
(2) H2O
Br2 / NaOH(aq)
E
97 2B 7 b iii 7b Identify J, K, L, M and N in the following reactions. iii O CH3
CH3
C
CHC OCH3
L
CH3
CH3
C
CHCH2OH + CH3OH
98 1B 8 b ii iii 8b 20.0 g of 4-nitrobenzoic acid (C7H5NO4) reacted with PCl5 to give a product which reacted exothermically with NH3 to give T. After treatment with Br2(l) and NaOH(aq), T gave a solid. Crystallization of the solid from ethanol gave 9.3 g of U (C6H6N2O2). ii Give the structures of T and U. iii Briefly give three reasons which could explain why the yield of U in the above preparation is not quantitative (i.e. 100%). 98 2B 7 a iii 7a Briefly explain why each of the following proposed reactions cannot lead to the target molecule. In each case, using the same starting molecule, outline a feasible synthetic route to obtain the target molecule, with no more than three steps. In each step, give the reagent(s) used, the conditions required and structure of the product. iii O O + CH3CH2 C OH
LiAlH4, H3O
CH3CH2 C H
99 2B 5 a iv 5a Identify D, E, F, G and J in the following reactions. iv CO2NH2 CH3O
Br2 / NaOH heat
G
Page 7
3
Reaction Mechanism
Unit 14
Page 1
Topic
Reaction Mechanism
Unit 14
Reference Reading
19.3 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg.
Syllabus
Reaction of nitrile Partial and complete hydrolysis of nitrile Dehydration of amide Reduction of nitrile
Notes
c)
220–221, 228–231, 234–237, 239–244, 251, 254–258, 259–263, 265–267 Organic Chemistry, Solomons, 5th Edition pg. 680–682, 691–693, 701–707, 724–726, 790–793, 845–846 Organic Chemistry, Fillans, 3rd Edition pg. 225–226, 229–238, 240–243, 247–248, 251–259, 262, 275 Organic Chemistry, Morrison Boyd, 6th Edition pg. 669–671, 677–680, 683–684, 697–698, 724, 737–740, 755–778, 797–799, 802–807,830–831 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 236–237, 254–255, 505 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 519 Organic Chemistry, 6th Edition, Solomons, pg. 711, 825-827, 918
Reactions of nitrile C N is quite similar to carbonyl group
Nitrile
O
with the N replacing the O.
C
(1) Hydrolysis of nitrile Nitrile can be hydrolysed to carboxylic acid or carboxylate by boiling with aqueous acid or alkali, with the amide intermediate. By employing an suitable reaction condition, amide intermediate can be isolated from further hydrolysis to carboxylic acid. e.g. Heating with dilute sulphuric acid will yield carboxylic acid while heating with concentrated sulphuric acid will only yield amide. Or, amide can be obtained by hydrolysis using cold dilute acid instead of a hot one. Partial hydrolysis of Nitrile to Amide H
H
isoamide H
O H
O H
O H
O
H R C N δ+ δ−
amide
H
H O+ R C N
R C N H
R C N H
O H tautomerization
Hydrolysis of Amide to Carboxylic acid O R C
L
O R C
Nu
Temperatur e of hydrolysis Complete Partial Partial
O R C Nu + L
L
Nu
Concentration of sulphuric acid Low High Low
Degree of hydrolysis High Low High
Hydrolysis of amide to carboxylic acid follows the regular pattern of the nucleophilic addition-elimination reaction of acid derivatives. O R C NH2 H O H
O H R C N H H O + H
O
O
R C
H N+ H
R C
H N+ H
H O
H
H O
H
O R C O H
NH3
O R C O-
NH4+
Reaction Mechanism
Unit 14
Page 2
(a) Dehydration of amide With the appropriate dehydrating agent, phosphorus pentoxide (P2O5, with molecular formula P4O10) or boiling ethanoic anhydride (CH3CO)2O or thionyl chloride (SOCl2), amide can be converted back to nitrile. O
P2O5 or (CH3CO)2O or SOCl2
R C NH2
R C N
(2) Reduction of nitrile Similar to the carbonyl group, nitrile group is susceptible to the reduction by hydride ion through nucleophilic addition pathway. For carbonyl compound, the carbonyl group is reduced to alcohol. A nitrile is reduced to amine by addition of hydride. from LiAlH4 H-
H H R C N -Al H Li+
H Al H Li+
R C N δ+ δ− nitrile
hydrolysis of nitrile
Past Paper Question
96 1A 3 f i 97 2B 6 a iii 98 2B 6 d ii 99 1A 5 b i
dehydration of amide
H3O+
H H R C N H
H
H
H
Glossary
H H R C N -Al H Li+
H primary amine
reduction of nitrile
96 2C 9 a i 97 2B 7 b iv 99 2B 7 b i
96 1A 3 f i 3f i Give the structures of compounds D and E in the following organic synthesis :
2
O C
HCN
H
+
D
H3O heat
OH D:
E
OH E:
CN
COOH
1 + 1 marks (Deduct ½ mark for each minor mistake in the structures.) 96 2C 9 a i 9a Each of the following conversions can be completed in not more than three steps. Use equations to show how you would carry out each of these conversions in the laboratory. For each conversion, give the reagent(s), conditions, and structure of the intermediate(s). (For answers with more than 3 steps, deduct 1 mark for each extra step.) i CH2Br CH2CH2NH2
(1) NaCN
CN (1)
C
(1) (? LiAlH4 or H2 / cat (Pd / Pt / Ni) heat (?
3 marks Some candidates erroneously used HCN in the first and LiAlH4/H in the last step. Acid hydrolysis should be followed by an alkaline work-up as the target molecule is an amine and not a salt. +
3
Reaction Mechanism Unit 14 97 2B 6 a iii 6a Each of the following conversions can be completed in not more than three steps. Use equations to show how you would carry out each conversion in the laboratory and for each step, give the reagent(s), conditions, and structure of the product. iii O CH3CH2C N
CH3CH2
C
CH3CH2
C
O O
97 2B 7 b iv 7b Identify J, K, L, M and N in the following reactions. iv O P2O5 heat
CH3CH2CH2 C NH2
M
98 2B 6 d ii 6d Identify K in the following reaction. ii CONH CN 2
K
99 1A 5 b i 5b Consider the following reactions : CHO
i
HCN
D
(1) LiAlH4 (2) H3O
+
E
Give the structures of D and E.
99 2B 7 b i 7b With no more than four steps, outline a synthetic route to accomplish each of the following transformations. In each step, give the reagent(s) used, the conditions required and the structure of the product. i O CH3CHCH2 C OCH2CH3 OH
CH3CHCH2CO2H CO2H
Page 3
Reaction Mechanism
Unit 15
Page 1
Topic
Reaction Mechanism
Unit 15
Reference Reading
20.0 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 102–110 Organic Chemistry, Solomons, 5th Edition pg. 342–349, 351–356, 360, 366–370 Organic Chemistry, Fillans, 3rd Edition pg. 127–141 Organic Chemistry, Morrison Boyd, 6th Edition pg. 317–318, 327–342, 357–362 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 423 Organic Chemistry, 6th Edition, Solomons, pg. 327-333
Syllabus
Electrophilic addition
Notes
Electrophilic reaction is a kind of reaction initiated by electrophile. It includes electrophilic addition and electrophilic substitution. VII. Electrophilic addition When an electrophile is attracted towards an negative, electron rich, centre, a bond may be formed between the electrophile and the substrate. A. Addition of HBr to alkene + δ Because bromine is more electronegative than hydrogen, the H–Br bond is polarized, δ H Br . -
Upon heterolytic bond fission, a H+ and a Br- will form. In this case, H+ ion is the electron deficient species and is an electrophile. It is attracted towards the electron rich double bond by the π electrons with the formation of a carbocation intermediate.
H+ C
H
r.d.s.
C
C
XH X
+ C
C
C
The reaction is completed by addition of the Br-, a nucleophile, to the carbocation. 1.
Markownikoff's rule.
For an unsymmetrical alkene, addition of hydrogen halide may have two possible orientation. e.g. H Cl H H C H 1 C H
2 C
C H
H
H H H 2-chloropropane
H
H H propene
C C
+
H Cl Cl H H H C
C C
H
H H H 1-chloropropane
If the H+ ion is attached to the C-1, the product will be 2-chloropropane. If the H+ ion is attached to the C-2, the product will become 1-chloropropane. It is found that in this reaction, 2-chloropropane is the major product and 1-chloropropane is the minor product. This can be explained by the difference in activation energy involved in the two different pathway.
Reaction Mechanism
Unit 15
Page 2
In 2-chloropropane formation, the rate determining step is the attachment of the H+ ion to the C-1 to form a 2º carbocation. While for the 1-chloropropane, the H+ ion is attached to the C-2 to form a 1º carbocation. As 2º carbocation is more stable than the 1º carbocation (refer to SN1 reaction), it involves a lower activation energy of formation and forms at a faster rate.
Formation of 2-chloropropane Cl-
H+ H H
H C
C
C H
H r.d.s.
H C
H + C C H
H H H 2° carbocation (more stable)
H H
H Cl H H C
C C H
H H H 2-chloropropane
Formation of 1-chloropropane Cl-
H+ H H
H C
C
C H
H H
r.d.s.
H H + H C C C H H H H 1° carbocation (less stable)
Cl H H H C
C C H
H H H 1-chloropropane
This observation is generalized and is called Markownikoff's rule. Markownikoff’s rule states that “in the addition of hydrogen halide to an alkene, the hydrogen atom adds to the carbon with more hydrogen atoms.” or “if an unsymmetrical alkene combines with a hydrogen halide, the halide ion adds to the carbon atom with the fewer hydrogen atoms.” To be more general, the electrophile joins to the carbon will more hydrogen atoms to yield a more stable carbocation.
Reaction Mechanism 2.
Unit 15
Page 3
Reactivity of alkene towards electrophilic addition
Because the rate of electrophilic addition is depending on the stability of carbocation intermediate. Alkene has the general order of reactivity towards electrophilic addition as follows. H C
H C
<
C H
H
CF3
H
H C
<
C H
H
H
R
R C
<
C H
R
R
H C
<
C R
H C H
–F is an electron withdrawing group which makes the carbocation intermediate less stable. –R is an electron donating group which make the carbocation intermediate more stable. H
H C C
is particularly reactive because the carbocation involved is stabilized by resonance.
H H H C
+ C
H H
Glossary
electrophilic addition
Past Paper Question
91 1A 1 a 92 2C 9 a iv 93 2C 9 b 94 1A 3 b i ii 95 1A 3 c i ii 96 2C 9 b 98 2B 5 c i
+
H H C
C
H H
Markownikoff’s rule
H H C
C
H H
+
Reaction Mechanism 91 1A 1 a CH3 1a H
Unit 15
Page 4 4
C CH2 E
Using E, give reagents and a mechanism to explain what is meant by each of the following : A reaction obeying Markownikoff’s rule. A polymerization reaction. Markownikoff’s rule The answer must involve “the addition of acid and explain and show the correct direction of addition”. Z
H H
Z
Me
Me
Me
+ ZMe
Me
Acid; could be HI, HCl, H2SO4, H3O+, HX but not HNO3 Mechanism: Correct use of arrows or H+ in the first step Intermediate Carbonium ion (draw or write) Correct product
½ mark ½ mark ½ mark ½ mark
X
e.g. but use of HX (max. ½ mark) If wrong product Me The mechanism is Electrophilic Addition Polymerization reaction Reagents: can be radical R· or ion either cationic polymerization (e.g. above using an acid) ½ mark or anionic polymerization - using a base e.g. KNH2 Mechanism: Correct intermediate or arrows or indicate chain initiation, propagation and termination. 1 mark Direction of addition is not important. Product: must clearly show a polymer ½ mark Radical e.g. R
R
CH2 CH
R
R
CH3 n
Cationic e.g. Me H+ CH2
Me CH2
H
etc. H
Me Me
CH2 CH
Me
CH3
CH CH2 CH
n
Anionic e.g. NH2- CH2
C
Me H
Me CH2
etc.
H
Me
Me
NH2 CH2 CH CH2 CH
CH2 CH CH3 n
Whilst Markownikoff seemed to present few problems, the polymerisation was obviously more difficult problems with reagents, mechanism and product.
92 2C 9 a iv 9a Give the structural formula(e) of the major organic product(s) from each of the following reactions : iv H3C excess HBr/CCl CHC CH
2
4
H3C CH3 CH3
Br Br CH3
2 marks
Reaction Mechanism Unit 15 93 2C 9 b 9b Outline the mechanism for the reaction between but-1-ene and HBr to give bromobutane. Explain why 2bromobutane is the major product, rather than 1-bromobutane. H Br
Br-
3
BrCH3CH2CH2
CH3CH2CH CH3
CH3CH2CH CH2
Page 5
CH2
or CH3CH2CH2
CH3CH2CH CH3
CH2 Br
Br
1-bromobutane
2-bromobutane
2-bromobutane is the major product because
CH3CH2CH CH3 2° carbocation
is more stable than
CH3CH2CH2 CH2 1° carbocation
.
(2 marks for mechanism, 1 mark for explanation) Many candidates simply stated the Markownikov's rule without offering any explanation.
C
94 1A 3 b i ii 3b i Give the major product from the reaction: CH3CH=CH2 + HBr → CH3CHCH3 Br
ii
1
2-bromopropane or isopropyl bromide
1 mark
Outline the mechanism of this reaction. Another product is also formed. Give its structure and account for the fact that it is only the minor product. Mechanism: Br
BrCH3 CH CH2
H Br δ+ δ-
CH3
C
CH3
CH3
C
CH3
H
H
1 mark 1 mark
Minor product is CH3CH2CH2Br It is a minor product because its intermediate The secondary carbocation
CH3
C H
CH3
CH3 CH2 C H
H
is less stable than the secondary carbocation
is more stable than the primary carbocation
CH3 CH2 C H
H
CH3
C
CH3
H
because the two
electron releasing methyl groups disperse the charge more and so stabilise the positive ion more effectively. 1 mark Or The carbocation formed exists as an equilibrium CH3
C
CH3
H more stable
C
e
CH3 CH2 C
3
H
H less stable
Mechanism – 1 mark Minor product – 1 mark Explanation – 1 mark ½ bonus mark – The electron release may be considered to be due to the inductive effect and the hyperconjugative effect. In the mechanism of the reaction, the wrong direction of the curly arrows was a common mistake.
.
Reaction Mechanism Unit 15 95 1A 3 c i ii 3c i Give the structure of the major product formed from the following reaction : CH3
Page 6 1
+ HBr
CH3 Br
ii
1 mark Outline a mechanism for the above reaction. (Movement of electron pairs should be indicated by curly arrows.)
2
CH3 H or CH3
C
CH3 Br
CH3
H Br δ+ δ-
Br-
2 marks Generally well answered. Some did not use complete arrows to indicate the breaking of covalent bonds.
96 2C 9 b 9b Give the structure of the major product in the following reaction and outline the mechanism of the reaction. (Movement of electron pairs should be indicated by curly arrows.)
3
CH2 + HBr
Br CH3
Major product :
1 mark Mechanism : -
Br
(?
CH2 (? (?
C
CH3
H Br
Br CH3
(?
2 marks
Candidates were poor in writing curly arrows. Many used H+ instead of HBr as the electrophile.
98 2B 5 c i 5c Give the structure of the major organic product, G, in (i) below. Outline a mechanism for the formation of the major product in the reactions. i H3C CH3 C
H3C
HBr
C
H
G
7
Reaction Mechanism
Unit 16
Page 1
Topic
Reaction Mechanism
Reference Reading
20.1–20.2 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 102–110 Organic Chemistry, Solomons, 5th Edition pg. 342–349, 351–356, 360, 366–370 Organic Chemistry, Fillans, 3rd Edition pg. 127–141 Organic Chemistry, Morrison Boyd, 6th Edition pg. 317–318, 327–342, 357–362 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 421–422, 424–425, Organic Chemistry, 6th Edition, Solomons, pg. 336-339, 346-351, 422-425
Syllabus
Electrophilic addition
Notes
B. Other relevant reaction 1.
Unit 16
Addition of Br2 to alkene
Bromine in CCl4 Normally, Br–Br molecule breaks symmetrically by homolytic fission. But under the influence of electron rich double bond, Br–Br bond will be polarized when the molecule is approaching the double bond. The polarization weakens the Br–Br bond and makes it break unsymmetrically.
Owing to the presence of lone pair, halogen atom is capable to form a single bond with the positive centre of the carbocation formed. Br
δ−
Br δ+ C
C
BrBr C
Br C +C
+ C
carbocation (does not form indeed)
Br-
Br
Br
C
C
bromonium ion
Bromonium ion intermediate is more stable than ordinary carbocation intermediate because the positive charge is dispersed over the 3 membered ring. In ordinary carbocation, the positive charge is concentrated on one carbon atom only.
Finally, the bromide ion Br- attacks the C atom from another side of the molecule and yield a 1,2-disubstituted bromoalkane. Overall reaction Br2 + R C C R R
R
in dark
Br Br R C
C
R R
R
Reaction Mechanism
Unit 16
Page 2
Bromine water In the presence of water, halohydrin will be formed instead of 1,2-dihalide. This is because H2O is a better nucleophile than Br-. H H O
Br C +C
H O
H
C
Br
O
Br
C
C
C
bromonium ion
halohydrin
H
H2O attacks the bromonium ion to form halohydrin. Anyway, bromine water can also be used to test for double bond or triple bond. N.B. 2.
Hydrin means a hydroxyl compound.
Addition of H2SO4 to alkene
Alkene doesn't reacts with dil. H2SO4(aq) normally. However, it reacts with conc. H2SO4(l) to give an addition product. OH O C
C
H O
S
H O H
C
O
O + C
O
-
O
S
O H
O
S
H
O
C
C
O
alkyl hydrogensulphate
sulphuric acid molecule
This reversible reaction can be used to extract alkene from a mixture with alkane. By bubbling impure alkene through conc. H2SO4(l), only alkene will dissolve in it. Conc. H2SO4(l) can be separated by separating funnel. By heating the conc. H2SO4(l), alkene will evaporate and can be recovered. a) Preparation of alkanol from alkene Since H2SO4 is a strong acid, HSO4- would be a very weak base, therefore, HSO4- is a very good leaving group. By adding water into alkyl hydrogensulphate, it can be converted to alkanol through nucleophilic substitution.. OH O
S
H
O
C
C
O
H nucleophilic substitution
H
H
O H
H O
C
C
C
C
O H alkyl hydrogensulphate
H
alkanol
Reaction Mechanism 3.
Unit 16
Page 3
Hydration of alkene
An alkene can also be converted to alkanol by direct hydration using the catalyst of aqueous sulphuric acid or phosphoric acid. Hydration of alkene proceeds through an ionic mechanism and obeys the Markownikoff’s rule. N.B.
Water is a nucleophile which does not react with electron rich double bond directly.
e.g. Hydration of propene O
+ H O H O
H H H
H
H C
C
C H
r.d.s.
H C
H
H
H
H
H C
C
C
H
H
H
H H H 2º carbocation (more stable)
H H
H
H
H H
+ C C
H
+O
H H O H
H H
H C
C C
H H H propan-2-ol
H
+ H O H H regenerated
In the above mechanism, H3O+(aq) is consumed in the first step and regenerated in the last step. It only acts as a catalyst. Since propan-2-ol can be prepared by direct hydration of propene while propan-1-ol cannot be, propan-2-ol is much cheaper than propan-1-ol. The cheap propan-2-ol is commonly used as the rubbing alcohol. For those alkanols involving formation of less stable 1º carbocation, e.g. ethanol, a more vigorous reaction condition is required. CH2 CH2 + H2O
H3PO4 300 ºC
CH3 CH2 OH
Comparing with H2SO4(l), H3PO4(l) is not an oxidizing agent with will not oxidize CH3CH2OH to CH3COOH.
Reaction Mechanism 4.
Unit 16
Page 4
Ozonolysis of alkene
The term ozonolysis means breaking down (-lysis) by ozone (ozon-). Ozone reacts with an alkene vigorously to form a very unstable intermediate initial ozonide which rearranges immediately to ozonide. Since the product is potentially explosive, it is carried out at a low temperature.
Upon treatment with zinc and water, the ozonide will be reduced to 2 carbonyl compounds, aldehyde or ketone, depending on the original structure of the alkene.
Since there is a direct relationship between the structure of the carbonyl compound obtained and the structure of the original alkene, ozonolysis is useful in locating the position of the double bond in an alkene.
CH3 CH3CHCH CH2 3-methylbut-1-ene
N.B.
(1) O3, CH2Cl2, -78 ºC (2) Zn / H2O
CH3 O CH3CH C H
O +
2-methylpropanal
H C H methanal
CH2Cl2 is only the solvent.
In the presence of the reductive zinc metal, the aldehyde formed will not be oxidized to carboxylic acid. If the alkene is a cyclic one, there will be only one product molecule instead of two. (1) O3, CH2Cl2, -78 ºC (2) Zn / H2O cyclohexene
O H C
O CH2 CH2 CH2 CH2 C H hexane-1,6-dial
These characteristics of ozonolysis reaction makes it particularly useful in determining the structure of an organic molecule. N.B.
The carbonyl compounds formed from ozonolysis can be identified by melting point determination of the oxime and 2,4-dinitrophenylhydrazone derivatives.
Reaction Mechanism 5.
Unit 16
Page 5
Preparation of ethane-1,2-diol
A diol can be prepared by oxidation of a double bond using cold dilute basic / acidic potassium permanganate or osmium tetroxide. N.B.
Basic potassium permanganate is more oxidative than acidic potassium permanganate and the reduction product would be brown MnO2(s) instead of colourless Mn2+(aq).
