Alcohol, Phenol & Ether_Chemistry

AISM-09/C/APE

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SUBJECT – CHEMISTRY

TOPIC – ALCOHOL, PHENOL AND ETHER

COURSE CODE – AISM-09/C/APE

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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.

C CH

H

H H

H

OH Contain only one - OH group

Dihydric

C COH

H

H H

H

OH Contain tw o - OH group

Polyhydric

C CH

H

OH OH

H

C

H

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 CH2OH

Secondary

R C

OH

R

H

Tertiary

R C

OH

R

R

or 20 or 30Alcohol

Illustration 1:

Classify the following into primary, secondary and tertiary

alcohols:

(a)

OH

CH3

(b)

CH3

OH

(c)

OH

Solution:

(a) Tertiary, (b) Secondary, (c) Tertiary

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PREPARATION OF ALCOHOLS

1. From Alkanes

Alkanes having tertiary carbon on oxidation with cold alkaline KMnO4 give

tertiary alcohol.

R C

H

R

R

KMnO 4/OHR C

OH

R

R

2. From Alkenes

Alkenes can be converted into alcohol by the following reactions:

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R CH CH2

HOH/H 2SO4

(i) Conc. H 2SO4

(ii) HOH

(i) Hg(OCOCH 3)2/HOH

(ii) NaBH 4

(i) BH 3/THF

(ii) H 2O2/OH

CO H2

Co(CO) 4

R CH2 CH2 CHO

H2/Pt

R CH2CH2CH2OH

R CH2 CH2OH

R CH CH3

OH

R CH CH3

OH

R CH CH3

OH

(anti markonikov's product)

[Oxo process]

3. From alkyl halides

Alkyl halides give alcohol with KOH/NaOH or with moist Ag2O.

R X

HOH/NaOH

moist Ag 2O

R OH

R OH

4. Reduction of aldehydes and ketones

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(a) Reduction by reducing agents

(i) Aldehyde gives primary alcohol

R C H

O

[H]

Reducing agentR CH2OH

(ii) Ketone gives secondary alcohol

R C R

O

[H]

Reducing agentR CH

OH

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) H2/Ni

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NaBH4 and aluminium isopropoxide reduces only carbonyl

group and has no effect on any other group.

4NaBH

3 3 2CH CH CH CHO CH CH CH CH OH

Reduction with aluminium 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.

H2C CH C CH3

O

LiAlH 4CH3 CH2 CH CH3

OH

H5C6 CH CH CHOLiAlH4

H5C6 CH2 CH2 CH2OH

(b) Reduction by Grignard reagents

Addition followed by hydrolysis

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R Mg X

H C H

O

R' C H

O

R' C R''

O

R C H

O MgX

H

R C H

OMgX

R'

R C R''

OMgX

R'

H2O/H

H2O/H

H2O/H

R C H

OH

H

Mg

OH

X

R C H

OH

R'

Mg

OH

X

R C R'

OH

R''

Mg

OH

X

30 alcohol

20 alcohol

10 alcohol

Methanol cannot be prepared by this method.

5. Reduction of carboxylic acid, Acid chlorides and esters:

(a) Reduction by LiAlH4

R C G

O

LiAlH 4R CH2OH H G

G = OH (acid)

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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 BH3.

R C

O

OH3

2

i BH / THF

ii H O/HRCH 2OH

R C

O

O R' 3

2

i BH /THF

ii H O/HR CH2OH R'OH

(c) Bouveault – Blanc reaction

R C

O

OR'Na/C2H5OH

RCH 2OH R'OH

Illustration 2:

(CH2)n

COOH

COOCH3LiAlH4

(i) BH3/THF

(ii) H2O/H

A

B

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Solution:

(CH2)n

COOH

CH2OH

A =

If compound has COOH as well as ester group then reactivity of ester is

more than acid towards LiAlH4.

(CH2)n

COOCH 3

CH2OH

B =

Acid is more reactive than ester towards BH3.

Exercise 1:

Find A and B.

(i)

CH3

2

2

Hg OAC

THF / H OA 4NaBH

OHB

(ii) 2 4CuO CuCr O

3 2 3CH CH COOCH A B

6. 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.

CH3 CH2 CH2 NH2

NaNO 2/HClCH3 CH2 CH2OH CH3 CH

OH

CH3

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.

H2C CH2

O

R Mg X H2C CH2 R

OMgX

HOH/H

HO CH2 CH2 R

δ

δ δ δ

Illustration 3:

(a) Find A, B, C, D, E.

PhBrMg

dry etherA

O

B

O

HC

(b) 3

2

i CH MgBr

3 2 5 ii H O / HCH COOC H D E

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Solution:

(a) A = PhMgBr B = Ph

OC =

Ph

OH

(b)

CH3 C

CH3

CH3

OH C2H5OH

8. Fermentation of carbohydrates

Sucrose

InvertaseC6H12O6 C6H12O6

glucose fructose

Zymase

C12H22O11 H2O

2C2H5OH 2CO2

Illustration 4:

(i)

C2H5MgBr

(i) C CH2

CH3

CH3

O

(ii) HOH/H

A

(ii)

CH C

CH3

CH3CH3

CH CH2 CH3Oxymercuration

DemercurationB

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Solution:

(i)

CH3 C

CH3

OH

CH2 C2H5 ( 2 methyl 2 pentanol)

(ii)

CH3 CH

CH3

C

CH3

OH

CH2 CH2CH3

(30 alcohol)

Exercise 2:

(i) Identify A, B & C

RCO2Et + (CH2)5

MgBr

MgBr 3+H O

A 2

Δ

-H OB 2H /Ni C

(ii)

H C

O

CH2 CH2 OH RMgX RH H C

O

CH2 CH2 OMgX

( 1 eq)

H C

O

CH2 CH2 OH RMgX H C(OH)

R

CH2 CH2 OH

( 2 eq)

Account the reason for the above reactions.

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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) 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.

(A) Reaction involving breaking of carbon – oxygen bond

Order of reactivity of alcohol. 3 > 2 > 1

(i) SN reaction

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R OH

HCl/Anhy ZnCl2

Δ

NaX/H2SO4

PCl5 or PCl3

P/Br2 or PBr3

Δ

SOCl2/Py ridine

SOCl2/Ether

or

R Cl

R X NaHSO 4 H2O

R Cl

R Br

R Cl SO2 HCl

(ii) Dehydration of alcohol

Dehydration of alcohol to give alkene.

(a) Dehydrating agents are

Conc H2SO4/ , KHSO4/ , H3PO4/ , Anhyd Al2O3/ , Anhyd PCl5/ ,

Anhyd ZnCl2/ , BF3/ , P2O5/ .

(b) Reactivity of alcohols. (Ease of dehydration)

3 > 2 > 1

(c) Product formation always takes place by saytzeff rule.

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CH3 CH2 CH

OH

CH32 4Conc. H SO

ΔCH3 CH CH CH3

(Major)

CH3 CH2 CH CH2

(Minor product)

* Alcohols on acetylation gives acetyl derivative which on pyrolytic

elimination always gives Hofmann product.

CH3 CH2 CH

OH

CH33 2

CH CO O /Py CH3 CH2 CH CH3

OCOCH 3

(Major)

CH3 CH2 CH CH2

(Minor)

Δ

CH3 CH CH CH3

Mechanism in presence of acidic medium

E1 mechanism: follow saytzeff‘s rule.

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CH3 C

CH3

CH3

CH2 O H H 2 4Conc.H SOCH3 C

CH3

CH3

CH2 O H

H

CH3 C

CH3

CH3

CH2

1,2methyl shiftCH3 C

CH3

CH2 CH3

CH3 C

CH3

CH CH3

Exercise 3:

Write mechanism

CH3

CH CH3

OH

H

CH3

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.

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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

23

33R OH Al RO Al H

2

Aluminium alkoxide

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(ii) Esterification (With carboxylic acid)

R' OH R C OH

O

H

R C O

O

R' H2O

It is reversible acid catalysed reaction. It follow SN1 mechanism.

