ncert chemistry xii 11

11 Alcohols, Phenols and Ethers

Intext QuestionsIntext Questions

Question 1. Classify the following as primary, secondary and tertiary

alcohols.

(i) CH C

CH

CH

CH OH23

3

3

(ii) H C CH CH OH2 2==

(iii) CH CH CH OH3 2 2 (iv)

(v) (vi)

If CH OH2 group is present, alcohol is 1° (primary), if CHOH group ispresent is 2° (secondary) and if COH group is present, alcohol is 3°. Soclassify on this basis.

Solution. (i) Primary alcohol (ii) Primary alcohol

(iii) Primary alcohol (iv) Secondary alcohol(v) Secondary alcohol(vi) Tertiary alcohol

Question 2. Identify allylic alcohols in the above examples.

Following groups are included in allylic alcohols.

CH CH CH OH2 2Primary (1 )allyl alcohol

== °

CH ==CH C OH2

Secondary (2 )allyl alcohol

°

H

C

CH CH2

Tertiary (3 )allyl alcohol

==

°

C

C

C

OH

Solution. (ii) H C==CH CH OH2 2

(vi)

Alcohols,�Phenols and Ethers 303

CH—CH3

OH

CH —CH—CH2 3

OH

CH CH—C—OH——

——

CH3

CH3

CH CH—C—OH

CH3

CH3

Chemistry-XII Alcohols, Phenols and Ethers

Question 3. Name the following compounds according to the IUPAC

system.

(i) CH CH CH CH

CH OH

CH CH

2

3 2 3

CH Cl2

CH3

(ii) H C CH

CH

CH CH

OH

CH

CH OH

CH3

2

3

2 3

(iii)

(iv) H C==CH CH

OH

CH CH CH2

2 2 3

(v) CH C==C

CH Br

CH OH23

3

Prefix according to the C-chain; for alcohol add suffix —ol; for doublebond —ene; tell the position of substituents, double bond and functionalgroup (s).Then, given the name in the following manner.Name of substituent with position (in alphabetical order) + parentC-chain name + primary suffix (antiene etc.) + secondary suffix withposition (for main functional group).

Solution.

(i) CH CH CH

CH Cl

CH CH

CH

CH

CH OH

5 4 3

2

1

3-chl

3 2

3

3

2

2

oromethyl-2-isopropylpentan - 1-ol

12CH OH

(ii) CH CH CH CH CH

CH OH

6

2,5 dimethylhexan-1,3-

3

5 4

2

3 2

3

di-ol

CH3

304 NCERT Class�XII Chemistry�Solutions

OH

Br


(iii)

(iv) H C==CH CH

OH

CH CH CH2

1 2

Hex-1-en-3-ol

3 4

2

5

2

6

3

(v) CH C==C CH OH

CH Br

4

2

2-bromo-3-methylbut-2-en-

3

3 2 1

3

1-ol

Question 4. Show how are the following alcohols prepared by the

reaction of a suitable Grignard reagent on methanal?

(i) CH CH

CH

CH OH3 2

3

(ii)

HCHO + Mg CH OH Mg(OH)H O2

2R R XX → ++H. Form here, it is clear that

HCHO gives CH OH2 groups, so R of Grignard reagent is the remainingpart of given alcohols. Thus, select the suitable Grignard reagent bysubstituting the value of R.

Solution. (i) HCHO + (CH CH MgBrMethanal

Isopropylmagnesiumbromide

3)2 →Dry ether

[ ) ](CH CH CH OMgBr3 2

Addition compound2 ⋅

(CH CH OH3

Isobutyl alcohol

)2 2CH

or 2-methylpropan-1-ol


3-bromocyclohexanol

OH

Br

12

3

CH OH2

–Mg(Br)OH H O/H2+


Question 5. Write structures of the products of the following reactions.

(i) CH CH==CH3 22

+H O/H →

(ii)

(iii) CH CH CH

CH

CHO3 24NaBH

→

3

(i) Acidic hydration of propene i.e., addition of H O2(according to Markownikoff’s rule) result in the formation of alcohol.

(ii) Reduction of C O== group (not in ester group) into —OH.

(iii) Reduction of —CHO group into — CH OH2 .

Solution.

(i) CH CH== CH3 22

Propene Markownikoff'srule

H O/H →+

CH CH CH1 2 3

3 3

Propan-2-ol

OH


O

CH —C—OCH2 3

O

NaBH4

O

CH —C—OCH2 3

O

NaBH4

Reduction

22 1

1

methyl-2-oxocyclohexylethanoate

(ii)

OH

CH —C—OCH2 3

O

Methyl (2-hydroxycyclohexyl)ethanoate

HCHO +

MgBr

Dry ether

Methanal Cyclohexylmagnesiumbromide

CH OMgBr2

Intermediateproduct

H O/H2

+

CH OH2

Cyclohexylmethanol

–Mg(OH)Br

(ii)