By cold dilute basic / acid potassium permanganate
By osmium tetroxide
Overall reaction
a) Oxidative Cleavage of double bond If an alkene is treated with hot basic potassium permanganate instead of a cold one, the product will not be a diol. The hot basic potassium permanganate is strong enough to break the double bond by oxidation. CH3CH2
C C
CH3CH2
CH3
(1) KMnO4, OH-, hot (2) H+
CH3
O
O H3C C
CH3CH2 C CH2CH3 + pentan-3-one
3-ethyl-2-methypent-2-ene
CH3
propanone
The reaction is similar to ozonolysis of alkene. The difference is that any aldehyde formed will be oxidized to carboxylic acid immediately.
(1) KMnO4, OH-, hot (2) H+ cyclohexene
O HO C
O CH2 CH2 CH2 CH2 C OH hexane-1,6-dioic acid
And any terminal =CH2 group will be oxidized completely to carbon dioxide and water. CH3 CH3CHCH CH2 3-methylbut-1-ene
(1) KMnO4, OH-, hot (2) H+
CH3 O CH3CH C OH + CO2
+ H2O
2-methylpropanoic acid
The products obtained from the oxidation can also be used to deduce the structure of the alkene.
Reaction Mechanism
Unit 16
Page 6
b) Comparison of ozonolysis and oxidative cleavage Ozonolysis 1) O3, CH2Cl2, -78ºC 2) Zn / H2O
Reagent used =CR2 (disubstituted terminal group)
Ketone
=CHR (monosubstituted terminal group)
Oxidative cleavage KMnO4, OH-, heat
O R C
Aldehyde
Ketone
R
R C
R
O R C O
=CH2 (non-substituted terminal group) 6.
Methanal
O H C
R
Carboxylate ion / carboxylic acid
O H C
O
O -
R C OH
CO2
H
Oxymercuration of alkyne (Preparation of ketone from alkyne)
Reagent : water, catalyst – Hg2+ ion (HgSO4) and strong acid (H2SO4) H R C C H
HgSO4(aq) H2SO4(aq)
R C
C H
O
keto-enol tautomerization
H R C C
H
O H
H Markovinkov orientation
keto form
In the presence of acid and mercuric ion (Hg2+) catalyst, water can be added to a triple bond to form an enol. Enol is then rearranged to form a ketone. The addition of the water to the alkyne follows the Markownikoff’s rule.
Glossary
bromonium ion hydrin diol osmium tetroxide
Past Paper Question
90 2C 8 a ii 91 2C 7 a ii 94 1A 3 a ii 96 2C 9 a ii iv 97 1A 5 a iv 98 1A 4 c i iii
alkyl hydrogensulphate phosphoric acid ozonolysis oxidative cleavage oxymercuration mercuric ion
97 2B 6 a i 98 2B 6 d iv
97 2B 7 a i 98 2B 7 a iv
ozonide
97 2B 7 b ii
90 2C 8 a ii 8a It is suggested that the structure of a compound having the molecular formula C12H11ClO4 is either A or B. COOH Cl
CH C
COOH CH3 CH3
COOH A
ii C
CH C
CH3 CH2Cl
COOH B
How would you show the presence of a carbon-carbon double bond in the side chain? Decolourization of bromine water / permanganate solution. 2 marks Some candidates proposed very clumsy ways to detect the presence of a carbon-carbon double bond e.g. ozonolysis.
2
Reaction Mechanism Unit 16 Page 7 91 2C 7 a ii 7a State with explanations, what you would observe in each of the following experiments, and write equations for the reactions. ii Propene is bubbled into aqueous alkaline potassium manganate(VII). 2½ The purple colour of KMnO4 turns green due to manganate(VI) ion MnO42-. As the ions are further reduced, the 1½ mark green turns to a brown suspension of MnO2. CH3 CH CH2
MnO4-
CH3CH CH2
OH OH
+ MnO2
1 mark Very few candidates recognized that MnO4- is reduced to MnO2 in alkaline medium; hence they gave wrong observations.
C
94 1A 3 a ii 3a Consider the two compounds: CH3CH2CHCH3
CH3CH2CH2NH2 and
OH C
ii
D
Dehydration of C gives 3 products, E, F, and G all with the formula C4H8. On treatment with ozone followed hydrolysis, E gives methanal among other products, but F and G do not give methanal. Give structures for E, F and G and an equation for the ozonolysis reaction involving E. CH3
F and G : E:
C
C
H CH3
CH3
C
CH3
CH3
cis-but-2-ene
H
and
H C
H
C
CH3 trans-but-2-ene
CH2 CH CH2 but-1-ene
CH2 CH CH2
O3, -78°C Zn / H2O
1 mark each 1 mark
O
O
H C
H + CH3CH2C
H
1 mark
Some candidates produced very poor drawings of the cis- and trans- isomers of but-2-ene.
96 2C 9 a ii iv 9a Each of the following conversions can be completed in not more than three steps. Use equations to show how you would carry out each of these conversions in the laboratory. For each conversion, give the reagent(s), conditions, and structure of the intermediate(s). (For answers with more than 3 steps, deduct 1 mark for each extra step.) ii Br O
C(CH3)3
H C
O
(CH3)3
3
C C(CH3)3
(? (1) O3 (2) (CH3)2S or Zn or Et3N or Ph3P (?
- +
(CH3)3CO K or alc. KOH, heat (1)
(1)
3 marks Some candidates omitted heating in the first step and wrongly used KMnO4 in the last step. Some used O3 and Zn together as one step instead of as two steps.
C iv
3
OH
OH
OH conc. H2SO4 or H3PO4 or P2O5 or Al2O3 (? heat (?
MnO4 / OH (? cold (?
(1)
C
4
3 marks Heating was omitted in the dehydration step. NaOH or NaOEt was wrongly used as the dehydrating agent. Most candidates did not give the correct conditions (cold and weakly alkaline) for the dihydroxylation of the alkene.
Reaction Mechanism Unit 16 97 1A 5 a iv 5a Give a structure for each of the compounds D, E, F. G and H: iv (1) O3 C8H16 G
(2) Zn / CH3CO2H
2
97 2B 7 b ii 7b Identify J, K, L, M and N in the following reactions. ii OH
5
CH2Br
98 2B 6 d iv 6d Identify M in the following reaction. iv CH (1) O3 (2) Zn, H2O
NaOH
3 1
4
M
98 2B 7 a iv 7a Briefly explain why each of the following proposed reactions cannot lead to the target molecule. In each case, using the same starting molecule, outline a feasible synthetic route to obtain the target molecule, with no more than three steps. In each step, give the reagent(s) used, the conditions required and structure of the product. iv Br CH3CH2CH CH3
12
K
98 1A 4 c i iii 4c On treatment with dilute H2SO4(aq), E gives mainly two isomeric compounds, F and G, both of which have the formula C4H8. On treatment with bromine, both F and G give a product H with formula C4H8Br2. i Draw structures for F, G and H. iii Outline the mechanism for the formation of H from either F or G.
3
9
3 3
97 2B 7 a i 7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Each test should include the reagent(s), the expected observation with each compound and the chemical equation(s). i and
CH2
5
butanal only
97 2B 6 a i 6a Each of the following conversions can be completed in not more than three steps. Use equations to show how you would carry out each conversion in the laboratory and for each step, give the reagent(s), conditions, and structure of the product. i (CH ) CCH CH (CH ) CC CH 3 3
Page 8
H3CC
CCH3
12
Reaction Mechanism
Unit 17
Page 1
Topic
Reaction Mechanism
Unit 17
Reference Reading
21.0 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 133–135, 183, 205, 208, 274, 276 Organic Chemistry, Solomons, 5th Edition pg. 636–644, 652, 842–843, 939–940, 946 Organic Chemistry, Fillans, 3rd Edition pg. 157–160, 205 Organic Chemistry, Morrison Boyd, 6th Edition pg. 517–518, 525–527, 529, 832, 898 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 639, 641–647 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 440–441, 445 Organic Chemistry, 6th Edition, Solomons, pg. 617-618, 655-658
Syllabus
Electrophilic substitution Aromaticity Arenium ion
Notes
VIII.
Electrophilic substitution
Benzene is another molecule containing multiple bond besides alkene and alkyne. With the presence of π electrons, it is also an ideal substrate for the electrophilic attack by an electrophile.
H
E+
H
H
H H
H
However, benzene reacts with bromine less readily than alkene or alkyne does. And, even if there is reaction, substitution product is the only product obtained. Br H
H
H
H
+ HBr H
H H
H
H
H
substitution product Br
Br
X
H
Br H
Br
H H
H
H H
addition product
The difference is caused by the aromaticity of the benzene ring.
N.B.
Aromaticity –
the extra stability associated with the benzene ring through resonance. Or to be more precise, any ring system with (4n + 2) number of π electron will have this kind of extra stability, where n is a natural number. For benzene, n = 1.
Reaction Mechanism
Unit 17
Page 2
When a Br–Br molecule is added to a benzene ring, the hybridization of 2 carbon atoms are change from sp2 hybridization to sp3 hybridization. The aromaticity of the benzene ring is lost. This involves a very high activation energy and is unfavorable. On another hand, the bromobenzene formed from substitution is still aromatic where the activation energy involved is relatively lower. H
H
Br sp2 H
H H
addition
+ Br
Br
sp3
H H
+ H
H H the aromaticity is retained
substitution
H
H
H Br
Br
Br H
H H
H H breaking down of aromaticity
Although the substitution product is aromatic, inevitably, the aromaticity of a benzene ring is lost temporarily when an electrophile is added to the ring. This makes substitution reaction of benzene occur only in the presence of a catalyst or under a very rigorous condition. E H
E+
H
H H
+
H
H
E
H
H
H
H
H
H
H
H
H
H H
H
r.d.s.
E
H
H
E H
H
H
Aromaticity is retored
H
+
H
+
H + H
H
Arenium ion (temproray lost in aromaticity)
H
H
H
H
H + H+ H
H
Arenium ion
The positive charge possessed by the electrophile is dispersed over the ring of arenium ion. And by losing a hydrogen ion, the aromaticity is restored. The reaction conditions and mode of reaction can be summarized as follows : Reaction condition Normal circumstance With Lewis acid catalyst In the presence of uv or radical initiator
Mode of reaction no reaction electrophilic substitution radical addition
Reaction Mechanism
Unit 17
Page 3
A. Representation of arenium ion The arenium ion intermediate can be represented by the drawings of several resonance structures or a single resonance hybrid. H E
H E +
+
H E
H E
OR
+
+
resonance hybrid
resonance structures
Glossary Past Paper Question
electrophilic substitution
electrophile
aromaticity
arenium ion
Reaction Mechanism
Unit 18
Page 1
Topic
Reaction Mechanism
Reference Reading
21.1.1–21.1.4 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 133–135, 183, 205, 208, 274, 276 Organic Chemistry, Solomons, 5th Edition pg. 636–644, 652, 842–843, 939–940, 946 Organic Chemistry, Fillans, 3rd Edition pg. 157–160, 205 Organic Chemistry, Morrison Boyd, 6th Edition pg. 517–518, 525–527, 529, 832, 898 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 639, 641–647 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 441–444, 447–448 Organic Chemistry, 6th Edition, Solomons, pg. 658-664, 930-931, 974
Syllabus
Electrophilic substitution
Notes
B. Other relevant reaction 1.
Unit 18
Sulphonation of benzene
Benzene reacts with fuming sulphuric acid (H2S2O7 = SO3 + H2SO4) readily at room temperature or with concentrated sulphuric acid (H2SO4) slowly. If necessary, moderate heating may be employed (≈ 80ºC) to accelerate the reaction. Indeed, the electrophile is not SO42- ion since it is negatively charged. It is the sulphur trioxide molecules, SO3. It is either present in the fuming sulphuric acid or from the self dissociation of concentrated sulphuric acid molecules. H2SO4 + H2SO4 d H3O+ + HSO4- + SO3
+
H2SO4
SO3H
+ H2O
Overall reaction benzenesulphonic acid
Since the reaction is reversible, benzenesulphonic acid can be desulphonated by heating to a temperature over 100ºC, or by diluting the sulphuric acid, or by passing steam through the sulphonic acid. Sodium salt of sulphonic acid can be used to prepare phenol by fusion with solid sodium hydroxide.
Reaction Mechanism
Unit 18
Page 2
a) Preparation of phenol SO3-Na+
O-Na+
350°C
+ 3 NaOH
+ Na2SO3 + H2O
sodium benzenesulphonate
O-Na+
OH + HCl
+ Na+Clphenol
2.
Nitration of benzene
Concentrated nitric acid does not react with benzene at all. Ordinary concentrated nitric acid contains only H+ ion and NO3- ion. NO3- ion is a nucleophile which does not react with benzene. In the presence of concentrated sulphuric acid, concentrated nitric acid reacts with it to produce nitronium ion / nitryl cation / (NO2+) which is an electrophile. Since sulphuric acid is a stronger acid than nitric acid, it protonates the hydroxyl group of the nitric acid and converts it into a better leaving group. Without sulphuric acid, the concentration of NO2+ ion in conc. nitric acid is rather low. 2HNO3 e NO2+ + NO3- + H2O Formation of nitronium ion (nitryl cation) O
+ O H O N O
O
+ + O H O N O- + HSO 4 H
S O
O
H
H
a better leaving group + + O H O N OH
+ O N O + H2O nitronium ion
Electrophilic substitution by nitronium ion O H
H H
+ O N O
O
H H
H
N+
r.d.s.
H
H
-
H H
+ H
H
Arenium ion O
N+ O
H H HSO4
-
+ O O N
H H
+
-
H H + H2SO4
H
-
H
H
Arenium ion
H
H
Reaction Mechanism
Unit 18
Page 3
Since nitro group –NO2 is an electron withdrawing group, it decreases the electron density of the benzene ring. Therefore, further nitration of nitrobenzene is more difficult and requires prolonged heating. O + O N
O + O N
O + O N
+ +
Formation of trinitrobenzene takes days of prolonged heating.
Reaction Mechanism 3.
Unit 18
Page 4
Halogenation of benzene
At room condition, halogen does not react with benzene at all. X2
no reaction
The reaction is found to be catalyzed by Lewis acid, e.g. FeCl3, FeBr3 and AlCl3. The Lewis acid accepts an electron pair from the halogen molecule. This polarizes the halogen molecule and promotes the formation of positive halogen ion. X X
FeX3 Lewis acid (an electron acceptor)
X+ + FeX4electrophile
Halogen X2 and iron can also be used instead because the halogen reacts with the iron readily to produce the Lewis acid e.g. FeX3 catalyst. 2 Fe + 3 X2 → 2 FeX3 Example of bromination of benzene is illustrated as follows :
Reaction Mechanism 4.
Unit 18
Page 5
Alkylation of benzene (Friedel-Crafts alkylation)
Similar to halogenation of benzene, benzene can also be alkylated in the presence of anhydrous AlCl3 and haloalkane. This method is named after the inventors, Charles Friedel and James M. Crafts. AlCl3 is a Lewis acid which catalyzes the reaction by promoting the formation of carbocation (an electrophile). Cl R Cl
R+ + Cl carbocation (an electrophile)
Al Cl Cl
Cl Al Cl Cl
Then, the carbocation attacks the benzene ring through an electrophilic substitution pathway. Eventually, the benzene ring is alkylated (replacing a hydrogen by an alkyl group) and AlCl3 is regenerated. H
H +
R
H
H H
R
H r.d.s.
H
H
H H
+ H
H
Arenium ion H Cl
Cl Al
Cl
H
Cl
R
H
+ H
R
H
H
Arenium ion
H H + HCl + AlCl3
H H
H
Alkylbenzene
Summary of different electrophilic substitution reaction of benzene
Glossary
Reaction Sulphonation
Reagent conc H2SO4(l) / H2S2O7(l)
Catalyst Nil
Condition 80 ºC
Electrophile SO3
Nitration
conc. HNO3(aq)
conc. H2SO4(l)
55 ºC
NO2+
Halogenation
X2
FeCl3 / AlCl3
X+
Alkylation
R–Cl
FeCl3 / AlCl3
R+
sulphonation fuming sulphuric acid benzenesulphonic acid fusion nitration nitronium ion / nitryl cation benzenamine / aniline halogenation Lewis acid Friedel-Crafts alkylation
Reaction Mechanism Past Paper Question
Unit 18
92 2B 6 Ab iii 96 2C 8 c iii 97 2B 7 b v
Page 6
92 2B 6 Bb iii
92 2B 6 Ab iii Bb iii What are the products of the reactions between iii concentrated nitric(V) acid and concentrated sulphuric(VI) acid? 2HNO3 + H2SO4 d NO2+ + H3O+ + NO3- + HSO46Bb What are the products of the reactions between iii concentrated sulphuric(VI) acid and concentrated nitric(V) acid? 2HNO3 + H2SO4 d NO2+ + H3O+ + NO3- + HSO4C The products of this reaction were not widely known. 6Ab
1 1 mark 1 1 mark
96 2C 8 c iii 8c Identify K, L, M, N and P in the following reactions : (Deduct ½ mark for each minor mistake; max. Deduction for the whole question = 2 marks) iii NO2
1
NO2
M CH3
CH3
C
NO2
M : conc. HNO3, conc.H2SO4 (heat) 1 or 0 mark HNO3 / H2SO4 or HNO2 or HNO3 for M. Both nitric(V) and sulphuric(VI) acids used should be concentrated and the abbreviation conc. should be used instead of c;
97 2B 7 b v 7b Identify J, K, L, M and N in the following reactions. v (1) conc. H2SO4 (2) NaOH(aq)
N
5
Reaction Mechanism
Unit 19
Page 1
Topic
Reaction Mechanism
Unit 19
Reference Reading
21.1.5–21.1.6 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 133–135, 183, 205, 208, 274, 276 Organic Chemistry, Solomons, 5th Edition pg. 636–644, 652, 842–843, 939–940, 946 Organic Chemistry, Fillans, 3rd Edition pg. 157–160, 205 Organic Chemistry, Morrison Boyd, 6th Edition pg. 517–518, 525–527, 529, 832, 898 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 639, 641–647 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 476–477, 521, 523–524 Organic Chemistry, 6th Edition, Solomons, pg. 930-931, 974
Syllabus
Electrophilic substitution
Notes
5) Diazocoupling of diazonium ion NH2 NaNO2 HCl, 0-5 蚓
benzenamine (an aromatic primary amine)
+ N N
diazonium ion
When benzenamine is treated with HNO2(aq) (or NaNO2(s) / HCl(aq)) at 0–5 ºC, diazonium ion is formed. Beside the substrate of nucleophilic substitution, diazonium ion is also an electrophile which is capable to accept electron form a benzene ring. Nucleophilic substitution of diazonium ion (SN) (The high leaving ability of N2 makes diazonium ion a good substrate of nucleophilic substitution)
+ N N
Nu
-
+ N2
Nu
diazonium ion
Electrophilic substitution of diazonium ion (SE) (The positive charge and the multiple bond on the azide group makes diazonium ion a good electrophile)
+ N N
H OH
N
-
OH
N H diazonium ion
naphthalen-2-ol
OH
+
N N
+ H2O OH
azo dye (an orange red ppt.)
Phenol compound (e.g. naphthalen-2-ol) is particularly reactive to electrophilic substitution because the electron density of the benzene ring is enhanced by the hydroxyl group. H
H
H
O
O+
O+
-
-
If napthalen-2-ol is used, the substitution product formed is an orange red ppt. The colour of the ppt. is so intense that it can be used as a dye. Since the compound contains nitrogen, it is called azo dye.
Reaction Mechanism
Unit 19
Page 2
a) Colour of a substance Normally, most simple organic molecules are colourless because they absorbs ultra-violet and infra-red only. A substance is coloured only if it absorbs light from the visible region. The colour of azo dye is due to the presence of the extensive π system. This makes the difference between the energy levels smaller. Therefore, the molecule will absorb visible light during electron transition. For the same reason, some transition metal ions are coloured in the presence of water. (This will be discussed in the section of transition metal : d-d transition). H
H
O N N
N N
Lewis structure of azo dye
6.
O
Orbital representation of azo dye
Bromination of phenol
Through resonance, the lone pair on the oxygen is donated to the benzene ring. This makes the ortho (2,6) and para (4) positions of the benzene more electron rich than the meta (3,5) position does. Therefore, ortho and para positions of the benzene ring is more vulnerable to the attack of an electrophile. The hydroxyl group is said to be ortho-para directing. H
H
H
H
O
+O
+O
O -
-
-
Phenol reacts with aqueous bromine solution readily to form white precipitate of 2,4,6-tribromophenol. The formation of the ppt. can be served as a test for phenol. Since the yield of the reaction is almost 100%, the reaction can also be used in quantitative analysis (e.g. titration). H
H
O
O + 3 Br2
Br
Br
+ 3 HBr
Br 2,4,6-tribromophenol (a white precipitate)
Because phenol is found to be more reactive than ordinary benzene, hydroxyl group is also known as an activating group in electrophilic substitution of benzene.
Glossary
diazocoupling meta position
napthalen-2-ol dye azo dye ortho position ortho-para directing activating group
para position
Reaction Mechanism
Unit 19
90 1B 4 b ii 91 1B 5 b 94 2C 9 a iv 95 1A 3 d ii 96 1B 4 b 97 1B 8 a i ii 98 2B 7 a ii
Past Paper Question
Page 3
91 2C 7 a iii 96 2C 8 c ii
90 1B 4 b ii 4 A mixture contains equal amounts of X, Y and Z. OH
X b.p. 182°C
4b ii
CH3
NH2
Br Z b.p. 184°C
Y b.p. 184°C
Outline tests involving observable colour changes to show that: Y is a primary aromatic amine, and Add NaNO2 and dil. HCl to Y at 5ºC followed by 2-naphthol (β-naphthol). An orange ppt. denotes presence of a primary aromatic amine. 1 mark
91 1B 5 b 5a
CH3
CH3
NH2 X
5b
NO2 Y
The amine X, C7N9N, may be prepared from Y, C7H7NO2, by reaction with excess hot granulated tin and concentrated hydrochloric acid. Describe simple chemical tests which would enable a distinction to be made between X and phenylmethylamine (C6H5CH2NH2). Your answer should include chemical equations, reaction conditions and a description of any visible change. For X CH3 NH2
0°C
OH
N N Cl+ no observable change
C
NH2
HONO/HCl
4
N N Ar
CH3
HONO/HCl
OH
orange or red or yellow ppt.