CH3 C

O

O H

CH3 C

O

OH

H

CH3 C

OH

OH

R O H

CH3 C

OH

OH

O RCH3 C

OH

O

OR

H H

CH3 C

OH

ORH2O

CH3 C

O

OR

H

Increasing the size of alkyl group on alcohol part decreases the

nucleophilic character because steric hindrance increases.

1Reactivityα

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 OH N O

(nitrous acid)

R O N O

(Alkyl nitrite)

H2O

(b)

R O H OH S OH

O

O

RO S OH

O

OAlkyl hydrogen sulphate

(iv) Alkylation of Alcohol

3 42

3 2 3

CH SO /NaOH

3orCH I/K CO

R O H R O CH

Methylation is mainly used for determination of hydroxyl groups in an

unknown compound.

Molecular weight of methylated ether prodcued molecular weight of reac tantNo. of hydroxyl groups

14

Exercise 4:

Arrange the following in increasing order of acidic strength.

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(a)

CH3 CHOH

CH3

(i)

CH3 CH2OH

(ii)

CH3 OH

(iii)

CH3 C

CH3

CH3

OH

(iv)

(b) Arrange the following in increasing order of esterification:

MeCOOH EtCOOH (Et)2CHCOOH

(i) (ii) (iii)

(C) Oxidation of alcohol

Oxidation of alcohol is dehydrogenation reaction which is 1,

2–elimination reaction.

R C

O

H

H

R'1, 2 elimination

R C

O

R' H2

So oxidation of alcohol numbers of - hydrogen atom.

(a) 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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R CH2OH[O]

RCHO

R CH R'

OH

[O]R C

O

R'

CH3 C OH

CH3

CH3

[O]no reaction

Note:

PCC (Pyridinium chloro chromate) is a selective reagent which converts

1 alc to aldehyde.

(b) With strong oxidising agent

Oxidising agents are

(i) 4KMnO /OH /

(ii) 4KMnO /H /

(iii) 2 2 7K Cr O /H /

(iv) 3Conc. HNO /O

2n carbonn carbon

RCH OH RCOOH

R C

H

OH

R[O]

R C

O

R

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(D) Dehydrogenation with Cu/573K or Ag/573K

(a) 1 alcohol aldehyde

Cu/ 573K

2R CH OH RCHO

(b) 2 alcohol ketone

Cu/573KR CHOH R R CO R

(c) 3 alc undergo dehydration to form alkene.

CH3 C

CH3

CH3

OHCu/ 573 K

CH3 C

CH3

CH2

H2O

Reduction

R O HHI/Red P

R H

(E) Miscellaneous reactions of mono hydric alcohol

(i) Methylation with CH2N2 in presence of BF3.

2 2

3

CH N

3 2BFR O H R O CH N

(ii) Haloform reaction

Ethyl alcohol and 2 methyl alcohol gives haloform reaction.

C C

H

H

H

H

H OH 2

2

CaOCl

H O(HCOO) 2Ca CHCl 3

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Exercise 5:

(i) Out of these compound which gives iodoform test:

(a) CH3 CH2 CHOHCH3

(b) PhCH2CHOHCH3

(c) PhCHOHCH3

(d) CH3CH2OH

(e) CH3COCH2 COOC2H5

(ii) O

O

ONaBH4

CH3OH

LiAlH4

(A)

(B)

Distinguishing 1 , 2 , 3 alcohol

Test 1 alc 2 alc 3 alc

(I) Lucas test

[ZnCl2 + HCl]

No reaction at

room

temperatur

e

White turbidity

after 5-

10 min.

HCl

ZnCl 2

R CH R

Cl

H2O

RCH(OH)R

White turbidity

instantan

eously

R3C OH HCl

ZnCl 2

R3C Cl

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(II) Victor Meyer test

(P/I2,AgNO2,

HNO2, NaOH)

Red colour Blue colour Colourless

RCH 2OH

P/I 2

RCH 2I

AgNO2

RCH 2NO 2

HONO

R C

NOH

NO 2

Nitrollic acid

NaOH

R C

NO Na

NO 2

Sodium nitrolate (red )

CHOH

R

RP/I 2

CHI

R

RAgNO2

CHNO 2

R

RHNO 2

C

R

R

NO 2

N O

NaOH

(Pseudo nitrole)

Blue

R3C OH

P/I 2

AgNO 2

R3C NO 2

HNO 2

R3C I

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 (HIO4). The

reaction breaks carbon carbon bonds and produced carbonyl

compounds (aldehyde, ketones or acids)

H C

H

OH

CCH3 OH

H

HIO 4 HC O

H

HIO 3 H2OCH3CHO

It takes place through a cyclic intermediate.

C OH

C

CH3

H

CH3

OHCH3

IO 4

C O

C

CH3

H

CH3

OCH3

I O

O

O

CH3 C

CH3

O H C

CH3

O IO 3

Other examples

R C

O

C

O

R' HIO 4RCOOH R'COOH

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OH

R' OH R

OH

2HIO 4 RCHO HCOOH R'CHO

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:

3 3 2 3 3 3 43 3 3 3CH CBr, CH C CH ; CH COCOCH , CH CO Na ,CH , no reaction .

Illustration 6:

Write products

(a)

H C

H

C

OH

O

CH OH

H

IO4

(b)

H C

H

CH2

CH3

CH OH

H

IO4

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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 of

dehydration.

6 5 2 6 5 3 6 5 33 2C H CCH OH, C H CHOHCH , C H COHCH

(b) Find product

(a) CH2OCH3

CHOH

CH2R

IO4

(b) OH OH

OH

OHOH

OH IO4

(c) R CHOH

CHR' NH2

PINACOL – PINACOLONE REARRANGEMENT

Action of H2SO4 on 1, 2 diols.

Ditertiory – 1, 2 diols convert in to ketones on treatment with H2SO4.

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R C

OH

R

C

OH

R

RH

R C

R

R

C

O

R

Mechanism

CH3 C

OH

CH3

C

O

CH3

CH3

H

HCH3 C

OH

CH3

C CH3

CH3CH3 C

OH

C CH3

CH3

CH3

δ δ

δ

Transition state

CH3 C

OH

C CH3

CH3

CH3

Methyl shift

CH3 C

O

CH3

C CH3

CH3

HCH3

-HCH3 C

O

C CH3

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.

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Solution:

Hydration, hydroboration and oxymercuration – demercuration of alkenes.

Hydration:

CH3 C

CH3

CH3

CH CH22H O

HCH3 C

CH3

CH3

CHCH 3

OH

Hydroboration

CH3 C

CH3

CH3

CH CH23

2 2

i BH

ii H O , OHCH3 C

CH3

CH3

CH2CH2OH

Oxymercuration – demercuration

CH3 C

CH3

CH3

CH CH222

4

i Hg OAC , THF, H O

ii NaBHCH3 C

CH3

CH3

CHCH 3

OH

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Illustration 8:

Which of the following alcohols would react fastest with Lucas

reagent?

CH3CH2CH2CH2OH , CH3CH2CH

OH

CH3 , CH3CH

OH

CH2OH , CH3 C

CH3

CH3

OH

Solution:

3 3CH COH , it being a tertiary alcohol.

Exercise 7:

Find out

(i)

CH3 CH3

Ph

OH

Ph

OH

HA

(ii)

CH3 Ph

CH3

OH

Ph

OH

B

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Exercise 8:

Give a possible structure for the substance C5H10O2 behaving in the following

manner.

C5H10O26 5 2C H NHNH

O

2I ,NaOH

Fehling

Solution

ETHER

* R O R Alkoxy alkane (Di alkyl ether)

* R = R Symmetrical ether.

