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SUBJECT – CHEMISTRY TOPIC – CARBOXYLIC ACIDS & ITS DERIVATIVE COURSE CODE – AISM-09/C/CAID Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Contents :- CARBOXYLIC ACIDS & ITS DERIVATIVE Introduction……………………………………………………………………………...3 Nomenclature ………………………………….……………………………………...5 General methods of preparation……………………………………………...11 Physical properties…………………………………………………………………..24 Chemical reactions…………………………………………………………………..27 Reactions of RCOOH………………………………………………………………..28 Salt formation………………………………………………………………………….31 The mechanism of easterification reaction………………………………35 Carboxylic acid derivatives……………………………………………………….44 Chemical reactions of acid derivatives……………………………………..47 Base promoted hydrolysis of esters: Saponification………………….52 Claisen Condensation……………………………………………………………….55 Amides……………………………………………………………………………………..57 Answers to excersies…………………………………………………………………66 Miscellaneous…………………………………………………………………………..75 Solved examples……………………………………………………………………….83 IIT level questions………………………………………………………………….….87
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INTRODUCTION Carboxylic acids are characterized by the presence of carboxyl group. The –COOH group which itself is made up of a carbonyl group (C=O) and a hydroxyl group (
OH) is called carboxyl group (carb from carbonyl and oxyl
from hydroxyl)
O
O
C
O
Carbonyl
H
C
Hydroxyl
O
H
Carboxyl
Carboxylic acids may be aliphatic or aromatic
O R
O
C
Ar O
H
O
Aliphatic carboxylic acid
of
resonating
H
Aromatic carboxylic acid
(Where R = H or any alkyl group)
Comparison
C
(Where Ar is any aryl group)
structures
of
carboxylic
carbonyl group Carbonyl group has two resonance structures (I and II)
C
O
C
(I)
O
(II)
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group
and
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However, for a carboxyl group, three resonance structures (A, B and C) can be written.
O
O
C
O
C O (A)
H
C O (B)
H
O
H
(C)
In both structures (A) and (C), the C – atom and the two O – atoms have eight electrons in their respective valence shells while in structure (B), C – atom has only six electrons. Therefore, structure (B) is less stable than structure (C), in other words the two important resonance structures of carboxyl group are structures (A) and (C). In both these structures, carboxyl carbon is electrically neutral. However in case of aldehydes and ketones, only one structure i.e. I is electrically neutral. As a result, the carboxyl carbon of the resonance hybrid is less positive and hence less electrophilic than the carbonyl carbon of aldehydes and ketones. However, it may be noted that like carbonyl group, carboxyl group is also polar due to resonance structures (B) and (C).
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NOMENCLATURE The aliphatic carboxylic acids are commonly known by their initial names, which have been derived from the source of the particular acid. Examples:
HCOOH Formic acid [Latin: Fermica = ant]
CH3COOH Acetic acid [Latin: acetum = Vinegar]
CH3–CH2–COOH Propionic acid [Greek: Proton = First; Pion = Fat]
CH3(CH2)2COOH Butyric acid [Latin: Butyrum = Butter]
CH3(CH2)3COOH Valeric acid
CH3(CH2)14COOH Palmitic acid
CH3(CH2)16COOH Stearic acid
Alternative system of nomenclature is naming the acids as the derivatives of acetic acids. The only exception being formic acid.
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Example: CH3 – CH2 – COOH Methyl acetic acid (CH3)3C – COOH Trimethyl acetic acid According to the IUPAC system of nomenclature, the suffix of the monocarboxylic acid is ‘oic acid’, which is added to the name of the alkane corresponding to the longest carbon chain containing the carboxyl group, e.g. HCOOH methanoic acid CH3 – CH2 – CH2 – COOH butanoic acid The positions of side-chains (or substituents) are indicated by numbers, the numbering to be started from the side of the carboxyl group.
5
4
3
2
1
1 CH3 – CH – CHCH2 – COOH | | CH3 CH3
3, 4-dimethylpentanoic acid
NAMING OF ACYL GROUPS, ACID CHLORIDES AND ANHYDRIDES: The group obtained from a carboxylic acid by the removal of the hydroxyl portion is known as an acyl group. The name of an acyl group is created by Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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changing the - ic acid at the end of the name of the carboxylic acid to –yl, examples:
O || H –C –
O || H –C – O – H Formic acid
Formyl group
O || C – OH
O || C–
Benzoic acid
Benzoyl group
Acid chlorides are named systematically as acyl chlorides.
O H3C Cl acetyl chloride
An acid anhydride is named by substituting anhydride for acid in the name of the acid from which it is derived.
O
O
OH
O O OH succinic acid
O dihydrofuran-2,5-dione
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O COOH
O
COOH
O 2-benzofuran-1,3-dione
phthalic acid
Naming of Salts and Esters: The name of the cation (in the case of a salt) or the name of the organic group attached to the oxygen of the carboxyl group (in the case of an ester) precedes the name of the acid.
O H3C O
O -
O Na 1-phenylethanone
+
CH3 ethyl 4-methylbenzoate
Sodium benzoate
Naming of Amides and Imides: The names of amides are formed by replacing –oic acid (or –ic acid for common names) by amide or –carboxylic acid by carboxamide.
O O H3C NH2 acetamide
NH2 cyclohexanecarboxamide
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If the nitrogen atom of the amide has any alkyl groups as substitutents, the name of the amide is prefixed by the capital letter N; to indicate substitution on nitrogen, followed by the name(s) of alkyl group(s).
O O 2N N
CH3
CH3 N-ethyl-N-methyl-4-nitrobenzamide
If the substituent on the nitrogen atom of an amide is a phenyl group, the ending for the name of the carboxylic acid is changed to anilide
O H3C
NH NH
N-phenylacetamide
O N-phenylbenzamide
Some dicarboxylic acids form cylic amides in which two acyl groups are bonded to the nitrogen atom. The suffix imide is given to such compounds.
O
O
H N
NH O
pyrrolidine-2,5-dione
O 1N-isoindole-1,3(2H)-dione
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Illustration 1: Give
the
IUPAC
names
and
common
names
of
compounds (i)
(ii)
COOH COOH
(iii)
iv)
OH
COOH HOOC
HOOC
OH
H3C COOH
HO
COOH
Solution: IUPAC Name
General name
(i)
Ethanedioic acid
Oxalic acid
(ii)
Butanedioic acid
Succinic acid
(iii)
2-Hydroxy propanoic acid
Lactic acid
(iv)
2, 3 Dihydroxybutanedioic acid
Tartaric acid
Illustration 2: Which is not a hydroxy acid? (A)
lactic acid
(B)
tartaric acid
(C)
citric acid
(D)
succinic acid
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the
following
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Solution: (D)
GENERAL METHODS OF PREPARATION 1.
Oxidation of alcohols, aldehydes and ketones Refer AEP & AK
Illustration 3: How will you prepare CH3CH2COOH from CH3 CH=CH2 Solution:
CH3
CH
BH3 / THF
CH2
CH3
H2O2 / OH
CH2
CH2
OH
MnO4 / H
CH3
Illustration 4: Bring about the following transformations CH 2
OH
MnO2
A
Ag NH3 OH 2 H
B
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CH2
COOH
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Solution:
O A:
2.
B:
CHO
C
OH
Oxidation of alkyl benzenes Although benzene and alkane are quite unreative towards the usual oxidizing agents (KMnO4, K2Cr2O7 etc). The benzene ring renders an aliphatic side chain quite susceptible to oxidation. The side chain is oxidised down to the ring and only a carboxyl group ( to indicate
the
position
of
the
original
side
COOH) remains
chain. Potassium
permanganate is generally used for this purpose, although potassium dichromate or dilute nitric acid can also be used. (Oxidation of a side chain is more difficult, however, than oxidation of an alkene and requires prolonged treatment with hot KMnO4)
CH 2CH 2CH 2CH 3 hot KMnO
n - butyl benzene
COOH 4
CO 2 Benzoic acid
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This reaction is used for two purposes (a) synthesis of carboxylic acids and (b) identification of alkyl benzenes. 3.
Carbonation of Grignard reagents The Grignard synthesis of a carboxylic acid is carried out by bubbling gaseous CO2 into the ether solution of the Grignard reagent or by pouring the Grignard reagent on crushed dry ice (solid CO 2). In the latter method dry ice serves not only as reagent but also as cooling agent. The Grignard reagent adds to the carbon – oxygen double bond of CO2 just as in the reaction with aldehydes and keotnes. The product is the magnesium salt of the carboxylic acid, from which the free acid is liberated by treatment with mineral acid.
O R
MgX
Grignard's reagent
C
R COO Mg X
H
R
COOH
Mg
2
X
O
The Grignard’s reagent can be prepared from primary, secondary, tertiary or aromatic halides. The method is limited only by the presence of other reactive group in the molecule. The following synthesis illustrate the application of this method.
