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SUBJECT – CHEMISTRY TOPIC – ALCOHOL, PHENOL AND ETHER COURSE CODE – AISM-09/C/APE Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Contents :- ALCOHOL, PHENOL AND ETHER Introduction, Aliphatic hydroxyl compound………………………………………………………3 Preparation of Alcohols………………………………………………………………………………….….5 Physical properties, chemical properties ………………………………………………………….15 Periodate oxidation…………………………………………………………………………………………..28 Pinacol, pinacol- rearrangement ………………………………………………………………………30 Ether …………………………………………………………………..……………………………………………34 Preparation of ethers………………………………………………………………………………………..35 Properties of ether…………………………………………………………………………………………….39 Chemical reaction ……………………………………………………………………………………………..40 Phenol……………………………………………………………………………………………………………….47 Physical properties …………………………………………………………………………………………...50 Chemical properties…………………………………………………………………………………………..52 Mercuration………………………………………………………………………………………………………59 Answers to exercises……………………………………………………………………………….…………82 Miscelleneous exercises…………………………………………………………………………………….90 Solved problems………………………………………………………………………………………………..98 IIT level questions……………………………………………………………………………….…………….102 Objective problems……………………………………………………………………………….………….109 Fill in the blanks…………………………………………………………………………………………………117 Assignment problems………………………………………………………………………………………..122
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ALCOHOL, PHENOL, ETHER INTRODUCTION Hydroxy compound can be classified in the following three categories. Aliphatic hydroxy compound Monohydric i.e.
H
H
H
C
C
H
H
OH Contain only one - OH group
Dihydric
HO
H
H
C
C
H
H
OH Contain tw o - OH group
Polyhydric
H
H
H
H
C
C
C
OH
OH
OH
H
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Contain three and more than three hydroxyl group. Monohydric alcohols are of three types. Monohydric Alcohols
Primary or 10 R
CH 2OH
Secondary or 20 OH R
C
R
Tertiary or 30 Alcohol OH R
H
C
R
R
Illustration 1: Classify the following into primary, secondary and tertiary alcohols: CH3 OH
HO
OH H3C
(a)
(b)
(c)
Solution: (a)
Tertiary,
(b)
Secondary,
(c)
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Tertiary
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PREPARATION OF ALCOHOLS 1.
From Alkanes
Alkanes having tertiary carbon on oxidation with cold alkaline KMnO 4 give tertiary alcohol. H R
C
OH R
KMnO 4/OH
C
R
R
R
2.
R
From Alkenes Alkenes can be converted into alcohol by the following reactions:
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OH HOH/H 2SO 4
R
CH
CH 3
OH (i) Conc. H 2SO 4
R
CH
CH 3
(ii) HOH OH R
CH
CH2
(i) Hg(OCOCH (ii) NaBH
3)2/HOH
R
CH
CH 3
4
(i) BH 3/THF
R
(ii) H 2O 2/OH
CH 2OH
(anti markonikov's product)
CO H2 Co(CO)
CH 2
R
CH 2
CH 2
4
H2/Pt
[Oxo process] R
3.
CHO
CH 2CH 2CH 2OH
From alkyl halides Alkyl halides give alcohol with KOH/NaOH or with moist Ag 2O. HOH/NaOH R
4.
X
moist Ag 2O
R
OH
R
OH
Reduction of aldehydes and ketones Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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(a)
Reduction by reducing agents (i)
Aldehyde gives primary alcohol O R
(ii)
C
H
[H] Reducing agent
R
CH 2OH
Ketone gives secondary alcohol
OH
O R
C
R
[H] Reducing agent
R
CH
R
Reducing agents (i)
LiAlH4
(ii)
NaBH4
(iii)
Na/C2H5OH
(iv)
Metal (Zn, Fe or Sn)/Acid (HCl, dil H2SO4 or CH3COOH)
(v)
(a) Aluminium isopropoxide/isopropylalcohol, (b) H2O
(vi)
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NaBH4 and aluminium isopropoxide reduces only carbonyl group and has no effect on any other group.
CH3
CH
Reduction
CH CHO
with
NaBH4
aluminium
CH3
CH
CH CH2OH
isopropoxide
is
known
as
Meerwein–Ponndorf Verley (MPV) reduction. LiAlH4 has no effect on double and triple bonds but if compound is
- aryl,
,
- unsaturated carbonyl compound
then double bond also undergoes reduction. O H2C
H5C6
(b)
CH
CH
C
CH
OH CH3
LiAlH 4
CHO
LiAlH4
H3C
CH 2
H5C6
Reduction by Grignard reagents
Addition followed by hydrolysis
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CH
CH2
CH3
CH2 CH2OH
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O H
C
O H
R
C
MgX H
OH H2O/H
R
C
H
X H
Mg OH
H 10 alcohol
O R
Mg
R'
X
C
OMgX H
R
C
OH
H
H2O/H
R
C
R'
X H
Mg OH
R' 20 alcohol
O OMgX R'
C
OH
R'' R
C
R''
H2O/H
R
C
R'
X R'
Mg
R'' 30 alcohol
Methanol cannot be prepared by this method. 5.
Reduction of carboxylic acid, Acid chlorides and esters: (a)
Reduction by LiAlH4
O R
C
G
LiAlH 4
R
CH 2OH
H
G
G = OH (acid)
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OH
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G = Cl (acid chloride) G = OR (ester) (b)
Reduction by BH3
Carboxylic acids and esters are reduced in to primary alcohol by BH 3.
O R
C
i BH3 / THF
OH
RCH 2OH
ii H2 O / H
O R
(c)
C
O
i BH3 / THF
R'
ii H2O/H
R
CH 2OH R'OH
Bouveault – Blanc reaction
O R
C
OR'
Na/C2H5OH
RCH 2OH R'OH
Illustration 2:
COOCH 3
LiAlH4
A
(CH2)n COOH
(i) BH3/THF (ii) H2O/H
B
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Solution:
CH 2OH A = (CH 2)n COOH
If compound has
COOH as well as ester group then reactivity of ester is
more than acid towards LiAlH4. CH 2OH B = (CH 2)n COOCH
3
Acid is more reactive than ester towards BH 3.
Exercise 1: Find A and B. CH3
Hg OAC
(i) (ii)
6.
2
THF / H2 O
NaBH 4
A
OH
CH3CH2COOCH3
B
CuO CuCr2 O4
A B
From aliphatic primary amines It react with nitrous acid to give alcohol.
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Nature of alcohol depends on the nature of carbon having
NH2
group. Reaction proceeds through carbocation hence rearranged alcohol is obtained. H3C
CH 2 CH 2 NH2
NaNO 2/HCl
H3C
CH 2 CH 2OH
H3C
CH
CH3
OH
7.
From Oxiranes
Oxiranes react with Grignard reagent to give mono hydric alcohol. Nature of G.R is basic hence it attack on less hindered carbon of oxirane ring. δ CH 2
H2 C
δ R
δ Mg
H2 C
X
O δ
CH 2 R
OMgX HOH/H HO
CH 2 CH 2 R
Illustration 3: (a)
Find A, B, C, D, E. O Mg dry ether
PhBr
(b)
CH3COOC2 H5
A
i CH3 MgBr ii H2O / H
B
H
C
D E
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Solution: (a)
A = PhMgBr (b)
O
B =Ph
C = Ph
CH3 H3C
C
OH
C2H5OH
CH3
8.
Fermentation of carbohydrates
H2O Invertase
C12H22O11 Sucrose
C6H12O6 glucose
C6H12O6 fructose
Zymase
2C2H5OH
2CO 2
Illustration 4: (i)
H3C (i) H3C
C2H5MgBr
C CH2 O
A
(ii) HOH/H
(ii) H3C
CH3 CH
C
CH
CH2
CH3
Oxymercuration Demercuration
B
H3C
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OH
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Solution: (i)
CH3 H3C
C
CH 2 C2H5 ( 2 methyl 2 pentanol)
OH
(ii)
CH3 CH3 H3C
CH
C
CH 2 CH 2CH 3
OH 0
(3 alcohol)
Exercise 2: (i)
Identify A, B & C MgBr
RCO2Et + (CH2)5
H 3O +
MgBr
(ii)
A
Δ -H2 O
H 2 /Ni
B
O H
C
C
O CH2
CH2
OH
RMgX
RH
H
C
CH2
CH2
OM gX
H
C(OH) CH2 CH2 OH
( 1 eq)
O H
C
CH2
CH2
OH
RM gX ( 2 eq)
R
Account the reason for the above reactions. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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PHYSICAL PROPERTIES Solubility Alcohols are soluble in water due to formation of H – bonding between water & them. As the molecular mass increases, the alkyl group become larger which resists the formation of H – bonds with water molecules and hence the solubility goes an decreasing. Boiling Point Intermolecular H – bonding is present between alcohol molecules. This makes high boiling point.
H
O R
H
O R
H
O R
Amongst the isomeric alcohols, the order of boiling point is 1
> 2 > 3
alcohol. CHEMICAL PROPERTIES Chemical properties of alcohols can be discussed under following categories:
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(A)
(A)
Reaction involving breaking of carbon – oxygen bond.
(B)
Reaction involving breaking of oxygen – hydrogen bond.
(C)
Oxidation of alcohols.
(D)
Dehydrogenation of alcohols.
(E)
Some miscellaneous reactions of monohydric alcohol.
Reaction involving breaking of carbon – oxygen bond Order of reactivity of alcohol. 3 > 2 > 1 (i)
SN reaction
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HCl/Anhy ZnCl 2
Δ NaX/H2SO 4
R
OH
PCl5 or PCl3
P/Br2 or PBr3
Δ SOCl2 /Py ridine or SOCl2 /Ether
(ii)
R
Cl
R
X NaHSO 4 H2O
R
Cl
R
Br
R
Cl
SO 2 HCl
Dehydration of alcohol
Dehydration of alcohol to give alkene. (a)
Dehydrating agents are
Conc H2SO4/ , KHSO4/ , H3PO4/ , Anhyd Al 2O3/ , Anhyd PCl 5/ , Anhyd ZnCl2/ , BF3/ , P2O5/ . (b)
Reactivity of alcohols. (Ease of dehydration) 3 >2 >1
(c)
Product formation always takes place by saytzeff rule. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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OH H3C
CH 2 CH
CH3
Conc. H2 SO 4 Δ
H3C
CH CH (Major)
CH3
H3C
CH 2 CH CH2 (Minor product)
* Alcohols on acetylation gives acetyl derivative which on pyrolytic elimination always gives Hofmann product.
OH H3C
CH 2 CH
OCOCH CH3
CH3 CO 2 O / Py
H3C
CH 2 CH
3
CH3
Δ
H3C
CH 2 CH (Major)
CH2 H3C
Mechanism in presence of acidic medium E1 mechanism: follow saytzeff‘s rule.
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CH CH (Minor)
CH3
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CH3 H3C
C
CH3 CH 2 O
H
Conc.H2 SO4
H
H3C
CH3
CH 2 O
CH3
CH3 H3C
C
C
H
H
CH3 CH 2 CH3
1,2methyl shift
H3C
C
CH 2
CH3
CH3 H3C
C
CH
CH3
Exercise 3: Write mechanism
CH3 OH CH
CH3 CH3
H CH3
(B)
Reactions
due
to
breaking
of
oxygen
hydrogen
bond.
(Reactions due to acidic character of alcohols) (a) Alcohols are acidic in nature because hydrogen is present on electro negative oxygen atom. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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(b)
Alcohol is weaker acid R
O
H
H R
O
acidity
stability of acid anions. Acidity of 1 > 2 > 3 Alcohols give following reactions due to breaking of oxygen – hydrogen bond. (i)
Reaction with metal
R
O
H M
R
O M 1/2 H2 Metal alkoxide
M = 1st group metal. M = Al, Mg, Zn
3R OH Al
RO
3
Al
3 H2 2
Aluminium alkoxide
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(ii)
Esterification (With carboxylic acid)
O
O H
R'
OH
R
C
OH
R
C
O
R'
H2O
It is reversible acid catalysed reaction. It follow SN1 mechanism.
O
O
H
OH
H H3C
C
H3C
O
C
H
H3C
OH
C OH R
OH H3C
C
OH OR
H2O
H3C
C
H
OH OR
H3C
O H
O
C
O
R
OH H
O H3C
C
OR
Increasing the size of alkyl group on alcohol part decreases the nucleophilic character because steric hindrance increases.