R 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

Oxirane

(Epoxide)

Oxetane

(Oxacyclo butane)

O

O

Tetra hydro furan

(Oxacyclo pentane)

PREPARATION OF ETHERS

(i) From 1 alcohol

(a) With H2SO4

2 4

02 3

H SO

140 C or Al O /525KR O H R O H R O R symmetrical ether

Order of dehydration 1 > 2 > 3 alcohol

(b) With diazomethane

2 2

2 5 3

CH N

3 2C H O AlR O H R O CH N

(c) 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.

CH3 CH2 CH2 CH2 OH6 64

0

Pb OAC hν /C H

80 C

OTetrahydro furan

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H2C (CH2)3

H

CH2OH6 64

0

Pb OAC hν /C H

80 C

O CH3

2 methyl tetra hydro furan

Williamson’s synthesis

NS 2 reaction of a sodium alkoxide with alkyl halide, alkyl sulphonate or alkyl

sulphate is known as Williamson synthesis of ethers.

2SNR ONa R'L R O R' NaL

2 2L X, SO R'' , O SO 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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CH3Br Na O C

CH3

CH3

CH3 C

CH3

CH3

CH3OCH3

Sodium ter. butoxide ter. butyl methyl ether

Illustration 9:

Write the product

(i)

C

CH3

CH3

O NaCH3CH3Cl

(A)

(ii)

C

CH3

CH3

BrCH3C2H5ONa

B C

Solution:

(i)

C

CH3

CH3

OCH3 CH3

(ii)

C

CH3

CH2

CH3 C2H5Br

Exercise 9:

Find product

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(i)

Br

CH2OH NaOH

p XyleneX

(ii)

20

NaOH / H O

25 CY

OH

Cl

(3) From Alkane

(a)

R CH CH22

4

i Hg OAC /R'OH

ii NaBH /OHR CH CH3

OR'

C

CH3

CH3

CH232

4

i Hg OAC / CH OH

ii NaBH / OHCH3 C

OCH 3

CH3

CH3

(b)

CH3 C

CH3

CH2 H OCH 32 4H SO

CH3 C

CH3

O CH3

CH3

(4) From Grignard reagent

Higher ethers can be prepared by treating - halo ethers with suitable

reagents.

CH3 O CH2Cl CH3MgI Dry ether CH3 O CH2CH3 MgCl

I

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(5) From Alkyl halide

2dry

2RI Ag O R O R 2AgI

PROPERTIES OF ETHERS

Dipole nature of ether

Ethers have a tetrahedral geometry i.e. oxygen is sp3 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.

O

R

R

O

R

R

net μ1100C

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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R O R

δ δ

Conc. H2SO4

1 mole warm

Conc. H2SO4

2 mole warm

H O H / Δ / h igh pressure

H

HI ( 1 mole)

Cold

2HI

Cold

3 mole HI/Red P

Δ

R C Cl

O

δ

δ/Anhy ZnCl 2

R' C O

O

C R'

O

5PCl /ΔR Cl R Cl POCl 3

R O C

O

R' R O C

O

R'

R Cl R O C

O

R

R H R H

R I R I

R I

2R O H

R OH

R O SO3H R O SO3H

R OH R O SO3H

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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2°R + 3°R More sterically hindered SN1

1°R + 3°R

Nature of mechanism decoded by

nature of solvent.

Aprotic orNon polar

Proticpolar

SN2 SN1

Methylcation is stabler than phenylcations

(B) Dehydration with H2SO4/ and Anhy Al2O3/

(i) When both alkyl groups has -hydrogen.

CH3 CH2 O CH CH2 CH3

CH3

α β

2 4Conc. H SO /Δ

CH2 CH2 H3C CH CH CH3 H2O

(ii) When only alkyl group has - hydrogen.

CH3 C

CH3

CH3

O CH32 4Conc. H SO

ΔCH2 C

CH3

CH3

CH3OHβ α

α

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Hot conc. H2SO4 react with secondary and tertiary ethers to

give a mixture of alcohols and alkenes.

(CH3)3CO—CH3 42Conc.H SO

hot (CH3)2—C=CH2 + CH3OH

(CH3)3CO—

hot

SOH.conc 42 (CH3)2—C=CH2 + HO

Cyclohexyl tert butyl ether isobutene

cyclohexanol

(C) Miscellaneous reactions

(1) Halogenation:

Monohalogenation takes place at carbon (with small amount)

(a)

CH2 O CH2 CH3CH3

Excess

2Cl /hν

CH3 CH O CH2

Cl

CH3

(b) CH2 O CH2 CH3CH3

Small

2Cl excess /hνCl 5C2 O C2Cl 5

(2) Reaction with CO: give ester

0

3

125 180 C

200 atm, BFR O R CO RCOOR

Illustration 10:

Explain why sometimes explosion occurs while distilling ethers.

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Solution:

It is due to formation of peroxide

CH3CH2—O—CH2CH3 + O2 h

CH3—CH—O—CH2CH3

OOH

Illustration 11:

The basicity of the ethers towards BF3 has the following order, explain.

O

> (CH3)2O >(C2H5)2O >[(CH3)2CH]2O

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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CH2BrKF

CH2F

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.

OO

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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H3CO C2H5

HBr

Br C2H5

3CH OH

(A)

H3CO

HBr

OH

3CH Br

(B)

PHENOL

These are organic compounds a hydroxyl group attached directly to a

benzene ring.

OH

Phenol or carbolic acid

OH

CH3

(o , p , m)

Cresol

Preparation

Industrial Method

(i) From chloro benzene (Dow’s process)

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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

NaOH/ 623K

300 atm p

ONa

H

OH

(ii) Cumene Process

Cumene obtained from propene & benzene cumene on air oxidation followed

by acidification with H2SO4 gives phenol & acetone.

0

3 4

250 C

H PO

HC

CH3

CH3

H3C CH CH2

Cumene

O2

95 - 1350

C O OH

CH3

CH3

CH3 C CH3

O

OH

Cumene hydroperoxide

H, H 2O 50-900C

(iii) From benzene sulphonic acid

It is fused with NaOH gives sodium salt of phenol.

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2H O /HNaOH

6 5 3 6 5 3 6 5 6 5FusionC H SO H C H SO Na C H ONa C H OH

(iv) From benzene diazonium chloride

This gives Ar SN1 reaction with H2O to form phenol.

N N Cl

2H O/H

Δ

OH

N2

Illustration 14:

Starting from 1-methyl cyclohexene, prepare the following:

(a) CH3

H

OH

OH

(b) CH3

OH

Solution:

(a) CH3

H

OH

OH

CH3

4

2

i KMnO

ii H O

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(b) CH3

OH

CH3

2H O/H

Exercise 12:

Starting from 1-methyl cyclohexene, prepare the following:

(a) CH3

H

OH

OH

(b) CH3

OH

OH

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:

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O O O O

(I)(II) (III)

O

(V)(IV)

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

NO2

(a)

OH

CH3

OH OH

, , ,

(b)

OH

NO2

OH

NO2

OH

NO2

NO2

OH

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 H R O H

O H O

H

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Reactions of phenol due to breaking of O H bond are given

below:

O H

Alcoholic

FeCl3

NaOH

(CH3)2SO4

NaOH

O Fe

3

blue or violet colour (test for phenolic group).

ONa

dry ether

CH2N2

O CH3

H2O

Anisole

O CH3

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Acylation

OH

R C Cl

O

or

R C O

O

C R

O

O C R

O

/ Pyridine

Phenyl esterpyridine

Fries rearrangement

Phenolic esters are converted in to o– and p–hydroxy ketones in the

presence of anhydrous AlCl3. Generally low temperature favours the

formation of p–isomer and higher temperature favour the o-isomer.

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O C CH3

O1600C

600C

OH

COCH 3

OH

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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OH

NH2

NH3/Anhy ZnCl2

3000C

Cl

SH

PCl / Δ5

2P S / Δ

5

O P

3

(C) EAS in Phenol:

It is strong activating group.