Basic reaction with HONO/0ºC for both X and phenylmethylamine No observable change in the first step Formation of diazonium salt Use of 2-naphthol Formation of orange or red or yellow ppt. For phenylmethylamine Ph
1
1 mark ½ mark ½ mark ½ mark ½ mark
OH + N2
Ph
0°C
Formation of alkanol ½ mark Nitrogen gas bubbles observed. ½ mark No coloured ppt. with β-naphthol (2-naphthol). Many cases of use of aryl sulphonic acids suggest some very bad teaching – as the use of sulphonic acid to distinguish amines is not in the syllabus.
91 2C 7 a iii 7a State with explanations, what you would observe in each of the following experiments, and write equations for the reactions. iii Aqueous phenol is treated with aqueous bromine. 2 The reddish brown solution of bromine is decolourized. Tribomophenol appears as a white precipitate. 1 mark OH
OH
Br
Br
+ Br2 Br
C
1 mark Most candidates did not give full observations. They only mentioned decolorisation of bromine or formation of white precipitate, but not both.
Reaction Mechanism Unit 19 94 2C 9 a iv 9a Give the structural formula(e) of the major organic product(s) (P to V) formed in each of the following reactions. NH2 iv NaNO / HCl 2
< 5°C
+ N N
S:
S
OH
Page 4
2
T
N N
T:
1 mark each
OH
C
Candidates were casual / careless in writing structural formula, particularly in putting the number of H atoms to each C atom. Common mistakes included : OH
for T (wrong position of coupling)
N N
95 1A 3 d ii 3d ii Write the structure of the product formed in the diazocoupling of naphthalen-2-ol with benzenediazonium chloride.
1
N N OH
C
(the diazocoupling must at the correct position) 1 mark The structure of the diazo-coupled product was generally poorly drawn, with no indication of the correct regiochemistry and the correct valence of nitrogen.
96 1B 4 b 4b Describe the experimental procedures of a chemical test that you would employ in order to demonstrate that a liquid-sample is an aromatic primary amine. State the observation of your test. Add NaNO2/HCl (or HNO2) in a test tube. 1 mark 1 mark Cool the mixture to 0 - 5ºC / ice temperature. Add napthalen-2-ol (β-naphthol) / phenol(aq). 1 mark 1 mark Observation : bright red ppt. (with naphthalen-2-ol) / bright orange ppt. (with phenol) (Award ½ mark for coloured precipitate) C Fairly well answered. Most candidates correctly described the procedures involved in the chemical test. But the name 'naphthalen-2-ol' was spelt wrongly in many cases. Some candidates gave chemical tests which were non-specific, e.g., heating the intermediate diazonium salt to give bubbles of nitrogen gas could be used to show that the given liquid sample was an aromatic primary amine. 96 2C 8 c ii 8c Identify K, L, M, N and P in the following reactions : (Deduct ½ mark for each minor mistake; max. Deduction for the whole question = 2 marks) ii + CH3
4
1
N N Cl
L
N N HO
L : HO
CH3
97 1B 8 a i ii 8a i Give the reaction conditions and the necessary chemicals/reagents required for the formation of benzenediazonium ion from phenylamine. ii Benzenediazonium ion reacts with naphthalen-2-ol to give compound T. Give the structure of T.
1 mark
3
Reaction Mechanism Unit 19 98 2B 7 a ii 7a Briefly explain why each of the following proposed reactions cannot lead to the target molecule. In each case, using the same starting molecule, outline a feasible synthetic route to obtain the target molecule, with no more than three steps. In each step, give the reagent(s) used, the conditions required and structure of the product. ii OH
NH2
heat
N N
OH
Page 5 12
Reaction Mechanism
Unit 20
Topic
Reaction Mechanism
Reference Reading
22–23 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 305–306 Organic Chemistry, Solomons, 5th Edition pg. 262–263, 267–274, 361–366 Organic Chemistry, Fillans, 3rd Edition pg. 127, 141–142 Organic Chemistry, Morrison Boyd, 6th Edition pg. 46–49, 351–354, 356–357, 388–389 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 910, 912–914 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 407–409, 426–431, 445–446 Organic Chemistry, 6th Edition, Solomons, pg. 334-335, 366-379, 393-397
Syllabus
Free radical Substitution (chlorination of methane) Free radical Addition (polymerization of alkene)
Notes
IX. Free radical Reaction
Page 1
Unit 20
Free radical is a species containing 1 or more unpaired electrons. Halogen atom and alkali metal atom are two examples of them. Oxygen molecule may also be considered as a free radical. This is because it contains two unpaired electron and is paramagnetic.
Examples of free radicals As free radical does not have a fulfilled octet, it is an electron deficient species. If possible, it tends to react with an electron rich species e.g. an double bond. Most of the free radicals are very reactive and cannot be isolated.
Reaction Mechanism
Unit 20
Page 2
A. Formation of free radical Free radical is formed in 3 ways : 1) Homolytic fission of a bond
A
B
Cl
Cl
A + B hυ or heating
heat
R O O R
Cl
+
Cl
chlorine free radical R O
+
O
R
The energy required to break the bond is furnished by exposing the molecule to light (photolysis, hυ) or by heating of the molecule. A radical forms more readily if atom A and B have similar electronegativity and the A–B bond is weak. Peroxide molecule containing weak O–O bond is usually used as a radical source. 2) By reaction of a molecule with another free radical
X A
A B
+ B X
a new free radical R O
R O H + Br
H Br
When a free radical gets in contact with another molecule, it may gain an electron from the molecule in order to attain octet. The molecule which loses an electron to the free radical becomes another free radical. Since the product is still a free radical, the reaction may continue. 3) By oxidation-reduction process H CH3 CH2
Br
Na
CH3 C
+ Br- + Na+
H
Radical can be formed from a molecule by a single electron transfer.
Reaction Mechanism
Unit 20
Page 3
B. Free radical Substitution 1.
Chain reaction e.g. chlorination of methane Most free radicals reacts through a chain mechanism. A chain mechanism consists of three steps : 1. chain initiation, 2. chain propagation, 3. chain termination. Besides free radical reaction, nuclear fission of 235U in nuclear bomb is also a chain reaction. Chlorination of methane is a typical example of a chain reaction.
a) Chain initiation (chain initiating step) Cl
Cl
hυ
Cl
+
Cl
Cl· radical is generated.
In the presence of light or heat, Cl–Cl bond breaks symmetrically and yields 2 Cl· radical. b) Chain propagation (chain propagating step) H Cl
H C
H H
Cl
H +
H
C
Cl
Cl· radical is consumed.
H
H Cl
H
C
H H
Cl
+ Cl
H
C
H
Cl· radical is regenerated.
H
Depending on the amount of the reactants present, mono-substituted chloroalkane may undergo further substitution. Cl Cl
H C
Cl H
Cl
H +
H
C
Cl Cl
Cl
H
Cl· radical is consumed.
H
C H
Cl H
Cl
+ Cl
C H
H
Cl· radical is regenerated.
Reaction Mechanism c)
Unit 20
Page 4
Chain termination (chain terminating step) Cl
+
Cl
Cl
Cl H
H Cl
C
Cl
H
C
H
H
H H
H
H C
C
H
H
+
Energy
radical is destroyed.
H H H
H C C H H H
When two free radicals meet each other, they may join together to form a molecule. However, free radical is very reactive, the molecule formed will also be very energetic. It breaks apart instantaneously and forms two free radicals again. The molecule formed will become stable only if the energy of the molecule is dissipated immediately. This happens when the molecule collides with the wall of the container once it is formed and transfers the energy to the wall. Since the occurrence of the chain termination is rather rare, it is estimated that a single Cl· radical may lead to chlorination of thousands of methane molecule. Apparently, there is more than 1 way of chain propagation and chain termination, the presence of a mixture of products is always a characteristic of radical reaction. 2.
Reaction between H2(g) and Cl2(g)
Reaction between H2(g) and Cl2(g) is also a free radical substitution reaction where the rate of reaction is very sensitive to the presence of light. Under controlled condition, H2(g) burns in Cl2(g) smoothly to produce HCl(g), from which HCl(aq) is prepared. However, if a mixture of H2(g) and Cl2(g) is exposed to strong light, the result may be explosive. Chain initiation hυ → Cl· + Cl· Cl–Cl
Chain propagation Cl· + H–H → H–Cl + H· H· + Cl–Cl → H–Cl + Cl· Chain termination H· + H· → H–H + energy Cl· + Cl· → Cl–Cl + energy
Reaction Mechanism
Unit 20
Page 5
C. Free radical Addition Besides the reaction with saturated compound, free radical reacts with unsaturated compound more readily because radical is an electron deficient species. 1.
Chain reaction e.g. polymerization of alkene
In polymerization of alkene, an organic peroxide is usually added into the alkene as a radical initiator. Diacylperoxide is commonly used. Chain initiating step O
O
R C O O C
O
heat
R
2R
2 R C O
+ 2 CO2
alkyl radical
Diacylperoxide
When diacylperoxide is heated, it dissociates into alkyl radical and carbon dioxide consequently. Chain propagating step H R H
H H
H C
R C
C H
H H
H
H H H H
H C
R C C
C
H
H H
C
H H
R C C C H
C
etc.
long chain polymer
H H H H
The electron deficient alkyl radical reacts with double bond of the alkene repeatedly and results in a very long chain polymer. Chain terminating step
R
H H
H H
(CH2 CH2)n C C H H
C C
H H R
(CH2 CH2)n C
C
H H
(CH2 CH2)n R
combination
H H H H R
H H
H H C C
(CH2 CH2)n R
(CH2 CH2)n C
C
C C
(CH2 CH2)n R
H H H H
disproportionation
H H R
(CH2 CH2)n C
H H oxidation product
H H
C
H C C
H
H H
(CH2 CH2)n R
reduction product
There are 2 possible terminating steps : combination and disproportionation In combination, two radicals combine together to form a molecule. In disproportionation, a hydrogen atom is transferred from a radical to another. Thus, the one lost the hydrogen is oxidized and the other radical is reduced.
Reaction Mechanism 2.
Unit 20
Page 6
Anti-Makownikoff orientation H Cl H H C H 1 C H
H 2 C C H
C C
H
H H H 2-chloropropane +
H Cl Cl H H
H H propene
H C
C C
H
H H H 1-chloropropane
In the electrophilic addition of HCl to propene, 2-chloropropane is found to be the major product because its formation involves a lower activation energy. The mechanism is also known as ionic addition of alkene where a hydrogen ion is added to the propene molecule first. -
Cl
+
H H H
H C
C C H
r.d.s.
H H C
H Cl H
H + C C H
H C C C
H
H H H 2-chloropropane
H H H 2?carbocation (more stable)
H H
However, in the presence of an peroxide (a radical initiator), 1-choropropane will become the major product. It is obvious that the two reactions have 2 different reaction pathways. When an peroxide is added, the reaction proceeds through an radical mechanism. Similar to a carbocation, a radical is also an electron deficient species. The influence of the substituent on the stability of a radical is similar to that of a carbocation. Relative Stability of Free radical
H C H benzyl radical
R >
R C R 3 radical
R >
R C H
H
R >
H C H
>
H C H methyl radical
Reaction Mechanism
Unit 20
Page 7
Chain initiation heat
R O O R R O
R O
+
O
R
R O H + Cl
H Cl
In the chain initiation step, the free radical formed from homolysis of the organic peroxide subtracts an hydrogen atom from the H–Cl molecule. An Cl· radical is formed first. Chain propagation H Cl Cl H H
H C
C
C H
H H
r.d.s.
Cl
H
H C
C C
H H H 2º radical (more stable)
Cl H H H
H C
C C
H + Cl
H H H 1-chloropropane
The Cl· radical is added to the propene molecule and a 2º radical is formed. This accounts for the fact that 1chloropropane is the major product of the reaction. The main difference between the ionic mechanism and the radical mechanism is that in ionic mechanism a H+ ion is added to propene first while in radical mechanism a Cl· radical is added to propene first. The effect of the peroxide on the orientation of the addition is also known as “peroxide effect”.
Glossary
free radical species paramagnetic homolytic fission photolysis free radical free radical substitution chlorination of methane chain reaction chain initiation chain propagation chain termination free radical addition polymerization of alkene diacylperoxide radical initiator combination / coupling disproportionation anti-Makownikoff’s rule peroxide effect
Past Paper Question
91 2C 7 a i 94 2C 9 b ii 97 1A 4 c ii 98 2B 6 d iii 99 1A 5 a i ii iii
91 2C 7 a i 7a State with explanations, what you would observe in each of the following experiments, and write equations for the reactions. i A mixture of pentane and bromine in tetrachloromethane is exposed to sunlight. 3½ The reddish brown bromine turns colourless. The reaction is a free-radical chain reaction. Depending on the amount of bromine, a variety of substitution products is obtained. 2½ mark CH3CH2CH2CH2CH3 + Br2 → CH3CH2CH2CH2CH2Br + CH3CH2CH2CH2CHBr2 etc. 1 mark
Reaction Mechanism Unit 20 94 2C 9 b ii 9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC) respectively. PVC is more rigid and durable than PE, but incineration of PVC causes a more serious pollution problem. ii Use equations to show a mechanism of the polymerization of ethene. The polymerisation of CH2=CH2 is a free-radical addition reaction. An organic peroxide is added as an initiator. Initiation R O O
heat
R
O
2R O
O
R C O O C
O
heat
R
alkyl radical
Diacylperoxide OR Propagation H R H
R C
C H
H H
C
H H
H
R C C
H H H H
H C
C
H
H H
1 mark
H H
H C
+ 2 CO2
2R
2 R C O
R C C C H
C
etc.
long chain polymer
H H H H
1 mark
Termination H H R
(CH2 CH2)n C
H H
C
C C
H H
H H
H H R
(CH2 CH2)n C
(CH2 CH2)n R
C C
H H
H H
H H H H R
(CH2 CH2)n C
C
C C
(CH2 CH2)n R
H H H H
H H
C
combination
(CH2 CH2)n R
disproportionation
H H R
(CH2 CH2)n C oxidation product
OR
H H
C
H C C
H
H H
(CH2 CH2)n R
reduction product
1 mark C
Mistake in using '
' instead of '
' to describe one-electron shift in free radical reaction.
97 1A 4 c ii 4c ii Outline a free radical mechanism for the conversion of ethene to poly(ethene). Your answer should include appropriate arrows to show how the new bonds are made. 98 2B 6 d iii 6d Identify L in the following reaction. iii H2C
CH Cl
L
CH2 CH Cl
n
99 1A 5 a i ii iii 5a Under certain conditions, methane reacts with chlorine to give chloromethane as the major product. i State the conditions for the reaction. ii Outline the mechanism and name the mechanistic steps of the reaction. iii Is the reaction of methane with chlorine an appropriate method for the preparation of dichloromethane ? Explain.
Page 8
3
Amino acids I.
Amino acids A. B.
Zwitterion Polypeptides
Amino acids
Page 1
Topic
Amino acids
Reference Reading
24 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 278–279 Organic Chemistry, Solomons, 5th Edition pg. 1096 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 524–525 Organic Chemistry, 6th Edition, Solomons, pg. 1146-1148
Syllabus
Amino acid Zwitterion Polypeptide
Notes
I.
Amino acids
Amino acid is a bifunctional compound containing amino group (–NH2) and carboxyl group (–COOH). NH2 O
O H2N CH2 C O H α
CH3 CH C O H α β
aminoethanoic acid
2-aminopropanoic acid
Both aminoethanoic acid and 2-aminopropanoic acid are also called α-amino acid because the amino group is attached to the α carbon. All the naturally occurring amino acids are α-amino acids, the only difference between them are the –R attached to the α carbon. Amino acid R = R O H2N Cα C O H H
H
Glycine
CH3
Alanine
CH3
Valine
CH CH3
α-amino acid
CH2
OH
Tyrosine
A. Zwitterion stronger acid weaker acid
Acid CH3COOH CH3NH3+
pKa 4.76 10.6
Conjugate base CH3COOweaker conjugate base CH3NH2 stronger conjugate base
Since CH3COOH is a stronger acid than CH3NH3+, thus CH3NH2 is a stronger base than CH3COO-, the carboxyl group is capable to react with the amino group by intramolecular proton migration. H R O + H N C C O H H H Cationic form (Predominant at low pH)
+
-H + +H
H R O + H N C C O H H Zwitterion (Dipolar ion)
+
-H + +H
H R O H N C C O H Anionic form (Predominant at high pH)
At low pH, the concentration of H+(aq) is high, the cationic form of amino acid will be predominant. At high pH, the concentration of H+(aq) is low, the anionic form of amino acid will be predominant. At medium pH, beside cationic form and anionic form, amino acid may also exist in form of dipolar ion (also known as zwitterion. Because of the of the formation of zwitterion, amino acids has 1. high melting point 2. high solubility in water but low solubility in organic solvent 3. large dipole moment..
Amino acids
Page 2 B. Polypeptides Amine is capable to condense with carboxylic acid through nucleophilic pathway to give an amide. (Refer to Mechanism of nucleophilic addition) H
O
R N H + amine
O H
R' C OH
R' C N
carboxylic acid
R +
H2O
amide group (or peptide linkage)
Amino acid is a bifunctional compound which is capable to undergo successive condensation (polymerization) with elimination of water. H H2N C
COOH
successive condensation H2N
CONH R
R
CONH R
CO2H R
polypeptide (a tripeptide consists of 3 monomers)
amino acid
The polymer formed is also known as polypeptide (protein) since the amino acids (monomers) are linked together by peptide linkage (amide group). Moreover, polypeptide can also be hydrolyzed back to amino acid in the presence of enzyme or acid / alkali catalyst. e.g. reflux with 6M HCl(aq) for 24 hours.
Glossary
amino acid polypeptide
Past Paper Question
91 1A 1 c 93 1A 3 e i iii 94 1A 3 c ii 98 1C 10 99 1A 6 a
bifunctional polymerization
a-amino acid zwitterion (dipolar ion) amide group (peptide linkage)
predominant
98 2B 6 a ii iii 99 2B 5 a v
91 1A 1 c 1c Give the structure of a dipeptide. Give a reagent that could cleave your dipeptide into smaller units. Draw the zwitter-ionic form of one such smaller unit R O
R O
R'
R'
NH3CHCNHCHCO2
NH2CHCNHCHCO2H
4
or R, R’ can be H, Me, Et, Ph, etc. (must be specified, if no -½) Note: must be from α-amino acids. (If an amide from NON-α-amino acid, with –CO2H and –NH2 1 mark) (If just an amide, e.g. RNHCOR -1½ mark) dil. H2SO4 or enzyme by acid hydrolysis (any acid) (Alkali causes racemisation, therefore ½ mark only)
2 marks
1 mark
R NH3CHCO2
C
1 mark Zwitter ion: This was considered by many to be very short and easy for 4 marks. However it was not very well answered especially since many misused R in such dipeptide structures as NH2RCO2H.
Amino acids 93 1A 3 e i iii 3e A dipeptide, W, yields only valine and glycine on hydrolysis. CH3
CH CH CO2H
H2NCH2CO2H
CH3 NH2 valine
i
glycine
How many different structural isomers are possible for W? Draw one of these structures. 2 Structural isomers are possible for W. O C
NH2 O
Page 3
H
H N
CH3CHCH C N CH2CO2H
2 ½ mark
CH2 NH2 O
CH3CHCH C OH
CH3
CH
3 or 1½ mark iii Draw the structure of the predominant form in which glycine exists in aqueous solution: with acidic conditions and under basic conditions.
CH3
CH CH CO2H
under acidic conditions
CH3 NH3 + CH3
under basic conditions
1 mark
CH CH CO2CH3 NH2
1 mark
94 1A 3 c ii 3c Hydrolysis of a protein gives rise to a number of amino acids. ii Draw the structure of an amino acid in its zwitter-ionic form.
1
N+H3 CH3
CH CO2α
1 mark
No mark for non-ionic form 97 2B 5 b i ii 5b The following equation represents the acid hydrolysis of a dipeptide D to produce compounds E and F. one of which is a chiral compound. CH3 O H2N CH C NHCH2CO2H D
i ii
H3O
+
E + F
Name all functional groups in D . Give one structure for E and one for F. Draw a suitable representation for the chiral product.
98 2B 6 a ii iii 6a Consider the structures of the two synthetic polymers shown below. CH2CH2
n
HN(CH2)6NHCO(CH2)4CO
poly(ethene)
n
nylon-6.6
ii Briefly explain why aqueous acids can more readily attack nylon-6,6 than poly(ethene) inducing degradation. iii Apart from acidic conditions, state one other condition under which nylon-6,6 degrades more readily than poly(ethene). 99 1A 6 a 6a Consider the amino acids, F and G. H + H3N C CO 2 + H3NCH2CO2
F
G
Draw three-dimensional structures of all dipeptides formed from F and G. 99 2B 5 a v
2
Amino acids 5a Identify D, E, F, G and J in the following reactions. v O +
NH
H3O heat
J
(Hint: J is a polymer.)
Page 4
Oxidation and Reduction I.
II.
Oxidation A.
Combustion of alkane
B.
Oxidation of alkanol and aldehyde
C.
Oxidation of aromatic side chain
Reduction A.
Reduction of nitrobenzene
B.