Reactivity α
1 Steric hindrence in RCOOH/ ROH
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Order of reactivity of alcohols CH3OH > 1 alc > 2 alc > 3 alc (iii) Ester formation with proton acid having –OH group: to give inorganic ester. (a) R
O
H
HO
N
O
(nitrous acid) (b)
R
O
H
HO
S
N
O H2O
(Alkyl nitrite)
O R
O O
OH
O
RO
S
OH
O Alkyl hydrogen sulphate
(iv) Alkylation of Alcohol
R O H
CH3
2
SO4 / NaOH
or CH3I / K 2 CO3
R O CH3
Methylation is mainly used for determination of hydroxyl groups in an unknown compound.
No. of hydroxyl groups
Molecular weight of methylated ether prodcued molecular weight of reac tan t 14
Exercise 4: Arrange the following in increasing order of acidic strength. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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(a)
CH3 H3C
CHOH
H3C
CH2OH H3C
OH H3C
CH3
C
OH
CH3
(i)
(ii)
(iii)
(iv)
(b) Arrange the following in increasing order of esterification:
(C)
MeCOOH
EtCOOH
(Et)2CHCOOH
(i)
(ii)
(iii)
Oxidation of alcohol Oxidation
of
alcohol
is
dehydrogenation
reaction
2–elimination reaction.
R
O
H
C
R'
O 1, 2 elimination
R
C
R'
H2
H
So oxidation of alcohol (a)
numbers of
- hydrogen atom.
With mild oxidising agents like: (i)
X2
(ii)
Fenton reagent [FeSO4/H2O2]
(iii)
Jones reagent / CH3COCH3 [CrO3/dil. BaSO4]
(iv)
K2Cr2O7/H+ cold
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which
is
1,
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R
[O]
CH2OH
RCHO
OH R
CH
O [O]
R'
R
C
R'
CH3 H3C
C
OH
[O]
no reaction
CH3
Note: PCC (Pyridinium chloro chromate) is a selective reagent which converts 1 alc to aldehyde. (b)
With strong oxidising agent Oxidising agents are (i)
KMnO4 / OH /
(ii)
KMnO4 /H /
(iii)
K 2Cr2O7 / H /
(iv)
Conc. HNO3 /
RCH2OH
O
RCOOH n carbon
n carbon
H R
C
O R
[O]
R
C
R
OH
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(D)
Dehydrogenation with Cu/573K or Ag/573K (a)
1 alcohol Cu / 573K
R CH2OH
(b)
aldehyde
2 alcohol
ketone Cu / 573K
R CHOH R
(c)
RCHO
3 alc
R CO R
undergo dehydration to form alkene. CH3
H3C
C
CH3 OH
Cu/ 573 K
H3C
CH3
C
H2O
CH2
Reduction
R
(E)
O
H
HI/Red P
R
H
Miscellaneous reactions of mono hydric alcohol (i)
Methylation with CH2N2 in presence of BF3. CH2N2 BF3
R O H
(ii)
R O CH3
N2
Haloform reaction Ethyl alcohol and 2 methyl alcohol gives haloform reaction.
H
H
H
C
C
H
H
OH
CaOCl2 H2 O
(HCOO) 2Ca
CHCl
3
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Exercise 5: (i)
Out of these compound which gives iodoform test: (a)
CH3
CH2
(b) (c)
CHOHCH3
PhCH2CHOHCH3
PhCHOHCH3 (d)
(e)
CH3CH2OH
CH3COCH2
(ii)
O
COOC2H5
NaBH4 CH3OH
O
LiAlH4
(A)
(B)
O
Distinguishing 1 , 2 , 3 alcohol 1 alc
Test (I) Lucas test [ZnCl2 + HCl]
No
reaction
2 alc
3 alc
at White turbidity White turbidity
room
after
temperatur
10 min.
e
RCH(OH)R
5-
HCl
instantan eously R3C
ZnCl 2 R
CH
R
ZnCl 2 H2O
Cl
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OH
R3C
Cl
HCl
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(II) Victor Meyer test Red colour
Blue colour
Colourless
(P/I2,AgNO2, HNO2, NaOH) RCH 2OH
R
R 3C OH CHOH
P/I 2
P/I 2
R P/I 2
RCH 2I
R 3C I
R
AgNO 2 RCH 2NO 2
CHI R
AgNO 2
AgNO 2
R
R 3C NO 2
HONO CHNO
NO 2 R
C
NOH Nitrollic acid NaOH NO 2 R
C
R
2
HNO 2
R C
NO 2
R N O (Pseudo nitrole) NaOH
NO Na Sodium nitrolate (red )
Blue
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HNO 2 No reaction (colourless)
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PERIODATE OXIDATION Compounds that have hydroxyl group on adjacent atoms undergo oxidation cleavage when they are treated with aq. Periodic acid (HIO 4). The reaction
breaks
carbon
carbon
bonds
and
produced
carbonyl
compounds (aldehyde, ketones or acids)
H H H
C
OH
H3C
C
OH
HIO 4
HC
O CH 3CHO HIO 3 H2O
H
It takes place through a cyclic intermediate.
CH3
CH3
H
C
OH
H3C
C
OH
IO 4
H
C
O
H3C
C
O
I
O
CH3
CH3
O H3C
C
O
O CH3
CH3
Other examples
O R
C
C
R'
HIO 4
RCOOH
R'COOH
O
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H
C
O
IO 3
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HO
OH 2HIO 4
R'
OH
RCHO
HCOOH
R'CHO
R
This oxidation is useful in determination of structure. Illustration 5: Write the products of the reaction of t-butyl alcohol with PBr3, conc. H2SO4, CH3COCl, Na, CH3MgBr, Na2Cr2O7/H2SO4. Solution:
CH3
3
CBr, CH3
3
C
CH2 ; CH3
3
COCOCH3 , CH3
3
CO Na ,CH4 , no reaction .
Illustration 6: Write products (a)
(b)
H H
H
C
OH
C
O
C
OH
H H
IO4
C
CH3 IO4
CH2 H
H
C
OH
H
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Solution: (a)
2HCHO + CO2
(b)
No reaction (as it is not a vicinal diol)
Exercise 6: (a)
Arrange
the
following alcohols in order
of
ease
dehydration.
C6 H5 (b)
3
CCH2OH, C6 H5 CHOHCH3 , C6 H5
COHCH3
Find product
CH (a)2OCH3 CHOH
2
HO (b)
IO4
OH
HO
OH
IO4
CH2R HO
R (c)
CHOH
R'
CH
OH
NH2
PINACOL – PINACOLONE REARRANGEMENT Action of H2SO4 on 1, 2 diols. Ditertiory – 1, 2 diols convert in to ketones on treatment with H2SO4. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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OH OH R
C
C
R
R
R
H
R
R
O
C
C
R
R
Mechanism
H3C
OH O
H
C
CH3
C
OH H
H3C
C
C
Methyl shift
CH3
CH3 CH3 H3C
δC
CH3
C
C
CH3
CH3
OH
CH3 CH3
H3C
OH
CH3
δC
CH3
δ CH 3 Transition state O H3C
C
CH3 C CH3
CH3
-H
H3C
O
H
C
C
CH3 CH3
Migratory preference of the group Migration depends on the stability of Transition state. In general migration of C6H5 > alkyl Illustration 7: List three methods with chemical equations for the preparation of alcohols from alkenes. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Solution: Hydration, hydroboration and oxymercuration – demercuration of alkenes. Hydration:
CH3 H3C
C
CH3 CH
CH2
H2 O H
H3C
CH3
C
CHCH 3
CH3 OH
Hydroboration
CH3 H3C
C
CH3 CH
CH2
i BH3 ii H2 O2 , OH
H3C
CH3
C
CH 2CH 2OH
CH3
Oxymercuration – demercuration
CH3
CH3 H3C
C CH3
CH
CH2
i Hg OAC 2 , THF, H2O ii NaBH4
H3C
C
CHCH
3
CH3 OH
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Illustration 8: Which of the following alcohols would react fastest with Lucas reagent?
CH3 CH3CH2CH2CH2OH , CH3CH2 CHCH3 , CH3 CH CH2OH , H3C OH
OH
C
OH
CH3
Solution:
CH3
3
COH , it being a tertiary alcohol.
Exercise 7: Find out (i)
Ph
(ii)
Ph
H3C
CH3
OH
H
A
OH
CH3 H3C
Ph
OH
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Ph
OH
B
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Exercise 8: Give a possible structure for the substance C 5H10O2 behaving in the following manner.
O
C6 H5 NHNH2
C5H10O 2
I2 ,NaOH
Fehling Solution
ETHER *
R
*
R = R Symmetrical ether. R
O
R Alkoxy alkane (Di alkyl ether)
R Unsymmetrical or mixed ether.
‗O‘ is to be counted with least number of C atom. Example: CH3
O
C2H5 Methoxy ethane
CH3
O
C6H5 Methoxy benzene
There are various types of cyclic ethers also.
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O
O
O Oxirane
Oxetane
(Epoxide)
(Oxacyclo butane)
Tetra hydro furan (Oxacyclo pentane)
PREPARATION OF ETHERS (i)
From 1 alcohol (a)
With H2SO4
R O H R O H
H 2SO4 1400 C or Al2O3 / 525K
R O R symmetrical ether
Order of dehydration 1 > 2 > 3 alcohol (b)
With diazomethane R O H
(c)
CH2N2 C2H5O 3 Al
R O CH3 N2
Alcohol having at least one hydrogen at fourth carbon gives five membered cyclic ether with Pb(OAC)4. The reaction is free radical reaction which is initiated by heat or light. H3C
CH 2 CH 2 CH 2 OH
Pb OAC 4 hν / C6H6 800 C
O
Tetrahydro furan
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H2C
(CH 2)3
Pb OAC 4 hν / C6H6
CH 2OH
800 C
O
CH3 2 methyl tetra hydro furan
H
Williamson’s synthesis SN 2 reaction of a sodium alkoxide with alkyl halide, alkyl sulphonate or alkyl
sulphate is known as Williamson synthesis of ethers.
R ONa R'L
L
X, SO2R'' ,
SN2
R O R' NaL
O SO2
OR'
In this reaction alkoxide may be alkoxide of primary, secondary as well as tertiary alcohol.
Alkyl halide must be primary.
In case of tertiary alkyl halide, elimination occurs giving alkenes
With a secondary alkyl halide, both elimination and substitution products are obtained.
R
X
Na . O
R
R
O
R'
Na X
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CH3 CH 3Br
Na O
C
CH3 CH3
H3C
O
CH3
C
CH3
CH3
Sodium ter. butoxide
ter. butyl methyl ether
Illustration 9: Write the product (i)
CH3 H3C
C
O Na
CH 3Cl
Br
C2H5ONa
(A)
CH3
(ii)
CH3 H3C
C
B
C
CH3
Solution: (i)
(ii)
CH3 H3C
C
O
CH3
CH3 H3C
CH3
C
C2H5Br
CH2
Exercise 9: Find product
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(i)
(ii)
OH NaOH / H2 O
CH 2OH
25 0 C
NaOH p Xylene
X
Cl
Br
(3)
Y
From Alkane (a)
OR' R
CH
i Hg OAC
CH2
2
/ R'OH
R
ii NaBH4 / OH
H3C CH2
i Hg OAC
2
/ CH3 OH
H3C
ii NaBH4 / OH
C
H3C
CH3
CH3
CH3
CH3 H3C
CH3
OCH 3 C
(b)
CH
CH2 H
C
OCH 3
H2SO4
H3C
C
O
CH3
CH3
(4)
From Grignard reagent
Higher ethers can be prepared by treating
- halo ethers with suitable
reagents. Cl H3C
O
CH 2Cl
CH 3MgI
Dry ether
H3C
O
CH 2CH 3
Mg I
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(5)
From Alkyl halide
2RI Ag2O
R O R 2AgI
dry
PROPERTIES OF ETHERS Dipole nature of ether Ethers have a tetrahedral geometry i.e. oxygen is sp 3 hybridized. The C–O–C bond angle in ether is 110 . Because of the greater electronegativity of oxygen than carbon, the C
O bonds are slightly polar and are
inclined to each other at an angle of 110 C, resulting in a net dipole moment.