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OH

OH

NO 220 % HNO3

250C

OH

Br Br

Br

3Conc. HNO

2 2Br /H O

2 4H SO / Δ

2 3or Br / CH COOH

2 2Br / CS

2 4or Br / CCl

OH

SO3H

2 4H SO

Δ

OH

R3R X Anhy AlCl

Δ

OH

SO3H

OH

Br

OH

Br

OH

O2N NO 2

NO 2

2, 4, 6, tri nitro phenol

OH

NO 2

o-nitro phenol

p-nitro phenol

(picric acid)

2, 4, 6, tri bromo phenol

p-bromo phenol

o-hydroxy benzene sulphonic acid

p-hydroxy benzene sulphonic acid(Major)

o-alkyl phenol

o-bromo phenol

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OH

OH

NO

HNO2

OH

CHO

3i CHCl / alc KOH

2ii H O /H

4ii CCl / alc KOH Δ

2ii H O /H

OH

COOH2i CO /NaOH

ii H

OH

CH2OH

2CH O

OH

COOH

O

N OH

OH

R

2 4ROH/H SO

Coupling reaction OH mild

ReimerTiemann reaction

ReimerTiemann reaction

0120 C, 7 atm

OH

CH2OH

OH

COOH

Major

Salicylic acid

Major product

OH

CHO

N NCl

OHN N

p hydroxy azo benzene Azodye

p nitroso phenol

Kolbe 's reaction

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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

3 2Hg OCOCH

Reflux

OH

HgOCOCH 3

Aq. NaCl

OH

HgCl

OH

I

I2/CHCl 3

Miscellaneous reaction

(i) Reaction with Zn dust

OH

Zn Δ ZnO

Dust

(ii) Oxidation

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OH

O

O

p - benzoquinone

(brow n in colour)

OH

OH

K2S2O8/OH

Elbs oxidationp - quinol

h ν

(iii) Condensation with phthalic anhydride

C

C

O

O

O

2 4

2

Conc. H SO

Δ H O

C

C

O

O

OHOH

OH

H

OH

H

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

3i CHCl , OH

ii H

CHO

OH

The electrophile is the dichloro carbene: CCl2, formation of carbene is an

example of -elimination.

(i)

OH H C Cl

Cl

Cl

-HClCCl 2

(ii) OH

OH

O

2CCl

Ar SE

O

CCl 2

H

O

CHCl 2

o (dichloro methyl) Phenoxide ion

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(iii) O

HC

Cl

Cl

OH

O

C

H

O

Cl

H

O

C

O

HH

OH

CHO

2. Kolbe’s reaction

OH

NaOH CO2

0i 120 C, 7 atm

ii H

OH

COOH

Mechanism

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OH O Na

NaOH

C O

O

(More reactive for Ar SE)

O

C

O

O Na

H

OH

C O Na

O

OH

COOH

Salicylic acid

H2O/H

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

02i NaOH/ CO , 120 C

4 7 atm ii H

OH

COOH3 2

CH CO O

H

O

COOH

COCH 3

Aspirin

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(ii) OH

COOH

OH

CO O

OH

H

Salol

OH


7 atm P ii H

(iii) OH OH

COOH3CH OH

H

OH

COOCH 3

Oil of w inter green


7 atmP ii H

(iv) OH OH

COOHZn dust

COOH02i NaOH/ CO , 120 C

7 atm P ii H

Illustration 16:

What product would you expect in the following reaction? Explain.

OH

CH3

3CHCl , KOH?

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Solution:

3 2CHCl aq. KOH : CCl

O

CH3

2CCl

O

CH3CCl 2

2H O

O

CH3CHCl 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 FeCl3 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

O

H

OH

CH3

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.

Exercise 17:

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How will you effect the following conversion?

(a) OH O

(b) C2H5OH CH3CH2CH2OH

Exercise 18:

How will you effect the following conversion?

(a) OH

(b) CH CH2 CH2CH2OH

(c) OH

CH3 CH3 (d) OH

OH

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Exercise 19:

Write mechanism

CH2OH

H Cyclohexene

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.

3CuO ZnO CrO

2 3573 673K, 200 300KCO 2H CH OH

Uses:

It is largely used as:

(a) a solvent for paints, varnishes and celluloids.

(b) for manufacturing of formaldehyde.

(c) for denaturing ethyl alcohol, i.e. to make it unfit for drinking

purpose. Denatured alcohols is called methylated spirit.

(d) in manufacture of perfumes and drugs.

(ii) 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.

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Uses:

It is largely used as an

(a) antiseptic.

(b) 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) As an important beverages.

(e) 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) As an antiseptic and disinfectant in soaps and lotions.

(ii) 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

dyes.

(v) As a preservative for ink.

Ethylene Glycol: Ethane 1, 2 diol

Preparation: Lab preparation by hydroxylation.

(i) CH2

CH2

34 2

Cold dilute alkaline

2KMnO 4H O

CH2OH

CH2OH

22MnO 2KOH3

Manufacture

CH2

CH2

2O / Ag

575K

H2C

H2CO

2H O/473K

hydrolysis

CH2OH

CH2OH

Ethylene epoxideor 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

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Reaction with sodium

(i)

2

Na, 323K

1H2

CH2ONa

CH2OH

CH2OH

CH2OHMonosodium glycolate

2

Na, 433K

1H2

CH2ONa

CH2ONa

Disodium glycolate

(ii) 52PCl

CH2Cl

CH2Cl

CH2OH

CH2OH

32POCl 2HCl

(iii) 22SOCl

CH2Cl

CH2Cl

CH2OH

CH2OH

22SO 2HCl

Ethylene dichloride

(iv) 2

HCl, 433K

H O

CH2Cl

CH2OH

CH2OH

CH2OH

Ethylene chlorohydrin

2

HCl, 473K

H O

CH2Cl

CH2Cl

Ethylene dichloride

(v)

p toulenesulphonic acid

RCH O

H2C

H2C

CHRCH2OH

CH2OH

2H O

Cyclic acid

(vi) Oxidation: Ethyelene glycol upon oxidation gives different

products with different oxidising agents. For example.

(a) O

CH2OH

CH2OH

CHO

CH2OH

O

COOH

CH2OH

Glycolaldehyde Glycollic acid

CHO

CHO

COOH

CHOGlyoxal

O

Glyoxalic acid

O

COOH

COOH

Oxalic acid

(b) With periodic acid HIO4 or lead tetra acetate.

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4HIO

CH2OH

CH2OHIodic acid

2 3H O HIOHCHO

HCHO

Also called malapride reaction.

3 4

2 CH COO Pb

CH2OH

CH2OH

3 3 22CH COOH CH COO Pb

HCHO

HCHO

Dehydration

(a) 773K

H2C

H2C

OCH2OH

CH2OH

2H O

(b) 2

2

ZnCl

H O

CH2

CHOH

CH2OH

CH2OH

Vinyl alcohol

Tautomerize

CH3

CHO

(c) 2 4conc. H SO

distill

CH2

CH2

O

CH2

CH2

O

CH2OH

CH2OH

2 22H O

1, 4 dioxane

(d) 3 4conc. H PO

distill

CH2

CH2

O

CH2OH

CH2OHCH2OH

CH2OH

2 2H O

Diethylene glycol

Uses

(a) As a solvent.

(b) Antifreeze in the radiators of cars and aeroplanes.

(c) In manufacture of terylene and other polyester.

Exercise 20:

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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.

H2C

HC

H2C

O

OCOR 2

OCOR 3

COR 1

3NaOH

CH2OH

CHOH

CH2OH

R1COONa

R1COONa

R1COONa

Sodium salt of fatty acid (soap)

(ii) From Propylene

2 32

2 2

aq.Na COCl , 773K

3 2 2 2 2 2HCl 423K, 12 atmAllyl chloride Allyl alcohol

HOCl Cl OH aq. NaOH

2 2 2 2trans Cl H O NaClglycerol

glycerol m

CH CH CH Cl CH CH CH HO CH CH CH

HOCH CHOH CH Cl CH OH CHOH CH OH

onochlorohydrin

(iii) Synthesis from its elements

3 3

2 4

Na in liq. NH CH IElectric ore

2 Berthelets synthesis 196 K NaI

H /Pd BaSO

3 3 2Lindlar 's catalystPr opyne Pr opylene

2C H CH CH Na C CH

H C C CH CH CH CH

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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.