Catalytic hydrogenation (Hydrogenation of alkene)
Oxidation and Reduction
Page 1
Topic
Oxidation and Reduction
Reference Reading
25 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 95, 144–145, 178, 243–244 Organic Chemistry, Solomons, 5th Edition pg. 669, 454–456, 458, 714 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 475–476, 518 Organic Chemistry, 6th Edition, Solomons, pg. 165-166, 473-475, 690-691, 915-916
Objectives
Oxidation and reduction Hydrogenation of double bond
Notes
Oxidation and Reduction I.
Oxidation
A. Combustion of alkane y y CxHy + ( x + 4 ) O2 → x CO2 + 2 H2O y z y CxHyOz + ( x + 4 - 2 ) O2 → x CO2 + 2 H2O B. Oxidation of alkanol and aldehyde Reagent : any strong oxidizing agent – e.g. KMnO4 / H+(aq) or K2Cr2O7 / H+(aq) Both 1º and 2º alkanol can be oxidized by strong oxidizing agent. 1º alkanol can be oxidized to aldehyde and then carboxylic acid easily. 2º alkanol will only be oxidized to ketone. Since 1º and 2º alkanol turns K2Cr2O7/H+(aq) green, K2Cr2O7 / H+(aq) cannot tell which is 1º alkanol and which is 2º alkanol. 1º and 2º alkanol can only be distinguished by Luca's test. 3º alkanol can easily distinguished from 1º and 2º alcohol since it has no reaction with common oxidizing agent. H R C O H
[O]
[O]
R C
H 1° alcohol
H
aldehyde
R R C O H
O
[O]
R C O H carboxylic acid
O R C
H 2?alcohol
O
R
ketone
R R C O H R 3?alcohol
[O]
Resistant to oxidation
In the oxidation of the primary alkanol, aldehyde can be prevented from further oxidation by distilling it off the reaction mixture once it is formed. This is because aldehyde has a lower boiling point than both alkanol and carboxylic acid.
Oxidation and Reduction
Page 2
C. Oxidation of aromatic side chain Reagent : a very strong oxidizing agent – hot KMnO4 / OH-(aq) or hot H2CrO3 (chromic acid) N.B. chromic acid is the same as K2Cr2O7(aq) in acidic medium. benzylic carbon
benzylic hydrogen H
O
H C H
C OH +
3 [O]
1) KMnO4/OH- (aq), heat
+ H2O
2) H+(aq)
benzenecarboxylic acid (benzoic acid)
Benzenecarboxylic acid can be prepared by oxidation of aromatic side chain using a very strong oxidizing agent. The oxidation involves the abstraction of benzylic hydrogen by the oxidizing agent. i.e. all alkyl groups containing benzylic hydrogen can be oxidized to carboxyl group. benzylic hydrogen H R
C
R
O R
C OH 1) KMnO4/OH- (aq), heat 2) H+(aq)
Oxidation of aromatic side chain is not limited to alkyl group. Alkenyl, alkynyl and acyl groups are also oxidized by hot alkaline potassium permanganate in the same way.
R
C
no benzylic hydrogen R 1) KMnO4/OH- (aq), heat 2) H+(aq)
resistant to oxidation
Oxidation and Reduction
Page 3
II. Reduction A. Reduction of nitrobenzene In organic synthesis, nitrobenzene can be converted to benzenamine / aniline by reduction. Fe in dil. HCl(aq) or Sn in conc. HCl(aq) can be employed for this purpose. The ammonium salt –NH3+ obtained, is then neutralized by OH(aq). NO2
NH2 (1) Sn / conc. HCl (2) OHbenzenamine / aniline
nitrobenzene
In Sn / conc. HCl, Sn is oxidized to Sn2+ by H+(aq) first. Since Sn2+ is a very strong reducing agent which is oxidized to Sn4+ readily, it can reduce the nitro group to amino group. High concentration of Cl- ions form complex [SnCl4]22+ 2+ (aq) with Sn . This makes Sn a little bit more stable and suitable to be used as a reagent in the laboratory. B. Catalytic hydrogenation (Hydrogenation of alkene) H2C=CH2(g) + H2(g) → CH3–CH3(g)
(Ni catalyst, 400ºC)
As hydrogen is adsorbed to the catalyst surface, the formation of hydrogen–metal bonds provides sufficient energy to dissociate H–H bonds. Hydrogen atoms are thereby made available on the catalyst surface for the reaction with the π bond. As hydrogen atoms are removed, the catalyst surface is regenerated and becomes available to adsorb more hydrogen molecules. The ethene molecule may also be adsorbed on the metal surface. This weakens the π bond in the molecule as well.
Glossary
chromic acid
benzylic hydrogen
alkenyl group
alkynyl group
Oxidation and Reduction
Past Paper
91 1B 5 a ii 5a
91 1B 5 a ii 91 2C 9 a iv 93 1A 3 a iii 95 2C 9 b iii 97 1A 5 b 98 2B 6 d i 99 2B 5 a ii iii
CH3
97 1A 6 a i
97 2B 7 a iv
CH3
NH2
NO2
X
ii
Page 4
Y
The amine X, C7N9N, may be prepared from Y, C7H7NO2, by reaction with excess hot granulated tin and concentrated hydrochloric acid. If 6 g of X is obtained from 10 g of Y, what is the percentage yield of the reaction? (Relative atomic masses : H, 1.0; C, 12.0; N, 14.0; O, 16.0) relative molecular mass of X, C7H9N = 107 relative molecular mass of Y, C7H7NO2 = 137 107 × 10 g for 100% yield Theoretical yield = 137 107 Percentage yield = 6 ÷ ( × 10) × 100% = 76.8% (76.7 - 76.9%) (-½ mark for wrong no of sig. fig.) 2 marks 137
91 2C 9 a iv 9a Outline chemical tests which would allow you to distinguish between the compounds in the following pairs. Describe what you would observe in each case. OH H iv CH3CH2CH2 C
CH3
2
2
CH3CH2CH2 C CH2OH
CH3
CH3
Warm the compound with acidic Na2Cr2O7. The primary alcohol will turn the solution from orange Cr(VI) to green Cr(III). No change in colour for the tertiary alcohol. Primary alcohol easily oxidized, not tertiary alcohol. H CH3CH2CH2 C CH2OH CH3
C
[O] Na2Cr2O7
H CH3CH2CH2 C COOH CH3
Many candidates incorrectly suggested the use of iodoform test to distinguish between the tertiary and the primary alcohols. Some candidates applying the Lucas test provided the inaccurate observation that tertiary alcohol gave white precipitate, instead of turbidity, on reaction with Lucas reagent.
93 1A 3 a iii 3a Consider the following compound, X:
iii Upon catalytic hydrogenation, one mole of X reacts with two moles of H2. Draw the structure of the product and give its systematic name. CH3CH2CH2CH2CO2H Pentanoic acid or n-pentanoic acid
2 1 mark 1 mark
95 2C 9 b iii 9b Identify K, L, M, N, P, R and S in the following reactions: iii M CH2
C
CHCO2CH3
1
CH3CH2CO2CH3
M : Pd or Ni or Pt / H2 (½ mark for H2 / cat. only) Common mistakes included : LiAlH4 or H2 without catalyst for M;
1 mark
Oxidation and Reduction 97 1A 5 b 5b Consider the following reactions:
Page 5
conc. H2SO4 heat CH3CH2COCl
CH3CH2CH2OH
HBr
J
L
Na2Cr2O7
NHNH2 NO 2
NO 2
M
+
H3O
K
N (a red precipitate)
Give structures for J, K, L, M and N. 97 1A 6 a i 6a i Vegetable oils (e.g. peanut oil) can undergo the following chemical reactions to give useful solid products P and Q. excess H2 / Pt
P
Vegetable oil NaOH(aq) heat
Q
Give one use each for P and Q . Suggest a possible structure for P . 97 2B 7 a iv 7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Each test should include the reagent(s), the expected observation with each compound and the chemical equation(s). iv O O C2H5
C
C2H5
and
CH3(CH2)3
98 2B 6 d i 6d Identify J in the following reaction. i J
(1) Sn / conc. HCl (2) NaOH
C
H
NH2
H3C
99 2B 5 a ii iii 5a Identify D, E, F, G and J in the following reactions. ii CH2CH3 CO2H E O2N
O2N
iii
CH2CH2OH +
Na2Cr2O7 / H3O heat
HO CH3
F
Uses of Different Compounds I.
Uses of halogeno-compounds A. B.
II.
Use as solvent Manufacture of polymer 1. Preparation of vinyl chloride 2. Physical properties of PVC and Teflon
Uses of alcohols A. B. C. D.
III.
Use as solvent Alcoholic drink Blending agent Ethan-1,2-diol
Uses of carbonyl compounds A. B.
IV.
Preparation of urea-methanal Use of propanone
Uses of carboxylic acids and their derivatives A. B. C.
V.
Food preservatives Manufacture of nylon and terylene Use of ester
Uses of amines and their derivatives A. B.
Azo compounds as dyes in dyeing industries Amine derivatives as drugs
Uses of compounds with different functional groups
Page 1
Topic
Uses of compounds with different functional groups
Reference Reading
26
Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 452, 466–467, 525–526, 575 Reading Syllabus
Uses of halogeno-compounds Uses of alcohols Uses of carbonyl compounds Uses of carboxylic acid and their derivatives Uses of amine and their derivatives
Notes
I.
Uses of halogeno-compounds A. Use as solvent Chlorinated hydrocarbons are good solvents for oils and greases, so they are widely used in dry-cleaning industry. Trichloroethene, CCl2=CHCl and tetrachloroethene, CCl2=CCl2 are two examples of drycleaning fluids. They are relatively non-flammable, easily removed because of their volatility and have little or no structural effect on fabrics. These properties make them an ideal choice for use as dry-cleaning fluids. B. Manufacture of polymer Some halogeno-compounds are important raw materials in the plastic industry. Poly(chloroethene) [Polyvinyl chloride, PVC] is produced by addition polymerization of chloroethene in the presence of a catalyst. n CHCl=CH2 → –[CHCl–CH2]n– Poly(tetrafluoroethene) [PTFE, "Teflon"] is produced by addition polymerization of tetrafluoroethene under pressure in the presence of a catalyst. n CF2=CF2 → –[CF2–CF2]n– 1.
Preparation of vinyl chloride
Vinyl chloride can be prepared by chlorination of ethene followed by elimination in the laboratory. CH3CH2O H
H C C
H
H
Cl2 Electrophilic addition
-
H H
alcoholic NaOH
H C C H Cl Cl
H
H C
Cl
C H
Industrially, vinyl chloride is prepared by heating 1,2-dichloroethane at 500 ºC. H H H C C H Cl Cl
Heat, 500蚓
H
H C
Cl
C H
Uses of compounds with different functional groups
Page 2
2. Physical properties of PVC and Teflon PVC is one of the largest consumers of plasticizers. Pure PVC is rigid, easily cracked and broken, and cannot be readily processed. The rigidity is due to the intermolecular force which occurs between the slightly negative chlorine atoms on one polymer chain and the slightly positive hydrogen atoms on an adjacent polymer chain. By the addition of suitable plasticizers, interaction between the polymer chains can be reduced and a flexible form of PVC can be produced. Modified PVC is suitable for raincoats, garden hose, and seat covers for automobiles, etc. Comparing with PVC, molecule of Telfon is non-polar. And owing to the high electronegativity and low polarizability of F, the polarizability of Telfon molecule is also very low. This makes the intermolecular forces acting amount the molecules very weak and makes Telfon to make the non-sticky surface coating.
F
F
F
F
F
C
C C
C
C
F
F
F
F
F
II. Uses of alcohols A. Use as solvent Many hydroxy compounds are good solvents, for example, methanol and ethanol are commonly used as solvents in laboratory and in industry. Industrial methylated spirit contains 95% ethanol/water mixture and 5% methanol. B. Alcoholic drink Ethanol is an essential component in alcoholic beverages. The ethanol in alcoholic drinks is produced by the fermentation of sugar or starch. C. Blending agent Ethanol can also be used as a motor fuel blending agent. As ethanol contains oxygen atom in its molecular structure, adding it to petrol helps the fuel to burn more efficiently. As a consequence, emission of carbon monoxide is reduced. D. Ethan-1,2-diol Ethane-1,2-diol is miscible with water in all proportions and has a high boiling point (hence not easily vaporized). These properties make it an ideal choice for use as an anti-freeze. Ethane-1,2-diol is also a raw material in the manufacture of a polyester known as Terylene (or Dacron). The polyester is produced by condensation polymerization of ethane-1,2-diol and benzene-1,4-dicarboxylic acid. n HOCH2CH2OH + n HO2C
CO2H
O CH2CH2
O
O
O C
C
n
+ (2n -1) H2O
Uses of compounds with different functional groups
Page 3
III. Uses of carbonyl compounds A. Preparation of urea-methanal Carbonyl compounds play an important role in the manufacture of plastics. Urea-methanal is produced by condensation polymerization between urea and methanal under heat and pressure. O
H O H H
n H2N C NH2 + n H C
N C N C
O
H
H
+ (n - 1) H2O n
In the presence of excess methanal, further heating causes cross-linkage to form between the polymer chains and hence a rigid structure is produced.
B. Use of propanone Propanone is the raw material in the production of methyl 2-methylpropenoate which is the monomer of poly(methyl 2-methylpropenoate) [Perspex].
Propanone is also an important solvent used in industry and laboratory as it can dissolve a variety of organic compounds.
IV. Uses of carboxylic acids and their derivatives A. Food preservatives Benzoic acid and sodium benzoate are commonly used as food preservatives in non-alcoholic beverages, fruit juices, margarine, ketchup, salads, jams and pickled products. B. Manufacture of nylon and terylene Nylon 6.6, –[CO(CH2)4CONH(CH2)6NH]n– , is an example of polyamides. It is the synthetic fibre for making ropes, threads, cords, lady's stockings, underwear and dresses. Industrially, nylon 6.6 is prepared by heating hexane-1,6-diamine and hexanedioic acid under heat and pressure. In laboratory preparation, hexanedioyl dichloride is used instead of hexanedioic acid to increase the yield. However, it is also more expensive. Terylene,
O CH2CH2
O
O
O C
C
n
is an example of polyesters. It is the synthetic fibre for making wash-
and-wear garments. C. Use of ester Liquid esters are good solvents and are used in all-purpose adhesives, nail varnish removers and as thinners for paints. Volatile esters have characteristic sweet fruity smells and are therefore generally used as artificial flavourings in food and drinks.
Uses of compounds with different functional groups
Page 4
V. Uses of amines and their derivatives A. Azo compounds as dye in dyeing industries Primary aromatic amine is used as a starting material for the manufacture of azo dyes. It reacts with nitric(III) acid to form diazonium salt which can undergo coupling reaction to form azo-compound. HNO2 0-5蚓
Ar NH2
Ar N2
+
Ar'
H
Ar N N Ar' azo compound
As azo-compounds are highly coloured, they are widely used in dyeing industry. Some examples of organic dyes are shown below. A recognition of the azo group, –N=N–, is all that is required. N.B. :
Students are NOT expected to reproduce the structures of these dyes.
Direct brown 138 (brown dye for fabrics) NH2 H2N
N N
Methyl orange (orange dye for fabrics)
NH2 NH2
N N
+-
Na O3S
N N -
+
SO3 Na
CH3 CH3
Ponceau (red dye for food product)
HO Na O3S
N
NH2
Sunset yellow FCF (orange-yellow dye for food product)
+-
N N
+-
Na O3S
N N
N N
+-
Na O3S -
+
SO3 Na -
+
SO3 Na
B. Amine derivatives as drugs Students should be made aware of the use of amine derivatives as drugs. A recognition of the amino group or derived functional group of amine is all that is required. Some examples of these amine drugs are shown below. N.B. :
Students are NOT expected to reproduce the structures of individual drugs.
Chlorpheniramine – an antihistamine that helps to relief allergic disorders due to cold (runny nose, watery eyes), hay fever, itchy skin, insect bites and stings, etc. It is present in some over-the-counter drugs such as Coltalin, Coricidin, Dristan, and Piriton. Chlorpromazine – a tranquillizer that sedates without inducing sleep. It is used to relieve anxiety, excitement, restlessness or even mental disorders.
CH2CH2N(CH3)2 N
CH
chlorpheniramine Cl
CH2CH2CH2N(CH3)2 N
chlorpromazine
S
Acetaminophen – (also known as paracetamol, or p-acetaminophenol) an analgesic that relieves pains such as headaches. It is believed to be less corrosive to the stomach and is an alternative to aspirin. Acetaminophen is present as the active ingredient in some over-the-counter pain-killers such as Panadol, Saridon, Tylenol and Fortolin.
O HO
NH C CH3 acetaminophen
Uses of compounds with different functional groups
Glossary
PVC Telfon Terylene / Darcron
Past Paper Question
90 1A 1 c ii 91 1A 1 e 92 1A 1 a ii 95 2C 9 b iv 99 2B 5 c i ii iii
90 1A 1 c ii HOOC 1c
Page 5
plasticizer polarizability urea-methanal perspex
blending agent
anti-freeze
CH2CH2CH2CH2 COOH E
ii
Show how E, by reaction with suitable reagents, may be converted into an important commercial product. any diamine E
H2N
NH2
CO2NH3R
NH(CH2)6NH2CO(CH2)4CO
∆
(CH2)4 ∆
Polyamide
CO2NH3R
2
n
1 mark for process, 1 mark for product
2 marks
OR via an acid chloride COCl E
SOCl2
(CH2)4 COCl
NH2
H2N
NH(CH2)6NH2CO(CH2)4CO
∆
Polyamide
n
∆ ∆ Other nylons are possible e.g. Nylon 6,6. Important is salt salt → amide or acid chloride → amide .
91 1A 1 e 1e Give all reagents and show how 1,4-dimethylbenzene may be converted into useful polyester. COCl CO2H
CH3 oxidation
SOCl2
Reagents: KMnO4 CH3
C
O HOCH2CH2OH
COCH2CH2O
COCl CO2H
C
HOCH2CH2OH H+ or pressure/heat
O
n
3 marks ½ mark for KMnO4 ½ mark for dicarboxylic acid ½ mark for the correct diol ½ mark for the reaction conditions 1 mark for the correct product Many candidates failed to give a polyester, giving amides instead, or the correct ester (many guessed quite odd alcohols : HOCH2(CH2)4CH2OH for example.)
92 1A 1 a ii 1a There are several isomers of benzenedicarboxylic acid. ii One of these isomer is used to make terylene. Outline the reaction involved. CO2H
CO2H (I)
CH2 CH2 OH
( ) CO2
OH +
various methods
CO
(II)
O
n
Terylene COCl
(II)
C
3
methods (i) SOCl2 → COCl → (ii) Heat (I) and (II) under pressure (iii) H+ or acid or H2SO4 or conc H2SO4 (I) ½ mark (II) 1 mark Methods ½ mark Terylene formula ½ mark Few gave details of reagents for condensation reaction.
2½
Uses of compounds with different functional groups 95 2C 9 b iv 9b Identify K, L, M, N, P, R and S in the following reactions: iv O O N
HOCH2CH2OH
C
Page 6
1
C OCH2CH2O n
Cl
N: C
O
O
O
C
C
Cl or HO C
O C OH
1 mark
Common mistakes included : benzoyl chloride for N;
97 1A 4 b ii 4b ii Upon oxidation, one of the isomers of dimethylbenzene produces a compound with the formula C8H6O4. This compound on condensation with ethane-1,2-diol gives a useful textile material B . Give the structure of B. 99 2B 5 c i ii iii 5c Dacron is the most common of the group of polymers known as polyesters. A segment of the polymer chain is shown below.
i ii
O
O
O
C
C OCH2CH2O C
O C OCH2CH2O
Suggest two types of material which can be made from polyesters. Draw the structures of the two monomers used in manufacturing Dacron. Name the type of polymerization involved. iii How can Dacron be degraded in the environment ?
Determination of Structure I.
II.
Determination of empirical formula and molecular formula A.
Different kinds of formula
B.
Determination of empirical formula
C.
Determination of molecular formula
Degree of unsaturation A.
Determination of degree of unsaturation
B.
Meaning of degree of unsaturation
III.
Sodium fusion test
IV.
Test for different functional groups by wet chemistry
V.
Introduction to IR and NMR spectroscopy A.
Use of infra-red (IR) spectrum in the identification of functional groups 1.
More examples
Structure Determination
Unit 1
Page 1
Topic
Structure Determination
Unit 1
Reference Reading
27.1–27.2 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 281–382
Syllabus
Determination of empirical formula and molecular formula Degree of unsaturation
Notes
I.
Determination of empirical formula and molecular formula
A. Different kinds of formula Formulas are frequently used in chemistry to represent a particle. Like equation, there are many different kind of formulas depending on the usage. Empirical formula - showing the simplest whole number ratio of atoms present (applicable to all kind of compound). e.g.
sodium chloride (ionic compound) NaCl
water (molecular compound) ethane (molecular compound) CH3 H2O
Ionic formula - showing the simplest whole number ratio of ions present (only applicable to ionic compound) e.g.
sodium chloride (ionic compound) Na+Cl-
Molecular formula - showing the actual number of different atoms in a molecule e.g.
water (molecular compound) H2O
ethane (molecular compound) C2H6
Structural formula - showing the connection of the atoms in a molecule e.g.
water (molecular compound)
ethane (molecular compound) H H
H
O
H C C H
H
H H
B. Determination of empirical formula The word empirical means "based on observation or experiment, not the theory". By determining the relatively amount of each kind of element in a compound experimentally, the empirical formula can be determined. Example : A compound X containing only carbon, hydrogen and oxygen burned completely in air to form carbon dioxide and water as the only products. 2.43 g of X gave 3.96 g of carbon dioxide and 1.35 g of water. Find the empirical formula of X. (Given : Atomic mass of H = 1.0, C = 12.0 and O = 16.0) Answer : 12.0 Since all the C is converted to carbon dioxide, the amount of C present = 3.96 g × 12.0 + 16.0 × 2 = 1.08 g 1.0 × 2 Similarly, all the H is converted to water, the amount of H present = 1.35 g × 1.0 × 2 + 16.0 = 0.15 g The remaining mass must from O, therefore the amount of O present = 2.43 g - 1.08 g - 0.15 g = 1.20 g
Structure Determination
Unit 1
Page 2
Because empirical formula shows the relative number of atoms present, the mass has to be converted to the number of atoms (no. of mole of atoms) first.