R
R O
1100C
R
O
net μ
R
The bond angle is slightly greater than the tetrahedral angle due to repulsive interaction between the two bulky groups.
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Chemical Reaction Dialkyl ethers reacts with very few reagents other than acids. The only active site for other reagents are the C
H bonds of the alkyls. Ethers
has ability to solvate cations (electrophile) by donating an electron pair from their oxygen atom. These properties make ether as solvents for many reactions. On standing in contact with air, most aliphatic ethers are converted slowly into unstable peroxides. Ether gives following reactions: 1.
Nucleophilic substitution reactions
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Conc. H2SO 4 1 mole warm Conc. H2SO 4 2 mole warm H O H / Δ / h ig h p re s s u re
HI ( 1 mole) Cold
R
R
O
3 mole HI/Red P
Δ
O
SO 3H
SO 3H R
O
R
2HI Cold
δ R
OH R
2R
H
δ O
R
O
SO 3H
H
I R
OH
R
I
R
I
R
H R
H
O
O
δ R
C
δ
Cl
/Anhy ZnCl 2
C
Cl R
O
O R'
R
O
C
O
C
R
O R'
PCl5 / Δ
R
O
R
Cl R
O
C
R'
R
Cl POCl
O
C
R'
3
Note: Type of ethers also make a difference in the mechanism followed during the cleavage of C—O by HI/HBr. Combinations
Mechanism follows
1°R + 2°R
Less sterically hindered
SN2
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SN1
More sterically hindered
2°R + 3°R
Nature
of
mechanism
decoded
nature of solvent. 1°R + 3°R Aprotic or Non polar
Protic polar
SN2
SN1
Methylcation is stabler than phenylcations (B)
Dehydration with H2SO4/ and Anhy Al2O3/ (i)
When both alkyl groups has -hydrogen. H3C
β CH 2 CH3
α CH
CH 2 O
CH3 Conc. H2 SO 4 / Δ
H2C
(ii)
CH 2
H3C
CH
CH
CH3 H2O
When only alkyl group has
- hydrogen.
CH3 H3C β
C α
CH3 O
CH3
Conc. H2 SO4 Δ
H2C
C
CH 3OH
α
CH3
CH3
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Hot conc. H2SO4 react with secondary and tertiary ethers to give a mixture of alcohols and alkenes. (CH3)3CO—CH3
Conc.H2SO4 hot
(CH3)2—C=CH2 + CH3OH
conc.H2SO 4 hot
(CH3)3CO—
(CH3)2—C=CH2 + HO isobutene
Cyclohexyl tert butyl ether
(C)
cyclohexanol
Miscellaneous reactions (1)
Halogenation: Monohalogenation takes place at
carbon (with small amount)
(a)
Cl H3C
(b) H C CH 3 2
(2)
Cl2 / hν
CH 2 O CH 2 CH3 Excess
O CH 2 CH3 Small
H3C
Cl2 excess / hν
CH
O
Cl 5C2
O
CH 2 CH3
C2Cl 5
Reaction with CO: give ester
R O R CO
125 1800 C 200 atm, BF3
RCOOR
Illustration 10: Explain why sometimes explosion occurs while distilling ethers. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Solution: It is due to formation of peroxide OOH CH3CH2—O—CH2CH3 + O2
h
CH3—CH—O—CH2CH3
Illustration 11: The basicity of the ethers towards BF 3 has the following order, explain.
> (CH3)2O > (C2H5)2O >[(CH3)2CH] 2O O
Solution: There are steric effects in the Lewis acid-Lewis base complex formation between BF3 and the respective ethers. Illustration 12: What are crown ethers? How can the following reaction be made to proceed?
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CH2F
CH2Br KF
KBr
Solution: Crown ethers are large ring polyethers and are basically cyclic oligomers of oxirane which may have annulated rings. They are designated according to ring size and the number of complexing oxygen atoms, thus 18-crown – 6 denotes an 18-membered ring with 6-oxygens. The molecule is shaped like a ―doughnut‖, and has a hole in the middle. O
O
O
O O
O
These are phase transfer catalysts. This is a unique example of ―host-guest relationship‖. The crown ether is the host, the cation is the guest. The cavity is well suited to fit a K + or Rb+ which is held as a complex. Interaction between host and guest in all these complexes are mainly through electrostatic forces and hydrogen bonds. The reaction can be made to process by using catalytic amount of crown ether, 18-crown-6.
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Illustration 13: Explain why
O
is much more soluble than furan in water.
Solution: THF is more soluble than furan. In THF, in contrast to furan the electron pairs are available for H-bonding with water which makes it more soluble in water. Exercise 10: What chemical methods can be used to distinguish between the following pairs of compounds? (a)
Ethoxy ethanol and methyl isopropyl ether.
(b)
Butyl iodide and butyl ethyl ether.
(c)
Ethyl propyl ether and ethyl allyl ether.
Exercise 11: Ether A cleaves much faster than B with conc. HI. Explain.
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HBr
H3CO
CH3OH Br
C 2 H5
C2 H5
(A)
HBr
H3CO
CH 3 Br
HO
(B)
PHENOL These are organic compounds a hydroxyl group attached directly to a benzene ring.
OH
OH
CH 3 Phenol or carbolic acid
(o , p , m) Cresol
Preparation Industrial Method (i)
From chloro benzene (Dow’s process) Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Chlorobenzene is heated with NaOH at 673 K and under pressure of 300 atm to produced sodium phenoxide which on acidification yields phenol. Cl
ONa
NaOH / 623K 300 atm p
(ii)
OH
H
Cumene Process
Cumene obtained from propene & benzene cumene on air oxidation followed by acidification with H2SO4 gives phenol & acetone. CH3 CH3 H3C
CH
CH2
HC
250 0 C H3 P O 4
CH3
C
O2 95 - 1350
O
CH3 Cumene hydroperoxide
Cumene H, H 2O 50-900C OH O H3C
(iii) From benzene sulphonic acid It is fused with NaOH gives sodium salt of phenol.
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C
CH3
OH
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NaOH Fusion
C6H5 SO3H
C6H5 SO3Na
C6H5 ONa
H2 O / H
C6H5OH
(iv) From benzene diazonium chloride This gives Ar SN1 reaction with H2O to form phenol. OH N
N Cl H2O/H Δ
N2
Illustration 14: Starting from 1-methyl cyclohexene, prepare the following: (a)
(b)
CH3
CH3
OH
OH
OH H
Solution: (a)
CH3
CH3 OH
i KMnO4 ii H2 O
OH H
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(b)
CH3
CH3 H2 O / H
OH
Exercise 12: Starting from 1-methyl cyclohexene, prepare the following: (a)
CH3
(b)
CH3
OH
OH
OH
OH
H
OH
PHYSICAL PROPERTIES Phenol is needle shaped solid, soon liquefies due to high hygroscopic nature. It is less soluble in water, but readily soluble in organic solvents. Phenol has high boiling point due to presence of hydrogen bonding. Acidity of phenol Phenol is weak acid. It reacts with aqueous NaOH to form sodium phenoxide, but does not react with sodium bicarbonate. The acidity of phenol is due to the stability of the phenoxide ion, which is resonance stabilized as shown below: Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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O
O
(I)
(II)
O
(III)
O
O
(IV)
(V)
In substituted phenols, the presence of electron withdrawing groups at ortho and para positions such as nitro group, stabilizes the phenoxide ion resulting in an increase in acid strength. It is due to this reason that ortho and para nitro phenols are more acidic than phenol. On the other hand, electron releasing groups such as alkyl group, do not favour the formation of phenoxide ion – resulting in decrease in acid strength. For example: (cresol are less acidic then phenol) Exercise 13: Arrange each group of compounds in order of decreasing acidity:
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OH
OH
(a)
,
OH
,
NO 2
OH
,
CH3
OH
OH
OH
OH NO2
(b)
,
,
,
NO2 NO2
NO2
Cl
CHEMICAL REACTIONS (A)
Reaction due to breaking of O – H bond
Phenol is more reactive than alcohol for this reaction because phenoxide ion is more stable than the alkoxide ion.
R O
O H
H
R
O
H
O
H
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Reactions of phenol due to breaking of
O
H bond are given
below: O
Alcoholic FeCl3
Fe blue or violet colour (test for phenolic group). 3
ONa
O
H
NaOH
H2O
O
CH2 N 2
CH3
Anisole
dry ether
O
CH3
(CH3 ) 2 SO 4 NaOH
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Acylation
O OH
O
C
R
O R
C
Cl
pyridine Phenyl ester
or O R
C
O O
C
R
/ Pyridine
Fries rearrangement Phenolic esters are converted in to o– and p–hydroxy ketones in the presence of anhydrous AlCl 3. Generally low temperature favours the formation of p–isomer and higher temperature favour the o-isomer.
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OH
O O
C
COCH
1600C
3
CH3
OH
600C
COCH 3
(B)
Reactions due to breaking of carbon- Oxygen bond Nucleophilic substitution reaction Phenols are less reactive than aliphatic compound because: (i)
–OH group is present on sp2 hybridised carbon. This makes C–O bond stronger.
(ii)
‗O‘ is more electronegative than halogens. This also makes C–O bond stronger than C–X.
(iii)
There is some double bond character between carbon and oxygen due to the resonance. This also makes C–O bond stronger.
However it give SN under drastic condition.
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NH2
NH3/Anhy ZnCl2 3000C
OH Cl
O
P
PCl5 / Δ
3 SH
P S5 / Δ 2
(C)
EAS in Phenol: It is strong activating group.
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OH
OH
NO 2
20 % HNO3 250C
o-nitro phenol NO 2
OH Conc. HNO3
p-nitro phenol
O 2N
NO 2
H2 SO 4 / Δ
2, 4, 6, tri nitro phenol (picric acid) OH NO 2
Br2 / H2 O
OH
Br
Br 2, 4, 6, tri bromo phenol
or Br2 / CH3 COOH
OH
OH Br
Br2 / CS 2
Br p-bromo phenol
or Br2 / CCl4
o-bromo phenol
OH
OH Br
H2 SO 4
SO 3H
Δ
o-hydroxy benzene sulphonic acid OH
R
SO 3H (Major) p-hydroxy benzene sulphonic acid R
X Anhy AlCl3
Δ
o-alkyl phenol
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OH
O
HNO2
N
NCl
NO
N
OH
p nitroso phenol
N Coupling reaction OH
N
OH
mild
p hydroxy azo benzene Azodye OH
OH
CHO i CHCl3 / alc KOH ii H2 O / H ReimerTiemann reaction
OH
OH
Major product
CHO ii CCl4 / alc KOH Δ
COOH Major
OH ii H2 O / H
Salicylic acid
ReimerTiemann reaction
OH COOH COOH
i CO2 / NaOH 1200 C, 7 atm ii H Kolbe ' s reaction
OH
OH CH 2OH
CH2
O
OH
CH 2OH
ROH / H 2 SO 4
R
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MERCURATION Mercuric acetate cation. [HgOAC]+ is a weak electrophile which substitutes in ortho and para position of phenol. Usually donating product is O– acetoxy mercuriphenol. The mercuric compound can be converted to iodophenol.
OH
OH
Hg OCOCH3
HgOCOCH 2
Re flux
OH
OH
3
I
HgCl
Aq. NaCl
I2/CHCl
Miscellaneous reaction (i)
Reaction with Zn dust OH
Zn Dust
(ii)
Δ
ZnO
Oxidation
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O
p - benzoquinone (brow n in colour)
hν
OH
O OH
K2S2O 8/OH
p - quinol
Elbs oxidation
OH
(iii)
Condensation with phthalic anhydride OH
OH
OH
H
OH
H O C C O C
Conc. H2SO4 Δ H2O
O C
O O Penolphthalein (Acid base indicator)
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Exercise 14: Discuss the product formed in the bromination of p-phenol sulfonic acid. Mechanism of some important reactions 1.
Reimer Tieman reaction
OH
OH
CHO
i CHCl3 , OH ii H
The electrophile is the dichloro carbene: CCl2, formation of carbene is an example of -elimination. (i)
Cl OH
H
C
Cl
-HCl
CCl 2
Cl (ii)
OH
O
O
O H
OH
CCl2 Ar SE
CHCl
2
CCl 2 o (dichloro methyl) Phenoxide ion
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(iii)
O
O O
H
C
Cl
Cl HC Cl
OH
H
OH
O O CHO
C
H
2.