CH2OH

CHOH

CH2OH

2

Na, room temperature

1H2

CH2ONa

CHOH

CH2OH

Na, room temperature

CH2ONa

CHOH

CH2ONa

monosodium glycerolate , 'disodium glycerolate

(ii)

CH2OH

CHOH

CH2OH

CH2Cl

CHOH

CH2OH

CH2OH

CHCl

CH2OH

CH2Cl

CHCl

CH2OH

CH2Cl

CHOH

CH2Cl

HClgas

2383K, H O 2gas, 383 K, 2H O

Excess of HCl

Glycerol

monochlorohydrin

Glycerol

monochlorohydrin

Glycerol

, dichlorohydrin

Glycerol

, dichlorohydrin

Excess of dry HCl gas, 383 K

2H O

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To replace the third hydroxyl group in either of two dichlorohydrins, PCl5 or

PCl3 is fused.

(iii) CH2OH

CHOH

CH2OH

23H O3HI

CH2I

CHI

CH2I

Unstable

2I

CH2

CH

CH2I

HI

Markonikof 's addition

CH3

CHI

CH2I

1, 2 diiodopropane

2I

CH3

CH

CH2

HI

CH3

CHI

CH3

Isopropyl iodide

3. Reaction with concentrated nitric acid:

H2C

HC

H2C

OH

OH

OH

HO

HO

HO

NO2

NO2

NO2

Glycerol

3 2 4Conc HNO Conc. H SO

283 298K

H2C

HC

H2C

O

O

O

NO2

NO2

NO2

3H2O

Glyceryl trinitrate Noble'soil

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

C

C

C

H

H

OH

H OH

H

OH

Glycerol

4

2

KHSO , 473 503K

2H OC

CH2

CHOH

(unstable)

Tautomerises CH

CH2

CHO

Acrolein

or Prop- 2-en-1-al

5. Oxidation.

CH2OH

CHOH

CH2OH

Glycerol

[O]

CHO

CHOH

CH2OH

Glyceraldehyde

[O]

COOH

CHOH

CH2OH

Glyceric acid

[O]

CH2OH

C

CH2OH

O

Dihydroxyacetone

[O]

COOH

CO

COOHMesoxalic acid

COOH

CHOH

COOH

Tartromic 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 (H2O2 + FeSO4) give a mixture

of glyceraldehyde and dihydroxyacetone. The mixture is called

glycerose.

(v) With periodic (HIO4) acid.

CH2OH

CHOH

CH2OH

42HIO

Glycerol

HCHOFormaldehyde

HCOOHFormic acid

HCHO

Formaldehyde

3 2Iodic acid

2HIO H O

(vi) With acidified potassium permanganate.

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CH2OH

CHOH

CH2OH

4

Acidified

KMnOO

Glycerol

COOH

COOH

Oxalic acid

2 2CO 3H O

6. Reaction with phosphorous halides.

CH2OH

CHOH

CH2OH

53PCl

Glycerol 1, 2, 3 - trichloropropane

CH2Cl

CHCl

CH2Cl

(Glyceryl trichloride)

33HCl 3POCl

7. Reaction 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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CH2O.COCH 3

CHOH

CH2OH

Glycerol monoacetate

CH2O.COCH 3

CHOH

CH2O.COCH 3

Glycerol diacetate

CH2O.COCH 3

CHO.COCH 3

CH2O.COCH 3

Glycerol triacetate

8. Acetylation. When treated with acetyl chloride, glycerol forms

glycerol triacetate.

CH2OH

CHOH

CH2OH

33CH COCl

Glycerol Glycerol triacetate

CH2OCOCH 3

CHOCOCH 3

CH2OCOCH 3

3HCl

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9. Reaction with oxalic acid

(i) H2C

CHOH

CH2OH

OH

Glycerol

H OOC COOHOxalic acid

2

383K

H O

H2C

CHOH

CH2OH

OOC COO H

Glyceryl monoxalate

2

Heat

CO

H2C

CHOH

CH2OH

OOCH

Glyceryl monoformate

HOH, hydrolysis

From water of crystallization

CH2OH

CHOH

CH2OH

Glycerol

HCOOH

Formic acid

(ii) CH2O H

CHO H

CH2OH

2

503K

2H O

Glycerol

CO

CO

HO

HO

Oxalic acid

H2C

HC

CH2OH

OOC

OOC

Glyceryl dioxalate

(Dioxalin)

2

503K

2H O

CH2

CH

CH2OH

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. As a preservative for fruits and other eatables.

8. 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) CH3 OH

HgOAC

HA =

CH3 OH

B =

(ii) A = CH3CH2CH2OH

B = CH3OH

Exercise 2:

(i)

RCO 2Et + OH

byfollowed

3MgBr MgBr

R

OH

RR

-H2O

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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second case you have two moles of RMgX that completes the

reaction because there are two R– for two different sites.

Exercise 3:

CH3

CH CH3

OH

H

CH3

CH CH3 ring

Expansion

CH3

CH3

H

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

OH

O

(A) =

OH

OH

OH

(B) =

Exercise 6:

(a) 6 5 2 6 5 3 6 5 33 2C H CCH OH C H CHOHCH C H C(OH)CH

(b) (i) No reaction

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(ii) 6HCOOH

(iii) RCHO + R CHO + NH3

Exercise 7:

(i)

A = C C

CH3

O

Ph

CH3

Ph

(ii)

A = C C

CH3

O

Ph

CH3

Ph

Here phenyl migration takes

place.

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.

CH3 C

O

CH2CH2CH2OH and CH3 C

O

CH2CH2OH

CH3

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Exercise 9:

X =

O

, w – halo alcohols undergoes

intramolecular Williamson

ether.

Y = O

Synthesis in the presence of NaOH

by NS 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.

O

I 1 2Attack byeither SN or SN

H CH3

(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

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Exercise 12:

(a)

CH3

H

OH

OHCH3

4

3

i OsO

ii NaHSO

(b) CH3

O

CH3

6 5 2C H CO OH 2H O

H

OHCH3

OH

H

Exercise 13:

OH

NO 2

(a)

OH

CH3

OH OH

> > >

(b)

OH

NO2

OH

NO2

OH

NO2

NO2

OH

Cl

> > >

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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

SO3H

23Br

OH

Br

BrBr

2 43HBr H SO

Exercise 15:

(a) 2-Pentanol responds to the Iodoform test

(b) Phenol gives coloration with FeCl3 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 by

resonance.

(f)

Ethanol contains the H C

H

CH3

OH grouping

(g) A tertiary alcohol form a stable tertiary carbonation.

Exercise 17:

(a) OH OH

2H ,

Na2 2 7Na Cr O

H ,

O

(b) 5 2HPCl HNOKCN

2 5 2 5 2 5 2 5 2 2 2 5 2C H OH C H Cl C H CN C H CH NH C H CH OH

Exercise 18:

(a) 2 4

0

H SO

160

SO3H

NaOH

SO3 Na

H

OH

(b) HBr aqueous OH

6 5 2 6 5 2 2 6 5 2 2PeroxideC H CH CH C H CH CH Br C H CH CH OH

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(c) CH3

OH OH

CH3

2H /Ni

2

H ,

H O

CH3

(d) OH

H HOCl

OH

Cl

2

H ,

H O

Cl

aq. KOH

OH

Exercise 19:

CH2OH

2

H ,Δ

H O

CH2

ring expansion H

Exercise 20:

3 4O NaBH

2 6HO CH 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?

(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?

Exercise 3: How is glycerol prepared from fats and oils? Why it is

highly viscous? Give its uses?