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CH3 H3C
C
CH3 OH
HCl
H3C
CH3
C
CH3
Mg
Cl
H3C
C
CH3
CH3 MgCl
CO 2
H
H3C
CH3
C
COOH
CH3 Trimethylacetic acid
Illustration 5: (i)
COOH H3C
CH3 Br 2
A
Fe
CH3
H3C Mg Ether
B
H
CO 2
CH3
CH3
Write A and B (ii)
COOH
Br
Mg Ether
H3C
CO 2
A
B
H
H3C
CH
CH C2H5
C2H5 p-bromo-sec-butyl benzene
p-sec-butyl benzoic acid
Write A and B Solution: (i)
Br H3C
MgBr
CH3
CH3
H3C Br 2
H3C Mg
COOH CH3
H3C
CO 2
H
CH3
Ether CH3 Mesitylene
CH3 Bromomesitylene
CH3
CH3 Mesitoic acid ( 2, 4, 6 - trimethyl - benzoic acid)
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(ii)
Br
H3C
CH
CH
H3C
CH
H3C
C2H5
C2H5
C2H5 p-bromo-sec-butyl benzene
COOH
H
CO 2
Mg Ether
H3C
COOMgBr
MgBr
C2H5 p-sec-butyl benzoic acid
Illustration 6: Bring about the following transformations. Br
O C
OH
Solution:
O Br
Mg / Ether
MgBr
CO2
CO 2MgBr
H3O
C
OH
Bring about the following transformations Exercise 1: (i)
O
Br
(ii)
OH
OH HC
CH
CH O
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Exercise 2: (i)
Why we can not get anhydrous formic acid by fractional distillation of the Fatty acids?
(ii)
4.
CH2=CH—CH=CH2
(i) HBr (ii) Mg/ether (iii) CO2 /H3O+
A. Identify A?
Hydrolysis of nitriles Aliphatic nitriles are prepared by treatment of alkyl halides with sodium cyanide in a solvent that will dissolve both reactants. In dimethyl sulfoxide (DMSO), reaction occurs rapidly and exothermically at room temperature. The resulting nitrile is then hydrolysed to the acid by boiling with aqueous alkali or acid.
H R Cl NaCN
DMSO
R
C
N
H2 O
OH
Illustration 7: (i)
NaCN n - C 4H9Br n - butyl bromide
A
C NH4
B
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R
COOH
NH 4
R
COO
NH 3
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(ii)
CH 2Cl NaCN
70% H2SO 4, reflux
A
B
NH 4
Benzyl chloride
Write A and B. Solution: (i)
NaCN n - C 4H9Br n - butyl bromide
n - C 4H9CN n - voleonitrile (pentanenitrile) (A)
aq. alc. NaOH, reflux
n - C 4H9COO NH 3 (B) H n - C 4H9COOH
NH 4
n - valeric acid (pentanoic acid) (C)
(ii)
CH 2Cl
70% H 2SO 4, reflux
NaCN
Benzyl chloride
CH 2COOH
CH 2CN
Phenyl acetonitrile
NH 4
Phenyl acetic acid
Exercise 3: Identify the missing reagents or products
O HCN/OH -
H3C
(A)
H 2 O/H +
(B)
conc. H 2 SO4
(C)
CH3 i) BH3/THF
i)BH 3 /THF ii)CH 3 COOH
ii) H2O2/OH
(E)
-
(D)
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Exercise 4:
OH
Conc. H 2SO 4
A
Cl2/H2O
B
CN
C
H3O
D
Cr 2 O72 - /H +
E
Exercise 5: A pleasant smelling optically active ester (A) has molecular weight 186. it does not react with Br2 in CCl4. Hydrolysis of (A) gives two optically active compounds B and C. Compound (C) gives positive iodoform test and on warming with conc. H 2SO4 gives D (saytzeff product) with no geometrical isomers. (C) on treatment with benzene sulphonyl chloride gives (E) which on treatment with NaBr gives optically active F. When Ag2+ salt of (B) is treated with Br2, racemic F is formed. Give structures of A to F. 5.
Use of alkoxide RO Na
CO
CH3 O Na CO
CH3OH CO
6.
dark
RCOONa
H
RCOOH
CH3COOH 2100 C, pressure Co CO 8
CH3COOH
Carbonylation of alkenes CH2
CH2
CO H2O
H3PO4 300 4000 C
CH3 CH2COOH
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CH3 CH
7.
CH2
CO H2O
H3PO4
CH3
300 4000 C
CH COOH | CH 3
Oxidative cleavage of alkenes, alkynes and cyclo alkenes RC
CH3C RCH
i O3 or KMnO4 ii H2O
CR'
i O3 ii H2O
CCH3 CHR
RCOOH R'COOH
2CH3COOH
i Alkaline KMnO4 ii H /
KMnO4 / Δ /H
2RCOOH
CH 2COOH CH 2COOH
Illustration 8: Compound A, C5H8O3, when heated with soda lime gives B which reacts with HCN to give C, C reacts with PCl 5 to give D which reacts with KCN to form E. E, on alkaline hydrolysis gives a salt which is isolated and heated with soda lime to produce n-butane. A, on careful oxidation with K2Cr2O7 gives acetic acid and malonic acid. Give structural formulae of A to E. Solution:
(C5H8O3)
Soda lim e
HCN
Compound B
Compound (C) PCl5
Careful oxidation with K2Cr2O7
COOH CH3COOH + CH2 COOH
Compound E
KCN
Compound D,
Alk. hydrolysis Salt
soda lim e
n-butane
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On Oxidation only two –COOH groups can be introduced, i.e., one to each carbon undergoing C–C fission, but in the resulting products we have three –COOH groups. Hence, one –COOH group is already there in compound A, the remaining portion C4H7O–, resembles with Keto substituted alkyl group
O
C3H7
This indicates that the given organic compound A is Keto substituted acid. To assign position to Keto group in carbon chain, we know that keto acids on careful oxidation undergo C–C bond fission at a place O ||
where - C - is situated, further Keto group is also converted into –COOH and remains with acid having small number of carbon atoms. From the above discussion it is clear that acetic acid is formed from:
O || CH3
C
arrangement
O ||
C3H7CO–has
CH3
C CH2
CH2
structure
and
O || CH3
C CH2
CH2
COOH
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compound
A
is
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A, with soda-lime undergoes decarboxylation to give CH3 – CO – CH2– CH3 (B) B, being Ketone will give addition product with HCN to form
OH | CH3 – C – CH2 – CH3 | CN PCl5
(C)
Cl | CH3 – C – CH2 – CH3 | CN KCN
(D)
CN | CH3 – C – CH2 – CH3 | Alkaline hydrolysis CN COONa | CH3 – C – CH2 – CH3 | COONa
Soda lim e
CH3 – CH2 – CH2 – CH3 (n-butane)
Illustration 9: How will you affect the following conversions?
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(i)
CH3
COOH
Cl
(ii) COOH
Solution: (i)
CH3
COOH
KMnO 4
COOH
Fe/Cl 2
OH Cl
(ii)
O
H2O/H 2+
Hg
CH3
NaOH I2
COOH
Illustration 10: An organic compound A (C4H8O3), acidic in nature, is oxidised to give B which on gentle heating produces C, (C3H6O) and CO2. A on heating yields D (C4H6O2) having an acid neutralization equivalent of 86. What are A, B and C? Solution:
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A on heating produces D (C4H6O2). It clearly tells us that it is dehydration reaction and A must be a
- hydroxyl acid.
OH |
So A is CH3 HC CH2 COOH and D is CH3
CH = CH
O
OH CH 3C
COOH
CH 2
COOH
Oxidation
H3C
C (B)
CH 2
COOH
H
Δ decarboxylation H3C
C
CH3
CO 2
O
(Because if a carbonyl group is present
to carboxylic group, the
compound undergoes decarboxylation on heating) Illustration 11: Formic acid is obtained when (A)
calcium acetate is heated with conc. H 2SO4
(B)
glycerol is heated with oxalic acid
(C)
acetaldehyde is oxidised with K2Cr2O7 and H2SO4
(D)
calcium formate is heated with calcium acetate
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Solution: (B) Exercise 6: A neutral organic compound A, C3H6O2 on hydrolysis gives a monobasic acid B and a neutral compound C. The acid B reduces HgCl 2 solution. The compound C gives iodoform test. Write structures of A, B and C and give equations to explain the reactions involved. PHYSICAL PROPERTIES Some important physical properties of carboxylic acids are given below, 1.
Solubility As the size of the alkyl group increases, the solubility of the acid decreases and polarity is reduced.
2.
Boiling points Due to intramolecular hydrogen bonding dimerization of acid takes place and boiling point of carboxylic acid is higher than expected.
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O R
H
C
C O
3.
O
H
R
O
Melting points The melting points of aliphatic carboxylic acids do not show a regular pattern. The first ten members show a alteration effect, i.e. the melting point of an acid containing even number of carbon atoms is higher than the next lower and next higher homologues containing odd number of carbon atoms.
Illustration 12: On the basis of H-bonding explain that the second ionization constant K2 for fumaric acid is greater than for maleic acid. Solution: We know that H-bonding involving acidic H has an acid weakening effect and H-bonding in conjugate base has an acid strengthening effect. Both
dicarboxylic
acids
have
two
ionisable
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hydrogen
atoms.
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O
O C
O
H
H
C O
C
C
H C H
OH C
C H
O C
O
O
Fumarate monoanion (no H-bonding)
Maleate monoanion (H-bonding)
Since the second ionisable H of the Maleate ion participates in Hbonding more energy is needed to remove this H because the H-bond must be broken. The maleate mono anion is, therefore, the weaker acid. Exercise 7: Explain the following (i)
Carbon-oxygen bond length in formic acid are 1.24 Å and 1.36 Å but in sodium formate both the carbon-oxygen bonds have same value i.e.1.27 Å.
(ii)
Acetic acid in the vapour state shows a relative molecular weight of 120.