H
Kolbe’s reaction OH
OH
NaOH CO 2
i 1200 C, 7 atm ii H
Mechanism
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COOH
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O Na
OH
O C
H
O
C
O
NaOH
O Na
O (More reactive for Ar SE)
OH OH O COOH C
H2O/H
O Na
Salicylic acid
Illustration 15: How will you convert? (i)
phenol to aspirin
(ii)
phenol to salol.
(iii)
phenol to oil of winter green.
(iv)
phenol to benzoic acid.
Solution:
(i)
OH
OH
O
COOH
i NaOH/ CO2 , 1200 C
COOH
CH3 CO 2 O
4 7 atm ii H
H
Aspirin
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COCH 3
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(ii)
OH
OH
OH
OH
COOH
i NaOH / CO2 , 1200 C 7 atm P ii H
CO
O
H
Salol
(iii)
OH
OH
OH
COOH
i NaOH / CO2 , 1200 C
COOCH
CH3 OH
7 atmP ii H
3
H
Oil of w inter green
(iv)
OH
OH
COOH
i NaOH / CO2 , 1200 C
Zn dust
7 atm P ii H
Illustration 16: What product would you expect in the following reaction? Explain. OH
CHCl3 , KOH
?
CH3
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COOH
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Solution:
CHCl3
aq. KOH
: CCl2
O
O
CCl 2
CH3
O
H2 O
H3C
CCl 2 H C 3
CHCl
2
It is an ‗abnormal‘ product formed in the Reimer-Tiemann reaction when the dienone cannot tautomerize to regenerate a phenolic system. Illustration 17: How would you distinguish between the following pairs? (a)
Phenol and cyclohexanol
(b)
Ethyl alcohol and methyl alcohol
Solution: (a)
Phenol gives coloration with FeCl 3 solution (b)
Ethyl alcohol responds to the iodoform test
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Exercise 15: How would you distinguish between the following pairs? (a)
2-Pentanol and 3-pentanol
(b)
1-Propanol and phenol
Exercise 16. Offer explanation for the following observations: (a)
Why is phenol unstable in the keto-form?
(b)
The following dehydration is extremely facile: CH3
CH3 H
O
OH
O
(c)
Why does thionyl chloride provide alkyl chlorides of high purity?
(d)
2-Methyl -2- pentanol dehydrates faster than 2 – methyl – 1 – pentanol.
(e)
Phenol is acidic but ethyl alcohol is neutral.
(f)
Ethanol responds to Iodoform test but tert- butanol does not.
(g)
A tertiary alcohol reacts faster than a primary alcohol in the Lucas test.
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How will you effect the following conversion? (a)
(b)
OH
O
C2H5OH
CH3CH2CH2OH
Exercise 18: How will you effect the following conversion? (a)
OH
(b)
(c)
CH
CH2CH2OH
OH
CH3
CH3
(d)
CH2
OH
OH
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Exercise 19: Write mechanism CH2OH
Cyclohexene
H
Some Commercially Important Alcohols And Phenols (i)
Methanol: Methanol is also called wood spirit since originally it was obtained by destructive distillation of wood. Now a days it is prepared by catalytic hydrogenation of water gas. CO
2H2
CuO ZnO CrO3 573 673K, 200 300K
CH3 OH
Uses: It is largely used as:
(c)
(a)
a solvent for paints, varnishes and celluloids.
(b)
for manufacturing of formaldehyde.
for denaturing ethyl alcohol, i.e. to make it unfit for drinking purpose. Denatured alcohols is called methylated spirit. (d)
(ii)
in manufacture of perfumes and drugs.
Ethanol: Ethanol is mainly prepared by hydration of ethene formation of carbohydrates gives only 95% alcohol the rest being water. This is called rectified spirit. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Uses: It is largely used as an (a) (b)
antiseptic.
solvent
for
paints,
lacquers,
varnishes,
dyes,
cosmetics,
perfumes, tinctures, cough syrups etc. (c)
As an important starting material for manufacture of ether, chloroform, Iodoform etc. (d)
(e)
As an important beverages.
As power alcohol a mixture of 20% absolute alcohol and 80% petrol (gasoline) with benzene or tetralin as a co-solvent. (f)
As an antifreeze in automobile radiators.
(iii) Absolute alcohol: Absolute alcohol is 100% ethanol prepared from rectified spirit 95.5% alcohol as follows: In laboratory absolute alcohol is prepared by keeping the rectified spirit in contact with calculated amount of quick lime for few hours and then refluxing and distilling it. Phenol or Carbolic Acid Uses (i) (ii)
As an antiseptic and disinfectant in soaps and lotions.
In manufacture of drugs like, aspirin, salol, salicylic acid, phenacetin. (iii)
In the manufacture of bakelite.
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(iv)
In the manufacture of picric acid, phenolphthalein, azo
(v)
As a preservative for ink.
dyes.
Ethylene Glycol: Ethane 1, 2 diol Preparation: Lab preparation by hydroxylation. (i)
CH2 3 CH2
2KMnO
4 Cold dilute alkaline
CH 2OH
4H2O
3
2MnO2 2KOH
CH 2OH
Manufacture
CH2
H2C O2 / Ag 575K
CH2
CH 2OH O
H2 O / 473K hydrolysis
H2C
CH 2OH
Ethylene epoxide or oxirane Physical properties: It is highly viscous because of the presence of two OH bond it undergoes extensive intermolecular H-bonding. Same reason owes to high solubility in water and high boiling point. Chemical properties Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Reaction with sodium (i)
CH 2OH CH 2OH
(ii)
Na, 323K 1 H2 2
CH 2ONa
CH 2OH Monosodium glycolate
2POCl3
2PCl 5
2HCl
CH 2Cl
CH 2Cl
CH 2OH
2SO2 2HCl CH 2Cl Ethylene dichloride
2SOCl 2
CH 2OH
(iv)
CH 2ONa Disodium glycolate
CH 2Cl
CH 2OH CH 2OH
(iii)
CH 2ONa
Na, 433K 1 H2 2
CH 2OH
CH 2Cl HCl, 433K H2 O
CH 2OH
CH 2OH
CH 2Cl HCl, 473K H2 O
CH 2Cl
Ethylene chlorohydrin Ethylene dichloride
(v)
CH 2OH
H2C RCH
O
p toulene sulphonic acid
CH 2OH
CHR
H2 O
H2C Cyclic acid
(vi) Oxidation:
Ethyelene
glycol
upon
oxidation
gives
different
products with different oxidising agents. For example. (a) CH 2OH CH 2OH
CHO
COOH
O
O
CH 2OH Glycolaldehyde
CH 2OH Glycollic acid
CHO
COOH O
CHO Glyoxal
COOH O
CHO Glyoxalic acid
COOH Oxalic acid
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CH 2OH
HCHO HIO4
HCHO
CH 2OH
H2O HIO3 Iodic acid
Also called malapride reaction. CH 2OH
HCHO 2 CH3COO 4 Pb
2CH3COOH
CH3COO 2 Pb
HCHO
CH 2OH
Dehydration (a) CH 2OH
H2 C
O H2O
773K
H2 C
CH 2OH
(b) CH 2OH
CH 2 ZnCl 2 H2 O
Tautomerize
CHOH
CH 2OH
CH 3 CHO
Vinyl alcohol
(c)
CH 2OH 2
conc. H2 SO4 distill
CH 2 CH 2 O
CH 2OH
O 2H2 O
CH 2 CH 2 1, 4 dioxane
(d)
CH 2OH 2
conc. H3 PO4 distill
CH 2 CH 2OH O
CH 2OH
H2 O
CH 2 CH 2OH Diethylene glycol
Uses (a)
As a solvent.
(b)
Antifreeze in the radiators of cars and aeroplanes.
(c)
In manufacture of terylene and other polyester.
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How would you convert cyclohexane to 1, 6 – hexanediol? Glycerol (Propane 1, 2, 3 triol) One of the most important trihydric alcohol. Preparation: (i)
By Saponification of oils and fats. H2 C
O
COR 1
HC
OCOR 2
H2 C
OCOR 3
3NaOH
CH 2OH
R1COONa
CHOH
R1COONa
CH 2OH
R1COONa Sodium salt of fatty acid (soap)
(ii)
From Propylene CH3
CH
CH2
Cl2 , 773K HCl
Cl CH2
HOCl Cl OH trans Cl2 H2O
CH
CH2
Allyl chloride
HOCH2
aq.Na2 CO3 423K, 12 atm
CHOH CH2Cl
HO CH2
CH
aq. NaOH NaCl
glycerol
CH2OH CHOH CH2OH glycerol
monochlorohydrin
(iii) Synthesis from its elements
2C H2
Electric ore Berthelets synthesis
H3C C Pr opyne
CH CH
CH
Na in liq. NH3 196 K
H2 / Pd BaSO4 Lindlar ' s catalyst
CH3
Na C CH
CH CH2
Pr opylene
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CH2
Allyl alcohol
CH3I NaI
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Physical Properties Highly viscous due to three –OH group due to which it undergoes extensive intermolecular H-bonding. Chemical Properties (i)
It undergoes reaction of both secondary and primary alcoholic group. CH 2OH CHOH
CH 2ONa Na, room temperature 1 H2 2
CH 2OH
CH 2ONa Na, room temperature
CHOH CH 2OH
monosodium glycerolate
(ii)
CHOH CH 2ONa , 'disodium glycerolate
CH 2OH Excess of HCl HCl gas CHOH 383K, H2O gas, 383 K, 2H2O CH 2OH
CH 2Cl
CH 2OH
CH 2Cl
CH 2Cl
CHOH
CHCl
CHCl
CHOH
CH 2OH Glycerol monochlorohydrin
CH 2OH Glycerol monochlorohydrin
CH 2OH Glycerol , dichlorohydrin
CH 2Cl Glycerol , dichlorohydrin
Excess of dry HCl gas, 383 K H2 O
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To replace the third hydroxyl group in either of two dichlorohydrins, PCl 5 or PCl3 is fused. (iii)
CH 2OH CHOH
CH 2I 3HI
3H2 O
CH 2OH
CHI CH 2I
CH 2 I2
CH
CH 3 HI Markonikof 's addition
CH 2I
CHI
I2
CH 2I
Unstable
3.
CH 3 CH CH 2
1, 2 diiodopropane
CH 3 HI
CHI CH 3
Isopropyl iodide
Reaction with concentrated nitric acid:
H2C
OH
HO
NO 2
HC
OH
HO
NO 2
H2C Glycerol
OH
HO
NO 2
Conc HNO3 Conc. H2 SO4 283 298K
H2C
O
NO 2
HC
O
NO 2
3H2O
H2C O NO 2 Glyceryl trinitrate Noble's oil Nitroglycerine
A mixture of glyceryl trinitrate and glyceryl dinitrate absorbed on Kieselguhr is called dynamite.
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4.
Reaction with KHSO4 – Dehydration.
H H
C
OH
HO
C
H
H
C
OH
CH2 KHSO4 , 473 503K 2H2 O
CH2 Tautomerises
C
CHOH (unstable)
H
CH CHO Acrolein or Prop- 2-en-1-al
Glycerol
5.
Oxidation.
CHO
COOH [O]
CHOH CH 2OH
Glyceraldehyde
CH 2OH
CHOH
COOH [O]
CH 2OH Glyceric acid
CHOH COOH Tartromic acid
[O] CHOH CH 2OH Glycerol
CH 2OH C
O
COOH [O]
CH 2OH Dihydroxyacetone
CO COOH Mesoxalic acid
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(i)
With dil. HNO3, a mixture of glyceric acid and tartronic acid is produced.
(ii)
With conc. HNO3, mainly glyceric acid is obtained.
(iii)
With bismuth nitrate, only mesoxalic acid is formed.
(iv)
Mild oxidising agents like bromine water, sodium hypobromite (Br2/NaOH) and Fenton‘s reagent (H 2O2 + FeSO4) give a mixture of glyceraldehyde and dihydroxyacetone. The mixture is called glycerose.
(v)
With periodic (HIO4) acid.
HCHO Formaldehyde
CH 2OH 2HIO4
CHOH CH 2OH
Glycerol (vi)
HCOOH Formic acid
2HIO3 H2O Iodic acid
HCHO Formaldehyde
With acidified potassium permanganate.