Exercise 4: Discuss the following reactions:

(a) Reimer and Tiemann reaction

(b) Kolbe‘s reaction

Exercise 5: Complete the following by putting structures of A, B, C, D.

3 3KOH als.PBr NHHBr

3 2 2CH CH CH OH A B C D

Complete the following.

Exercise 6: Compete the following by drawing correct structures.

SO3H2 4 2

i fuse withNaOHfu min g

H SO ii H O / H.................. ..................

Exercise 7: How the sugar is converted to ethanol?

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Exercise 8: Write the structure of organic compounds A to F in the

following sequence of reactions:

OH

3 32 2 2

2 4

HNO H OBr H / V Pt HNOZn

H SO HClA B C D E F

Exercise 9: Predict the products of the following reactions:

(a)

2 5C H O Na

Br

(b) HBr

3 2 6 55CH CH O C H

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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.

4 2 2CO Co H /Ni

2 2 2 3 2 3 2 2High temp.Ethene Propanal 1 propanoland pressure

CH CH CO H CH CH CHO CH CH CH OH

(b) 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.

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2 3Al O

3 2 2 2 2600KEtheneEthanol

CH CH OH CH CH H O

(b) Diethylether is formed.

2 4Conc. H SO

2 5 2 5 2 5 2413KEthoxy Ethane

2C H OH C H OC H H O

OH

Phenol

3Acetyl chloride

CH COCl

O COCH 3

Phenyl acetate

HCl

(c) 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).

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H2C

HC

H2C

O

O

O

COR

COR'

COR"

3NaOH

H2C

HC

H2C

OH

OH

OH

RCOONa

RCOONa

RCOONa

(Oil and fats) 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

Phenol

4CCl

340K

OH

COONa

2H O/H

OH

COOH

2-Hydroxybenzaldehyde (Salicyldehyde)

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OH

Phenol

4CCl /NaOH

340K

OH

COONa

2H O/H

OH

COOH small amount of 4-hydroxy benzoic acid is also formed

2-Hydroxy benzoic acid (Salicylic acid)

If CCl4 is used in place of CHCl3 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

Phenol

4 7 atm

2 400KCO

OH

COONa

2H O/H

OH

COOH

Salicylic acid(Major product)

Exercise 5:

3KOH alc.PBr

3 2 2 3 2 2 3 21 Propanol 1 Bromopropane Propene B

A

CH CH CH OH CH CH CH Br CH CH CH

HBrCH3 CH

Br

CH3

NH3CH3 CH CH3

NH2

2-Propanamine(D)

2-Bromopropane(C)

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Exercise 6:

SO3H2 4

fuming

H SO

SO3H

SO3H

2

i fuse with NaOH

ii H O/H

OH

OH1, 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.

Invertase Zymase

12 22 11 2 6 12 6 6 12 6 2 5 2Sugar Glucose Frucot se Ethanol

C H O H O C H O C H O 2C H OH 2CO

Exercise 8:

OH

Zn 3

2 4

HNO

H SO

NO 2

2Br

NO 2

Br

Sn /HCl

Phenol Benzene

[A]

Nitrobenzene

[B]m-Nitro

(bromobenzene)[C]

NH2

Brm-Bromo

Aniline[D]

N2Cl

Br

m-BromoBenzenediazonium

[E]

3H O 2HNO

HCl

OH

Brm-Bromophenol

[F]

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Exercise 9:

(a) OC2H5

(b) 3 2 2 6 54CH CH CH Br C H 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 HBF4.

(v) 2 methyl propene is heated with methanol in the presence of dil.

H2SO4.

Solution:

(i) Diethyl ether is formed.

2 30

Al O

3 2 2 3 3 2 2 3 2350 CCH CH OH H O CH CH CH CH O CH CH H O

(ii) C2H5 Br

C2H5 Br

Ag2O H5C2 O C2H5 2AgBr

diethyl ether

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(iii) Ethyl methyl ether is formed.

CH3 O C2H5Mg

Cl

Br

dry

etherH3C O CH2Cl CH3MgBr

(iv) Ethyl methyl ether is formed

4HBF

3 2 2 2 3 2 3 2dim ago methane

CH CH OH CH N CH CH O CH N

(v) Tertiary – butyl methyl ether

CH3 C

CH3

CH2 H O CH3

H2SO4 CH3 C

CH3

CH3

O CH3Methyl 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 C

CH3

CH3

Br Na O C

CH3

CH3

CH3CH3 C

CH3

CH2 CH3 C

CH3

CH3

OH NaBr

30 bromide tert- butoxide

Problem 4:

Identify the product in the following reaction

OH

OH 2

3

ClCH COCl

POClA 3 2CH NH B 4i KMnO / OH

ii HC Zn dust D

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Solution:

A =

OH

COCH 2Cl

OH

B =

OH

COCH 2NHCH 3

OH

C =

OH

COOH

OH

D =

COOH

Problem 5:

Write the various product when ethanol reacts with sulphuric acid in suitable

conditions.

Solution:

C2H5OH 2 4H SO

383KC2H5HSO 4

C2H5OC2H5

CH2 CH2

H2SO4/ 413 K

H2SO4/ 443 K

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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

OH

CH2 C6H5

(b) OH

C

H

OH

Solution:

(a) C O

H

H

H C

O

CH2 C6H5

(b)

O

CH O

Problem 7:

Effect the following conversion

(a) CH2OH CH2OH

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(b) CH3OH

CH2CH2OH

Solution:

(a) CH2OH CH2

H ring

Expansion

H2O

OHK2Cr2O7

O

Ph 3P CH2

CH2

HBO

CH2OH

(b)

CH3OH 5PClCH3Cl

Anhy AlCl 3

CH3

NBS

CH2Br

Mg dry ether

CH2MgBr

3

HCHO

H O

CH2CH2OH

Problem 8:

Postulate a mechanism for the following reactions:

CH3

OH

H

O

Solution:

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OH

OH

H

-H2OOH

ring expansion

O H

-H

O

Problem 9:

Conversion

(a) H

OH

H

D

(b) CH2 CH2 O O

Solution:

(a) H

OH

H

Br

PBr 3

Pyridine

Mg/dry ether

H

MgBr

D2O

H

D

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(b) CH2 CH2

MnO 4

KOHHOH 2C CH2OH

H

-H2OOH CH2

CH2

O

CH2CH2OH

2H , Δ, H OIntrmoleculadehydration

O O

Problem 10:

Identify the products in the following reaction sequence.

CH3

OH

3i CHCl

ii KOH, HA 4LiAlH

B 3CH COOH

HC

Solution:

A =

CH3

OH

CHO

B =

CH3

OH

CH2OH

C =

CH3

OH

CH2 O C CH3

O

Problem 11:

Distinguish between each pair of compounds by a simple test.

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(a) (CH3)2CHOH and (CH3)2CHSH

(b) CH3CH2CH2OH and (CH3)2CHOH

Solution:

(a) The thiol gives a precipitate with heavy metal cations like

2 2 2Hg , Pb and Cu

(b) (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

H3O

O

Solution:

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OH

OH

O

O H

O

H

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 I2/Na2CO3 followed by

acidification. Identify the compound A, B, C and D. Give the reasons.

Sol.

A = O

CH3

CH3

C CH3

CH3

H

B =

CH3OH

CH3

C = CHI3 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

N O

H

O

o - nitro phenol

O

C O

H

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) OH

CHO

(B) OH

CHO

(C) CHOOH

(D) OH

COOH

Sol.

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(A) Salicyldehyde

Problem 3:

CH3 C CH2

CH3

O

182H O

HA, A is

(A)

CH3 C

CH3

OH

CH2

OH18

(B)

CH3 C

CH3

OH

CH2

OH18

(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

CH3 CH23

2 2

i BH / THF

ii H O ,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

3CH INaOH NaOH

2 5 aqC H Br A B C will be

(A) propane (B) ethyl iodide

(C) butanol (D) methoxy ethane

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Solution:

3Na/CH INaOH

2 5 2 5 2 5 3aq William son's synthesisC H Br C H OH C H OCH

(D)

Problem 7:

Phenol can be distinguished from alcohol with

(A) Tollen‘s reagent (B) Schiff‘s base

(C) Neutral FeCl3 (D) HCl

Solution:

Phenol gives violet colour with neutral FeCl3.