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CHEMICAL REACTIONS The characteristic chemical behavior of carboxylic acids is, of course, determined by their functional group, carboxyl, –COOH. This group is made up of a carbonyl group (C = O) and a hydroxyl group (–OH). As we shall see, it is the –OH that actually undergoes nearly every reaction. Loss of H +, or replacement by another group but it does so in a way that is possible only because of the effect of the C = O.
H R
O
C C (d) (c) H
(b)
(a) O H
Carboxylic acids can also show various types of reactions. (i)
Removal of H+ (due to cleavage of O
H bond) by reaction with a
base (at ‘a’ above) (ii)
C
O bond breaks at b (by PCl 5, PCl3, SOCl2, NH3/ )
(iii)
Nucleophilic attack at point (c) in carboxyl carbon (Ester formation) Reaction in which OH is replaced by O R
C
NH2, Cl is SN type
O OH NH 2
R
C
NH2
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(iv)
Halogenation at
–C by P/Br2 at point d (Hell – Volhard Zelinsky
reaction) (v)
Oxidation of
- methylene group by SeO2 O SeO2
RCH2COOH
||
R C COOH H2O Se keto acid
Some reactions are summarized below:
RCOONH
RCONH
4
RCOOCH H2 RCOONa
X
O R
CH 2N2
C
CO 2
NaHCO
LiAlH 4 R'OH/H
RCOOR'
RCOONa
NH3 , Δ
NH 3 RCH 2OH
2
3
N3H Δ S o d a lim e, Δ
OH Y
CO2
RH
3
NaOH
Na
PCl 5
PCl 3
SOCl 2
ROCI
P 2O 5 (RCO) 2O
Reactions of RCOOH 1.
N2
Acidity and salt formation
Acidity of Carboxylic Acids Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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RCOONa
R NH2
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The acidity of a carboxylic acid is due to the resonance stabilization of its anion.
+
O
O
H
H
H R
R
R O (I)
O
O
H
O (II)
R O (III)
O (IV)
Because of the resonance, both the carbon oxygen bond in the carboxylate anion have identical bond length. In the carboxylic acid, these bond lengths are no longer identical. The acidity of carboxylic acid depends very much on the substituent attached to – COOH group. Since acidity is due to the resonance stabilization of anion, substituent causing stabilization of anion increases acidity whereas substituent causing destabilization of anion decrease acidity. For example, electron withdrawing group disperses the negative charge of the anion and hence makes it more stable causing increase in the acidity of the corresponding acid, on the other hand, electron-releasing group increases the negative charge on the anion and hence makes it less stable causing the decrease in the acidity. In the light of this, the following are the orders of a few substituted carboxylic acids. (a)
Increase in the number of Halogen atoms on
-position increases the
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Cl3CCOOH > Cl 2CHCOOH > ClCH2COOH > CH3COOH (b)
Increase in the distance of Halogen from COOH decreases the acidity e.g.
Cl
H3C COOH
Cl
H3C COOH
Cl
COOH
This is due to the fact that inductive effect decreases with increasing distance. (c)
Increase in the electronegativity of halogen increases the acidity. FCH2COOH > BrCH2COOH > ICH2COOH
Illustration 13: Which one of the following would be expected to be most highly ionised in water? (A)
CH2Cl–CH2–CH2COOH
(B)
CH3CHCl –CH2–COOH
(C)
CH3–CH2–CHCl–COOH
(D)
CH3CH2–CCl2 –COOH
Solution: (D)
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SALT FORMATION Carboxylic acids are weak acids and their carboxylate anions are strong conjugate bases and are slightly alkaline due to the hydrolysis of carboxylate anion compared to other species, the order of acidity and basicity of corresponding conjugate bases are as follows: Acidity
RCOOH > HOH > ROH > HC
Basicity
RCOO– < HO– < RO– < HC
1.
CH > NH3 > RH C– < NH2 < R–
The carboxylic acids react with metals to liberate hydrogen and are soluble in both NaOH and NaHCO3 solutions. For example. 2CH3COOH + 2Na
2CH3COO–Na+ + H2
CH3COOH + NaOH
CH3COO–Na+ + H2O CH3COO–Na+ + H2O + CO2
CH3COOH + NaHCO3 Examples:
2CH3 COOH Zn
CH3 COO 2 Zn H2
Acetic acid CH3 CH2
Zinc acetate
10
COOH NaOH
lauric acid
CH3 CH2
10
COO Na
H2O
Sodiumlaurate COOH
COO Na
NaHCO Benzoic acid
3
CO 2
H2 O
Sodium benzonate
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Illustration 14: Arrange the following illustration in order of increasing acidity (i)
HCOOH, ClCH2COOH, CH3COOH
(ii)
CH3COOH, (CH3)2CHCOOH, (CH3)3CCOOH
(iii)
ClCH2COOH, Cl 2CHCOOH, Cl3CCOOH
(iv)
ClCH2COOH,
CH3CH2COOH,
ClCH2CH2COOH,
(CH3)2CHCOOH,
CH3COOH (v)
CH3COOH, Cl2CHCOOH, CH3CH2COOH, Cl3CCOOH, ClCH2COOH
Solution: (i)
CH3COOH
HCOOH
ClCH2COOH
(ii)
(CH3)3CCOOH
(iii)
ClCH2COOH < Cl 2CHCOOH < Cl 3CCOOH
(iv)
(CH3)2CHCOOH
(CH3)2CHCOOH CH3CH2COOH
CH3COOH CH3COOH
ClCH2CH2COOH
ClCH2COOH (v)
CH3CH2COOH CH3COOH
ClCH2COOH
Cl2CHCOOH
Cl3CCOOH
Illustration 15: Acetic acid reacts with chlorine in the presence of catalyst, anhydrous FeCl3 to give (A)
acetyl chloride
(B)
methyl chloride
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(C)
tri-chloro acetic acid
(D)
chloral hydrate
Solution: (C) 2.
Conversion into functional derivatives O R
O
C
R OH
(a)
(Z = -Cl, -OR', -NH
C
2)
Z
Conversion into acid chlorides O R
C OH
O
SOCl 2 PCl 3 PCl 5
R
C
Cl Acid chloride
Examples: COOH
1000 C
PCl 5
Benzoic acid
Benzoyl chloride
n C17H 35 COOH n SOCl2 Stearic acid
CH3 COOH PCl3 Acetic acid
COCl POCl
reflux
Thionyl chloride
500 C
n C17H 35COCl SO2
HCl
Stearoyl chloride
3CH3 COCl H3PO3 Acetyl chloride
Illustration 16: Benzoyl chloride is prepared from benzoic acid by Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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3
HCl
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(A)
Cl2, h
(B)
SO2Cl2
(C)
SOCl2
(D)
Cl2, H2O
Solution: (C) (b)
Conversion into esters Conversion into Esters (Esterification): Carboxylic acid on
reacting with alcohols in presence of
dehydrating agent (H2SO4 or dry HCl gas) gives esters. The reaction is known as esterification.
O
O R–C
+ R OH OH
H+
R –C
+ H2O OR
Acid
This reaction is reversible and the same catalyst, hydrogen ion, that catalyzes the forward reaction, esterification, necessarily catalyzes the reverse reaction hydrolysis. The
equilibrium is particularly unfavourable when
phenols
(ArOH) are used instead of alcohol; yet if water is removed
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during the reaction, phenolic esters [RCOOAr] are obtained in high yield. The presence of bulky group near the site of reaction, whether in alcohol or in the acid, slows down esterification (as well as its reverse, hydrolysis). Reactivity CH3OH > 1° > 2° > 3° In esterification HCOOH > CH3COOH > RCH2COOH > R2CHCOOH > R3CCOOH The Mechanism of the Esterification Reaction: The step in the mechanism for the formation of an ester from an acid and an alcohol are the reverse of the steps for the acid-catalyzed hydrolysis of an ester, the reaction can go in either direction depending on the conditions used. A carboxylic acid does not react with an alcohol unless a strong acid is used as a catalyst, protonation makes the carbonyl group more electrophilic and enables it to react with the alcohol, which is a weak nucleophile.
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O || C
+ H+ O–H
CH3
CH3
+ O–H || C O–H
Protonation of the carbonyl group
CH3
+ O–H || C O–H
O–H | CH3 – C – O – H
-BH
+O
O CH3
O–H | CH3 – C – O – H
CH3
H
O CH3
H
Nucleophilic attack at the carbonyl group
B Deprotonation of the intermediate
Tetrahedral intermediate being protonated at another site
H+
CH3
O H – B+ || C O – CH3
CH3
O–H | CH3 – C – O + –H
+ O–H B || C O – CH3
CH3
O
+ H
O
H
Deprotonation of the carbonyl group
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H
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Examples: H COOH
CH 3OH
COOCH
3
H2 O
Methanol Benzoic acid
Methyl benzoate O H
CH 3COOH
H3C
CH 2OH
C
O
CH 2
Acetic acid Benzyl alcohol
CH3
CCOOH
SOCl2
CH3
3 Trimethyl acetic acid
3
CCOCl
C2H5 OH
CH3
CCOOC H5
2 3 Ethyl trimethyl acetate
Illustration 17: Assign a structure to each compound indicated by a letter in the following equations. O HO
O
C
(CH2) 9
C
OH
CH 3OH (excess) H2 SO4 , Δ
A
Solution:
O
A=
O
O
(CH 2)9 OCH 3
CH3
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Assign a structure to each compound indicated by a letter in the following equations. O OH CH 3CH2 -OH HCl(g), Δ
A
SOCl2 Δ
B
OH
(c)
Conversion into amides
O R
O SOCl 2
C OH
R
C
O NH 3
R
Cl An acid chloride
C NH2 An amide
Example: C6H5 CH2COOH Phenyl acetic acid
SOCl2
C6H5CH2COCl Phenyl acetyl chloride
NH3
C6H5CH2CONH2 Phenyl acetat amide
Illustration 18: Discuss the reason for that a characteristic reaction of aldehydes and Ketones is one of nucleophilic addition while Acyl compounds yield Nucleophilic substitution product. Solution:
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R
Nu C=O
R
C–O R
Nu–
H – Nu
Nu C–O–H R
R
R
Aldehyde or Ketone Nucleophilic addition
H Nu+
R
Nu – H
R
Nu C–O
C–O
C=O L
Nu
R
L
C = O + HL R
HL+
An acyl compound
Another acyl compound
Nucleophilic addition (L = –OH, –Cl, –NH2, OR etc.)