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CH 2OH COOH O
CHOH
Acidified KMnO4
CO2
3H2O
COOH Oxalic acid
CH 2OH Glycerol 6.
Reaction with phosphorous halides.
CH 2OH CHOH
CH 2Cl 3PCl 5
CH 2OH Glycerol
7.
Reaction
CHCl
3HCl 3POCl 3
CH 2Cl 1, 2, 3 - trichloropropane (Glyceryl trichloride) with
monocarboxylic
acids.
Glycerol
reacts
with
monocarboxylic acids to form mono-, di- and tri- ester depending upon the amount of the acid used and the temperature of the reaction. An excess of the acid and high temperature favour the formation of tri-esters. For example, with acetic acid, glycerol monoacetate, diacetate and triacetate may be formed.
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CH 2O.COCH
3
CHOH
CH 2O.COCH
3
CHOH
CH 2OH Glycerol monoacetate
8.
CH 2O.COCH
CHO.COCH
CH 2O.COCH 3 Glycerol diacetate
3
3
CH 2O.COCH 3 Glycerol triacetate
Acetylation. When treated with acetyl chloride, glycerol forms glycerol triacetate.
CH 2OH CHOH
CH 2OCOCH 3CH3COCl
CH 2OH Glycerol
CHOCOCH
3
3
3HCl
CH 2OCOCH 3 Glycerol triacetate
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9.
Reaction with oxalic acid (i)
H2C
OH
H
OOC COOH Oxalic acid
CHOH
H 2C
383K H2 O
CH 2OH
OOC
COO H
H2C Heat CO2
CHOH
CH 2OH Glyceryl monoxalate
Glycerol
OOCH
CHOH CH 2OH Glyceryl monoformate
CH 2OH HOH, hydrolysis From water of crystallization
CHOH
HCOOH Formic acid
CH 2OH Glycerol
(ii) CH 2O H CHO H
HO CO HO CO Oxalic acid
CH 2OH
503K 2H2 O
H2C
OOC
HC
OOC
CH 2 503K 2H2 O
CH
CH 2OH Glyceryl dioxalate (Dioxalin)
Glycerol
CH 2OH Allyl alcohol
Uses: Glycerol is used: 1.
In the preparation of nitroglycerine used in making dynamite. Nitroglycerine is also used for treatment of angina pectoris. 2.
As an antifreeze in automobile radiators.
3.
In medicines like cough syrups lotions etc.
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4.
In the production of glyptal or alkyl resin (a cross – linked polyester
obtained
by
the
condensation
polymerization
of
glycerol and phthalic acid) which is used in the manufacture of paints and lacquers. 5.
In making non-drying printing inks, stamp colours, shoes polishes etc.
6.
In the manufacture of high class toilet soaps and cosmetics since it does not allow them to dry due to its hydroscopic nature. 7.
8.
As a preservative for fruits and other eatables.
As a sweetening agent in beverages and confectionary.
Exercise 21: How does glycerol react with a. HI, b. (COOH) 2 and c. conc HNO3?
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ANSWER TO EXERCISES Exercise 1: (i)
H3C
H3C
OH
OH
HgOAC H
A=
(ii)
B=
A = CH3CH2CH2OH B = CH3OH
Exercise 2: (i)
R OH
followed by
RCO 2Et +
MgBr
MgBr
H3 O
-H2O R H2/Pd
(ii)
The R– (carbanion) pulls acidic hydrogen from OH group making it an acid-base reaction instead of usual nucleophilic addition reaction. On further hydrolysis it produces reactant. But in the
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R
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second case you have two moles of RMgX that completes the reaction because there are two R– for two different sites. Exercise 3:
CH3
CH3 OH CH
CH3
CH
H
CH3
CH3
CH3 H
ring Expansion CH3
CH3
H
Exercise 4: (a)
(iii) > (ii) > (i) > (iv)
(b)
(i) > (ii) > (iii)
Exercise 5: (i)
Only (c) and (d) gives iodoform test.
(ii)
O
O
(A) =
OH
OH
(B) =
OH
OH
Exercise 6: (a) (b)
C6H5
(i)
3
CCH2OH
C6H5 CHOHCH3
C6H5
2
C(OH)CH3
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(ii)
6HCOOH
(iii)
RCHO + R CHO + NH3
Exercise 7: (i)
(ii)
CH3 Ph A=
Here
C
C
O
CH3
A=
Ph
phenyl
CH3 Ph
migration takes
place.
C
C
O
CH3
Ph
Here methyl migration takes place.
Exercise 8: Considering the reactions of C5H10O2, it is likely that C6H5NHNH3, I2, NaOH will react only with a keto group while CH3COCl and oxidation are reactions of an alcohol. Therefore two isomeric structures are possible for the original compound.
O
O
CH 3 CCH 2CH 2CH 2OH and
CH 3 CCH 2CH 2OH CH3
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Exercise 9: ,
w
–
halo
alcohols undergoes
intramolecular
X=
O
Williamson
ether. Synthesis in the presence of NaOH O
Y=
by SN 2 reaction.
Exercise 10: (a)
Ethoxyethanol give the Iodoform test.
(b)
Butyl iodide with AgNO3 yields AgI precipitate.
(c)
Ethyl allyl ether decolorize bromine water.
Exercise 11: The cleavage reaction of an ether by HI is initiated by protonation of the ether oxygen.
I
Attack by either SN1 or SN 2
H O CH 3 (B)
In the second step the attack by Br has to take place by SN1 or SN2 process. Since the system B is rigid either attack is very slow. The ether A does not pose such a problem Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Exercise 12: CH3
CH3
(a)
OH
i OsO4 ii NaHSO3
OH H
(b)
CH3
CH3 C6 H5CO2 OH
O
OH CH3
H2 O H
OH H
Exercise 13:
OH
OH
(a)
>
OH
>
OH
>
CH3
NO 2 OH
OH
OH
OH NO 2
(b)
>
>
> NO 2
NO 2
Cl
NO 2
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Exercise 14: A
phenol
is
highly
reactive
towards
electrophilic
substitution.
Treatment of phenol with aqueous solution of bromine results in the replacement of every hydrogen ortho and para to the
OH group, and
it even cause the displacement of the sulfonic group to yield tribromophenol. OH
OH Br
Br 3HBr H2SO4
3Br2
SO 3H
Br
Exercise 15: (a)
2-Pentanol responds to the Iodoform test
(b)
Phenol gives coloration with FeCl 3 solution
Exercise 16: (a)
Phenol loses the aromatic stabilization in the keto form.
(b)
The alkene formed is more stable due to resonance.
(c)
Because the other products during the reaction of thionyl chloride with alcohols are gaseous.
(d)
2-Methyl-2-pentanol yield a stable alkene
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(e)
The
phenoxide
ion
formed
from
phenol
is
stabilized
resonance. (f)
H Ethanol contains the H
C
OH grouping
CH3
(g)
A tertiary alcohol form a stable tertiary carbonation.
Exercise 17: (a)
OH
OH
O
Na2 Cr2 O7
H2 , Na
(b) C2H5 OH
PCl5
H ,
C2H5 Cl
KCN
H
C2H 5 CN
C2H5 CH2NH2
HNO2
C2H5 CH2OH
Exercise 18: (a)
SO 3H
SO 3 Na NaOH
H2 SO4 1600
OH H
(b) C6H5 CH CH2
HBr Peroxide
C6H5 CH2CH2Br
aqueous OH
C6H5CH2CH2OH
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(c)
CH3
CH3
H2 / Ni
H , H2 O
OH
(d)
CH3
OH OH
OH HOCl
H
aq. KOH
H , H2 O
Cl
Cl
Exercise 19:
CH 2
CH 2OH
H ,Δ H2 O
ring expansion
H
Exercise 20:
O3
NaBH4
HO CH2
6
OH
Exercise 21: See the text.
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MISCELLANEOUS EXERCISES Exercise 1:
Write a note on: (a) Oxo process (b) absolute alcohol
Exercise 2:
What happens when?
Exercise 3:
(a)
ethanol vapours are passed over alumina at 600 K,
(b)
excess of ethanol is heated with conc.H2SO4 at 413K,
(c)
phenol is treated with acetyl chloride?
How is glycerol prepared from fats and oils? Why it is highly viscous? Give its uses?
Exercise 4:
Exercise 5:
Discuss the following reactions: (a)
Reimer and Tiemann reaction
(b)
Kolbe‘s reaction
Complete the following by putting structures of A, B, C, D. CH3CH2CH2OH
PBr3
A
KOH als.
B
HBr
NH3
C
D
Complete the following. Exercise 6:
Compete the following by drawing correct structures. SO 3H
Exercise 7:
fu min g H2 SO4
..................
i fuse with NaOH ii H2 O / H
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Exercise 8:
Write the structure of organic compounds A to F in the following sequence of reactions: OH
Zn
Exercise 9:
A
HNO3 H2 SO4
B
Br2
C
H2 / V Pt
D
HNO2 HCl
E
Predict the products of the following reactions: (a)
Br C2 H5O Na
(b)
CH3 CH2
5
O C6 H5
HBr
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ANSWER TO MISCELLANEOUS EXERCISES Exercise 1: (a)
Oxo process: It is used for the preparation of alcohols. When alkenes are treated with carbon monoxide and hydrogen in presence of octa-carbonyl dicobalt at high temperature and pressure, aldehyde are formed, which on reduction with H2/Ni give alcohols.
CH2
CH2
CO H2
Ethene
(b)
CO Co
4 2
High temp. and pressure
CH3
CH
2 Pr opanal
CHO
H2 / Ni
CH3
CH2
CH2
OH
1 propanol
Absolute alcohol: 100% ethanol is known as absolute alcohol. 95.6% alcohol and 4.4% water form azeotropic mixture and boil at same temperature. To prepare absolute alcohol, rectified spirit (95% alcohol) and benzene mixture is fractionally distilled. The first fraction obtained at 331.8 K contains the azeotropic mixture of total water (7.5%) some alcohol (18.5%) and maximum benzene (74%). The second fraction, which is an azeoptroopic mixture of remaining benzene (nearly 68%) and some alcohol (32%) is obtained at 341.2 K. The third and last fraction consists of absolute alcohol which is obtained at 351K.
Exercise 2: (a)
Dehydration with take place and ethylene is formed. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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CH3CH2OH Ethanol
(b)
Al2 O3 600K
CH2
CH2 H2O
Ethene
Diethylether is formed.
2C2H5 OH
Conc. H2SO4 413K
C2H5OC2H5 H2O
OH
Ethoxy Ethane
O
COCH 3 HCl
CH3COCl Acetyl chloride
Phenol
(c)
Phenyl acetate
Phenylethanoate is formed.
Exercise 3: Glycerol is prepared by the Saponification of oils and fats i.e., hydrolysis of oils and fats with NaOH. Fats and oils are the esters of glycerol and higher fatty acids i.e. glycerides of fatty acids. Glycerol is very viscous with extensive intermolecular hydrogen bonding. Glycerol molecule contains three
OH groups which form extensive
hydrogen bonding. Hence the intermolecular forces are very high and molecules are highly associated. Due to this reason it is very viscous and posses high boiling point (563K). Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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H2C
O
COR
HC
O
COR'
H2C
O
COR"
3NaOH
(Oil and fats)
H2C
OH
RCOONa
HC
OH
RCOONa
H2C
OH
RCOONa
Glycerol
Sodium salt of fatty acids (soaps)
Uses: (i)
Used
in
the
preparation
of
nitroglycerine
which
is
main
constituent of dynamite. (ii)
Used as antifreezing agent in automobile radiators.
(iii)
Used as a sweetening agent in confectionery and beverages.
Exercise 4: (a)
Reimer and Tiemann reaction: When phenol is refluxed with chloroform and sodium hydroxide at 340 K followed by hydrolysis, gives
o-hydroxy
benzaldehyde
(main
product)
and
p-hydroxy
benzaldehyde (minor product). This reaction is known as Reimer and Tiemann reaction.