Mass No. of mole of atoms The relative no of mole present The simplest whole no. ratio
C 1.08 g 1.08 g 12.0 gmol-1 = 0.090 mol 0.090 mol 0.075 mol = 1.2 1.2 × 5 = 6
H 0.15 g 0.15 g 1.0 gmol-1 = 0.15 mol 0.15 mol 0.075 mol = 2 2 × 5 = 10
O 1.20 g 1.20 g 16 gmol-1 = 0.075 mol 0.075 mol 0.075 mol = 1 1×5=5
The empirical formula of compound X is C6H10O5. C. Determination of molecular formula From the empirical formula and the molecular mass, molecular formula can also be determined. Example: Empirical formula of ethane is CH3 and the molecular mass is 30.0. Find the molecular formula. Answer : Empirical formula shows the simples whole no. ratio of atoms present but molecular formula shows the actual no. of atoms present. The different between them is only a whole number (an natural no.) factor. Formula mass of empirical formula × natural no. = Formula mass of molecular formula (12.0 + 1.0 × 3.0) × n = 30.0 15.0 × n = 30.0 30.0 n = 15.0 = 2 Molecular formula of ethane = (CH3)2 = C2H6. Example : Continue from the example given above. If the molecular mass of compound X (with the empirical formula C6H10O5) is 485, what will be its molecular formula ? Answer : Formula mass of empirical formula × natural no. = Formula mass of molecular formula (12.0 × 6 + 1.0 × 10 + 16.0 × 5 ) × n = 485 162.0 × n = 485 485 n = 162.0 = 2.99 n≈3 Molecular formula of X is (C6H10O5)3 = C18H30O15
Structure Determination
Unit 1
Page 3
II. Degree of unsaturation Open chain alkane has the general formula CnH2n+2. It has the maximum no. of hydrogen (2n+2) that can be joined to the C skeleton, it is said to be saturated (with H). For a compound with the general formula CnH2n-2, it has 4 H less than the maximum number. 2 moles of hydrogen (H2) is required to convert 1 mole of CnH2n-2 to a saturated compound. Therefore, the degree of unsaturation of CnH2n-2 is said to be 2. A. Determination of degree of unsaturation (or double bond equivalent) Halogen –
Both hydrogen (H) and halogen (X) atom form only 1 covalent bond, they can be treated as the same in the determination of degree of unsaturation. e.g. Both C4H10 and C4H9Br are saturated compounds with degree of unsaturation zero. H H H H H C
C C C
H H H H
Oxygen –
H H H H H
H C
C C
C
Br
H H H H
O atom forms 2 covalent bond, it uses up a bond to the C and a bond to the H in a saturated compound. It has no effect on the degree of unsaturation. e.g. Both C2H6 and C2H6O are saturated. H H
H H
H C C H
H C
H H
C O H
H H
e.g. Both C3H6 and C3H6O2 have 1 degree of unsaturation. H H H H C
C C
H
Nitrogen –
H H O H
H C
C C O H
H H
N atom forms 3 covalent bond, it uses up a bond to the C and 2 bonds to the H in a saturated compound. The presence of each N makes the molecule require one more H atom to become saturated. (Each N increases the degree of unsaturation by ½.) e.g. Both CH4 and CH5N are saturated. H H C
H H H
H
H C N H H
Degree of unsaturation can be interpreted as the difference between the no. of H required and the no. of H available. Degree of unsaturation =
(no. of C × 2 +2) - (no. of H + no. of X) + no. of N 2
Structure Determination
Unit 1
Page 4
B. Meaning of degree of unsaturation (or double bond equivalent) Degree of unsaturation cannot tell the actual structure of the molecule, but it provides some information about it. One degree of unsaturation may represent the presence of :
1. 2.
1 double bond, or a ring structure.
e.g. C4H8 H H H C C H
H H H H H C
C C C
H
H C C H H H
H H
e.g. C4H8O H H C
C C C
H H
H
O
H C
H
H C
C
H
H
H
H C
H H O H H
Two degree of unsaturation may represent the present of : 1.
2 double bonds, or 2. 1 double bond and a ring structure, or 3. 2 ring structures 4. 1 triple bond.
e.g. C6H10 H H H C H C C H
H H H H H H H C
C C C H
Glossary
C
C H H
H C C C H H H H
empirical formula ionic formula degree of unsaturation
H
H
H
H C H C
C
H
molecular formula
C
H
C H C H
H
H H H H C
C C C
H H H
structural formula
H C
C H H
Structure Determination
Past Paper Question
Unit 1
Page 5
92 2C 7 a 94 2C 7 b i ii 95 2C 9 a i ii 96 2C 8 a i 98 2A 1 b 99 2B 7 c i ii
92 2C 7 a 7 A carboxylic acid P, with a relative molecular mass less than 100, contains C, 55.8%; H, 7.0%; and O, 37.2% by mass . An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H2SO4 gave the desired ester Q but with much of the starting material P unchanged. (Relative atomic masses : H 1.0; C 12.0; O 16.0) 7a 2 Determine the molecular formula of P. 2 marks P : C4H6O2 94 2C 7 b i ii 7b Give, with explanation, THREE possible structures for each of the compounds (G,H and J) below. For (i) - (iii), 1 mark for the first correct structure, ½ mark for the second and ½ mark for the third structure. i G is a sweet smelling liquid which contains 54.4% carbon, 9.2% hydrogen, and 36.4% oxygen and has a relative molecular mass between 80 and 100. Mole ratio C : H : O 54.4 9.2 36.4 12.01 1.008 16.00 = 4.530 9.127 2.275 = 2 4 1 1 mark Empirical formula of G is C2H4O Since relative molecular mass of G is between 80 and 100, ∴ molecular formula of G is C4H8O2. 1 mark G is a sweet smelling liquid, ∴ it is an ester 1 mark Possible structure of G : O
O H C OCH2CH2CH3
C ii
CH3
CH3
O
O
C OCH2CH3 CH3CH2
C OCH3
(any 3)
Generally well-answered. Some candidates missed out the hint in the question for an 'ester'. H, C5H9Cl, is an acyclic chloroalkene which is optically active. H possesses a chiral carbon and a C=C bond Possible structure of H H H
C
CH3
H C OCH
* Cl
H
H * Cl
*
Cl
*
2 marks
3 1 mark
Cl
(any 3)
5
2 marks
A number of candidates did not realize the implication of optical activity on chemical structures. Many candidates considered a pair of enantiomers as two different structures.
Structure Determination Unit 1 95 2C 9 a i ii 9a An acyclic hydrocarbon G, with relative molecular mass of between 60 and 80, contains 85.6% carbon and 14.4% hydrogen by mass. i Determine the molecular formula of G. 85.6 Mole % of C = 12.01 = 7.13 14.4 Mole % of H = 1.008 = 14.29 C : H = 7.13 : 14.29 = 1 : 2 ∴ Empirical formula of G is CH2 1 mark Since m.w. is between 60 and 80 1 mark The molecular formula of G must be C5H10 ii Give all possible structures for the molecular formula determined in (i).
C
2
4
½ mark each for the 1st to 4th structure 1 mark each for the 5th and 6th structure Some candidates did not appear to understand the term 'acyclic'. Very few could give all six structures. Many did not realize that pent-2-ene can exist in cis- and trans-forms.
96 2C 8 a i 8a i An acyclic compound H of molecular formula C4H8O2 has a fruity smell. It does not produce a derivative with 2,4-dinitrophenylhydrazine nor with propanoyl chloride. Deduce the functional group(s) of H. Draw FOUR possible structures for H. H is not aldehyde / ketone ½ mark ½ mark because it does not form hydrazone derivative. or R
N NH
NO2
O2N
(½ mark)
H does not possess an –OH group / is not an alcohol because it does not form ester with propanoyl chloride. or
½ mark ½ mark
O
no R O C C2H5 formation
(½ mark)
H has a fruity smell, The functional group(s) in F is most likely an ester.
½ mark
Any FOUR of the following structures (½ marks for each structure)
2 marks
H
O O
C
Page 6
H
O O
O O
O O
O O
O
O
(Deduct ½ mark for each extra structure) Poorly answered. Many candidates did not appear to know the reaction of 2,4-dinitrophenylhydrazine and propanoyl chloride and hence the formation of the respective 2,4-dinitrophenylhydrazone and ester was omitted. Some did not know the reaction of ester with LiAIH4 or the structure of ethanoic anhydride.
98 2A 1 b 1b A gaseous compound D contains carbon and hydrogen only, and has density of 1.15 gdm-3 at 95.3 kPa pressure and 298 K. Assuming that D behaves ideally, calculate its molar mass and deduce its molecular formula. (1 kPa = 1 × 103 Nm-2)
4½
Structure Determination Unit 1 99 2B 7 c i ii 7c Compound M is a hydrocarbon isolated from oranges and lemons. i M contains 88.2% by mass of carbon and has a relative molecular mass of 136.2. Deduce the molecular formula of M. ii Based on the reactions given below, deduce the structure of M. M + 2H2
M
Pt
(1) O3 (2) Zn dust
CH3
CH(CH3)2
O CH3
O
C(CH2)2CHCH2C H + HCHO C O
CH3
Page 7
Structure Determination
Unit 2
Topic
Structure Determination
Reference Reading
27.3 27.4.1–27.4.4 Principles of Organic Chemistry, Peter R.S. Murray, 2nd Edition pg. 61 Organic Reactions at Advanced Level, D.G. Davies and T.V.G. Kelly, pg 5–6 Work Out Chemistry A-Level, David Burgess, pg 160–162
Page 1
Unit 2
Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 472, 487–489 Organic Chemistry, 6th Edition, Solomons, pg. 295-297 Reading Syllabus
Sodium fusion test Test for terminal alkyne Test for phenol
Notes
III. Sodium fusion test (Ignition test) Lassaigne sodium fusion test is a qualitative analysis used to test for any Nitrogen, Sulphur and Halogen in an organic compound. The organic compound is first decomposed by heating with sodium metal strongly in a ignition tube. The decomposed substance is dissolved in water and filtered. The filtrate is then examined for the presence of Nitrogen, Sulphur and Halogen. Nitrogen –
If nitrogen is present, cyanide ions are formed during the fusion. Iron(II) ions is added to form the complex hexacyanoferrate(II) ions. Fe2+(aq) + 6CN-(aq) → Fe(CN)64-(aq) A drop of FeCl3(aq) is added to react with hexacyanoferrate(II) ions to give a precipitate of Prussian blue. Fe(CN)64-(aq) + Fe3+(aq) → Fe[Fe(CN)6]- (blue ppt.)
Sulphur –
If sulphur is present, sulphide ions are formed during the fusion. In the presence of S2-(aq), a few drops of freshly prepared solution of sodium nitroprusside will turn the solution purple.
Halogen –
If halogen is present, halide ions are formed during the fusion. The filtrate has to be boiled with dil. nitric acid in fume cupboard to remove any CN- present because CN- will also form ppt. with acidified AgNO3(aq). The presence of any halide ions will form ppt. with acidified AgNO3(aq). Cl- – white ppt. of AgCl which turns purple grey under light. Br- – yellow ppt. of AgBr which turns yellowish grey under light. I- – yellow ppt. which is stable under light.
Structure Determination
Unit 2
Page 2
IV. Test for different functional groups by wet chemistry Criteria of a good test 1. Simple – If possible, a single step reaction is better than a multiple steps reaction. 2. Obvious change – The reaction must be accompanied with an obvious change e.g. colour change, evolution of bubble, formation of ppt.. For example, formation of an immiscible layer is usually not a very obvious observation. A reaction taking place doesn’t mean it can be served as a test. 3. Mild reaction condition – Under vigorous condition, there may be other side reactions. Test for terminal alkyne For R–C≡C–H, because the H on the terminal alkyne is exceptionally acidic, it can be displaced by Cu+ or Ag+ to form an ionic ppt. Two common reagents are used are diamminecopper(I) chloride [Cu(NH3)2]+Cl- and diamminesilver(I) nitrate [Ag(NH3)2]+NO3-. +
( NH ) → R–C≡C-Cu+(s) (red ppt.) R–C≡CH Cu 3 2
+
( NH ) R–C≡CH Ag → R–C≡C-Ag+(s) (white ppt.) 3 2
Test for phenol Phenol turns 'neutral' aqueous iron(III) chloride from yellow-orange to purple in colour. The phenoxide ion (or called phenolate ion) C6H5O- acts as a ligand with a lone electron-pair on the oxygen atom forming a dative bond with the iron(III) cation. [Fe(H2O)6]3+(aq) + 2C6H5O-(aq) → [Fe(H2O)4(C6H5O)2]+(aq) + 2H2O(l) It is important that the test be carried out in neutral or near-neutral conditions. If the conditions are alkaline, there will be plenty of C6H5O- ions but the iron(III) ions will precipitate as iron(III) hydroxide. In acidic medium, the concentration of C6H5O- ions is very minimal. The molecular phenol molecule C6H5OH is a very weak ligand which does not complex with iron(III) ions.
Structure Determination
Unit 2
Page 3
Test for 1º, 2º and 3º alcohol Both 1º and 2º alkanol can be oxidized by strong oxidizing agent. 1º alkanol can be oxidized to aldehyde and than carboxylic acid easily. 2º alkanol will only be oxidized to ketone. Since 1º and 2º alkanol turns K2Cr2O7/H+(aq) green, K2Cr2O7 / H+(aq) cannot tell which is 1º alkanol and which is 2º alkanol. 1º and 2º alkanol can only be distinguished by Luca's test. 3º alkanol can easily distinguished from 1º and 2º alcohol since it has no reaction with common oxidizing agent. H R C O H
[O]
H 1° alcohol
[O] H
aldehyde
R R C O H
O R C
[O]
O R C
H 2?alcohol
R
ketone
R R C O H R 3?alcohol
[O]
Resistant to oxidation
O R C O H carboxylic acid
Structure Determination
Unit 2
Page 4
D. Test for aldehyde Aldehyde is easily oxidized to carboxylic acid by any oxidizing agent. This characteristic can be used to test for the presence of the aldehyde. e.g. aldehyde turns acidified K2Cr2O7(aq) from orange to green. 1) KMnO4 / H+ or OH- (cold) 2) K2Cr2O7 / H+
O R C
H
aldehyde
3) HNO3 4) Pt catalyst / O2 5) Cu catalyst / O2
O R C OH carboxylic acid
6) Ag2O
1.
Silver Mirror Test / Tollen’s Test Tollen’s reagent, diamminesilver(I) hydroxide [Ag(NH3)2]OH is prepared by dissolving silver(I) oxide in excess NH3. It oxidizes aldehyde slowly to carboxylate. The silver(I) ion is reduced to silver metal and deposited on the surface of test tube to form a silvery surface. R–CHO(aq) + 2 [Ag(NH3)2]OH(aq) → RCOO-NH4+(aq) + 2 Ag(s) + H2O(l) + 3 NH3(aq) silver mirror The test tube used must be very clean otherwise the silver will not deposit on the test tube surface to form a silver mirror. Instead, it will deposit in the middle of the solution to form gray ppt. Potential hazard of using Tollen’s reagent The Tollen’s reagent is potentially explosive if allowed to evaporate to dryness.
2.
Fehling’s Test Fehling’s reagent consists of Solution A : CuSO4(aq) Solution B : Sodium potassium tartrate in excess NaOH(aq). Fehling’s solution is freshly prepared by addition of Solution B to Solution A until the blue ppt. just redissolves to give a deep blue solution due to the formation of a Cu(II) complex ion. The tartrate ion serves as the complexing agent in here. The presence of complexing agent keeps the Cu2+(aq) soluble even in alkaline medium. R–CHO(aq) + Cu2+(aq) + NaOH(aq) + H2O(l) → RCOO-Na+(aq) + Cu2O(s) + 4H+(aq) blue brick red ppt.
O C OH C OH HO C
H
C OO tartrate ion
Upon reduction, the Cu(II) ion is reduced to Cu(I) with the formation of brick red ppt. of Cu2O. 3.
Benedict’s Test Benedict’s solution is an aqueous solution of Na2CO3, CuSO4 and sodium citrate. Its action is similar to that of Fehling’s solution. But in Benedict’s solution, the complexing agent is the citrate ion instead of the tartrate ion. R–CHO + Cu2+ NaOH + H2O → RCOO-Na+ + Cu2O + 4H+ blue brick red ppt. N.B.
Cu+(aq) ion is unstable in aqueous medium. It disproprotionates spontaneously to Cu2+(aq) and Cu(s). 2Cu+(aq) d Cu(s) + Cu2+(aq) Therefore, the reacting medium is keep alkaline to prepcipitate any Cu+ formed to Cu2O(s)..
C
COO-
HO C
COO-
C
COO-
citrate ion
Structure Determination
Unit 2
Page 5
Characteristic behaviour of functional groups Class of compound Alkene
Tests 1. Br2 rapidly decolorised. 2. Brown ppt. with alkaline KMnO4.
Comments Direct addition. Brown MnO2(s) ppt.
Alkyne
1. 2.
Direct addition. Terminal alkyne group only.
Alkanoic acid
Liberates CO2 from aqueous NaHCO3.
Acid halide
Hydrolysis to give X- ; Test with AgNO3(aq).
Amide
Gives NH3 on warming with NaOH(aq).
Aliphatic 1º amine
Gives N2 bubbles with cold 'HNO2'.
Aromatic 1º amine
Does not give N2 bubbles with cold 'HNO2' but can be coupled to a phenol to give orange diazo dye.
1º alkanol
1. 2. 3.
Steamy fumes of HCl with PCl5. Ester with RCO2H. Acidified orange dichromate(VI) goes green on warming.
2º alkanol
1. 2.
Steamy fumes with PCl5. Acidified orange dichromate(VI) goes green on warming. Turns cloudy slowly with ZnCl2 in conc. HCl(aq).
3. 3º alkanol
1. 2. 3.
Br2 rapidly decolorised. White ppt. with [Ag(NH3)2]+NO3- and red ppt. with [Cu(NH3)2]+Cl-.
Steamy fumes with PCl5. No change with acidified orange dichromate(VI). Turns cloudy quickly with ZnCl2 in conc. HCl(aq).
Phenol
1. 2. 3.
No CO2 with NaHCO3 solution. White ppt. with bromine water. Violet complex with neutral Fe3+(aq).
Aldehyde
1.
Reduces Fehling's and Benedict's solutions to orange/red Cu2O. Ag mirror or grey ppt with Tollens' reagent. Alkaline KMnO4 gives brown ppt. DNPH gives orange ppt. White ppt. with saturated aqueous NaHSO3(aq).
2. 3. 4. 5. Ketone
Methyl ketone
1. 2. 3.
No reaction with alkaline KMnO4. DNPH gives orange ppt. White ppt. with saturated aqueous NaHSO3(aq).
Yellow ppt. with I2 / NaOH(aq).
Fruity smell, (H+ catalyst) Aldehyde and acid product.
Ketone product.
3º alcohol is resistant to oxidation.
Ketone is resistant to mild oxidation. Methyl ketone only. +ve result with methyl ketone, ethanol, ethanol and 2º alcohol with – OH on 2nd C.
Structure Determination
Unit 2
Page 6
Test Reagents for functional groups [Ag(NH3)2]+NO3-
Reagent Diamminesilver(I) nitrate solution – Tollens' reagent
Used to test for 1. Mirror or ppt. with aldehyde. 2. White ppt. with terminal alkynes.
Br2(aq)
Bromine water
1. 2.
Decolorised by alkenes, alkynes quickly and methane slowly under light. White ppt. with phenol.
Cr2O72-
Acidified aqueous potassium dichromate(VI)
Orange solution goes green with 1º and 2º alcohols and with aldehyde.
[Cu(NH3)2]+Cl-
Diamminecopper(I) chloride solution
Red ppt. with terminal alkynes.
DNPH
2,4-dinitrophenylhydrazine aqueous solution
Yellow / orange ppt with aldehydes and ketones
Fe3+
Neutral solution of iron(III) chloride
Violet complex with phenols
Fehling's or Benedict's solution
Both contain complexed Cu2+(aq)
Orange to red ppt. with aldehyde and reducing sugar.
'HNO2'
A mixture of HCl(aq) and NaNO2. Used below 5ºC.
N2 evolved with aliphatic 1º amines. Aromatic 1º amines doesn't give N2 but give diazo dyes when coupled with phenols.
I2 / NaOH
Iodoform test. Aqueous I2 in KI solution, made alkaline with NaOH.
Pale yellow ppt with methyl ketone, ethanol, ethanal and 2º alcohol with –OH on 2nd C.
MnO4-
Manganate(VII), i.e. KMnO4 solution made alkaline with Na2CO3. Used in the absence of water.
Brown ppt. with alkenes, alkyne and aldehydes.
NaHCO3
Saturated sodium hydrogencarbonate solution
Evolves CO2 with alkanoic acids.
NaHSO3
Saturated sodium hydrogensulphite solution
White ppt. with aldehyde and methyl ketone.
NaOH
Cold
Acids and phenols dissolves. NH3 evolved from NH4+ salts. Amines liberated from amine salts.
Na metal
PCl5 ZnCl2 in conc. HCl(aq)
Hot Phosphorous(V) chloride. Used as solid Luca's Test
Evolves H2 from –OH in alcohols and alkanoic acid.