Elimination
Nucleophilic substitution
The initial step in both reactions involves nucleophilic addition at the carbonyl carbon atom. It is after the initial nucleophilic attack has taken place that the two reactions differ. The tetrahedral intermediate formed from an aldehyde or ketone usually accepts a proton to form a stable addition product. By contrast, the intermediate formed from an acyl compound usually eliminates a leaving group: this elimination leads to regeneration of the carbon oxygen double bond and to a substitution
product.
The
overall
process
in
the
case
of
acyl
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mechanism. Acyl compounds react as they do so because they all have good leaving groups attached to the carbonyl carbon atom. 3.
Reduction
R COOH
LiAlH4
R CH2OH 10 Alcohol
Examples:
4(CH 3)3CCOOH
3LiAlH 4
Ether
[(CH 3)3CCH 2O] 4AlLi
2LiAlO 2
4H2
(CH 3)3CCH 2OH Neopentyl alcohol ( 2, 2 - dimethyl - 1 - propanol) CH 2OH
COOH
LiAlH 4
m - toluic acid
CH3
CH3 m - methyl benzyl alcohol
4.
Substitution in alkyl or aryl group
(a)
Halogenation
of
Aliphatic
Acids
(Hell-Volhard-Zelinsky
Reaction)
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In the presence of phosphorus, aliphatic carboxylic acids react smoothly with chlorine or bromine to yield a compound in which
-
hydrogen has been replaced by halogen.
CH3
COOH
Cl2 /,P
Cl CH2
Cl2 ,/ P
COOH
Cl2CH COOH
Cl2 ,/ P
Cl3CCOOH
The function of the phosphorus is ultimately to convert a little of the acid into acid halide so that it is the acid halide, not the acid itself, that undergoes this reaction. P + X2
PX3
R – CH2 – COOH + PX3
RCH2 – COX COX
R
COX
X2
R
CO 2H
HX
H2O
R
X
X
The halogen of these halogenated acid undergoes nucleophilic displacement and elimination same as it does in the simple alkyl halides. Halogenation is therefore the first step in the conversion of a carboxylic acid into many important substituted carboxylic acid.
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COOH R
COOH
NH3
R
Br
An
NH2
ha log enated acid
An
a min o acid
COOH R
COOH
NaOH
H
R
Cl
OH
An
hydroxy acid
Examples: CH 3COOH Acetic acid
Cl 2/P
Cl 2/P ClCH 2COOH Chloroacetic acid
Cl 2CHCOOH Dichloroacetic acid Cl 2/P Cl 3CCOOH Trichloroacetic acid
CH3
CH3
CH 3CHCH 2COOH Isovaleric acid
Br 2/P
CH 3CH CHCOOH Br α Bromoisovaleric acid
(b)
Ring substitution in aromatic acids: –COOH
deactivates
and
directs
incoming
electrophilic
position. Example:
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to
meta
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COOH
COOH
HNO 3/H 2SO 4/Heat NO 2 m - nitrobenzoic acid
benzoic acid
Illustration 19: Hydrolysis of a compound A (C7H3Cl5) gives an acid B of the formula C7H4Cl2O2. Decarboxylation of acid yields a neutral substance (C), the nitration of which forms only one mono derivative (D). Identify A, B, C and D. Solution:
Cl
Cl CCl 3
A=
COOH B=
Cl
Cl
Cl
Cl NO 2
C=
D=
Cl
Cl
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Illustration 20. Which of the following will not undergo HVZ reaction? (A)
2, 2-dimethyl propanoic acid
(B)
propanoic acid
(C)
acetic acid
(D)
2-methyl propanoic acid
Solution: (A)
CARBOXYLIC ACID DERIVATIVES There are four carboxylic acid derivatives. These are generally represented O ||
as R C Z , where Z is halogen (usually Cl), OCOR , OR or NH2 (or NHR or NR2 ). (a)
When Z is halogen (usually Cl), the derivatives are called as acid chlorides. O R
(b)
C
Cl
When Z is
OR , the derivatives are called as esters. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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O ||
(c)
When Z is O C R' , the derivatives are called carboxylic anhydrides.
(d)
Where Z is or
NH2, the derivatives are called amides. When Z is
NHR
NR2 they are called N – substituted amides.
Synthesis of acid derivatives Carboxylic acid derivatives are exclusively prepared from carboxylic acids. The preparation methods of carboxylic acid derivatives are already discussed under the chemical reactions of carboxylic acids. Illustration 21: (a)
PhLi
(b)
CH3
1. Excess CO2
A
2. H
3
COOH
2 eq. CH3Li
CH3
B
H
3
Ag2 O Br2 ,
COOH
C
Solution: (a)
O Ph
O
C
O
Ph
C
O O
H
Ph
C (A)
(b)
O ||
CH3
3
CH3
3
CCOOH CCOOH
CH3 Ag2 O Br2 ,
3
C C OCH3 B
CH3
3
CBr
CO2
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OH
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Illustration 22: Suggest a likely mechanism for each of the following reactions. (a)
OH
R
C
CX 3
H2O
RCOO
CHX 3
O
(b)
O C
NH2
O NH 2
C
NH 3
F
F
Solution: (a)
O OH
R
C
(b)
O CX 3
R
C
OH
O C
CX 3 O
NH 3
C
NH 2 F
F
NH2
O C
F
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NH2
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Exercise 9: (i)
O
O
AlCl3
A(C10H10O3)
Zn(Hg) HCl
B(C10H12O 2) SOCl 2
O H2SO 4
(C10H10) F
(C10H12O)E
LiAlH4
(C10H10O) D
AlCl3
C
NBS G
(ii)
alc. KOH
H(C10H8)
When heated with acid (e.g. conc. H 2SO4), O – benzoyl benzoic acid yields a product of the formula C14H8O2. What is the structure of this product?
CHEMICAL REACTIONS OF ACID DERIVATIVES (i)
Acyl chloride We have already seen that acyl chlorides are the most reactive of all acid derivatives. As a result, acyl chlorides are often selected as the starting material for the preparation any other acid derivative. Let us see how this is done. O R
C
O Cl
R
C
O O Na
R
C
O O
C
R'
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O R
O
C
Cl
R
R
OH
O R
C
Cl
NH 3
R
C
C
Cl
R'NH 2
NH2
R
C
NHR'
O
C
Cl
R'R"NH
C
R
C
NR'R" O
O H3C
R'
O
O R
O
O
O R
C
Cl
H2NC 6H5
C6H5NHC
CH3
HCl
Acetanilide O OH
O
C
CH3
CH 3COCl COOH
COOH Acetyl salicylic acid (aspirin)
Reaction of acetyl chloride with olefins Acetyl chlorides add on to the double bond of an olefin in the presence of a catalyst (AlCl 3 or ZnCl2) to form a chloro ketone which on heating, eliminates a molecule of hydrogen chloride to form an unsaturated ketone.
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O H3C
CH
CH2
ZnCl 2
CH 3COCl
H3C
CH
CH 2
C
CH3
Cl
Δ O H3C
CH
CH
C
CH3 HCl
Conversion into acids: Hydrolysis. O R
C
O Cl
H2O
R
C OH HCl An acid
Illustration 23: Acetyl chloride reacts with water more readily than methyl chloride. Explain. Solution: Alkyl halides are much less reactive than acyl halides in nucleophilic substitution because nucleophilic attack on the tetrahedral carbon of RX involves a hindered transition state. Also, to permit the attachment of the nucleophile a bond must be partly broken. In CH 3COCl, the nucleophile, attack on > C = O involves a relatively unhindered transition of acyl halides occurs in two steps. The first step is similar to addition to carbonyl compound and the second involves the loss of chlorine in this case. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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O C CH3
+H2O Cl
O-
O–
C
C
CH3 Cl +OH2
CH3 Cl +OH2
Transition state
Intermediate
–
O
O –
Cl
C CH3
+
–H
+ OH2
CH3
Transition state
+Cl–
C
OH
Product
Illustration 24: Hydrogenation of benzoyl chloride in the presence of Pd and BaSO4 gives (A)
benzyl alcohol
(B)
benzaldehyde
(C)
benzoic acid
(D)
phenol
Solution: (B) (ii)
Carboxylic acid anhydrides Carboxylic acid anhydrides can be used to prepare esters and amides.