(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:

HI

3 3 3 33 3CH C O CH CH OH CH C I

(A)

Problem 9:

Which of the following is the strongest acid?

(A) OH

NO2

(B) OH Cl

(C) NO2OH

(D) OH

NO2

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. H2SO4.

The product is

(A) Phenolphthalein (B) bakelite

(C) salicylic acid (D) fluorescein

Solution:

Phenolphthalein

(A)

Problem 11:

An organic compound X of molecular formula C4H10O undergoes

oxidation to give a compound Y of molecular formula C4H8O2. 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 ZnO-

Cr2O3.

(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. ZnCl2

(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 (CH2N2)

(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 ZnCl2

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.

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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) 3 2PBr ClNa

3 3Ether heatX Y CH CH Z

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(b)

X 2H O/HCH3 CH CH3

OH

Cu/ 573KY

(i) C2H

5MgI

(ii) H2O/H+

Z

6. How will you convert phenol to salol?

7. Name the reagents used in the following reactions:

(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.

8. 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) Pentan–1–ol, n–butane, pentanal, ethoxyethane.

9. 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.

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(b) Would you get any di-tert-butyl ether from this reaction?

Explain.

11. How the following conversion carries out?

(i) Propane Propan–2–ol

(ii) benzyl chloride benzyl alcohol

(iii) Ethyl magnesium chloride propanol–1

(iv) Methanol to Ethanol

12. Identify the major products in the following reactions.

(i) NO 2

OCH 3

3

2 4

HNO

H SOA

(ii) OH

CH3

2Br

waterB

13. Give one test to distinguish

(a) Ethanol and methanol

(b) Phenol & Benzyl alcohol

14. Compound (A) C7H8O is insoluble in water and dilute NaHCO3 but

dissolve in dilute NaOH solution. On treatment with Br2 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.

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Level – I

1. Write the product of the following reactions

CH3

3i BH

ii OxidationA

2. Write the product A & B

CH CH CHO

A

B

H2/Pt

NaBH 4

3. Write the product A & B

CH3

4

Cold

alkaline KMnOA 2 3Cr O

ACOHB

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

to

CH3

OH

Write mechanism of above conversion.

7. Identify compounds A – D in the following reaction.

CHC 6H5

3

2

i O

ii Zn/H OA B

dil. NaOH

2aq. ethanolA B C H O

3

2

i O

ii Zn/H OC D A

2H /PtD

OH

OH

8. Give mechanism of the following reaction

4

3

i CCl NaOH/Δ

ii H O

OH

COOH

OH

9. Identify the product in the given reaction

1 equiv

H CH3

BH3/THFA

H2O2 OHB

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10.

O

Conc. H2SO4

ΔX

H2OY

Identify X and Y.

11. Identify A & B

N

H

CHCl 3OH

A B

Pyrole

12. Give a simple test that distinguish between the compounds.

(a) Allyl and n – propyl alcohol.

(b) Benzyl methyl ether and benzyl alcohol.

13. 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

OH

Ph

OH

O C

Me

C

Ph

Me

Ph

(b) OH

OH

O

3. Give products

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CH3 C

CH3

CH3

O CH3

Anhydrous HI

ether I

Conc. HI

( 2 )

A B

C D

4.

CH(OH)Me

Me

HA [ C9H16]

A on ozonolysis gives nonane – 2, 8 dione. What is A and how is it

formed?

5. An organic compounds (P) having molecular formula C5H10O 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

NO 2

i OH

ii Et BrA

Zn /HClB

20

NaNO

HCl, 50 CC

C6H5OH

DLiAlH 4

(F)(E)

Soluble in NaOH

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7. (a) Diphenyl ether is not very easily prepared using Williamsons

sysnthsis. Provide an useful synthetic route for the same.

(b) 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 CCl4. p–bromo phenol is obtained. Explain.

8. Write mechanistic step of the following reaction?

CH2OH

H

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 AlCl3 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 C5H12O. 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)

O CH3

1mole

HI (A)

(b)

O CH3

Excess HI

Δ(B)

14. Find product

(a)

CH3 C

CH3

CH2 MeOHH2SO4

(A)

(b) OH

H(B)

OH

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(c)

RCOOEt (H2C)5

MgBr

MgBr

H3OC

H2/NiD

(d)

C

CH2OH

CH2OH

O 2IO 4 E F

15. (Z) C6H14O 2 4Δ /H SO(A) C6H12

HClB C

C6H13Cl

Balc. KOH

D (Isomer of A)

DO3/H2O

E

(Given negative test with Fehling solution

but responds to iodoform)

AO3/H2O

F G

(both give positive Tollen‘s test but do not

give iodoform test)

Conc. NaOHF G HCOONa a primary alcohol

Identify A to G and Z.

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Objective:

Level–I

1. 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

2. 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

3. A mixture of benzoic acid and phenol may be separate by treatment

with

(A) NaHCO3 (B) NaOH

(C) NH3 solution (D) KOH

4. In the lucas test of alcohols, appearance of cloudiness is due to the

formation of

(A) aldehyde (B) ketone

(C) acid chloride (D) alkyl chloride

5. The dehydration of 1 – butanol gives

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(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

6. Ethyl alcohol is obtained when ethyl chloride is boiled with

(A) alc. KOH (B) aq. KOH

(C) AlCl3 (D) H2O2

7. The number of methoxy groups in a compound can be determined by

treating with

(A) Na2CO3 (B) NaOH

(C) HI and AgNO3 (D) CH3COOH

8. Diethyl ether absorbs oxygen to form

(A) red coloured sweet smelling compound

(B) CH3COOH

(C) ether peroxide

(D) ether suboxide

9. 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. Order of reactivity of HX towards ROH is

(A) HI > HBr > HCl (B) HBr > HI > HCl

(C) HCl > HI > HBr (D) HI > HCl > HBr

11. 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

12. Ethyl iodide reacts with moist silver oxide to produce

(A) ethane (B) propane

(C) ethyl alcohol (D) diethyl ether

13. Reaction of tertiary butyl alcohol with hot Cu at 350 C produces

(A) butanol (B) butanal

(C) 2 – butene (D) 2 methyl propene

14. 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 H2SO4 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 – 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

2. The order of reactivity of phenyl magnesium bromide with the

following compound is

O

Me Me

(I)

O

Me H

(II)

O

Ph Ph

(III)

(A) II > III > I (B) I > III > II

(C) II > I > III (D) all react with the same rate

3. Which one is the stronger base?

(A) CH3CH2O (B) CF3CH2O

(C) both of equal strength (D) can not say

4. The acidic character of 1 , 2 , 3 alcohols H2O and RC CH is in the

order

(A) H2O > 1 > 2 > 3 > RC CH

(B) RC CH > 3 > 2 > 1 > H2O

(C) 1 > 2 > 3 > H2O > RC CH

(D) 3 > 2 > 1 > H2O > RC CH

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5. Choose the correct statement (s) for the reaction

OCOCH 33 2AlCl / CS

heatCOCH 3

OHOH

COCH 3

(a)

(b)

(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

6. Predict the major product

O

Excess HI

(A) HO CH2 CH2 CH2 CH2 I

(B) HO CH2 CH2 CH2 CH2 OH

(C) I CH2 CH2 CH2 CH2 I

(D) no reaction

7. Dipole moment of CH3CH2CH3, CH3CH2OH and CH3CH2F is in order

(I) (II) (III)

(A) I < II < III (B) I > II > III

(C) I < III < II (D) III < I < II

8. 3 – methyl – 3- hexanol can be prepared by

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(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