NH3 from NH4+ salts and amide. Steamy fumes of HCl from –OH in alcohols, alkanoic acids and phenols. Turns cloudy with 3º alcohol quickly. Turns cloudy with 2º alcohol slowly. Does not turn cloudy with 1º alcohol at all.
Structure Determination
Unit 2
Glossary
sodium fusion test sodium nitroprusside
Past Paper Question
90 1B 4 b i iii 91 2C 7 a iv 93 2C 8 a ii iv 94 1B 4 b 95 2C 7 a iii 96 2C 7 c i 97 1A 4 d 98 1B 8 a iii
ignition tube hexacyanoferrate(II) ion wet chemistry
Page 7 Prussian blue
97 1B 8 c i 98 1A 4 b i ii
90 1B 4 b i iii 4 A mixture contains equal amounts of X, Y and Z. OH
X b.p. 182°C
4b
C
NH2
Y b.p. 184°C
CH3
Br Z b.p. 184°C
Outline tests involving observable colour changes to show that: i X is a phenolic compound; Add alcoholic FeCl3 solution. A colour change to red or blue denotes presence of phenol group. 1 mark iii Z contains bromine. Carry out a sodium fusion to resulting solution add HNO3 and aqueous AgNO3. A pale yellow ppt. sparingly 1 mark soluble in dil. NH3(aq) denotes presence of Br. Few knew the Na fusion test. Some were able to devise a feasible procedure e.g. using the Grignard reagent and testing the inorganic product—showing initiative for which they were rewarded in full.
1 1
91 2C 7 a iv 7a State with explanations, what you would observe in each of the following experiments, and write equations for the reactions. iv Hexanal is treated with ammoniacal silver nitrate solution (Tollens’ reagent). 2 + A silver mirror is formed on the surface of the test tube or some silver formed. Ag is reduced by hexanal to Ag 1 mark metal and deposits on the glass surface. CH3CH2CH2CH2CH2CHO Ag(NH3)2 CH3CH2CH2CH2CH2C
O + Ag O-NH4+
1 mark
Structure Determination Unit 2 93 2C 8 a ii iv 8a Give a chemical test to distinguish one compound from the other in each of the following pairs. Your answer should include the reagents required, the observation expected, and the chemical equation(s) for each test. CH3CH2CH2CH2CHO and CH3CH2COCH2CH3 ii Reagent: Fehling’s reagent / Tollen’s reagent Observation: Red ppt. / Formation of silver mirror Equation: CH3CH2CH2CH2CHO + 2Cu2+ + 6OH- → CH3CH2CH2CH2COO- + 2Cu2O(red ppt.) + 3H2O or CH3CH2CH2CH2CHO + Ag(NH3)2+ + 2OH- → CH3CH2CH2CH2COO- + 2Ag(silver mirror) + H2O A few candidates incorrectly stated that Ag2O was the product formed in the " Silver Mirror" test. Some candidates could not give the formula of Tollen's reagent or Fehling's solution.
C iv
OH CH3 and
1 mark
3 1 mark
OH CH3
3
1 mark 1 mark
CH2OH
Reagent: Lucas reagent (ZnCl2 in conc. HCl) Observation:
Page 8
gives turbidity / cloudy / misty faster than
CH2OH
1 mark
Equation: OH CH3
C
H+
CH3
Cl + H2O
Cl-
CH3
1 mark Many candidates did not understand the chemistry involved in the Lucas test and the reason for observing turbidity. Many candidates incorrectly applied the Iodoform Test to distinguish the given pair of alcohols.
94 1B 4 b 4b You are provided with a sample of an aliphatic aldehyde. Describe, with experimental details, the silver mirror test (Tollens’ test) that you would perform, in order to establish that the sample is an aldehyde. What hazard will you face if you allow the reaction mixture to evaporate to dryness? Clean test tube ½ mark 1 mark Add excess NH3 to AgNO3 in test tube until initial ppt. dissolves Add aldehyde ½ mark Place the tube in a beaker of warm water (No direct gentle heating) ½ mark Hazard : an explosion may occur ½ mark C Candidates were reluctant to itemize the procedure, including experimental details. Frequent omissions included the need to use a clean test tube; the steps in making Tollens' reagent; and the use of a water bath.
3
Structure Determination Unit 2 95 2C 7 a iii 7a Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer should include the reagents used, the observation expected and the chemical equation(s) for each test. iii CH3 CH2OH and OH
Treat compound with MnO4-/H+ or Cr2O72-/H+ CH2OH
COOH
Cr2O72- / H+
½ + ½ mark CHO
or
½ mark
The solution changes colour from orange/yellow to green.
½ + ½ mark
CH3
No reaction with
OH .
½ mark
OR Treat compound with HCl and ZnCl2
½ + ½ mark
CH3 OH gives turbidity at a faster rate than
C
1 mark
CH3
CH3 OH
CH2OH
+ HCl
Cl
+ H2O
1 mark
The Lucas test was generally used, but most candidates gave wrong observation : precipitation instead of turbidity; they probably thought that the tertiary chloride was a solid. Some candidates gave incorrect reagents such as Zn instead of ZnCl2; ZnCl2(aq) instead of anhydrous ZnCl2(s); dilute HCl instead of concentrated HCl. Some gave the wrong formula ZnCl for zinc chloride. The simpler reagent, Cr2O72-/H+ only appeared occasionally. Some candidates did not give the correct colour change involving this common oxidizing agent.
Page 9
3
Structure Determination Unit 2 96 2C 7 c i 7c Suggest a chemical test to distinguish one compound from the other in each of the following pairs. Your answer should include the reagents used and the observation expected. i O CHO and
Page 10
2
Treat compound with Tollen's reagent / ammoniacal silver(I) oxide / ammoniacal silver nitrate(V) / Ag(NH3)2+ 1 mark CHO
gives a silver mirror / silver deposit
½ mark
O
No reaction with
½ mark
Or Treat compound with Fehling's reagent / a mixture of CuSO4,NaOH and sodium potassium tartarate
(1 mark)
CHO
produces a (bright) red / orange / reddish brown ppt.
(½ mark)
O
No reaction with
(½ mark)
Or Treat compound with Cr2O42-/H+ (cold MnO4-/H+)
(1 mark)
CHO
changes the colour of the solution from orange to green
(½ mark)
(changes colour of MnO4-/H+ from purple to colourless) O
No reaction with C
(½ mark)
Some candidates erroneously gave Brady's test for differentiation. One common mistake of the candidates was to take CuSO4 as Fehling's reagent.
97 1A 4 d 4d Suggest, with explanation, a chemical reaction which would enable 4-methylbenzoic acid and 4-methylphenol to be distinguished from each other.
2
97 1B 8 c i 8c For each of the following groups of compounds, suggest a chemical test which would enable each compound to be distinguished from the other(s). In your answer also give the changes that you would expect to observe for each compound. i CH3CH2CH2OH, (CH3)2CHOH and (CH3)3COH 2 98 1A 4 b i ii Alcohol E has the structure CH3CH(OH)C2H5. 4b i Draw the structures of three structural isomers of E, all of which are alcohols. ii Describe how the reagent Zn/concentrated HCl can be used to distinguish E from the three structural isomers.
1½ 1½
98 1B 8 a iii 8a Show how you would iii distinguish between C6H5COCl and C6H5COBr using a chemical test.
1
Structure Determination
Unit 3
Page 1
Topic
Structure Determination
Unit 3
Notes
27.4.0 Past Paper Questions of structural determination Organic Chemistry, 6th Edition, Solomons, pg. 359-360, 742
Past Paper Question
90 2C 7 a b c 91 2C 9 b i ii iii 92 1A 1 b 92 1A 1 f ii 93 2C 9 a i 94 2C 7 b iii 95 2C 7 b i ii 96 2C 7 a i ii 98 2B 6 c ii
92 2C 7 a b e 95 2C 9 a i iii iv
90 2C 7 a b c 7 Consider the following experimental information. I. X (C9H11NO) gives on hydrolysis, two compounds with formulae CH5N and C8H8O2. When the latter compound is added to sodium hydrogencarbonate solution, a colourless gas is evolved. II. Y (C9H10) gives on vigorous oxidation, an acid with formula C7H6O2. Y decolourizes a solution of bromine in tetrachloromethane. III. Z is an optically active ester containing 64.6% carbon, 10.8% hydrogen and 24.6% oxygen, and has a relative molecular mass between 120 and 140. Upon hydrolysis, Z gives propanoic acid. (Relative atomic masses: H, 1.0; C, 12.0; O, 16.0) 7a Draw TWO possible structures for each of the compounds X, Y and Z. X: Y:
9
O C NHCH3
O H
C NHCH3 CH3
C C
CH3
H
H
C C
H CH3
CH3
Z: molecular formula calculated C7H14O2 O
O
CH3
CH3CH2 C O C
CH2CH3
CH3CH2 C O C H CH2CH3
H
7b
7c
3 marks
CH3
1 mark for each structure For each pair of possible structures, name the type of isomerism involved. 3 X: Structural isomerism 1 mark Y: Geometric isomerism / cis-trans isomerism 1 mark Z: Enantiomerism / optical isomerism / stereoisomerism 1 mark Using only one of the possible structures for each of X, Y and Z, write equations for the reaction described above. 5 O
COOH
C NHCH3 hydrolysis CH3 H
+ CH3NH2 CH3
C C
CH3
2 marks H
COOH [O]
H
C C
CH3 H
Br Br2
Br
1 mark each O
CH3
CH3
CH3CH2 C O C
CH2CH3
H2O
CH3CH2OOH + HO C
C
CH2CH3
H
H
1 mark
The majority of the candidates could provide correct structures for X and Y, but not for Z. O C
CH2NH2
One common incorrect structure for X was : and the related reaction equation was incorrectly given as : O
O C CH2NH2
hydrolysis
C OH
+ CH3NH2
Common incorrect structures for Y were :
Structure Determination
Unit 3
Page 2
Although most candidates could work out the molecular formula for Z using the given analytical data, they failed to provide the correct structures which would display optical activity i.e. a chiral centre. Candidates had difficulty in drawing a 3 dimensional representation for Z properly. Many candidates could not distinguish between different types of isomerism in (b). It is suggested that terms like "chain isomerism", "functional group isomerism", "position isomerism" or "metamerism" should be dropped and the more general term "structural isomerism" should be used to describe different structures. One other common mistake in isomerism for Y was regarding structures with the double bond at different locations as geometric isomers, e.g. CH CHCH3
CH2CH CH2
91 2C 9 b i ii iii 9b Suggest a possible structure for each of the compounds J, K, L, M, and N below and explain briefly your deductions. i J, C4H6O, on oxidation gives K, C4H6O2.
ii
CHO
[O]
COOH
CHO
[O]
COOH
CHO
[O]
COOH
CHO
[O]
COOH
H
C
or or
OH
Cl
CH3
PCl5 H
CH3 + HCl
H
O3
CHO
O3
CHO
Zn/H2O
or
PCl5 H
CH3 + HCl
O3 O CHO
or
Zn/H2O
3
CHO CHO
M N 1 mark for M, 1 mark for N Ozonolysis of M without changing the carbon number → double bond forms part of a ring, i.e. cyclic alkene. 1 mark Many candidates could not account for the ring structure of M by relating the number of carbon atoms in M and N.
92 1A 1 b 1b Compound X with formula C10H14 is optically active. On treatment with potassium manganate(VII), X gives benzoic acid (C6H5COOH). Give the structure of X. Me or
C
3
Cl
CH3
or 1 mark L can exist as pair of enantiomers → chiral centre in L 1 mark L gives out HCl on reaction with PCl5 → -OH group present 1 mark A few candidates did not realize the relation between the existence of enantiomers and chiral centre; hence they gave an incorrect structure for L. iii M, C6H10, on ozonolysis gives N, C6H10O2. Zn/H2O
C
or
J K 1 mark for J, 1 mark for K Oxidation increase the number of O atom on J 1 mark L, C5H12O, can exist as a pair of enantiomers, and reacts with phosphorus pentachloride to give hydrogen chloride. OH
3
CH
Et
Few commented on the 3-D aspect of the structure.
2
Structure Determination 92 1A 1 f ii CO2H 1f
Unit 3
Page 3
CH3 CH NH3
ii
V Give the structure of the compound with a relative molecular mass of approximately 230, which on refluxing with dilute hydrochloric acid gives V as the only product. (Relative atomic masses: H 1.0; C 12.0; N 14.0; O 16.0) The structure of the compound is C9H17N3O4 with m.w. 231. CONH
CONH
1½
CO2H
NH2
Essential features (i) 2 peptide links: CONH (ii) tripeptide
1½ mark
92 2C 7 a b e 7 A carboxylic acid P, with a relative molecular mass less than 100, contains C, 55.8%; H, 7.0%; and O, 37.2% by mass . An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H2SO4 gave the desired ester Q but with much of the starting material P unchanged. (Relative atomic masses : H 1.0; C 12.0; O 16.0) 7a 2 Determine the molecular formula of P. 2 marks P : C4H6O2 7b Give the structures of four carboxylic acids having the molecular formula you determined in (a). Give the 6 systematic names for any one of the carboxylic acids and its methyl ester. COOH H
but-3-enoic acid or 3-butenoic acid
COOH
CH3
H
H
H
CH3
trans-but-2-enoic acid
COOH
cis-but-2-enoic acid
COOH
2-methylpropenoic acid
any 4 structures - 4 marks
COOH
C
cyclopropanecarboxylic acid ester : methyl but-3-enoate etc. Many candidates did not show the structure of cis and trans isomer clearly, e.g. H H CH3
C
C
COOH
CH3 H H C
C COOH
for cis isomer, and for trans isomer. CH3
7e
name of acid - 1 mark name of ester - 1 mark
CH2
Many candidates treated CH2 C COOH and CH3 C COOH as two different structures. Candidates were generally weak in naming; examples of common mistakes are : CH3 CH CH COOH butan-2-enoic acid CH3 CH CH COOCH3 methy but-2-enoate or methoxy but-2-enoate. Show, using equations, how you would convert Q to P. COOCH3
+ KOH
COO-K+
Heat
2
+ CH3OH
Q
H+ COOH P
2 marks
Structure Determination Unit 3 Page 4 93 2C 9 a i 9a A compound W of relative molecular mass 121 has the elemental composition: C, 79.3%; H, 9.1%; N, 11.6%. W reacts with nitric(III) acid to give a derivative X. X reacts with ethanoic anhydride to give a sweet-smelling liquid Y which contains no nitrogen. Oxidation of X with chromic(VI) acid yields Z, C8H8O. Z reacts with iodine and sodium hydroxide to produce a yellow precipitate and sodium benzoate. i Deduce the structures of W, X, Y and Z, using all the given data. Explain your reasoning. 10 C H N 79.3 9.1 11.6 9.1 11.6 79.3 No. of atoms 1.008 14.01 12.0 = 6.61 : 9.1 : 0.828 ≈ 8 : 11 : 1 2 mark Empirical formula of W is C8H11N Molecular formula is (C8H11N)n Molecular mass = 121 ∴n=1 1 mark X reacts with ethanoic anhydride to form a sweet-smelling liquid Y. ∴ X is an alcohol and Y is an ester 1 mark W reacts with HNO2 to give an alcohol ∴ W carries an –NH2 group / 1º amine 1 mark Z is obtained from oxidation of X (an alcohol) and it gives positive iodoform test. It is a methyl ketone. 1 mark NH2 C
W is
X is
H CH3
O
C O C
Y is C
OH
CH3
CH3
Z is
H
C
CH3
H O C
CH3
(4 marks for structure, 6 marks for deduction) A number of candidates were able to give correct structures for the unknowns; however, they were unable to show clearly and logically how these structures were deduced from the given information. Generally, candidates lacked the necessary technique in presenting an answer for this type of problem. Many candidates suggested X was a diazonium salt because X was formed from the reaction between an amine and HNO2. They had ignored the fact that X gave an ester on reacting with anhydride.
94 2C 7 b iii 7b Give, with explanation, THREE possible structures for each of the compounds (G,H and J) below. For (i) - (iii), 1 mark for the first correct structure, ½ mark for the second and ½ mark for the third structure. iii J, C10H14O, reacts with PCl5 to give hydrogen chloride. Vigorous oxidation of J gives benzene-1,2-dicarboxylic acid, but mild oxidation of J gives a ketone. Suggest one chemical test, giving the expected observations, which would allow you to eliminate one of your possible structures of J in (iii). ½ mark J reacts with PCl5 to give HCl, ∴ J carries a –OH group Stronge oxidation of J gives , J is a 1,2-disubstituted aromatic compound 1 mark Mild oxidation of J gives a ketone, J is a 2º alcohol ½ mark Possible structures of J OH
OH
OH
2 marks
Iodoform test / I2 in NaOH Observation : OH
OH
C
and
OH
will give a pale ppt., while
will not.
2 marks
Poorly-answered. Many candidates did not make use of all information to deduce that J was a secondary alcohol as well as a 1,2-disubstituted benzene derivative.
6
Structure Determination Unit 3 Page 5 95 2C 7 b i ii 7b An acidic compound D, C6H6O4, on heating to 140ºC dehydrates to give compound E, C4H2O3. D gives compound F upon hydrogenation. i Give the structures of D, E, F. 3 O H H
C C
O
C O H
H C C H C C
C O H
ii
C
HOOCCH2CH2COOH
O
O
C
O
(1 or 0) mark each D (must show the cis structure) E F Poorly answered. Few candidates gave the correct structure for D. Most candidates erroneously thought that dehydration only involved the elimination of water from alkanols to give alkenes. Some gave both trans- and cisisomers for D probably because they did not realize that only cis-butenedioic acid would dehydrate to form an anhydride at 140ºC. Most failed to appreciate geometrical isomerism and less than 5 % named D and E correctly. Give the systematic names for D and E. D : cis-butenedioic acid 1 mark E : butenedioic anhydride 1 mark Many candidates missed out the stereochemical prefix cis in the naming of D.
95 2C 9 a i iii iv 9a An acyclic hydrocarbon G, with relative molecular mass of between 60 and 80, contains 85.6% carbon and 14.4% hydrogen by mass. i Determine the molecular formula of G. 85.6 Mole % of C = 12.01 = 7.13 14.4 Mole % of H = 1.008 = 14.29 C : H = 7.13 : 14.29 = 1 : 2 ∴ Empirical formula of G is CH2 1 mark Since m.w. is between 60 and 80 1 mark The molecular formula of G must be C5H10 iii G, on ozonlysis, yield H and J which both produce positive iodoform reactions. Give the structures of G, H and J. O
∴ Structure of G : CH3 C iv C
C
C
2
3
O
Structures of H and J : CH3 C CH3 and CH3 C H CH3
2
1 + 1 mark
CH3 H
1 mark
Although structures for H and J were given correctly, many candidates could not deduce the structure for G. Some candidates could present G in (iii) but did not include it as one of the structures in (ii). Give the systematic name of G. 1 2-methylbut-2-ene 1 mark G was frequently but erroneously named as 3-methylbut-2-ene.
96 2C 7 a i ii 7a Two compounds F and G. with the same molecular formula C4H8O, give the following results in four chemical tests. Reagent F G H2NOH no reaction no reaction Cr2O72-/H+ colour changes from orange to green no reaction Br2 /CH3CCl3 decolorization no reaction I2/NaOH yellow precipitate no reaction i Give the structure for F. Explain your answer with reference to the result of each of the chemical tests. 1 mark F is CH2=CHCH(OH)CH3 F does not form any oxime. ∴ F is not aldehyde or ketone or, Aldehyde or ketone will give the following reaction
½ + ½ mark (½ mark)
5
Structure Determination O R C
R'
Unit 3
Page 6
N OH
+ H2NOH
R C
R'
(½ mark) ½ mark ½ mark (½ mark)
F can be oxidized by chromic(VI) acid ∴ F is a (primary / secondary) alcohol / bears an –OH group or, alcohols can be oxidized by chromic(VI) acid chromic acid
RR'CHOH
RR'C O
(½ mark)
F gives addition reaction with Br2. ∴ C=C / C≡C / unsaturation is present or, compounds with C=C / C≡C form addition product with Br2 C C
+ Br2
C
Br
C
Br
½ + ½ mark (½ + ½ mark)
(1 mark) F gives positive reaction in iodoform test ∴ H
C
or
C O
CH3
is present.
½ + ½ mark
-
R C OH
or,
CH CH3 OH
I2 / OH
-
RCOO + CHI3
(1 mark)
CH3
Candidates were weak in logical deduction. Many candidates missed the important hints and deduced the structure without explanation or without referral to related reactions. They should learn how to summarize information, reduce possibilities step by step, and then draw conclusions about the structure. Many candidates failed to give oxime as the derivative of carbonyl compound with NH2OH. Some erroneously attributed the colour change with chromic(VI) acid to the double bond. Some did not mention the words addition with Br2 and oxidation with chromic(VI) acid. Some wrongly concluded that ii
C
CH3
OH
was present instead
CH CH3 OH
for a positive iodoform
test. Suggest a structure for G and explain your suggestion. Any ONE of the following: O
O
O
O
OH
2
1 mark
Explanation: G does not possess
O C
,
C C
, 1º or 2º alcohol / G is 3º alcohol
½ mark
it can only be ether / tertiary alcohol ½ mark Some candidates erroneously thought that cyclobutanol is a tertiary alcohol and assigned it as compound G.
C
98 2B 6 c ii 6c 2-Ethanoyloxybenzoic acid (aspirin) is one of the most common substances used to relieve pain. CO2H OCOCH3
ii
2-ethanoyloxybenzoic acid Suggest how you could show the presence of the two functional groups in 2-ethanoyloxybenzoic acid.