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O R
C
O O
O
C
R'OH
R
R
C
O OR'
R
C
OH
NH 3 excess O R
C
O NH 2
R
Anhydrides can be hydrolysed to get back acids. O R
O
C
O
O
C
R
H2O
2R
C
OH
Illustration 25:
MeO
CHO
(X)
CH3 COONa H3 O
COOH
The compound (X) is (A)
CH3COOH
(B)
BrCH2 – COOH
(C)
(CH3CO)2O
(D)
CHO – COOH
Solution: (C) Exercise 10:
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C
O NH 4
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O
O
CH3 -CH2 -OH TsOH Δ
A
O
(iii) Esters Ester hydrolysis Acid catalysed esterification is an essentially reversible reaction. If you follow the backward course of reactions of esterification it gives you the mechanism for ester hydrolysis.
O R
C
O OR'
H2O H
R
C
OH R'OH
Base promoted hydrolysis of esters: Saponification Esters undergo base promoted hydrolysis also. This reaction is known as saponification, because it is the way most of the soaps are manufactured. Refluxing an ester with aqueous NaOH produces an alcohol and the sodium salt of the acid.
O R
C
O OR'
NaOH
H2O
R
C
O Na
R'OH
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This reaction is essentially irreversible because carboxylate ion is inert towards nucleophilic substitution. Mechanism
O R
C
O OR'
R
C
O OR'
R
C
OH R'O
OH
OH
O R
C
O
ROH
If an ester is hydrolysed in a known amount of base (taken in excess), the amount of base used up can be measured and used to calculate the saponification equivalent; the equivalent weight of the ester, which is similar to the neutralization equivalent of an acid. Transesterification Esters can also be prepared by transesterification (an alcohol displacing another from an ester)
O
O R"
C R
OR'
OH
HA heat
R'
C R
OH
OR"
The mechanism of this reactions is quite similar to esterification. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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O H2C
O
CHCOCH 3
CH 3CH 2CH 2CH 2OH
H2C
CHCOCH 2CH 2CH 2CH 3
Transesterification is an equilibrium reaction. To shift the equilibrium to right, it is necessary to use a large excess of the alcohol whose ester we wish to make or else to remove one of the products from the reaction mixture. Reduction of esters (i)
Catalytic hydrogenation O ||
CH3 CH2CH2 C O CH3
(ii)
H2 , CuO, Hg / H 1500 C high pressure
CH3CH2CH2CH2OH CH3OH
Chemical reduction O ||
CH3 CH2CH2 C O CH3
LiAlH4 / H or Na / C2H5 OH
CH3CH2CH2CH2OH CH3OH
Reaction of esters with Grignard reagents The reaction of carboxylic esters with Grignard’s reagent is a good method for the preparation of 3 alcohols.
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O R
C
O OR' R"MgX
R
OMgX R" R"MgX
C
R
C
OH
R" H
R"
R
C
R"
R"
Initially ketones are formed. However, as we know, ketones themselves readily react with Grignard reagent to yield teriary alcohols. Claisen Condensation When
ethyl
acetate
reacts
with
sodium
ethoxide,
it
condensation reaction. After acidification, the product is a
undergoes
a
- keto ester,
ethyl aceto acetate (commonly known as – aceto acetic ester)
O 2CH 3COC 2H5
NaOC 2H5
H3C
O
O
C
CH C
O OC 2H5
C2H5OH
HCl
O
CH 3 CCH 2 COC 2H5
Na
Condensation of this type is known as Claisen Condensation. For esters, it is the
exact
counterpart
of
the
Aldol
Condensation.
Like
the
Aldol
Condensation, the Claisen Condensation involves nucleophilic attack by a carbanion on an electron – deficient carbonyl compound. In the Aldol Condensation, nuclephilic attack leads to addition (the typical reaction of aldehydes and ketones). In the Claisen Condensation, nucleophilic attack leads to substitution (the typical reaction of acyl compounds)
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Mechanism
O Step- 1
R
CH
O
C
OC 2H5
OC 2H5
RCH
C
O OC 2H5
RCH
C
OC 2H5
H
O
O Step- 2 R
CH 2C
HC OC 2H5
C OC 2H5
RCH 2
O
R
O
C
CH
C
OC 2H5
OC 2H5
R
O RCH 2
O
C
CH
C
OC 2H5 C2H5O
R
O
H
O
Step- 3 RCH 2C
C
COC 2H5 C2H5O
O
R
RCH 2C
O C
C
OC 2H5
C2H5OH
R
This step is highly favourable and draws the overall equilibrium toward the product formation
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O Step- 4 RCH 2C
O C
C
O OC 2H5
H3 O rapid
RCH 2 C CH C OC 2H5
R
OH
O
RCH 2C
R Keto form
O CCOC 2H5 R
Enol form
When planning a claisen condensation with an ester it is important to use alkoxide ion that has the same alkyl group as the alkoxyl group of the ester. This is to avoid the possibility of transesterification. An intramolecular claisen condensation is called Dieckmann condensation. In general, the Dieckmann condensation is useful only for the preparation of five and six membered rings. Amides Preparation: In the laboratory amides are prepared by the reaction of ammonia with acid chlorides or, acid anhydrides. Basic Character of Amides: Amides are very feebly basic and form unstable salts with strong inorganic acids. e.g. RCONH2HCl. The structure of these salts may be I or II
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O–H
+
+ O –
R–C
Cl
or
–
R–C
Cl
NH2
NH3
I
II
Acidic Character of Amides: Amides are also feebly acidic e.g. they dissolve mercuric oxide to form covalent mercury compound in which the mercury is probably linked to the nitrogen. 2RCONH2 + HgO
(RCONH)2Hg + H2O
Exercise 11:
PhHC
CH
CH
CH3
OH
PhHC
O C
CH
CH
CH3
OH (Optically active)
O
Optically active
Exercise 12: An ester A (C4H8O2) on treatment with excess methyl magnesium chloride followed by acidification gives an alcohol B as the only Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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organic product. Alcohol B on oxidation with NaOCl followed by acidification, gives acetic acid. Deduce the structures of A & B. Show the reactions involved. (a)
Hydrolysis of amides Amides undergo hydrolysis when they are heated with aqueous acid or aqueous base. Acidic hydrolysis O R
C
O Δ H2 O
NH2 H3O
R
OH NH 4
C
Basic hydrolysis O
O R
(b)
C
Δ H2O
NH2 OH
R
C
O
NH 3
Reduction of amides Amides are reduced by Na/C2H5OH or by LiAlH4 to a primary amine. O R
(c)
C
NH2
4H
Na / C2H5 OH
R
CH 2 NH2
H2O
Reaction with P2O5
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When heated with P2O5 amides are dehydrated to cyanides. O R
C
NH2
P2 O5 H2 O
R
C
N
Amides may also be converted to cyanides by PCl 5. O R
(d)
C
NH2
PCl5
RCCl 2NH 2
2HCl
R
C
N
Hofmann rearrangement Amides with no substitutents on the nitrogen react with solutions of Br2 or Cl 2 in NaOH to yield amines through a reaction known as Hofmann rearrangement. O R
C
NH2 Br 2
4NaOH
RNH 2 2NaBr
Na 2CO 3
Mechanism
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2H2O
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O R
O
H
C
OH
N
O
H OH
R
C
H
N Br
H
R
C
N
Br
Br
Br O
R
N
C
O
-Br
R
C
N
Br
OH
H
OH R
N
C
R
N
C
R
O
O
N
C
O
Carbamate ion O
OH H
RNH2
CO2
OH
OH
HCO3
Illustration 26: Acetamide reacts with NaOBr in alkaline medium to form (A)
NH3
(B)
CH3NH2
(C)
CH3CN
(D)
C2H5NH2
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(B) Illustration 27:
A(C8H8O)
NH 2OH/HCl
B
and
C
H
H
D
E O
Alc. KOH
alc. KOH H3C
C
Cl
F (C7H6O2)
G (C6H7N)
Keeping in mind the fact that B, C, D and E are all isomers of molecular formula (C8H9NO), identify A to G. Solution:
O
Ph
C CH3
(A)
(B)
OH C
N
H3C Ph (C)
C H3C
O
N OH
(D)
Ph
C
NH
CH3
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O (E) H3C
C
(G)
NH2
Ph
NH
Ph
(F)
Ph
COOH
Illustration 28: Formic acid and acetic acid may be distinguished by reaction with (A)
sodium
(B)
dilute acidic permanganate
(C)
2, 4 dinitrophenylhydrazine
(D)
sodium ethoxide
Solution: (B) Exercise 13: (a)
CHCOOH H2C
CH
CH
CH2
A
(b)
CH 2 COOH
CHCOOH
Exercise 14: Identify A, B, C and D? (i)
R—COOH
ND3 /Δ
A
KOH Br2 /Δ
B
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2
NH 3 Δ
B
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(ii)
NH3 /Δ
R—COOH
C
D
KOD Br2 /Δ
Exercise 15: (i)
SOCl 2
O Ph
C
A
H2/Pd(BaSO4)
C
OH 1 eq. DIBALH
CH2N2
B H
Find out A, B and C (ii)
O ConvertPh
(iii)
(a)
O CH 3C
(b)
OH to PhNH2
C
OH
i 2 eq. CH 3 Li ii H +
A
O 1 eq. CH 3CH 2OH
C O
1. LiAlH4 2. H
C
B C
O
What are A, B and C? Exercise 16: An organic compound (A), C8H14O forms an oxime and gives positive haloform reaction. On ozonolysis it gives acetone and a compound (B), C5H8O2 (B) forms a dioxime and on subjecting to haloform reaction gives an acid (C), C4H6O4. On treatment with excess of Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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ammonia and strong heating (C) gives a neutral compound (D), C4H5O2N (D) on distillation with Zn dust forms pyrrole. Suggest structures for (A), (B), (C) and (D). Give the IUPAC names of compounds (A) to (D). Exercise 17: An acidic compound (A), C4H8O3 loses its optical activity on strong heating yielding (B), C4H6O2 which reacts readily with KMnO4. (B) forms a compound C with SOCl 2, which on reaction with (CH3)2NH gives (D). The compound (A) on oxidation with dilute chromic acid gives a compound (E) which on gentle heating readily gives (F), C3H6O. Give structures of (A) to (F) with proper reasoning.