OH
OH
OH COONa
CCl 4 340K
Phenol
COOH H2 O / H
2-Hydroxybenzaldehyde (Salicyldehyde)
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OH
OH
OH COOH small amount of 4-hydroxy benzoic acid is also formed
COONa CCl 4 / NaOH 340K
H2 O / H
2-Hydroxy benzoic acid (Salicylic acid)
Phenol
If CCl4 is used in place of CHCl 3 the main product will be salicylic acid. (b)
Kolbe’s Reaction: When sodium phenoxide is treated with carbon dioxide under pressure (4 to 7) and at 400 K, sodium salicylate is formed which on acidic hydrolysis gives salicylic acid. This reaction is known as Kolbe‘s reaction. OH
OH
OH COONa
CO2
4 7 atm 400K
Phenol
COOH H2 O / H
Salicylic acid
(Major product)
Exercise 5:
CH3CH2CH2OH
PBr3
1 Pr opanol
CH3CH2CH2 Br
KOH alc.
1 Bromopropane A
CH3CH
NH2 H3C CH CH3 2-Propanamine (D)
NH 3
H3C
CH
CH3
HBr
Br 2-Bromopropane (C)
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CH2
Pr opene B
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Exercise 6: SO 3H
SO 3H
OH
fu min g H2 SO4
i fuse with NaOH ii H2 O / H
SO 3H
OH 1, 3 - Dihydroxy benzene
Exercise 7: Molasses, which is the main source of sugar is treated with yeast. The enzymes present in yeast ferment the sugar and ethanol is obtained. Sugar is converted to glucose and fructose by the invertase enzyme. Glucose is further converted to ethanol by the zymase enzyme.
C12H22O11 H2O
Invertase
Sugar
C6H12O6 C6H12O6 Glucose
Zymase
2C2H5OH 2CO2
Frucot se
Ethanol
Exercise 8: OH
NO 2
HNO3 H2 SO4
Zn
Phenol
Benzene [A]
NO 2
NH2
Sn / HCl
Br2
Nitrobenzene [B]
Br m-Nitro (bromobenzene) [C]
OH
N2Cl
H3 O
Br m-Bromophenol [F]
m-Bromo Aniline [D]
Br
HNO2 HCl
Br m-Bromo Benzenediazonium [E]
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Exercise 9: (a)
(b)
O C2H5
CH3 CH2
4
CH2Br
C6H5 OH
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SOLVED PROBLEMS
Subjective:
Board Type Questions Problem 1: What happens when? (i)
Vapours of ethanol are passed over heated alumina at 523 K.
(ii)
Ethyl bromide is heated with dry silver oxide.
(iii)
Methyl magnesium bromide is treated with methoxy chloro methane.
(iv)
Ethyl alcohol is heated with diazomethane in presence of HBF 4.
(v)
2 methyl propene is heated with methanol in the presence of dil. H2SO4.
Solution: (i)
Diethyl ether is formed. CH3 CH2
OH H O CH2CH3
Al2 O3 3500 C
CH3 CH2
O CH2
(ii) C2H5 Br Ag2O C2H5 Br
H5C2
O
C2H5 2AgBr
diethyl ether
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(iii) Ethyl methyl ether is formed. Br H3C
O
CH 3MgBr
CH 2Cl
dry ether
H3C
C2H5 Mg
O
Cl
(iv) Ethyl methyl ether is formed CH3 CH2OH
(v)
HBF4
CH N
2 2 dim ago methane
CH3 CH2
O CH3
N2
Tertiary – butyl methyl ether CH3 H3C
C
CH3 CH2
H
O
CH3
H2SO 4
H3C
C
CH3
O CH3 Methyl ter - butyl ether
Problem 2: Why phenol has smaller dipole moment than methanol? Solution: In case of phenol, the electron withdrawing inductive effect of oxygen is opposed by electron releasing resonance effect. Hence, phenol has smaller dipole moment. In case of methanol only electron withdrawing inductive effect is operative. Hence, it has higher dipole moment.
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Problem 3: Why di tert – butyl ether can not be obtained by Williamson‘s method? Solution: In order to prepare the above ether the reagents to be used are tert–butyl bromide and tert–butoxide. Since the tertiary bromide prefers to under go elimination, therefore, major product of the reaction shall be iso butylene and not di tert – butyl ether. CH3 H3C
C
CH3
CH3 Br
Na O
CH3
C
H3C
CH3
C
CH3 CH2 H3C
CH3
CH3
30 bromide
OH NaBr
C
tert- butoxide
Problem 4: Identify the product in the following reaction
OH
OH
ClCH2 COCl POCl3
A
CH3 NH2
B
i KMnO4 / OH ii H
C
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D
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Solution:
OH
OH
OH
OH
B=
A=
COCH 2NHCH
COCH 2Cl
3
COOH
OH
OH
D=
C=
COOH
Problem 5: Write the various product when ethanol reacts with sulphuric acid in suitable conditions. Solution:
C2H5OH
H2 SO 4 383K
C2H5HSO 4
H2SO 4/ 413 K
C2H5OC 2H5
H2SO 4/ 443 K H2C
CH2
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IIT Level Questions Problem 6: Give products expected when each of the following diols are reacted with periodic acid HIO4 (a) HO
CH2
. CH
CH2
C6H5
(b)
OH
H C
OH
OH
Solution: (a)
O
H C
O
H
C
CH 2
C6H5
H
(b)
H
C
O
O
Problem 7: Effect the following conversion (a)
CH2OH
CH2OH
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(b)
CH2CH2OH
CH3OH
Solution: (a)
CH 2OH
CH 2 H
ring Expansion
H2O CH2
CH 2OH
Ph 3P
HBO
O CH2
K2Cr 2O 7
OH
(b) PCl5
CH 3OH
CH 3Cl
CH3 Anhy AlCl 3
CH 2Br
NBS Mg dry ether
CH 2CH 2OH
HCHO H3 O
CH 2MgBr
Problem 8: Postulate a mechanism for the following reactions: O CH3
H
HO
Solution:
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OH H -H2O
HO
HO ring expansion
O
O
H
-H
Problem 9: Conversion (a)
H
H
OH
D
(b) H2C
CH2
O
O
Solution: (a)
H H OH
PBr 3 Pyridine
H
Mg/dry ether
Br
MgBr D2O H D
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(b)
H2C
CH2
MnO 4 KOH
HOH 2C
H -H2O
CH 2OH
OH
CH 2 CH 2 O CH 2CH 2OH
Intrmoleculadehydration H , Δ, H O 2
O
O
Problem 10: Identify the products in the following reaction sequence.
OH i CHCl 3 ii KOH , H
A
LiAlH4
B
CH3COOH H
C
H3C
Solution:
H3C
C=
B=
A= CHO
OH
OH
OH
H3C
CH 2OH
H3C
O CH 2
O
C
CH3
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(a) (b)
(CH3)2CHOH and (CH3)2CHSH
CH3CH2CH2OH and (CH3)2CHOH
Solution: (a)
The thiol gives a precipitate with heavy metal cations like
Hg 2 , Pb
(b)
2
and Cu
2
(CH3)2CHOH (2 alc) gives a yellow precipitate of CHI3 with I2/OH (iodoform test)
Problem 12: Write mechanism of the following reaction OH
OH
H3 O
O
Solution:
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OH
OH
O
H
O H
O
Problem 13: Compound (A) C6H14O is insoluble in water and gives negative Lucas test. On treatment with conc. H2SO4 followed by hydrolysis it gives only one compound (B) (C3H8O). Compound (B) is soluble in water and gives red colour with ceric ammonium nitrate. (B) gives yellow precipitate (C) and compound (D) on treatment with I 2/Na2CO3 followed by acidification. Identify the compound A, B, C and D. Give the reasons. Sol.
H
H3C
HO
A=
O
H3C
C
CH3
B= CH3
CH3
C = CHI3
CH3
D = CH3COOH
Problem 14:
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Explain
the
low
a
boiling
point
and
decreased
water
solubility
by
o–nitro phenol and o–hydroxy benzaldehyde as compared with this m and p–isomers. Solution: Intramolecular H-bonding (chelation) in the o-isomers inhibits intermolecular attraction,
lowering
the
boiling
point
and
reduces
H-bonding with H2O, decreasing water solubility. Intramolecular chelation cannot occurs in m – and p – isomers.
O
O H
N
H
O
O o - nitro phenol
C
O H
o - hydroxy benzaldehyde
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Objective:
Problem 1: When ether is exposed to air for some time an explosive substance is produced (A)
peroxide
(B)
TNT
(C)
oxide
(D)
super oxide
Solution: (A)
Peroxide
Problem 2: What is the major product obtained when phenol is treated with chloroform and aqueous alkali? (A)
(B)
OH
OH
CHO CHO
(C)
(D) HO
OH
CHO COOH
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(A)
Salicyldehyde
Problem 3:
CH3 H3C
C
CH 2
H2 O18 H
A, A is
O
(A)
(B)
CH3 H3C
C
CH2
OH
OH
18
CH3 H3C
C
CH2
OH
OH 18
(C) Both
(D) none
Solution:
Stability of carbocation is 3 > 2 > 1 (A)
Problem 4: Organic acid without a carboxylic acid group is …………………….. (A)
ascorbic acid
(B)
vinegar
(C)
oxalic acid (D)
picric acid
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Solution: Picric acid is a phenol i.e. 2, 4, 6 tri nitro phenol (D) Problem 5: The product of the following reaction H3C
i BH3 / THF
CH2
ii H2O2 ,OH
(A)
1 – pentanol
(B)
2 – pentanol
(C)
pentane
(D)
1, 2 – pentane diol
Solution: Anti Markowvnikov product. (A) Problem 6: Product in the reaction C2 H5 Br
NaOH aq
A
NaOH
B
CH3 I
C will be
(A)
propane
(B)
ethyl iodide
(C)
butanol
(D)
methoxy ethane
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Solution:
NaOH aq
C2H5Br
C2H5OH
Na / CH3I William son's synthesis
C2H5OCH3
(D) Problem 7: Phenol can be distinguished from alcohol with (A)
Tollen‘s reagent (B)
Schiff‘s base
(C)
Neutral FeCl3
HCl
(D)
Solution: Phenol gives violet colour with neutral FeCl 3. (C) Problem 8: Tert – butyl methyl ether on heating with HI of one molar concentration gives (A)
CH3OH + (CH3)3C I
(B)
CH3I + (CH3)3COH
(C)
CH3I + (CH3)3CI (D)
none of these
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Solution:
CH3
3
C O CH3
HI
CH3 OH
CH3
3
C I
(A) Problem 9: Which of the following is the strongest acid? (A)
(B)
OH
HO
Cl
NO 2
(C)
(D) HO
OH
NO 2
NO 2
Solution: Due to greater electron withdrawing effect of NO2 group than Cl atom nitrophenols are stronger acids than p–chloro phenol among nitro phenols P nitro phenol is the strongest acid. (C)
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Problem 10: Phenol is heated with phthalic anhydride in presence of conc. H 2SO4. The product is (A)
Phenolphthalein (B)
bakelite
(C)
salicylic acid
fluorescein
(D)
Solution: Phenolphthalein (A) Problem 11: An organic compound X of molecular formula C4H10O undergoes oxidation to give a compound Y of molecular formula C 4H8O2. X could be (A)
CH3CH2CH2CH2OH
(B)
CH3CH2CH(OH)CH3
(C)
(CH3)2CHCH2OH
(D)
(CH3)3COH
Solution: (A) & (C)
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Problem 12: Which of the following methods is/are not the industrial methods to prepare methanol? (A)
catalytic reduction of carbon monoxide in presence of ZnOCr2O3.
(B)
reacting methane with steam at 900 C with a Ni catalyst.
(C)
reducing formaldehyde with lithium aluminium hydride.
(D)
reacting formaldehyde with aqueous sodium hydroxide solution.
Solution: (B), (C) & (D) Problem 13: Which of the following reagent (s) or test (s), can be used to distinguish 2 methyl propanol and 2-methyl propanol–2? (A)
I2/NaOH
(B)
HCl/anhyd. ZnCl 2
(C)
Victor-Meyer test
(D)
oxidation with Cu at 573 K
Solution: (B), (C) & (D)
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Problem 14: Phenol has higher pka value than (A)
acetic acid
(B)
p-methoxy phenol
(C)
p-nitrophenol
(D)
ethanol
Solution: (A) & (C) Problem 15: Which of the following reactions can be used to prepare methyl phenyl ether? (A)
reacting sodium phenoxide with methyl chloride
(B)
reacting phenol with diazomethane (CH 2N2)
(C)
reacting sodium methoxide with chlorobenzene
(D)
reacting C6H5OMgCl with chloromethane
Solution: (A) & (B)
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Fill in the Blanks Problem 16: Williamson‘s synthesis involves the reaction of an………………..with an………………. Solution: Alkyl halide, alkoxide Problem 17: An ether is more volatile than alcohol having the same formula due to absence of……………….. Solution: Intermolecular hydrogen bonding Problem 18: Anisole reacts with HI to form……………………
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Solution: Methyl iodide and phenol Problem 19: Power alcohol is a mixture of absolute alcohol and…………….in the ratio……………. Solution: Petrol, 20 : 80 Problem 20: Glycerol on reaction with potassium hydrogen sulphite and heat yields a product which…………to form acraldehyde. Solution: Tautomerise
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Problem 21: Phenol on exposure to air produces a………………..coloured product known as…………………….. Solution: Red, phenoquione True and False Problem 22: t-butyl alcohol reacts less rapidly with metallic sodium than the primary alcohol. Solution: True Problem 23: o-nitro phenol is more soluble in water than p-nitrophenol.
