Alcohol, Phenol & Ether_Chemistry

August 8, 2017 | Author: Yoshitha Kuntumalla | Category: Ether, Ester, Alcohol, Aldehyde, Ethanol
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AISM-09/C/APE

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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)

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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)



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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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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OH

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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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..................

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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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H3O

F

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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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H2O

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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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Zn dust

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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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. [email protected] Tel: +91-120-4224242, +91-120-4224248

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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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