4
Structure Determination
Unit 4
Topic
Structure Determination
Reference Reading
27.5 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 384–390 Organic Chemistry, 6th Edition, Solomons, pg. 541-547
Page 1
Unit 4
Assignment Reading Syllabus
Introduction to IR and NMR spectroscopy Interpretation of IR spectrum
Notes
V. Introduction to IR and NMR spectroscopy Other than wet chemistry, infra-red spectroscopy and nuclear magnetic resonance (NMR) are modern instrumental techniques in helping chemists to understand the structures of molecules. IR spectrum Infra-red spectroscopy depends on the vibrations of atoms with respect to each other in the molecule. The infra-red spectrum can tell us whether particular functional groups are present in a molecule
IR spectrum of octane N.B.
The actual structure cannot be deduced from the IR spectrum alone, other tests are required to confirm the result.
Structure Determination
Unit 4
Page 2
NMR spectrum Nuclear magnetic resonance depends on the magnetic properties of the atomic nuclei (e.g. hydrogen nuclei) in a molecule. The proton NMR spectrum can tell us the number of hydrogen nuclei present in the molecule and give information about the structural environment of the hydrogen.
NMR spectrum of 1,1,2-chloroethane
Structure Determination
Unit 4
Page 3
A. Use of infra-red (IR) spectrum in the identification of functional groups On an IR spectrum, different functional groups have characteristic absorption wavenumbers, therefore, the functional group(s) present in an organic compound can be identified from a given IR spectrum. The frequency of infra-red spectrum is usually measured in wavenumber (cm-1). This means the number of wavelength in each cm. The wavenumber is directly proportional to the frequency. An IR spectrum can be divided into several region : 1. 2. 3. 4.
4000 – 2500 cm-1 2500 – 2000 cm-1 2000 – 1500 cm-1 1500 – 500 cm-1
Absorption of single bonds formed by H and other elements. e.g. C–H, O–H, N–H Absorption of triple bonds. e.g. C≡C, C≡N Absorption of double bonds e.g. C=C, C=O Fingerprint region, consists of complicated bands and is unique to each compound.
IR spectrum of octane CH3(CH2)6CH3 In A-Level, we only focus on the identification of functional groups. Therefore, only the region from 4000 cm-1 to 1500 cm-1 will be studied. Characteristic group absorption wavenumbers of covalent bonds in organic molecules
Structure Determination Unit 4 Page 4 1 The following examples show how to make use of the characteristic absorption wavenumbers to identify the functional groups present in organic compounds. Fig. 1 shows a portion of the IR spectrum of ethanol. The peak at 2950 cm-1 corresponds to the absorption of C–H group and that at 3340 cm-1 corresponds to the absorption of O–H group. Hence, a O–H group is present in the ethanol.
Fig. 1 IR spectrum of ethanol
Fig. 2 shows a portion of the IR spectrum of methanoic acid. The peak at 1710 cm-1 corresponds to the absorption of C=O group and that at 3100 cm-1 corresponds to the absorption of O–H group. Hence. both the C=O and O–H groups are present in the methanoic acid. It should be noted that the absorption of the O–H group in alcohols and carboxylic acids does not usually appear as a sharp peak. Instead. a broad band is observed because the vibration of O–H group is complicated by the hydrogen bonding.
Fig. 2 IR spectrum of methanoic acid
Structure Determination
Unit 4
1.
More examples
a.
Hex-5-en-2-one H C C C C C C H
H H H H O H H H
H H H H
b.
Hex-1-yne H C C C C C C H H H H H
H
Page 5
Peak / cm-1 Bond 3080 C–H 1720 C=O 1635 C=C
Peak / cm-1 Bond 3300 C–H* 2900 to 2950 C–H 2200 C≡C * Absorption at 3300 cm-1 is due to C–H bond in terminal C≡C triple bond, which is weaker.
Structure Determination
Unit 4 H H H H
c.
O H
Butyl ethanoate H C C C C O C C H H H H H
d.
O CH3
3-methylbutan-2-one CH C C CH 3 3
H
Page 6 Peak / cm-1 Bond 2900 to 2950 C–H 1740 C=O
Peak / cm-1 Bond 2880 to 2980 C–H 1720 C=O
Structure Determination
Unit 4 H H H H H
e.
Butylamine H C C C C N H H H H H
H H
f.
Propanenitrile H C C C N H H
Glossary
infra-red (IR) spectroscopy nuclear magnetic resonance (NMR) spectroscopy wavenumber fingerprint region absorption peak
Page 7 Peak / cm-1 Bond 3300 to 3400 N–H 2850 to 2950 C–H
Peak / cm-1 Bond 2900 to 3000 C–H 2250 C≡N
spectrum / spectra
Structure Determination
Past Paper Question
Unit 4
Page 8
97 2B 6 c 98 1B 8 a ii 99 2B 6 a iii
Sample question 7b An organic compound Z, with relative molecular mass below 100, has the following composition by mass. C 66.7%, H 11.1% and O 22.2%. i Determine the molecular formula of Z. no. of mole of C : no. of mole of H : no. of mole of O 66.7 11.1 22.2 = 12.0 : 1.0 : 16.0 = 5.56 : 11 : 1.39 = 4.00 : 7.91 : 1.00 ≈4:8:1 Empirical formula of Z is C4H8O. Formula mass of C4H8O = 12.0 × 4 + 1.0 × 8 + 16.0 = 72. Since the molecular mass of Z is below 100, the molecular mass of Z should be 72. ii A portion of the infra-red (IR) spectrum of Z is shown below:
Using the IR spectrum and the result from (i), deduce two possible structures of Z, each belonging to a different homologous series. Given : Characteristic Group Absorption Wavenumbers of Covalent Bonds in Infra-red Spectra (stretching modes) Bond Characteristic Range Wavenumber / cm-1 C=C Alkenes 1610 to 1680 C=O Aldehydes, ketones, acids, esters 1680 to 1750 C–C Alkynes 2070 to 2250 C≡N Nitriles 2200 to 2280 O–H Acids (hydrogen-bonded) 2500 to 3300 C–H Alkanes, alkenes, arenes 2840 to 3095 O–H Alcohols, phenols (hydrogen-bonded) 3230 to 3670 N–H Primary amines 3350 to 3500 From the molecular formula C4H8O, the degree of unsaturation of the molecule is 1. The peak at 1700 cm-1 and absence of a board peak at 3100 cm-1 indicate the presence of C=O bond. Z may be butanal (an aldehyde) or butan-2-one (a ketone). H H H O H C
C C C
H H H
H H O H H
H C
C C C
H H
H
H
butanal butan-2-one iii Suggest one chemical test, giving the expected observation(s), that can be used to distinguish the two compounds having the structures as deduced in (ii) above. By iodoform reaction, when I2(aq) and NaOH(aq) are added, only butan-2-one will give yellow precipitate of CHI3(s) but butanal has no reaction. Sample question 1 The infrared spectra in figure 1 and 2 represent an ester and an alkyne. Identify the peaks which are starred on each spectrum and so decide which spectrum represents which type of compounds.
11
Structure Determination
2
Unit 4
Page 9
Figure 2 Figure 1 Figure 1 : an ester Figure 2 : an alkyne An alkene A (C5H10), on ozonolysis, gave two different compounds, B and C with molecular formulae C3H6O and C2H4O respectively. Compound B has an infrared spectrum as shown in figure 3 below.
Figure 3 IR spectrum of compound B When B and C were separately treated with acidified dichromate(VI), B did not react but C gave D with molecular formula C2H4O2, D gave effervescence with sodium hydrogencarbonate. Deduce possible structure for A, B, C and D based on above information. A : 2-methylbut-2-ene B : propanone C : ethanal D : ethanoic acid 97 2B 6 c 6c Compound H. C3H6O2, does not react with NaBH4 and displays the following infra-red spectrum. Deduce all possible structures of H.
98 1B 8 a ii 8a Show how you would ii distinguish between propan-2-ol and propanone using spectroscopy, 99 2B 6 a iii
Structure Determination 6a Cl
Unit 4 O NH2
4-chlorophenylamine
Cl
NH CCH3
N-(4-chlorophenyl)ethanamide
iii Suggest how to show the presence of the amide group in N-(4-chlorophenyl)ethanamide by (I) a chemical test and (II) a spectroscopic method.
Page 10
Organic synthesis I.
Retrosynthetic analysis
II.
Structural analysis A.
III. IV.
Chain length 1.
Carbon chain
2.
Nitrogen and Oxygen containing chain
B.
Degree of unsaturation
C.
Oxidation or Reduction
D.
Position of functional group
Systematic approach to organic synthesis Examples
Organic Synthesis
Page 1
Topic
Organic Synthesis
Reference Reading
28 Organic Chemistry, Solomons, 5th Edition pg. 157 Organic Chemistry, Stanley H. Pine, 5th Edition pg. 722–723
Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd., pg. 530–534 Reading Assignment Organic Chemistry, 6th Edition, Solomons, pg. 169-171
Syllabus
Organic Synthesis
Notes
Synthesis is the real test of our ability to use and control organic reactions. Before attempting any synthesis, you must be familiar with all kinds of conversion available first. Besides the knowledge about the conversion (organic reaction), the following analysis may help you to plan the synthesis. I.
Retrosynthetic analysis
Retro- means working backwards. 1.
It is better to begin with the target molecule and work backwards from it. The no. of ways of preparing a target molecule is usually fewer than the no. of way that a starting molecule can react. If we start from the starting molecule, the no. of possible synthetic path may be numerous. The molecule from which the target molecule is prepared is called precursor. Target molecule ⇒ 1st Precursor ⇒ 2nd Precursor ⇒ ⇒ Starting molecule For a beginner, it may be quite difficult to plan the complete path at once. The structural analysis may help you to identify some of the precursors (and reagents) required, therefore a long path can be broken into several shorter pathways.
2.
The no. of steps involved should be as few as possible. Try not to use path with more than 4 steps. If the yield of each step is 80%, a 4 steps conversion means merely (80%)4 = 41% overall yield.
3.
Use a precursor (or a reagent) which will give a higher yield. e.g. bromoalkane instead of alcohol in nucleophilic substitution acyl halide instead of carboxylic acid in nucleophilic addition-elimination
4.
Even a conversion seems to be simple may not be so simple. e.g. Direct conversion of acyl bromide (RCOBr) to bromoalkane (RCH2Br) is impossible. This is the feasible way : acyl bromide → carboxylic acid → alcohol → bromoalkane
Organic Synthesis
Page 2
II. Structural analysis (The following guideline is only applicable to A-Level where the choices of conversion are limited.) Before planning of any organic synthesis, it is important to know the difference between the starting molecule and the target molecule. A. Chain length 1.
Carbon chain If the no. of C is increased by 1 i. Substitution reaction with CN- (haloalkane precursor) ii. Addition reaction with CN- (carbonyl compound precursor) If the no. of C is decreased by 1 i. Haloform reaction (methyl ketone or methyl alcohol precursor) ii. Hoffman degradation (1º amide precursor)
2.
Nitrogen and Oxygen containing chain If the chain length is increased. i. Acylation by acylating agent e.g. RCOCl (nucleophile precursor e.g. RNH2 or ROH) ii. Substitution reaction with RI (nucleophile precursor e.g. RNH2 or ROH)
B. Degree of unsaturation (consider C=C and C≡C bond only) If the degree of unsaturation is increased → Elimination with strong base e.g. CH3CH2ONa+ (haloalkane or dihaloalkane precursor) If the degree of unsaturation is decreased → Catalytic hydrogenation (alkene or alkyne precursor) C. Oxidation or Reduction If it is oxidized to i. Carboxylic acid (methyl ketone, aldehyde, 1º alkanol or alkyl benzene precursor) ii. Ketone (2º alkanol precursor) If it is reduced to i. Alkanol (any acid derivatives except amide precursor) ii. Amine (amide, nitrile or aromatic nitro compound precursor) D. Position of the functional group If there is no change in the position of the functional group, it may be converted by direct conversion. If there is change in the position, the functional group may be converted by elimination followed by addition. III. Systematic approach to organic synthesis 1. 2. 3. 4. 5. 6. 7.
Name all the functional groups on the original and target molecule, must be specific e.g. 1º or 2º. Identify if there is any change in length of carbon chain. Identify if the molecule is oxidized or reduced. Identify if there is any change in degree of unsaturation. Figure out the possible intermediate (precursor). Repeat step 1 to 5 until all the intermediates are figured out. Assign reagent to each conversion.
Organic Synthesis
Page 3
IV. Examples Example 1 CH3CH2CH2COOH
Round 1 1 – Naming 2 – Chain length 3 – Redox 4 – Unsaturation 5 – Intermediate
CH3CH2CH2CH2COOH
Starting molecule : CH3CH2CH2COOH (carboxylic acid) Target molecule : CH3CH2CH2CH2COOH (carboxylic acid) The no. of C is increased by one. N.A. No double or triple bond involved. Since the chain is lengthened by 1, nitrile should be the intermediate which can be hydrolyzed to carboxylic acid. CH3CH2CH2COOH
Round 2 1 – Naming 2 – Chain length 3 – Redox 4 – Unsaturation 5 – Intermediate
2 – Chain length 3 – Redox 4 – Unsaturation 5 – Intermediate
CH3CH2CH2CH2OH
CH3CH2CH2CH2CN
Starting molecule : CH3CH2CH2CH2OH (1º alkanol) Target molecule : CH3CH2CH2CH2CN (nitrile) No change on the main C chain. N.A. No double or triple bond involved. –OH group is not a good leaving group. This makes alkanol not a good substrate for nucleophilic substitution by CN- ion. –OH has to be converted to a good leaving group first, e.g. –Br. CH3CH2CH2CH2OH
7
CH3CH2CH2CH2COOH
Starting molecule : CH3CH2CH2COOH (carboxylic acid) Target molecule : CH3CH2CH2CH2CN (nitrile) No change on the main C chain. Carboxylic acid is reduced. No double or triple bond involved. Carboxylic acid has to be reduced to alkanol first. CH3CH2CH2COOH
Round 3 1 – Naming
CH3CH2CH2CH2CN
CH3CH2CH2CH2CH2Br
CH3CH2CH2CH2CN
Overall conversion
CH3CH2CH2COOH
LiAlH4
CH3CH2CH2CH2OH
PBr3
CH3CH2CH2CH2CH2Br
KCN / H+
CH3CH2CH2CH2CN
H3O+
CH3CH2CH2CH2COOH
Organic Synthesis
Page 4
Example 2 +
(CH3)2CHCH2COOH
Round 1 1 – Naming 2 – Chain length 3 – Redox 4 – Unsaturation 5 – Intermediate
Starting molecule : (CH3)2CHCH2COOH (carboxylic acid) Target molecule : (CH3)2CHCH2CH2N+(CH3)3I- (quartenary ammonium iodide) No change on the main C chain. Carboxylic acid is reduced. No double or triple bond involved. Quartenary ammonium salt can only be prepared from alkylation of amine, therefore, the intermediate may be a 1º amine, (CH3)2CHCH2CH2NH2. (CH3)2CHCH2COOH
Round 2 1 – Naming 2 – Chain length 3 – Redox 4 – Unsaturation 5 – Intermediate
-
(CH3)2CHCH2CH2N (CH3)3I
+
(CH3)2CHCH2CH2 NH2
-
(CH3)2CHCH2CH2N (CH3)3I
Starting molecule : (CH3)2CHCH2COOH (carboxylic acid) Target molecule : (CH3)2CHCH2CH2NH2 (1º amine) No change on the main C chain. Carboxylic acid is reduced. No double or triple bond involved. Amine has 3 possible precursors : Nitrile (by reduction), amide (by reduction or Hoffman degradation) or haloalkane (by alkylation of ammonia) Since there is no change in the length of C chain, nitrile intermediate and Hoffman degradation of amide can be ruled out. (CH3)2CHCH2CH2 Br (CH3)2CHCH2CH2 NH2 1. (CH3)2CHCH2COOH (Remark : Only 1º amine can be prepared by alkylation of ammonia) O
2. Round 3 1 – Naming
(CH3)2CHCH2C NH2
(CH3)2CHCH2COOH
(CH3)2CHCH2CH2 NH2
Starting molecule : Target molecule :
2 – Chain length 3 – Redox 4 – Unsaturation 5 – Intermediate
(CH3)2CHCH2COOH (carboxylic acid) 1. (CH3)2CHCH2CH2–Br (haloalkane) 2. (CH3)2CHCH2CONH2 (1º amide) No change on the main C chain. 1. haloalkane – Carboxylic acid is reduced. 2. 1º amide – Carboxylic acid is neither oxidized nor reduced. No double or triple bond involved. 1. Carboxylic acid has to be reduced first before it can be converted to haloalkane. Therefore, the intermediate may be alkanol. (CH3)2CHCH2COOH
2.
(CH3)2CHCH2CH2 OH
(CH3)2CHCH2CH2 Br
Carboxylic acid can be converted to amide through acid chloride intermediate. Because ammonia is a base which forms a salt RCOO-NH4+ with carboxylic acid, it does not form amide with acid directly. O (CH3)2CHCH2COOH
7
O
(CH3)2CHCH2C Cl
(CH3)2CHCH2C NH2
Overall conversion
Path 1 (CH3)2CHCH2COOH
LiAlH4
(CH3)2CHCH2CH2 OH
PBr3
(CH3)2CHCH2COCl
NH3
(CH3)2CHCH2CH2 Br
excess NH3
(CH3)2CHCH2CH2 NH2
excess CH3I
+
Path 2 (CH3)2CHCH2COOH
Glossary
SOCl2
organic synthesis
O (CH3)2CHCH2C NH2
retrosynthetic analysis
LiAlH4
(CH3)2CHCH2CH2 NH2
target molecule
excess CH3I
precursor
-
(CH3)2CHCH2CH2N (CH3)3I
(CH3)2CHCH2CH2N+(CH3)3I-
structural analysis
Organic Synthesis
Page 5
90 2C 9 b i ii 91 2C 8 a b d 93 2C 7 b i ii 94 2C 8 a i ii 95 1A 3 e ii 98 2B 6 c i 99 2B 7 b ii
Past Paper Question
95 2C 8 b i ii iii
90 2C 9 b i ii 9b Using equations, show the reactions you would employ for the following conversions in the laboratory. Give the reagent(s) for each step and structures of all intermediate compounds. O i C
CH3
O C
COOH
1. I2, NaOH CH3
3
CH2OH
CH2OH
LiAlH4
2. H+(aq)
3 marks ii
CH2CH2Br
C
3
CH
Br Br2
NaOEt
CH2CH2Br
NaOEt
C
CH
Br
3 marks
91 2C 8 a b d 8 Using equations, show the reactions you would employ for the following conversions in the laboratory. Give the reagent(s) for each step and the structures of all intermediate compounds. OH 8a
NO2
conc. HNO3 conc. H2SO4
8b
OH
HNO2
3 marks
(CH3)2CHCH2COOH → (CH3)2CHCH2CH2COOH (CH3)2CHCH2COOH
C
NH2
[H+] Sn, HCl
LiAlH4
PBr3
(CH3)2CHCH2CH2OH
4
(CH3)2CHCH2CH2Br
KCN
→ RCN Some common mistakes were : ROH KCN
RX → RCN O
O
hydrolysis
(CH3)2CHCH2CH2COOH
4 marks
(wrong reagent; should use KCN) 3
CH3
CH3CH2CH2 C NH2
CH3CH2CH2 C NH2
(CH3)2CHCH2CH2CN
H2O, H+
(without going through RX)
HCN
8d
CH3CH2CH2CH2 N CH3 LiAlH4
CH3CH2CH2CH2NH2
C
CH3I
O
Some common mistakes were :
3
R C NH2
CH3 CH3CH2CH2CH2 N CH3
Zn / Hg HCl
RCH2NH2
3 marks (wrong reagent)
Organic Synthesis 93 2C 7 b i ii 7b Show how you would carry out the following conversions in the laboratory, giving the structures of intermediate compounds and the reagents for each step. O O i CH 3
CH3
C
O
Page 6
3
C
COOH
MnO4- / OH-, heat
COCl
PCl5
O C
O O
C
COOH
3 marks C
Common mistakes revealed in the candidates' synthetic conversions were : RCOOH →
O
heat
ii
CH3CH2CH2COOH
(without dehydrating agent);
R
3
CH3CH2CH2CH2COOH LiAlH4
CH3CH2CH2COOH
C
O
R C O C
CH3CH2CH2CH2OH
PBr3
CH3CH2CH2CH2CH2Br
KCN / H+
CH3CH2CH2CH2CN
+
H3O
CH3CH2CH2CH2COOH
3 marks
Common mistakes revealed in the candidates' synthetic conversions were : ROH → RCN KCN
(without going through RX)
RCOOH → RCHO LiAlH 4
94 2C 8 a i ii 8a Show how you would carry out the following multi-step conversions. For each step, give the structure of the intermediate compound and the necessary reagent(s). CH2CH2OH C CH i
CH2CH2OH
Br Br C C
Br2
conc. H2SO4
H
3
C CH KOH / EtOH
H H
3 marks C
Common mistakes of the candidates included : RCH2CH2OH → RCH=CH2 RCH=CH2 → RC≡CH directly alc . NaOH
ii
CH2OH
3
O
CH2OH
CH2CH2 O C
PBr3 (or PCl5)
Br
CH3
CN
KCN
O
O COOH
LiAlH4
OH
H3O+, heat
Cl
CH2CH2 O C
CH3
3 marks C
Common mistakes of the candidates included : ROH → RCN KCN or HCN
Organic Synthesis 95 1A 3 e ii 3e Use equations to show how you would carry out the following conversions in the laboratory. For each conversion, give the reagent(s), conditions and structure of the intermediate compound(s) formed. ii O O
Page 7
3
COCH2CH2OC
CH2=CH2 to
O CH2 CH2
MnO4-
CH2 CH2
OH-
OH OH
C
Cl
O
O
COCH2CH2OC
3 marks Some candidates did not state that concentrated sulphuric(VI) acid is needed for the acid-catalyzed esterification if benzenecarboxylic acid is used.