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Solution: False Problem 24: 2-methyl – 2- propanol gives cloudiness with HCl and ZnCl 2 immediately at room temperature. Sol: True Problem 25: The hydroboration oxidation process gives product corresponding to Markonikov‘s addition of water to the carbon – carbon double bond. Solution: False Problem 26: p-amino phenol has higher pka value than phenol. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Solution: True
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ASSIGNMENT PROBLEMS
Subjective:
Level–O 1.
Write all the canonical forms of phenol.
2.
Account for the following: (i)
When HI react with methyl phenyl ether, phenol and methyl iodide are formed and not methyl alcohol and iodobenzene?
(ii)
An ether would posses a dipole moment even if the alkyl groups present in it are identical.
3.
O – nitro phenol is less soluble in water than p – nitro phenol. Why?
4.
3, 3 – dimethyl butane 2 – ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetra methyl ethylene as a major product. Suggest a suitable mechanism.
5.
Complete the following sequences of reactions by supplying X, Y and Z: (a)
X
PBr3
Y
Na Ether
CH3 CH3
Cl2 heat
Z
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(b) X
H2O/H
H3C
CH
CH3
Cu / 573K
Y
(i) C2H5MgI (ii) H2O/H+
Z
OH
6.
How will you convert phenol to salol?
7.
Name the reagents used in the following reactions:
8.
(i)
Benzyl alcohol to benzoic acid.
(ii)
Phenol to 2, 4, 6 tribromo phenol.
(iii)
Ethanol to ethanol
(iv)
Ethene to ethane – 1, 2 diol.
Arrange the following set of compound in order of their increasing boiling points. (a)
Pentan–1–ol, butan–1–ol, butan–2–ol, ethanol, propan–1–ol, methanol.
(b) 9.
Pentan–1–ol, n–butane, pentanal, ethoxyethane.
Arrange the following compounds in increasing order of their acid strength: Propan–1–ol, 2, 4, 6–trinitrophenol, 3–nitrophenol, 3, 5–dinitrophenol, phenol, 4–methyl phenol.
10.
(a)
Explain why a non symmetrical ether is usually prepared by heating a mixture of ROH and R OH in acid. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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(b)
Would you get any di-tert-butyl ether from this reaction? Explain.
11.
12.
How the following conversion carries out? (i)
Propane
Propan–2–ol
(ii)
benzyl chloride
(iii)
Ethyl magnesium chloride
(iv)
Methanol to Ethanol
benzyl alcohol propanol–1
Identify the major products in the following reactions. (i)
(ii)
NO 2
OH CH3
HNO3 H2 SO4
Br2 water
A
B
OCH 3
13.
14.
Give one test to distinguish (a)
Ethanol and methanol
(b)
Phenol & Benzyl alcohol
Compound (A) C7H8O is insoluble in water and dilute NaHCO3 but dissolve in dilute NaOH solution. On treatment with Br 2 water it readily gives a precipitate of C7H5OBr3. Write down the structure of the compound.
15.
Give the mechanism of Reimer Tiemann reaction on phenol. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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Level – I 1.
Write the product of the following reactions i BH3 ii Oxidation
A
CH3
2.
Write the product A & B H2/Pt CH
3.
CH
CHO
A
NaBH 4
B
Write the product A & B CH3 Cold alkaline KMnO4
A
Cr2 O 3 ACO H
B
4.
Convert nitro benzene to m – nitro phenol.
5.
An unknown compound A (C4H10O2) reacts with sodium metal to liberate hydrogen gas. A is inert towards periodic acid, it react with CrO3 to form B (C4H6O3). Identify A and B.
6.
Convert
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OH CH3 OH
to
Write mechanism of above conversion. 7.
Identify compounds A – D in the following reaction. CHC 6H5 i O3 ii Zn/H2O
dil. NaOH aq. ethanol
A
B
C
i O3 ii Zn/ H2O
C
A
B
H2O
D A OH
H2 / Pt
D
OH
8.
Give mechanism of the following reaction OH
OH
COOH
i CCl4 NaOH/ Δ ii H3 O
9.
Identify the product in the given reaction BH 3/THF
1 equiv
H
A
H2O2 OH
B
CH3
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10. Conc. H2SO4
Δ
H2O
X
Y
Identify X and Y.
O
11.
Identify A & B CHCl
OH 3
A
B
N H
Pyrole 12.
13.
Give a simple test that distinguish between the compounds. (a)
Allyl and n – propyl alcohol.
(b)
Benzyl methyl ether and benzyl alcohol.
How will you synthesize? CH3 from phenol
14.
An organic compound (A) gives positive Libermann reaction and on treatment with CCl4/KOH followed by hydrogenation gives (B). B on reduction gives (C) C7H8O2. (B) react with (CH3CO)2O/CH3COOH give a pain reliever (D). Identify A to D.
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15.
Provide suitable mechanism CH3
OH CH3
Level – II 1.
An ester A (C4H8O2) on treatment with excess of methyl magnesium bromide followed by acidification gives an alcohol (B) as the sole organic product. Alcohol (B) on oxidation with NaOCl followed by acidification gives acetic acid. Deduce the structure of A and B and show the reactions involved.
2.
Propose a mechanism for the following reaction. (a)
Me
Ph
Me
Ph
O
Me
Ph
C
C
Ph
Me OH
OH
(b)
OH O
OH
3.
Give products
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Anhydrous HI ether I
CH3 H3C
C
O
A
B
CH3 Conc. HI
CH3
C
D
(2)
4. CH(OH)Me H
A [ C 9H16]
Me
A on ozonolysis gives nonane – 2, 8 dione. What is A and how is it formed? 5.
An organic compounds (P) having molecular formula C 5H10O treated with dilute H2SO4 gives two compounds Q & R both gives positive iodoform test. The reaction of C5H10O with dilute H2SO4 gives reaction 1015 times faster than ethylene. Identify organic compound of Q & R. Give the reason for the extra stability of P.
6.
Complete the following reaction. OH
i OH ii Et Br
A
Zn / HCl
B
NaNO2 HCl, 500 C
C
C6H5OH NO 2 LiAlH 4 (E) (F) Soluble in NaOH
D
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7.
(a)
Diphenyl ether is not very easily prepared using Williamsons sysnthsis. Provide an
(b)
useful synthetic route for the same.
When phenol is treated with Br2 presence of H2O. 2, 4, 6 tri bromo phenol is obtained whereas on treatment with Br2 in presence of CCl 4. p–bromo phenol is obtained. Explain.
8.
Write mechanistic step of the following reaction?
H
CH 2OH
9.
Compound (A) gives positive Lucas test in 5 minutes when 6.0 gm of (A) heated with Na metal, 1120 ml of H2 is evolved at STP. Assuming (A) to contain one of oxygen per molecule, write structural formula of (A). Compound (A) when heated with PBr3 gives (B) which when treated with benzene in presence of anhydrous AlCl 3 gives (C). What are (A), (B) and (C)?
10.
Carry out the following conversion. OH
to
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11.
Treatment of compound (A) C8H10O with chromic acid & pyridine gives (B) C8H8O. Treatment of (B) with two equivalent Br2 yields (C) C8H6OBr2
which
on
treatment
with
caustic
soda
followed
by
acidifications gives a compound (D) C8H8O3. (D) liberates CO2 on treatment with NaHCO3 and is resolvable. 12.
Compound X, Y and Z are isomeric alcohol with formula C 5H12O. X and Y react with chromic acid solution, Y forming an acid A. The three isomers react with HBr with decreasing relative rates of Z > X >> Y, all giving the same C5H11Br (B) in varying yields. X alone can be oxidised by I2/OH to C. Write the structure of X, Y, Z with proper explanations.
13.
Predict the product (s) and write the mechanism of each of the following reactions (a)
(b) 1 mole HI
O
14.
Excess HI Δ
(A) O
CH3
CH3
Find product (a)
CH3 H3C
(b)
C
CH2 OH
H2SO 4
MeOH
(A)
OH
H
(B)
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(c)
MgBr RCOOEt
H3O
(H2C) 5
C
H2/Ni
D
MgBr
(d)
CH 2OH C
O
2IO 4
E F
CH 2OH
15.
(Z) C6H14O
B
D
alc. KOH
O 3/H 2O
Δ / H2 SO 4
(A) C6H12
HCl
B C C6H13Cl
D (Isomer of A)
(Given negative test with Fehling solution
E
but responds to iodoform) A
O3/H 2O
F
G
(both give positive Tollen‘s test but do not give iodoform test)
F
G
Conc. NaOH
HCOONa
a primary alcohol
Identify A to G and Z.
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Objective:
Level–I 1.
2.
3.
Which one of the following has the maximum acidic strength? (A)
Phenol
(B)
o-nitro phenol
(C)
p-methyl phenol
(D)
o, p-dinitro phenol
The boiling point to isomeric alcohols follows the order (A)
primary > secondary > tertiary
(B)
tertiary > secondary > primary
(C)
secondary > tertiary > primary
(D)
all have same boiling point
A mixture of benzoic acid and phenol may be separate by treatment with
4.
(A)
NaHCO3
(B)
NaOH
(C)
NH3 solution
(D)
KOH
In the lucas test of alcohols, appearance of cloudiness is due to the formation of
5.
(A)
aldehyde
(B)
ketone
(C)
acid chloride
(D)
alkyl chloride
The dehydration of 1 – butanol gives Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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6.
7.
(A)
1 – butene as the main product
(B)
2 – butene as the main product
(C)
equal amounts of 1 – butene and 2 – butane
(D)
2 – methyl propane
Ethyl alcohol is obtained when ethyl chloride is boiled with (A)
alc. KOH
(B)
aq. KOH
(C)
AlCl3
(D)
H2O2
The number of methoxy groups in a compound can be determined by treating with
8.
9.
(A)
Na2CO3
(B)
NaOH
(C)
HI and AgNO3
(D)
CH3COOH
Diethyl ether absorbs oxygen to form (A)
red coloured sweet smelling compound
(B)
CH3COOH
(C)
ether peroxide
(D)
ether suboxide
Which of the following compounds is oxidised to prepare methyl – ethyl ketone? (A)
2 – propanol
(B)
1 – butanol
(C)
2 – butanol
(D)
2 methyl 2 propanol
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10.
11.
12.
13.
14.
Order of reactivity of HX towards ROH is (A)
HI > HBr > HCl
(B)
HBr > HI > HCl
(C)
HCl > HI > HBr
(D)
HI > HCl > HBr
Glycerol has (A)
one 1 and one 2 alcoholic groups
(B)
one 1 and two 2 alcoholic groups
(C)
two 1 and one 2 alcoholic groups
(D)
two 2 alcoholic group
Ethyl iodide reacts with moist silver oxide to produce (A)
ethane
(B)
propane
(C)
ethyl alcohol
(D)
diethyl ether
Reaction of tertiary butyl alcohol with hot Cu at 350 C produces (A)
butanol
(B)
butanal
(C)
2 – butene
(D)
2 methyl propene
1 alcohol can be converted to aldehyde by using the reagent (A)
pyridinium chloro chromate
(B)
potassium di chromate
(C)
potassium permanganate
(D)
all of above
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15.
Reaction of ethanol with H 2SO4 and suitable conditions can lead to the formation of (A)
C2H5HSO4
(B)
ethene
(C)
ethoxy ethane
(D)
all of them
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Level – II 1.
2.