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ANSWERS TO EXERCISES Exercise 1: (i)
O Mg / Ether
Br
OMgBr
MgBr O
H3O
OH
KMnO4 / H
OH OH
HC
(ii)
HBr
CH
CH2
CHBr
Mg / Ether
CH2
CH MgBr
i CO2 ii H3O
O
Exercise 2: (i)
Boiling points of formic acid (100.3°C) and water (100°C) are almost equal. CH2
(ii)
HBr
H2C
Mg / ether
Br
H3C
Br Mg
H3C
CO2
COOH
H3C
H3O
Exercise 3:
OH
A = H3C
CN
B=
OH
O
CH3
OH
H3C
CH3
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O
C=
O
OH
D=
H3C
HO
OH
CH3
CH2
H3C
O
H3C
OH
E=
Exercise 4:
OH
OH Cl
A=
B=
COOH
OH CN
C=
D=
Exercise 5: (A)
O (H3C) 2HC
CH
C
O
CH3
(B) H3C
CH
CH
CH
CH(CH 3)2
CH3 COOH
CH3 CH3
(C) H3C
CH
CH
OH
CH3
(D) H3CHC (E)
H3C
CH3
C(CH3)2 CH
CH(CH 3)2
OSO 2C6H5
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O
COOH E=
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(F)
H3C
CH
CH
CH3
Br
CH3
Exercise 6: Since neutral compound A on hydrolysis gives an acid B and neutral compound C, it must be an ester.
H
Ester
Acid Alcohol
A
B
C
Further since the acid B reduces mercuric chloride, it must be formic acid (HCOOH) which is the only reducing carboxylic acid. So A must be HCOOC2H5.
H
HCOOC2H5
HCOOH C2H5 OH B
A
HCOOH 2HgCl2 C2H5 OH C
NaOH I2
C
Hg2Cl2
2HCl CO2
HCOOH CHI3
Exercise 7: (i)
In formate ion resonance gives rise to identical bond lengths H—C—O– O
O H—C = O
H—C
O
–
O
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Whereas no such resonance is noticed in formic acid (H — COOH) and thus C—O bonds are different in HCOOH (ii)
Acetic acid undergoes intermolecular H–bonding to form dimeric state (CH3COOH)2 and thus results in double molecular weight.
Exercise 8: O O (b)
A=
CH3 OH O O CH3
B= Cl
Exercise 9: (i)
O C
O
O
CH 2 CH 2 CH 2 COH
CH 2 CH 2 COH B=
A=
O CH 2 CH 2 CH 2 C
ClD =
C=
O
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E=
F=
OH Br G=
(ii)
H=
O
O
C
C
C
O
C
OH
O (Friedel - crafts alkylation)
Exercise 10:
O
(c)
O
A=
CH3
OH O
Exercise 11:
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Ph
CH
CH
CH
Ph
CH3
CH
CH
OH
C
CH3
O
O Ph
CH
Ph
O
C
O
O O Ph
C
O O
Ph
CH
CH
CH
CH3
Ph
C
OH
Ph
CH
OH
A= H
C
CH3 O
OH
HC B = H3C
CH
CH3
CH3
Exercise 13: (a)
COOH (Diel's - Alder Reaction) COOH (A)
(b)
O H2C
C
H2C
C
CH O
Exercise 12:
O
CH
NH
O
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CH3
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Exercise 14:
(i)
R—COOH
ND3 ,
RCOND2 (A)
KOH Br2
RND 2 (B)
(ii)
R—COOH
NH3 ,
RCONH2 (C)
KOD Br2
R NH2 (D)
Exercise 15: (i)
O A = Ph
C
O B = Ph
Cl
C
OCH 3
O C = Ph
(ii)
C
H
O Ph
C
O OH
SOCl 2
Ph
O
C
NH 3
Cl
Ph
C
NH 2
Excess OH Br 2 Ph
NH 2
(Hoffmann rearrangement)
(iii)
(a)
O CH 3C
O OH
CH 3Li
H3C
C
OLi OLi
CH 4
CH 3Li
H3C
C
OLi
CH3 H O H3C
C
OH CH3
H3 C
C
(A) CH3
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OH
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(b)
O C
O
C
OH
CH 2
CH3
CH 2OH
CH 2OH
O (B)
(C)
Exercise 16: A is (CH3)2 C = CH–CH2–CH2–COCH3 6–methyl–5–hepten–2–one B is CH3–CO–CH2–CH2–CHO 4–oxopentanal C is HO2 CCH2 CH2 CO2 H 1, 4–butandioic acid D is succinimide
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CH2 – CH2 O=C
C=O N | H
Exercise 17:
OH |
A.
CH3
C H CH2
COOH
B.
CH3
CH
CH COOH
O
O
||
C.
CH3
CH
CH C Cl
O
O
C CH2
C OH
||
E.
CH3
||
D.
CH3
CH
CH C N CH3
O
||
||
F.
CH3
C CH3
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MISCELLANEOUS EXERCISES Exercise 1: Suggest a suitable oxidising agent for the conversion. CH 3
2
C = CHCOCH 3
CH 3
2
C = CHCO 2 H
Exercise 2: Arrange the following compounds in order of increasing acid strength: i CH 3CH 2 CH Br COOH,CH 3 CH Br CH 2 COOH, CH 3
(ii)
2
CHCO 2 H, CH 3CH 2 CH 2 COOH
Benzoic acid, 4-nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4methoxybenzoic acid.
Exercise 3: Why does benzoic acid not undergo Friedel-Crafts reaction? Exercise 4: Why carboxylic acids do not form oximes? Exercise 5:
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Arrange the following in decreasing order of the boiling points: CH3CH2CH2CH2OH, CH3CH2OCH2CH3.CH3CH2CH2COOH. Exercise 6: Why is an amide more acidic than amines? Exercise 7: Identify (A) and (B) in the following sequence of reactions. (i)
CH CH
(ii)
CH3 COOH
CH3COOH excess HgSO4 AIPO4 1075K
A
A
distil
CH3 COOH
B CH3CHO B
Exercise 8: Why excess amine is necessary in the reaction of an acyl chloride with an amine? Exercise 9: Write the structure of the products (a)
acetic acid + benzyl alcohol
(b)
propionic acid + isobutanol
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Complete the following: H 5 C6
OH
PCl5
O
H3C
heat
OK O
H2 O
Electrolysis
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ANSWERS TO MISCELLANEOUS EXERCISES Exercise 1: Alkaline KMnO4, acidified K2Cr2O7 or HNO3 can not be used since all of these will cleave the molecule at the site of double bond giving a mixture of ketone/acids. The most suitable reagent for this oxidation is NaOI (I2/NaOH) since methyl ketones on treatment with NaOH undergo Iodoform reaction to give Iodoform alongwith the Na salt of a carboxylic acid having one carbon atom less than the starting methyl ketone.
CH3 H3C
C
O CH
C
CH3 CH3 3NaOI
H3C
C
CHCOONa
CHI3
2NaOH
4 Methylpent 3 en 2 one H / H2O
(Mesityl oxide)
CH3 H3C
C
CHCOOH
3 Methylbut 2 en 1 oic acid ( ,
Dimethylacrylic acid)
Exercise 2: (i)
We know that +I, effect decreases while
I effect increases the acid
strength of carboxylic acids. Since +I effect of isopropyl group is more than that of propyl group, therefore, (CH 3)2CHCOOH is a weaker acid than CH3CH2CH2COOH. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Further
since
I
effect
decreases
with
distance,
therefore,
CH3CH2CH(Br)COOH is a stronger acid than CH 3CH(Br)CH2COOH. Thus, the overall acid strength increases in the order: CH3 H3C
CH
COOH < CH 3
< H3C
H2C
CH
CH 2 COOH
CH 2 COOH < H3C
CH 2 CH
Br
(ii)
COOH
Br
Since electron donating groups decrease the acid strength, therefore, 4–methoxybenzoic acid is a weaker acid than benzoic acid. Further since electron–withdrawing groups increase the acid strength,
therefore,
both
4–nitrobenzoic
acid
and
3,
4–dinitrobenzoic acids are stronger acids than benzoic acid. Further due to presence of an addition NO2 at m – position w.r.t COOH group, 3, 4–dinitrobenzoic acid is a little stronger acid than
4–nitrobenzoic
acid.
Thus
the
overall
acid
strength
increases in the order: 4 methoxybenzoic acid benzoic acid
4 nitrobenzoic acid
3,4 dinitrobenzoic acid
Exercise 3:
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Due to deactivation of the benzene ring by electrons withdrawing effect of the
COOH group.