9. 3H O

4 8 2 3 4 102 parts

Ester (A)C H O CH MgBr B C H O

Alcohol B reacts fastest with Lucas reagent. Hence A and B are

(A)

CH3 C

O

O C2H5 ,(CH3)3COH

(B)

H C

O

O C3H7 ,(CH3)2CHOH

(C)

CH3 C

O

O C2H5 ,(CH3)2CHOH

(D)

H C

O

O C3H7 ,(CH3)3COH

10. Vinyl carbinol is

(A) HO CH2 CH = CH2 (B) CH3C(OH) = CH2

(C) CH3 CH = CHOH (D) 3 2CH CH CH OH

11. The reaction of elemental sulphur with Grignard reagent followed by

acidification leads to the formation of

(A) mercaptan (B) sulphoxide

(C) thio ether (D) sulphonic acid

12. 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

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13. For the preparation of tert-butyl methyl ether by Williamson‘s method

the correct choice of reagents is

(A) methoxide and tert – butyl bromide

(B) methanol and 2 – bromobutane

(C) 2 – butanol and methyl bromide

(D) tert-butoxide and methyl bromide

14. Allyl alcohol is obtained when glycerol reacts with the following at

260 C

(A) formic acid (B) oxalic acid

(C) both (D) none

15. 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.

O CH3

H

step 1

O CH3

H

I

step 2

OH

CH3I

Phenyl carbocation is highly unstable and bond between C6H5–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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O

R

R

1100C Net dipole moment

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

N

H

O

O

4.

CH3 C

CH3

CH3

CH

OH

CH3

HCH3 C

CH3

CH3

CH

OH2

CH3

-H2O

CH3 C

CH3

CH3

CH CH3

1, 2 methyl shift

CH3 C

CH3

CH CH3

CH3

-HCH3 C

CH3

C CH3

CH3

tetra methyl ethylene

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5. (a) X = CH3OH, Y = CH3Br, Z = CH3CH2Cl

(b)

X = CH3CH = CH2, Y = CH3COCH3, Z =

CH3 C

C2H5

CH3

OH

6. OH

2

3

Kolbe 's reaction

i CO 400K, OH

ii H O

OH

COOH

OH

H

OH

COO

Phenyl salicylate (salol)

7. (i) Alkaline KMnO4

(ii) Br2 water

(iii) P.C.C

(iv) 1% alkaline KMnO4

8. (a) Methanol < ethanol < propan–1–ol < butan–2–ol < butan–1–ol

< pentan–1–ol.

(b) n–butane < ethoxy ethane < pentanal < pentan–1–ol.

9. 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.

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12. (i)

A =

O2N

NO 2

OCH 3

(ii)

A =

Br

OH

Br

CH3

13. (a) Iodoform test given by ethanol.

(b) Neutral FeCl3 test given by phenol.

14. OH

CH3

m - Cresol

15. CHCl 3

OHCCl 3

ClCCl 2

OH

OH

O

CCl 2

O

H

CCl 2

O

CHCl 2

O

CHO OH2

H2O

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Level – I

1.

HCH3

OH

H

2. A = CH2CH2CH2CHO B = CH CH CH2OH

3.

A =

CH3

OH

OH

B =

CH3

OH

O

4. NO 2

3

2 4

funing HNO

Conc. H SO , Δ

NO 2

NO 2

4 2NH S

NH2

NO 2

NaNO 2/H2SO4

0 - 50C

N2 HSO 4

NO 2

2H O

Δ

OH

NO 2

5.

CH3 CH

OH

CH2 CH2 OHA =

CH3 C

O

CH2 COOHB =

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6. OH

Hring

expansion

CH3

(Hydride shift)

CH3

OH

CH3

OH

-H2O

7.

C O

H

A =

O

B =

C

H

O

C =

O

O

D =

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8. OH

OH

O

CCl 3

Cl

O

CCl 3

H

O

CCl 3

OH

O

COOH

OH

O

CO 2

OHH3O

OH

COOH

9.

A = BH2

H

H

CH3

B = OH

H

H

CH3

10. OH O SO3H(X) =

OH OH(Y) =

11.

N

H

CHO

A = B =

N

Cl

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

Zn

dust

Cl

O

AlCl 3

CH3

O

CH3

Zn(Hg)/HCl

14.

A =

OH

Phenol

B =

OH

COOH

Salicylic acid

C =

OH

CH2OH

C7H8O2

2 hydroxy benzyl alcohol

D =

OCOCH 3

COOH

Aspirin

15. OH

CH3

OH2

CH3

-H2O

H

CH3

CH3

-H

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Level – II

1

. OH

CH3

CH3 H

H C

O

O C

CH3

H

CH3

(B)(A)

2

.

(a)

Me Ph

Me

OH

Ph

OH

H

Me

OH

Ph

Ph

Me

OH

Ph

Me

Ph

O

Me

H

Ph

Me

Ph

O

Me

Ph

Me

C-H

CH3 Ph

(b)

OH

OHH

OH2

OH

H-H2O

OH

O H-H

O

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3.

CH3 C

CH3

CH3

O CH3

Anhydrous HI

ether ICH3 C

CH3

CH3

OHCH3I

Conc. HI

( 2 )CH3OH CH3 C

CH3

CH3

I

4. A Nonane 2, 8 dione (No loss of C)

H2C

CH2

CH2 CH2

CH2

C

C

CH3

CH3

O3/H2OH2C

CH2

CH2 CH2

CH2

C

C

CH3

CH3

O

O

CH(OH)Me

Me

HCH Me

Me

20 carbocation

1, 2 alkyl shift

(ring expansion)

Me

Me

Me

Me

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5.

CH2 O CH3

CH3

(C5H10O)

C

CH3 O CH3

CH3

H

Highly stable carbocation

H2OCH3 O

CH3

(Q)

C2H5OH

(R)

C O

CH3

CH3

CH3

Stability by resonacne

6

.

A =

OEt

NO 2

B =

OEt

NH3 Cl

C =

OEt

N2Cl

D = N N

OEt

OH

F =

OH

NH2

E =

OEt

NH2

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7. (a) OH

NaOH

O

X

O

O

Se/Δ

(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.

CH2OH

2

H

H O

CH2

H

H

H

ring expansion

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9. (A) is secondary alcohol.

CnH2n+1OH = Molecular weight

Molecular weight of (A) = 6 11200

60 gm A1120

CnH2n+1OH = 60 (n = 3)

(A) C3H7OH

CH3 CH CH3

OH

CH3CH(OH)CH 3

(A)

PBr 3CH3CHBrCH 3

(B) Anhy AlCl 3

HC

CH3

CH3

(C)

10. OH

3

O

CrO

O

3 3Ph P CH Cl

t BuLi

CH2

2 2C H N , Δ

11. CHCH 3

OHA =

C CH3

O

B =

C CHBr 2

O

C =

CH COOH

OH

D =

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12.

Br C

CH2

CH3

CH3

CH3

(B)

Hydride shift

Methyl shift

SN' displacement

CH3 CHCH(CH3)2

OH

(X)

I2/OH(CH3)2CHCOOH

(CH3)2CCH2OH

(Y)

(CH3)3C.CH2OH

(A)

(CH3)2.C(OH)CH2CH3

Oxidation

OxidNo Reaction

(Z)

13.

O CH3

1 mole HI

O CH3

H

OH CH3

I

OH CH3

I

(A)

(a)

O CH3 OH CH3

IH

OH2 CH3

I

SN2

I

I CH3

I

H2O

(B)

(b)HI

14. (a)

CH3 C

CH3

CH3

OCH 3

(b) CH3

C CH3

O

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(c) R

(d)

R

E = 2HCHO + F = CO2

15. Z =

H2COH

D =

A =

E = CH3 C CH3

O

B =

CH3

CH3CH3

Cl CH3

F =

O

H

C =

Cl

G = HCHO

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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

LEVEL - II

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

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