C
95 2C 8 b i ii iii 8b Each of the following conversions can be completed in not more than three steps. Use equations to show how you would carry and each of these conversions in the laboratory. For each conversion, give the reagent(s), conditions, and structure of the intermediate compound(s) 1 mark for reagents 1 mark for intermediate C All the conversions could be accomplished in two steps. Many candidates did not state the experimental conditions for the transformations, e.g., reaction temperature, concentration of the acid or base etc. Common mistakes were as follows : i NO2 OH
3
CH3
CH3 NO2
NH2
Fe or Zn in dil. HCl Sn in conc. HCl LiAlH4
NaNO2, HCl or HNO2
Na or K in EtOH H2 with Pt or Ni
CH3
OH
0 - 5 °C CH3
CH3
3 marks HNO3 instead of HNO2 was given. Direct conversion of the nitrobenzene into the phenol without going through the amine stage was erroneously given.
C ii
3
CH3CHCH3
CH3CH2CH2OH
OH conc. H2SO4 or Al2O3
CH3CH2CH2OH
heat
CH3
CH CH2
H3O+ or H+ or H2O, heat or conc. H2SO4, H2O
CH3CHCH3 OH
3 marks NaOH was wrongly used as the dehydrating agent. Aqueous KOH or NaOH was wrongly used for the dehydrohalogenation, but alcoholic metal hydroxide for the substitution reaction. Nucleophilic displacement of secondary alkyl halides is not of high preparative value because the reaction is generally low-yielding due the competing yet facile elimination. Heating is necessary in the dehydration of alkanol with concentrated H2SO4 and in the hydrolysis of alkene. Some candidates wrongly used H+ as the reagent for acid hydrolysis instead of H2O/H+ or H3O+. They did not realize that the reactant for the hydrolysis is water and H+ is just a catalyst. The following examples should illustrate the point.
C
OH
+
CH3CH CH2
H2O / H
CH3CHCH3 +
CH3CH CH2
iii
CH3OH / H
OCH3 CH3CHCH3
3
OH
CHO
H C CO2H
NO2
NO2
Organic Synthesis
Page 8 OH
CHO NaCN / H+ or HCN NO2
OH
H C CN
H3O+ or H+ / H2O
H C CO2H
heat NO2
NO2
3 marks Again H+ was used as the reagent for acid hydrolysis of the nitrile group and the important condition, heating, was omitted.
C
98 2B 6 c i 6c 2-Ethanoyloxybenzoic acid (aspirin) is one of the most common substances used to relieve pain. CO2H OCOCH3
i
2-ethanoyloxybenzoic acid Outline a synthetic route for preparing 2-ethanoyloxybenzoic acid from 2-methylphenol.
99 2B 7 b ii 7b With no more than four steps, outline a synthetic route to accomplish each of the following transformations. In each step, give the reagent(s) used, the conditions required and the structure of the product. ii OH CH2NH2
C CH3 H CH3O
CH3O
Organic Laboratory Technique I.
II.
III.
Purification of organic compound A.
Solvent extraction
B.
Steam distillation
C.
Chromatography
D.
Recrystallization
E.
Filtration and Suction filtration
Use of quickfit apparatus A.
Handling of quickfit apparatus
B.
Different setup of quickfit apparatus 1.
Reflux setup
2.
Distillation setup
Testing for purity A.
Determination of melting point
B.
Determination of boiling point
Organic Laboratory Techniques
Unit 1
Topic
Organic Laboratory Techniques
Reference Reading
29.1.1 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.
Page 1
Unit 1
Assignment Reading Syllabus
solvent extraction
Notes
I.
Purification of organic compound
A. Solvent extraction Technique of solvent extraction is commonly used to purify organic compound. It has an advantage that no heating is required. This is particularly important to some heat sensitive compounds which decompose easily upon heating. Water and ether are the solvents commonly used. They are immiscible with each other. Moreover, ether is a very volatile solvent which can be evaporized by a warm water bath / at reduced pressure to obtain the pure solute dissolved in it. Precaution : Since ether is very volatile, a pressure may build up inside the separating funnel while being shaken. The pressure should be released occasionally by opening the tap, especially, at the beginning of the extraction process when there is a lot of air inside the flask. Theoretically, in solvent extraction, a solute will partition between the two solvents. Practically, the solute is usually more soluble in one solvent than another. For example, if a brown iodine solution in KI(aq) is shaken with ether. The brown aqueous layer will turn to pale yellow while the ethereal layer will turn to deep purple. This is because iodine is much more soluble in ether than in KI(aq). The following are the solubilities of some organic compounds : More soluble in aqueous (polar) solvent Salt of an organic acid e.g. sodium benzoate, sodium phenoxide Salt of an organic base e.g. benzenammonium chloride Small alkanol e.g. ethanol
More soluble in organic (non-polar) solvent Organic acid e.g. benzoic acid, phenol Organic base e.g. benzenamine Large alkanol e.g. phenol, decan-1-ol
Organic Laboratory Techniques
Unit 1
Page 2
Solvent extraction separates different components in a solution by their difference in solubility. For an aqueous solution containing ethanol(more water soluble) and benzoic acid(more ether soluble), benzoic acid can be extracted by shaking the solution with ether. Eventually, the ether layer is evaporated to obtain pure solid benzoic acid.
However, if the two solutes have similar solubility in both aqueous and organic solvent, the solutes must be converted to other derivatives with different solubility first.
For example, both benzoic acid and phenol are very soluble in ether but not very soluble in water. However, benzoic acid is more acidic than phenol, benzoic acid reacts with sodium carbonate but phenol does not. Benzoic acid and phenol in an ethereal solution can be separated by converting benzoic acid to sodium benzoate first. This can be done by shaking the ethereal solution with aqueous sodium carbonate solution. An insoluble organic acid can easily be converted to a soluble salt by reacting with an alkali. For a relatively strong acid, e.g. benzoic acid, a weak base, like sodium hydrogencarbonate, is usually used to neutralized the acid. For a very weak acid, e.g. phenol, a very strong base, like sodium hydroxide, must be used. After the separation have been done, the benzoic acid can be regenerated by adding hydrochloric acid to sodium benzoate. And the benzoic acid is then extracted with ether again.
Glossary
solvent extraction
Past Paper Question
90 1B 4 a 91 1B 5 a i 92 1B 5 b 94 1A 3 a i 96 1A 3 e 97 1B 8 b 98 1B 8 c 99 2B 6 a ii
partition
ethereal layer
aqueous layer
Organic Laboratory Techniques Unit 1 90 1B 4 a 4 A mixture contains equal amounts of X, Y and Z. OH
X b.p. 182°C
4a
CH3
NH2
Y b.p. 184°C
Br Z b.p. 184°C
Outline a chemical method to separate X, Y and Z from each other. Two methods are possible: Method I Dissolve the mixture in diethyl ether, place in a separating funnel and shake with dil. HCl and separate. The ether layer will contain X and Z, the aqueous layer will contain Y. 1 mark To aqueous layer add dil. alkali (NaOH), ether and separate - the ether layer on evaporation will yield the amine Y. 1 mark To ether layer containing X and Z, add dil. alkali, and shake, separate. Ether layer will contain Z. 1 mark Neutralize alkaline layer with acid (dil. HCl) and extract with ether which on evaporation will yield X. 1 mark
4
Method II Dissolve the mixture in diethyl ether using a separating funnel. Shake with NaOH and separate. 1 mark Acidify (HCl) the aqueous layer and extract with ether to obtain X. 1 mark Acidify (dil. HCl) the other layer (from NaOH extraction), shake and separate. The ether layer will contain Z. 1 mark Add NaOH or alkali to this aqueous layer and extract with ether. The ether layer contains Y. 1 mark Answers which separate by fractional distillation of X, Y, Z. (0 mark) If no solvent (-1 mark) If in doubt, ask ‘will separation succeed?”. If yes award mark, if maybe award ½ mark, if no then 0 mark. Some extraordinary answers were given (e.g. using Na or HNO2, or not using any solvent in the separation process) showing absolutely no training in this basic technique.
C
91 1B 5 a i 5a
CH3
CH3
NH2 X
i
C
Page 3
NO2 Y
The amine X, C7N9N, may be prepared from Y, C7H7NO2, by reaction with excess hot granulated tin and concentrated hydrochloric acid. If, in this reaction, some Y remains unreacted, suggest a procedure to isolate pure X from the reaction mixture. The reaction mixture would contain X, Y, HCl and Sn which need to be separated. 1 mark Firstly remove Sn by filtration and Y by extraction with ether. 1 mark 1½ mark Followed by basification with NaOH of the aqueous layer and extraction with ether (Instead of ether extraction, distillation / steam distillation may be used.) Evaporation of ether gives X. ½ mark This kind of basic organic process is still not well answered.
92 1B 5 b 5b An incomplete reaction of an amine with ethanoic anhydride gave a liquid mixture on cooling. Briefly outline a procedure, using a separating funnel and appropriate reagent(s), to isolate the ethanoylation product. The mixture contains phenylamine, ethanoic acid, ethanoic anhydride and N-phenylethanamide. Use acid/base separation technique and hydrolyse all anhydride to acid. Procedure: Shake mixture with dil. HCl and ether. Remove ether layer, it contains N-phenylethanamide and ethanoic acid. All ethanoic anhydride have been converted to ethanoic acid with phenylamine forming a salt which is water soluble. Shake resultant ether layer with dil. alkali (or Na2CO3 or NaHCO3). The alkali forms a salt with ethanoic acid which is water soluble. Remove ether layer which contains only N-phenylethanamide. In this way, N-phenylethanamide is isolated from other parts present in the mixture. ½ mark for each underlined C Use of separating funnel - very weak. This basic organic technique had obviously never been experienced by students. Every student should do at least one separation. Most answers had no solvent (ether) for example. Candidates can learn more about the separation of mixtures either through laboratory work or by the use of flow charts.
4
3
Organic Laboratory Techniques 94 1A 3 a i 3a Consider the two compounds: CH3CH2CHCH3
CH3CH2CH2NH2
C
C
D
Explain how you would separate a mixture of C and D making use of the difference in their chemical properties and solubility. Add dil.(aq) HCl and (diethyl) ether to C + D 1 mark Separate the layers (using a separating funnel ½ mark In aq. HCl layer – D 1 mark In ether layer – C ½ mark In place of HCl, H2SO4 can be used In place of ether, CHCl3, CH2Cl2 OR only 2 marks If D forms salts with HCl and soluble in H2O while C does not If D may be recovered by addition of alkali (extracting with ether, drying and evaporation) ½ bonus mark Quite a number of candidates did not realize the difference in the basicity of the two compounds. Some did not even know how to carry out a solvent extraction.
96 1A 3 e 3e The structures of compounds A and B are shown below : OH
OH
B
A
Outline a laboratory method, based on a difference in their chemical properties, to separate a mixture of A and B . mixture of A and B
shake with NaOH(aq) and ether / CHCl3 / CCl4 / CH3CCl3
aqueous layer
O-Na+
ether layer
add acid
OH
evaporate / distill ether OH
OH
A
C
B
(Accept answer in form of paragraph.) Generally well-answered, except that some candidates erroneously used NaHCO3 instead of NaOH for the separation, and some gave a test to distinguish between A and B instead of a method to separate them from a mixture.
97 1B 8 b 8b You are provided with a mixture of two liquids, hexan-1-amine (U) and ethyl ethanoate (W). Outline an experimental procedure, based on a solvent extraction process, to enable U to be separated from W. 98 1B 8 c 8c You are provided with a mixture of two liquids, heptanoic acid and hexan-3-one. Outline an experimental procedure, based on a solvent extraction process, to isolate pure heptanoic acid in good yield. 99 2B 6 a ii
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and
OH
i
Unit 1
3
3
Organic Laboratory Techniques 6a Cl
NH2
4-chlorophenylamine
ii
Unit 1 O Cl
NH CCH3
N-(4-chlorophenyl)ethanamide
Based on the fact that the amino group of 4-chlorophenylamine is more basic than the amide group of N-(4chlorophenyl)ethanamide, outline a procedure to separate a mixture of the two compounds.
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Organic Laboratory Techniques
Unit 2
Topic
Organic Laboratory Techniques
Reference Reading
29.1.2–29.1.5 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.
Page 1
Unit 2
Assignment Reading Syllabus
Steam distillation Chromatography Recrystallization Filtration and suction filtration
Notes
B. Steam distillation (Refer to Phase Equilibria – Unit 4)
Setup of steam distillation Steam distillation is used in the purification of organic compound which is insoluble in water. The presence of water lowers the boiling point of the mixture to avoid decomposition of the compound. C. Chromatography (Refer to Phase Equlibria – Unit 5) Paper chromatography is usually used to separate a mixture of compounds for identification purpose. Nevertheless, chromatography can also be used for separation purpose in larger scale. However, gas chromatography or column chromatography would be used instead.
Organic Laboratory Techniques
Unit 2
Page 2
D. Recrystallization A solute starts to crystallize from a solution if the solution is saturated. The amount of solute that can be dissolved in a solvent is depending on the amount of solvent and the temperature of the solvent. Therefore, it is possible to make a solution saturated by reducing the amount of solvent (by evaporation) or by lowering the temperature (lowering the solubility by cooling). The size of the crystal is depending on the speed of crystallization, not the method of crystallization. Although a large crystal are usually prepared by evaporation, indeed, if a saturated solution is allowed to cool down very slowly, a large crystal can also be obtained. Recrystallization enhances the purity of a crystal. For example, a crystal containing 95% sodium chloride and 5% sugar can be purified by using a minimium amount of hot water to dissolve the crystal and recrystallizing the crystal. The solution prepared would have a very high concentration of sodium chloride (95%) and a very low concentration of sugar (5%). Upon recrystallization, the solution will become saturated with sodium chloride but not with sugar. The sugar will remain in solution form. And the purity of the sodium chloride crystal obtained would be higher, say (99.5%). E. Filtration and Suction filtration Filtration is used to separate an insoluble substance from a liquid. However, the speed of filtration may be very slow if the solid to be filtered is very fine. The process can be speeded up by using suction i.e. suction filtration. The suction flask is connected to a filter pump to generate a partial vacuum. A filter paper is placed onto the Buchner funnel with a few drops of solvent sticking the filter paper in place. The pump is switched on and the mixture to be filtered is placed in the Buchner funnel.
Suction filtration
When the filtration is done, it is important to disconnect the filter pump from the suction flask before switching off the water tap. Otherwise, the partial vacuum in the suction flask may draw the water from the tap into the flask and contaminate the filtrate.
Glossary
steam distillation recrystallzation
chromatography suction filtration
gas chromatography column chromatography filter pump Buchner flask
Organic Laboratory Techniques
Past Paper Question
Unit 2
Page 3
92 1B 5 a iii 95 1B 4 b i
92 1B 5 a iii 5a Ethanolylation (acetylation) of phenylamine (C6H5NH2) may be carried out by refluxing for 1 hour with ethanoic anhydride (acetic anhydride) giving a product which is insoluble in cold water. (Ethanoic anhydride hydrolyses in water to give ethanoic acid. The b.p. of phenylamine and ethanoic anhydride are 184°C and 140°C respectively; the m.p. and b.p. of the product are 114°C and 304°C respectively.) iii Suggest one method for the isolation of the product from the reaction mixture. I Throw the mixture into ice-water. The product with m.p. 114ºC which is insoluble in water will precipitate. Filter and wash with cold water. 2 marks or II. Boil off all the reagent and by-product, leaving the high boiling (304ºC) product. 2 marks 95 1B 4 b i 4b i Describe the experimental procedure by which you would recrystallize a sample of impure benzoic acid Dissolve in minimium quantity of hot (boiling) water using boiling tube / conical flask. ½ + ½ mark ½ mark Filter the solution while it is hot through fluted filter paper into beaker / conical flask. ½ mark Leave for crystals to form (scratch side, seed..) ½ mark Filter crystals off and wash with minimium amount of cold water. ½ mark Dry between filter papers until not damp C The recrystallization of benzoic acid requires the minimum quantity of hot water. Many candidates . confused solvent extraction and fractional distillation with recrystallization.
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3
Organic Laboratory Techniques
Unit 3
Topic
Organic Laboratory Techniques
Reference Reading
29.2 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.
Page 1
Unit 3
Assignment Reading Notes
II. Use of quickfit apparatus Quickfit apparatuses of the same set have identical junction size. All joints are finished with ground glass. When different pieces are connected, the junctions are air tight. Names of different pieces quickfit apparatus 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.
Pear shaped flask Stillhead Liebig condenser Screwcap adapter Receiver adapter Thermometer Dropping funnel with Rotaflo tap Stopper Air leak / steam inlet tube Round bottom flask Air condenser / drying tube Sintered glass funnel Drying tube Pear shaped flask with angled side neck Air leak / steam inlet tube Adapter with 'T' connection Screwcap adapter
Organic Laboratory Techniques
Unit 3
Page 2
A. Handling of quickfit apparatus Cleaning See that the ground surfaces are absolutely free from dirt before use. Lubrication For many preparations ground glass joints can be used unlubricated, but for reactions at high temperature apply a small amount of lubricant to the top portion of the cone only to ensure easily disassembly of the setup. Methods of Heating Use direct heating with a small flame whenever possible. Adjust the bunsen flame to a height of about 2 cm. Keep the tip of the flame about 3 cm from the base of the flask so that heaing is, in effect, by an air bath. Smooth boiling can be induced, if necessary, by adding antibumping granules. Clamping Supporting rather than clamping should always be the principle with assemblies of glassware. The clamping should not be too tight since the glassware will expand upon heating. Where more than one clamp has to be used, a hint is always to adjust and tighted the bosshead on the clamp, and finally on the stand. It is advisable is fix the flask to be heated first because its position is depending on the position of the bunsen burner. Seizure – sticking of joints Strong alkali should never be allowed to remain in contact with a joint face. Its attack on the matt surface is quite rapid and it left between two ground surfaces sticking them together very effectively. Disassembly of the joints while apparatus is still warm and application of a little lubricant are recommended for prevention of seizure. If the joints are stuck together, they may be separated by i) rocking the cone in the socket, ii) tapping the socket flange on a wooden surface, iii) heating the socket and not the cone in a localized flame. C. Different setup of quickfit apparatus 1.
Reflux setup
2.
Distillation setup
Organic Laboratory Techniques
Glossary
quickfit apparatus
Past Paper Question
92 1B 5 a i
Unit 3 ground surface
seizure
92 1B 5 a i 5a Ethanolylation (acetylation) of phenylamine (C6H5NH2) may be carried out by refluxing for 1 hour with ethanoic anhydride (acetic anhydride) giving a product which is insoluble in cold water. (Ethanoic anhydride hydrolyses in water to give ethanoic acid. The b.p. of phenylamine and ethanoic anhydride are 184°C and 140°C respectively; the m.p. and b.p. of the product are 114°C and 304°C respectively.) i With the aid of a labelled diagram, show the laboratory apparatus required to carry out the above reaction.
C
Page 3
½ mark for each piece of apparatus Many did not know the difference between reflux and fractional distillation. Generally very poorly answered. Experimental knowledge and labelling poor. It is obvious that few have used the apparatus in a laboratory.
2
Organic Laboratory Techniques
Unit 4
Topic
Organic Laboratory Techniques
Reference Reading
29.3 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.
Page 1
Unit 4
Assignment Reading Syllabus
Melting point determination Boiling point determination
Notes
III. Testing for purity A. Determination of melting point A pure substance has a sharp melting point but a mixture does not melts sharply. According to this behaviour, the purity of a substance can be tested. Furthermore, the presence of impurity will also lower the melting point of the mixture. Example 1 Pure ice melts at 0ºC but a mixture of salt and ice melts at about -20ºC to 25ºC . This phenomenon is called depression of melting point by an impurity. Example 2 Pure octadecan-1-ol melts at 58ºC and pure napth-1-ol melts at 120ºC. Experimentally, pure octadecan-1-ol will start to melts at 57.5ºC and when the temperature reaches 58.5ºC, all solid will be melted. If it is mixed with a little napth-1-ol, the mixture will start to melts at 50ºC and will melts completely at 55ºC. By knowing whether a substance melts sharply or not, the purity of the substance can be tested. A pure substance only has a very narrow melting range.
Organic Laboratory Techniques
Unit 4
Page 2
B. Determination of boiling point Unlike melting point determination, a sharp boiling point may not represent a pure sample. A constant boiling liquid may be an azeotropic mixture of two substances. However, boiling point can be used as a kind of positive confirmation of the identity of an unknown sample. For example, if the boiling point of a liquid is 103ºC, the liquid must not be water.
Glossary
determination of melting point
Past Paper Question
95 1B 4 b i ii
depression of melting point
determination of boiling point
95 1B 4 c
95 1B 4 b i ii 4b i Describe the experimental procedure by which you would recrystallize a sample of impure benzoic acid Dissolve in minimium quantity of hot (boiling) water using boiling tube / conical flask. ½ + ½ mark ½ mark Filter the solution while it is hot through fluted filter paper into beaker / conical flask. ½ mark Leave for crystals to form (scratch side, seed..) ½ mark Filter crystals off and wash with minimium amount of cold water. ½ mark Dry between filter papers until not damp C The recrystallization of benzoic acid requires the minimum quantity of hot water. Many candidates . confused solvent extraction and fractional distillation with recrystallization. ii How would you test the purity of the product obtained in (i) ? Use melting point determination, compare with literature value ½ + ½ mark ½ + ½ mark or the crystals should have sharp melting point ½ + ½ mark or mix with pure benzoic acid, and determine the m.p. of the mixture ½ + ½ mark or carry out further recrystallization, until the products obtained have a consistent m.p.
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