Exercise 4: Due to resonance between lone pairs of electrons on the O-atom of the OH group and C = O, the carboxyl carbon is less electrophilic than carbonyl carbon in aldehyde and ketones. Therefore, nucleophilic addition of NH2OH to the C = O group of carboxylic acid does not occur and hence carboxylic acids do not form oximes. Exercise 5: CH3CH2CH2COOH > CH3CH2CH2CH2OH > CH3CH2OCH2CH3 Hint: Due to hydrogen bond Exercise 6:
O R
O NH2
R
NH 2
Firstly due to delocalization of lone pair of electrons of N atom over the C = O group, amino group gets a positive charge which makes N–H bond weak. Secondly the anion left after removal of a proton is stabilized by resonance. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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O
O
R
NH
R
NH
In amines no such stabilization is possible. Exercise 7: (i)
CH
CH3 COOH excess HgSO4
CH
Acetylene
(ii)
CH3 COOH Acetic acid
AIPO4 , 1075K H2 O
CH3
CH OOCCH3 Ethylidene diacetate
CH2
C
O
CH3 COOH
Ketone
distil 2
CH3CO 2 O CH3CHO Acetic anhydride
Acetaldehyde
CH3 CO 2 O Acetic anhydride
Exercise 8:
O
O
R NH2 R
Cl
HCl R
NHR'
The products of this reaction are an amide and HCl. HCl formed in this reaction protonates unreacted amine and since protonated amines are not nucleophiles, the reaction with acyl chloride stops.
R NH2
HCl
RNH3
Cl
No nucleophilic site
O R
Cl
RNH3
No reaction
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Hence, reaction must be carried out with twice as much as amine as acyl chloride. Exercise 9: (a)
CH3—COOCH2—Ph
(b)
CH3—CH2COOCH2—CH(CH3)2
Hint: Esterification Exercise 10: (a)
COCl
COOH
HCl
PCl5
(b)
CH3 COOK H2O
electrolysis
POCl3
CH3COO
K
At anode: CH3
2CH3COO
CH3
2CO2
2e
At cathode: 2e
2H2O
2OH
H2
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SOLVED PROBLEMS
Subjective:
Board Type Questions Problem 1: Write chemical reactions to effect the following transformation. (a)
Butan-1-ol to butanoic acid
(b)
Benzyl alcohol to phenylacetic acid
(c)
Bromobenzene to benzoic acid
(d)
p-methylacetophenone to benzene-1, 4-dicarboxylic acid
Solution: (a)
CH3
CH2
CH2
CH2
OH
Aq. KMnO4
CH3 CH2CH2COOK
Butan-1-ol
(b)
C6H5 CH2 OH
(c)
Bromo benzene
KCN
C6H5 CH2Br
C6H5 CH2 CN
C6H5MgBr
Dry ice
CH3CH2CH2COOH Butanoic acid
Hydrolysis H / H2 O
Benzyl bromide
Mg dry ether
C6H5Br
Pot. Butanoate
PBr3 or HBr
Benzyl alcohol
Dil. H2 SO4
C6H5 CH2 COOH Phenylacetic acid
C6H5COOMgBr
H3 O
Phenyl mag. bromide i.e. CO 2 (s)
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C6H5COOH Benzoic acid
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(d)
H3C
COCH 3
P-methylacetophenone
KMnO4 / KOH,
KOOC
COOK
Dil. H2SO4
Dipotassium benzene- 1, 4dicarboxylate
HOOC
COOH
Benzene-1, 4-dicarboxylic acid
Problem 2: Arrange the following compounds in increasing order of their boiling points: acetic acid, methyl formate, acetamide, propan-1-ol Solution: (i)
Out of all these compounds methyl formate does not undergo Hbonding therefore, its boiling point is the lowest.
(ii)
Amongst the remaining three compounds intramolecular Hbonding is most extensive in acetamide, followed by acetic acid and least in propan-1-ol. Therefore, their boiling points decrease in the same order i.e. acetamide>acetic acid>propan-1-ol. Thus, the overall increasing order of their boiling points is methyl formate ii > i
(C)
ii > i > iii
(D)
iii > i > ii
(iii) CH
CCOOH
Solution: sp hybridized carbon is most electronegative and it stabilizes the carboxylate ion. to the maximum extent. Problem 10:
OH COOH H3C
CrO3
A
Δ
B
A and B are (A)
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(B)
CH3CH(OH)CH3, CH3COCH3
(C)
CH3COCH3, CH3COOH
(D)
CH3CH = CHCOOH, CH3CH = CH2
Solution:
OH
O COOH
O COOH
CrO 3
Δ
(A)
(B)
(A) Problem 11:
*
CH3 CH2COOH
NaHCO3
A a gas
i Me MgX ii H3 O
B
*
B is C is C14 *
(A)
CH3 COOH
(C)
CH3CH2 COCH3
*
*
(B)
CH3COOH
(D)
CH3 COCH3
*
Solution:
*
*
CH3CH2COOH *
CO2
MeMgX
NaHCO3
H3 O
CH3CH2COONa H2O CO2 *
CH3 COOH
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Problem 12:
O
OC2H5
C2 H5O
A
CH2CO2C2H5
A is (A)
(B)
O
O
OC2H5
(C)
CO2C2H5
(D)
O
O O
O
Solution:
O
O
O
O C
OC 2H5
C2 H5 O
CH 2CO 2C2H5
O
OC 2H5 CH
C
O
C2H5
C2H5
O
Problem 13: The carboxylic acid which has maximum solubility in water is (A)
phthalic acid
(B)
succinic acid
(C)
malonic acid
(D)
salicylic acid
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Solution: Salicylic acid has intramolecular hydrogen bonding. So it is least soluble. Out of the first three acids malonic acid has smallest non – polar alkyl part. So it is most soluble. (C) Problem 14:
Br
i Mg/ ether ii CO2 iii H3O
O H CH3
O
The final product is (A)
(B)
COOH
MgBr
O
CHO H CH3
(C)
CH3
O
(D) None of these
COOH
CHO CH3
Solution:
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MgBr
Br
COO COOH CO2
Mg / ether
O
O H CH3
O H3O
H
CH3
O
H
CH3
O
O
CHO CH3
Problem 15: In the reaction CH3 CH2COOH
NH3
A
B
P2 O5
C,
Identify C. (A)
CH3CH2CN
(B)
CH3CH2CH2NH2
(C)
CH3CH2CH2OH
(D)
(CH3CH2CO)2O
Solution: O CH 3CH 2COOH
NH 3
H2CH 3C
C
O ONH 4
Δ
H3C
CH 2 C
NH 2
P2O 5 H3C
CH 2 CN
Fill in the blanks Problem 16: Benzoic acid has a………….pKa value than phenol. Solution: Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Lower Problem 17: The acid derivatives are hydrolysed…………..readily in either alkaline or acidic medium than in neutral medium. Solution: More Problem 18: O-nitrobenzoic acid is…………acid than p-nitrobenzoic acid. Solution: Stronger Problem 19: Malonic acid on heating with phosphorous pentoxide gives………………….. Solution: Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Carbon suboxide (C3O2) Problem 20: Cinnamic
acid
can
be
prepared
from
benzaldehyde
using…………….reaction and shows…………..isomerism. Solution: Perkin, geometrical True and False Problem 21: Methanoic acid like ethanoic acid can be halogenated in the presence of red phosphorous and Br2. Solution: False: Due to the absence of
- hydrogen, methanoic acid can not be
halogenated by using Br2/red P.
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Problem 22: As the number and size of the substituents around the
COOH group
increases the rate of esterification slows down. Solution: True Problem 23: The boiling point of propionic acid is less than that of n-butanol, an alcohol of comparable molar mass. Solution: False: Due to extensive intermolecular hydrogen bonding, and the cyclic dimer formation, propanoic acid has higher b.p. than n-butanol. Problem 24: The conjugate base of chloro acetic acid is more resonance stabilized than the conjugate base of acetic acid.
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Solution: True: Due to the presence of chlorine atom. Problem 25: The correct order of acid strength of the given acid is H 2 SO4 > CH3COOH > H 2O > HC ≡ CH > C2 H5 OH
Solution: False: The correct order is H2SO4 CH3CO2H H2O C2H5OH HC CH .
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ASSIGNMENT PROBLEMS
Subjective:
Level-O 1.
Describe the following: (a)
2.
Saponification
(b) Decarboxylation
How is acetic acid prepared from? (i) Acetylene
(ii) Ethyl alcohol
3.
How is Grignard reagent employed to prepare a carboxylic acid?
4.
Explain the following about acetic acid:
5.
(i)
Its boiling point is higher than that of n – propanol.
(ii)
It is weaker acid than formic acid.
(iii)
It is stronger acid than phenol.
Acid anhydrides have higher boiling points than the corresponding carboxylic acids.
6.
Give IUAPC name for each of the following compounds. (a) H3CHC CHCOOH
(b)
COOH
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(c)
(d) Cl2CHCOOH
CHCOOH CH3
(e)
O HO C(CH 2)3COOH
7.
HOCH2CH2CH2CH2COOH can be prepared from HOCH 2CH2CH2CH2Br in two ways. One is through Grignard synthesis and the other through nitrile synthesis. Which method would you prefer?
8.
Why the bond length of C = O in carboxylic acids is bit longer than in aldehydes?
9.
Acetic acid can be halogenated in presence of phosphorus and chlorine but formic acid cannot be halogenated in the same way.
10.
During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as fast as it is formed. Explain.
11.
Identify the products in the following reactions (a)
12.