2 – phenyl propene on acidic hydration gives (A)
2 – phenyl – 2 propanol
(B)
2 phenyl – 1 – propanol
(C)
3 – phenyl – 1 – propanol
(D)
1 – phenyl – 2 – propanol
The order of reactivity of phenyl magnesium bromide with the following compound is O
Me
O
Me
Me
(I)
3.
4.
O
H
Ph
(II)
Ph (III)
(A)
II > III > I (B)
I > III > II
(C)
II > I > III (D)
all react with the same rate
Which one is the stronger base? (A)
CH3CH2O
(B)
CF3CH2O
(C)
both of equal strength
(D)
can not say
The acidic character of 1 , 2 , 3 alcohols H2O and RC order (A)
H2O > 1 > 2 > 3 > RC
CH
(B)
RC
(C)
1 > 2 > 3 > H2O > RC
CH
(D)
3 > 2 > 1 > H2O > RC
CH
CH > 3 > 2 > 1 > H2O
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5.
Choose the correct statement (s) for the reaction OH
OH
OCOCH
3
AlCl3 / CS2 heat
COCH
3
(a) COCH (b)
6.
3
(A)
(b) is formed more rapidly at higher temperature
(B)
(b) is more volatile than (a)
(C)
(a) is more volatile than (b)
(D)
(a) is formed higher yields at lower temperature
Predict the major product Excess HI O
7.
(A)
HO
CH2
CH2
CH2
CH2
I
(B)
HO
CH2
CH2
CH2
CH2
OH
(C)
I
(D)
no reaction
CH2
CH2
CH2
I
Dipole moment of CH3CH2CH3, CH3CH2OH and CH3CH2F is in order (I)
8.
CH2
(II)
(III)
(A)
I < II < III (B)
I > II > III
(C)
I < III < II (D)
III < I < II
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9.
(A)
CH3MgI and 3 – hexanone, followed by hydrolysis
(B)
C2H5MgI and 2 – pentanone, followed by hydrolysis
(C)
C3H7MgI and 2 – butanone, followed by hydrolysis
(D)
any of the method above
Ester (A)C4H8 O2
CH3MgBr
H3 O
B C4H10 O
2 parts
Alcohol B reacts fastest with Lucas reagent. Hence A and B are (A) H3C
(C)
11.
C
O
C2H5 , (CH 3)3COH
O H
(D)
O H3C
10.
(B)
O
C
O
O
C3H7 ,(CH 3)2CHOH
O
C3H7 ,(CH 3)3COH
O H
C2H5 ,(CH 3)2CHOH
C
C
Vinyl carbinol is (A)
HO
(C)
CH3
CH2
CH = CH2
CH = CHOH
(B)
CH3C(OH) = CH2
(D)
CH3
CH CH2OH
The reaction of elemental sulphur with Grignard reagent followed by acidification leads to the formation of
12.
(A)
mercaptan
(B)
sulphoxide
(C)
thio ether
(D)
sulphonic acid
Conversion of chloro benzene into phenol by Dow‘s process is an example of (A)
free radical substation
(B)
nucleophilic substitution
(C)
electrophilic substitution (D) rearrangement Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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13.
For the preparation of tert-butyl methyl ether by Williamson‘s method the correct choice of reagents is
14.
(A)
methoxide and tert – butyl bromide
(B)
methanol and 2 – bromobutane
(C)
2 – butanol and methyl bromide
(D)
tert-butoxide and methyl bromide
Allyl alcohol is obtained when glycerol reacts with the following at 260 C
15.
(A)
formic acid
(B)
oxalic acid
(C)
both
(D)
none
The correct decreasing order of acidic strength is (A)
C6H5OH > C6H5CH2OH > C6H5COOH > C6H5SO3H
(B)
C6H5CH2OH > C6H5OH > C6H5SO3H > C6H5OH
(C)
C6H5COOH > C6H5CH2OH > C6H5OH > C6H5SO3H
(D)
C6H5SO3H > C6H5COOH > C6H5OH > C6H5CH2OH
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ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
Level – O 1.
Refer to text.
2.
(i)
In methyl phenyl ether. The Ist step is protonation of ether to form phenyl methyl oxonium ion. H O
CH3
H step 1
O
CH3
OH
I step 2
CH3I
Phenyl carbocation is highly unstable and bond between C 6H5–O is stronger than O–CH3 so I attacks on CH3 to form CH3I & phenol. (ii)
Oxygen of ether is sp3 hybridised, therefore ethers have tetrahedral geometry. Due to more electronegativity of oxygen than carbon, C–O bonds of ether are polar in nature.
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R O
1100C
Net dipole moment
R
The two dipoles do not cancel each other hence the molecule would posses dipole moment. 3.
In O – nitro phenol there is intramolecular hydrogen bonding. This inhibits its hydrogen bonding with water and reduces its solubility in water. O H O N O
4.
CH3 H3C
C
CH3 CH
CH3
H
H3C
CH3 OH
C
CH
CH3
CH3 OH 2 -H2O CH3 H3C
C
CH
CH3
CH3 CH3 H3C
C
C
CH3 1, 2 methyl shift
CH3 CH3 CH3
-H
H3C
C
CH
CH3
tetra methyl ethylene
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5.
(a)
X = CH3OH, Y = CH3Br, Z = CH3CH2Cl
(b)
C2 H5
X = CH3CH = CH2, Y = CH3COCH3, Z =
H3C
C
OH
CH3
6.
OH
OH
OH
OH
COOH Kolbe ' s reaction i CO2 400K, OH ii H3 O
COO H Phenyl salicylate (salol)
7.
8.
(i)
Alkaline KMnO4
(ii)
Br2 water
(iii)
P.C.C
(iv)
1% alkaline KMnO4
(a)
Methanol < ethanol < propan–1–ol < butan–2–ol < butan–1–ol < pentan–1–ol.
(b) 9.
n–butane < ethoxy ethane < pentanal < pentan–1–ol.
Propan–1–ol < 4 methoxyphenol < phenol < 3 nitro phenol < 3, 5– dinitrophenol < 2, 4, 6–trinitrophenol.
10.
(a)
A mixture of three ethers, R–O–R, R–O–R
and R –O–R
is
obtained. (b)
Not–butanol does not solvate the 3 carbocation readily because of steric hindrance. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
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12. (i)
(ii)
NO 2
OH
O 2N
Br
A=
CH3
A= OCH 3
Br
13.
(a)
Iodoform test given by ethanol.
(b)
Neutral FeCl3 test given by phenol.
14.
OH
m - Cresol CH3
OH 15. CHCl 3 OH
Cl
CCl 3 O
CCl 2 O H
OH
CCl 2
O
CCl 2
O CHCl
2
CHO 2 OH H2O
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Level – I 1.
OH H H3C
H
2. A=
CH 2CH 2CH 2CHO
3.
B=
CH
OH
OH
B=
O
OH
4.
NO 2
CH 2OH
CH3
CH3 A=
CH
NO 2
NH 2
NH4
funing HNO3 Conc. H2SO4 , Δ
2
S
NO 2
NO 2 NaNO 2/H 2SO 4
0 - 50C OH
N2
HSO
4
H2 O Δ
NO 2
5.
NO 2
OH A=
H3C
CH
CH 2 CH 2 OH
O B=
H3C
C
CH 2 COOH
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6.
OH CH3 ring exp ansion
H
-H2O
(Hydride shift)
CH3
CH3 OH
OH
7.
O H C A=
O B=
O
O H O C
C=
D=
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8.
OH
O
O
O H CCl 3
Cl OH
CCl 3
CCl 3
OH O OH
O
H3O
COOH OH
COOH
CO 2
9. A=
BH 2H
H
10.
OH H
B= H
CH3
(X) =
HO
O
(Y) =
HO
OH
11.
CH3
SO3H
Cl A=
B= N
CHO
N
H
12.
(a)
Add Br2/CCl4 – Allyl decolourises orange colour
(b)
Add small piece of Na. H2 releases from alcohol.
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13.
OH
O Cl
CH3 Zn dust
O
AlCl 3
Zn(Hg)/HCl
CH3
14.
OH
OH
COOH
A=
B=
Phenol
Salicylic acid
OH
OCOCH CH 2OH
C=
3
COOH D=
C7H8O 2
Aspirin
2 hydroxy benzyl alcohol
15.
CH3
H
OH 2
OH CH3
CH3
CH3 -H2O -H
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Level – II 1
O H . C
CH3 O
C
H3C OH
CH3 H3C
H (A)
2
(a)
H
(B)
Me
Me
Ph
.
H H3C
Ph
OH
Me
Ph
Me
Ph
Ph
OH
OH
Ph
OH
Me
Ph
Ph
Me
Me -H C Ph
O
(b)
Ph
Me
H
O
Me
H OH
-H2O
H
OH
OH OH
OH 2
O
-H
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O
H
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3.
CH3
CH3 H3C
C
O
Anhydrous HI ether I
CH3
CH3I
H3C
C
OH
CH3
CH3
CH3 Conc. HI
CH3OH
(2)
H3C
C
I
CH3
4.
A
Nonane 2, 8 dione (No loss of C) CH 2 CH 2 C
CH3
CH 2 CH 2 C O3/H2O
H2 C CH 2 CH 2 C
H2 C
CH3
CH3
O CH 2 CH 2 C
CH3
O Me CH(OH)Me Me
CH Me
H
Me
1, 2 alkyl shift (ring expansion)
Me
20 carbocation
Me
Me
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5.
CH3
CH3 H
H2C
O
Stability by resonacne
CH3
H3C
O CH3 Highly stable carbocation
CH3 C2H5OH H3C
H2O
O (Q)
6
OEt
OEt
OEt
. C=
B=
A=
D=
N2Cl
NH 3 Cl
NO 2
N
C
C
(C5H10O)
(R)
H3C
N
OEt OH OH
OEt
F=
E=
NH2
NH2
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O
H3C H3C
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7.
(a)
X
OH
O
O
NaOH
Se/ Δ
O
(b)
H2O ionises phenol into the most reactive phenoxide anion and hence tri substitution occurs. Moreover H2O can stabilizing the arenium ion intermediate formed as well as the electrophile Br+.
Hence
the
rate
of
reaction
is
fast
and
hence
tri – substitution results. Hence the rate of the reaction is slow and hence now substitution results. 8. H H2 O
CH 2OH
CH 2
ring expansion
H H H
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9.
(A) is secondary alcohol. CnH2n+1OH = Molecular weight Molecular weight of (A) =
6 11200 1120
60 gm A
CnH2n+1OH = 60 (n = 3) (A) C3H7OH H3C
CH
CH3
OH CH3 CH 3CH(OH)CH (A)
PBr 3 3
HC CH 3CHBrCH (B)
3
Anhy AlCl 3
CH3 (C)
10.
OH
CH2
O
11.
C H 2N 2 , Δ
Ph3P CH3 Cl t BuLi
O CrO3
O
CHCH
3
A=
C
OH
O C C=
CH3
B=
OH
CHBr 2
CH
COOH
D=
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12.
OH Hydride shift
H3C
CHCH(CH3)2 (X)
I 2/OH
(CH3)2CHCOOH
CH3 Br
C
Methyl shift
CH3
Oxidation
(CH3)2CCH2OH (Y)
(CH3)3C.CH2OH (A)
CH2 CH3 (B)
SN' displacement
(CH3)2.C(OH)CH2CH3 (Z)
Oxid
No Reaction
13. 1 mole HI
O
CH3
(a)
O
CH3
OH
H
CH3 I
I OH
CH3 (A)
HI
(b)
SN 2 I I
H I
O
CH3
OH
CH3
OH 2
H2 O I
CH3
I
CH3 (B)
14. (a)
(b)
CH3 H3C
C
CH3
OCH 3
C CH3
O
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CH3
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(c)
(d) R
R
E = 2HCHO + F = CO2
15.
H2COH Z=
D=
E = H3C
A=
C
CH3
O H3C
CH3
O H
B=
Cl
CH3
F=
CH3
Cl
G = HCHO
C=
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Objective:
LEVEL - I 1.
D
2.
A
3.
A
4.
D
5.
B
6.
B
7.
C
8.
C
9.
C
10.
A
11.
C
12.
C
13.
D
14.
A
15.
D
1.
A
2.
C
3.
A
4.
A
5.
C
6.
C
7.
A
8.
D
9.
A
10.
A
11.
A
12.
B
13.
D
14.
B
15.
D
LEVEL - II
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