HCOOCH3
i) CH3CH2MgBr / ether ii) H3O
(b) C6H5CHO CH3CH2CHO
dil. NaOH
Explain why benzamide is less easily hydrolysed than methyl benzoate.
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13.
There are two –NH2 groups in semicarbazide. However only one is involved in the formation of semicarbazones. Explain.
14.
Carbon-oxygen bond length in formic acid are 1.24 Å and 1.36 Å but in sodium formate both the carbon-oxygen bonds have same value i.e.1.27 Å.
15.
Acid and acid derivatives although contain >C=O group, do not undergo the usual properties of carbonyl group explain.
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Level – I 1.
Convert the following: (a)
Acetaldehyde
Butane 1, 3 diol
(b)
Acetaldehyde
But 2 enoic acid
2.
Discuss the action of heat on
- hydroxy carboxylic acids.
3.
Compare the Ka1 and Ka2 values of maleic and fumaric acids.
O
4.
R
||
NH2 are soluble in aqueous acids like HCl, H 2SO4 etc but R C NH2
are almost insoluble in those acids. Explain. 5.
What happens when: (i)
Dry chlorine is passed through acetic acid in presence of sunlight.
(ii) 6.
Formic acid is reacted with ammonical silver nitrate solution.
Complete the following equations by writing the missing A, B, C. CH3 CH2MgBr
7.
A
CH3 CH2CH2CH2OMgBr
H3 O
B
What is A and B? O
C4H8O3
A
CrO3
B
warm
H3C
CH3
CO2
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KMnO4
C
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O
8.
||
(a)
CH3
(b)
HO C CH2 || O
CH
CH CH2
C CH3
C OH 2
A
C OH || O
B
9.
Cite 2 reasons why malonic acid does not form a cyclic anhydride.
10.
CH CH
H2SO4 HgSO4
A
O
B
SOCl2
C
C6H6 anhy AlCl3
D
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Level – II 1.
A hydrocarbon A, (C8H10), on ozonlysis gives compound B (C4H6O2) only. The compound B can also be obtained from the alkyl bromide, C (C3H5Br) upon treatment with magnesium in dry ether, followed by CO2 and acidification. Identify A, B and C and also gives equations for the reactions.
2.
Predict the product of the reaction given below, given that it is formed under high – dilute conditions. EtO2C CH2 4 CO2Et
3.
EtO cat
Compound (A), C5H8O2, liberates CO2 on reaction with sodium bicarbonate. (A) exists in two forms neither of which is optically active. It yields compound (B), C5H10O2 on hydrogenation. Compound (B) can be separated into two enantiomorphs. Write the structural formulae of A and B.
4.
An organic compound (A) reacts with PCl 5 to form a compound (B). A also reacts with NaOH to form a compound (C). Both B and C react together to form a compound D (C6H10O3). What are A, B, C and D?
5.
O H3C a
Δ
CH3
H3C H
`
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O b
c
6.
B2H6 H
1 Ag2 O 2. Br2
H3C
C
O 2N
CH 2OH
Br
The active ingredient of an insect repellent is N, N–diethyl–m– toluamide. Outline a synthesis of this compound starting with benzene.
7.
A carboxylic acid (A) of unknown structure was found to contain only C, H and O. 154.5 mg of this acid required 11.9 mL of 0.22 N NaOH to reach equivalence point. Gentle heating of A led to evolution of CO2 and formation of new carboxylic acid (B) with nutralisation equivalent of 74. Suggest structures for A & B.
8.
Compound (A), C5H11NO is not soluble in cold dil. alkaline or acidic solutions. When (A) is refluxed with NaOH solution, a gas (B) is evolved and a salt (C) is formed. Acetyl chloride reacts with (B) to give (D), C4H9NO. (B) reacts with HNO2/HCl to give a yellow oil (E). Give structures of A to E.
9.
Compound (A) (C6H12O2) on reduction with LiAlH4 yielded two compounds B and C. The compound B on oxidation gave (D) which on treatment with aqueous alkali and subsequent heating furnished (E). The latter on catalytic hydrogenation gave C. The compound D was Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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oxidised further to give F which was found to be monobasic acid with molecular weight 60. Deduce the structures of A, B, C, D, E and F. 10.
Suggest a mechanism for each of the following tranforamations. CO 2Et CO 2Et
CO 2Et t - BuO t - BuOH
O
CO 2Et O
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Objective:
Level – I 1.
Which of the following will not give cyclic anhydride on heating? (A)
COOH
(B) H2C
COOH
H2 C
COOH
H2C COOH
(C)
(D)
CH 2COOH
COOH
H2C CH 2COOH
2.
COOH
In the reaction CH3 CH2COOH
Cl2 / red P
A
alc. KOH
B
the compound B is
3.
(A)
CH3CH2OH
(C)
CH2 = CH
COOH
(B)
CH3CH2COCl
(D)
CH3CHClCOOH
Give the structure of the product of the following reaction CH3
O
CH 3NH 2
O
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(A)
(B)
CH3
CH3
O
O
N
HO
CH3
(C) NH
C
3
(D) None
O H3C
NHCH
CH 2
CH 2
CH
CH3
OH
4.
5.
Malonic and succinic acids are distinguished by (A)
heating
(B)
NaHCO3
(C)
both A and B
(D)
none
Consider the following acids 1.
MeCH2COOH
2.
Me2CHCOOH
3.
Me3CCOOH
4.
Et3CCOOH
Correct order of rate of esterification.
6.
(A)
1>3>2>4
(B)
1>2>3>4
(C)
4>3>2>1
(D)
3>2>1>4
Which of the following compound on heating with caustic potash gives ammonia? (A)
CH3CONH2
(B)
CH3CH2NH2
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O
(C)
7.
(CH3)2C = NOH
(D)
||
CH2
C N CH3
2
Identify the final product in the following sequence of reactions O H2 C
CH 3CH 2MgBr
8.
9.
CH 2
X
H
Y
KMnO 4
Z
(A)
CH3CH2CH2COOH
(B)
CH3CH(CH3)COOH
(C)
CH3CH2COOH
(D)
CH3CH2CH2CH2OH
The typical reaction of carbonyl group is suppressed the most in (A)
CH3
CHO
(C)
CH3COCH3
(B)
CH3CONH2
(D)
CH3COOCH3
Which of the following carboxylic acids undergoes decarboxylation easily? (A) C6H5COCH2COOH
(B)
C6H5COCOOH
(C) C6H5 HC
(D)
H5C6
COOH
COOH
NH2
OH
10.
CH
Benzoic acid gives benzene on being heated with X and phenol gives benzene on being heated with Y. Therefore X and Y respectively are (A)
sodalime and copper
(B)
Zn dust and NaOH
(C)
Zn dust sodalime
(D)
sodalime and Zn dust
O
11.
X
P2 O5
||
R C NH2
Br2 OH
Y
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X and Y respectively are
12.
(A)
R
CN, RCH2NH2
(C)
RCN, RNH2
Al OC2H5
2CH3 CHO
(B)
RNH2, RCN
(D)
RCN, RCH2NH2
3
Product and the name of the reaction respectively are
13.
(A)
CH3COOCH2CH3, Tisckenko
(B)
CH3COOH, CH3CH2OH, cannizzaro
(C)
2CH3CH2OH, perkin
(D)
none
Which of the following reactions sounds correct? 1.
HOOC COOH
Br
Br 2
COOH Br
2.
COOH
HO COOH alk.KMnO COOH
COOH
4
Cold HO
COOH (meso)
3. HOOC
Br
COOH
Br
COOH
Br 2 COOH
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4.
HO HOOC
alk.KMnO
COOH
4
COOH HO
COOH (meso)
(A)
1 and 2
(B)
2&3
(C)
3&4
(D)
1&4
14.
O C
OH
CH 3ONa
(A)
(B)
O C
O
O C
CH3
O Na CH 3OH
(C)
(D)
O
O C
C
OH
CH3
O
15.
||
The reaction of R C NH2 with a mixture of Br2 and KOH gives R as a product. The intermediate involved in this reaction is O
(A)
||
R C NHBr
(B)
R N C O O
(C)
R
NHBr
(D)
||
R C NBr2
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NH2
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16.
Among the carboxylic acids: O
OH O
O
O
O H
CH3
NO 2 and
OH
COOH
OH
(a)
(b)
(d)
(c)
the, one or more than one undergoing decarboxylation readily is (are)
17.
(A)
Only a
(B)
Both a and b
(C)
a, b and d
(D)
a, b and c
Lactide
In the following reaction, hydroxy carboxylic acid
i) DIBALH ii) H+
A is
18.
(A)
lactone
(B)
(C)
a hemiacetal
(D)
Which of the following can form an anhydride on heating? O
O
(A) H
(B) OH
O
O
O
O
(D) HO
HO HO
(C)
19.
, - unsaturated acid
OH
O HO OH
O i) ND3 , ii) KOH, Br2 , H2O
Product is
OH
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NH2
ND 2
(A)
(B)
O
NH2
O
(C)
20.
ND 2
(D)
Acetamide is treated separately with the following reagents. Which one of these would give methylamine? (A)
PCl5
(B)
Sodalime
(C)
NaOH + Br2
(D)
None of these
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Level-II
O
1.
||
Ph C OH CH2N2
(A)
PhCOOCH2N2
(B)
PhCOOCH3
(C)
PhCH3
(D)
none of these
2.
A H
O
C
C B
3.
Pr oduct
H O
O
C D
H
O
(C
O) bond lengths designated by A, B, C, D are in order
(A)
A=C