(iii) CH CH C CHO3 2

2 1

2-methylbutanal

Reductio

4 3

3

→H

CH

n

NaBH4

CH CH CH CH OH4 2 1

3 2 2

2-methylbutan-1-ol

3

3CH

Question 6. Give structures of the products you would expect when

each of the following alcohol reacts with

(a) HCl ZnCl2 (b) HBr and

(c) SOCl2(i) Butan-1-ol (ii) 2-methylbutan-2-ol

Replacement of —OH by (a) —Cl (b) —Br (c) Cl

Solution. (i) Reactions of butan-1-ol

(a) CH CH CH CH OH4 3 2 1

3 2 2 2

Butan-1-ol(1 )Heat

→° (Lucas reagent)

HCl ZnCl2−

CH CH CH CH Cl + H O4 2 1

3 2 2 2 21-chlorobutane

3

(b) CH CH CH CH OH3 2 2 2Butan-1-ol

→ +Heat

HBr

1-bromobutanCH CH CH CH Br H O3 2 2 2 2

2 14 3

e

(c) CH CH CH CH OH3

SOCl

Heat →2 2 2

2

CH CH CH CH Cl SO Cl4 3 2 1

1-chlorobutane3 2 2 2 2 + + H

(ii) Reactions of 2-methylbutan-2-ol

(a) CH C

OH

CH

CH CH1 3

2-methylbutan-2-ol (3 )

HCl3

2

3

2

4

3

°

ZnCl

Lucas reagent

2 →

CH C

Cl

CH

CH CH H O1 3

2

2-chloro-2-methylbutan

3

2

3

2

4

3

+

e



(b) CH C

OH

CH

CH CH1 3

2-methylbutan-2-ol

HBr

H3

2

3

2

4

3

→

eat

2-bromo-2-methylbutane

CH C

Br

CH

CH CH1 3

3

2

3

2

4

3

+ H O2

(c) CH C

OH

CH

CH CH CH C

Cl

1 3 1SOCl

Heat

23

2

3

2

4

3 3

2

→

CH

CH CH3

2-chloro-2-methylbutane

3

2

4

3

+ +SO HCl2

Note *Butan-1-ol, being primary alcohol requires heating with Lucasreagent. 2-methyl butan-2-ol being tertiary alcohol reacts without heating.

*Lucas test is used to distinguish between primary, secondary and tertiaryalcohols.

Question 7. Predict the major product of acid catalysed dehydration of

(i) 1-methylcyclohexanol (ii) butan-1-ol

In acid catalysed dehydration process a molecule of water is liberatedaccording to Saytzeff’s rule i.e., results in the formation of moresubstituted alkene.

Solution. (i)

(ii) CH CH CH CH OH

Butan-1-ol

H , heat

H O

+

23 2 2 2 →

−

CH CH CH==CH CH CH==CH CHBut-1-ene But-2-ene(M

3 2 2 3 3 +

inor) (Major)

Note *OH from α-C atom�and H from β-C atom�is�eliminated.(i) According to Saytzeff’s rule, major product is1-methyl-cyclohexene,�since��it�is�more�substituted.


CH3

OH

1-methylcyclohexanol

H (Heat)+

–H O2

CH3

1-methylcyclohexene(Major)

CH2

1-methylenecyclohexane(Minor)

+


(ii) Major product is but-2-ene due to rearrangement to give secondarycarbocation,�which�is�more�stable.

Question 8. Ortho and para nitrophenols are more acidic than phenol.

Draw the resonance structures of the corresponding phenoxide ions.

Acidic strength depends on the relative stabilities of correspondingphenoxide ions based on resonance.

Solution. (i) Phenoxide ion


CH CH CH CH OH3 2 2 2 –H O2

H+CH CH CH CH3 2 2

H

+

shift

H ion–

CH CH CH CH3 3

1º carbocation(Less stable)

+

2º carbocation(More stable)

H

–H+

CH CH3 CH CH 3

But-2-ene

Butan-1-ol

O O O

�

�

�

O O

�

�


(ii) o-nitrophenoxide ion

(iii) p-nitrophenoxide ion

Due to R effect of NO2 group, o- and p-nitrophenoxide ionsare more stable than phenoxide ion. Consequently, both o- andp-nitrophenols are more acidic than phenol.

Note Structures in boxes have negative charge on that C-atom to whichelectron withdrawing NO2 group is attached. So, these structurescontribute�more�towards�the�acidic�character�than�the�others.


O

N

O

O O

N

O

O�

� � �

O

N

O

O

�

�

O

N

O

O

�

�

O

N

O

O+

�

�

O

N

O O

O

N

O O

O

N

O O

�

�

�

� � �

O

N

O O

O

N

O O� �

�


Question 9. Write the equations involved in the following reactions :(i) Reimer-Tiemann reaction

(ii) Kolbe’s reaction

(i) It is the reaction of phenol with chloroform in the presence of sodiumhydroxide. As a result, CHO group is introduced at o-position ofbenzene ring.

(ii) Reaction of phenol and NaOH, then with CO2, a weak electrolyte.Orthobenzoic acid is formed.

Solution. (i)

Question 10. Write the reactions of Williamson synthesis of2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

(i) Reaction of 3-methylpentan-2-ol with sodium.

(ii) Reaction of ethanol with hydrogen bromide.

(iii) Reaction of product formed in (i) reaction with the product of (ii).

Solution. (i) CH CH CH CH

CH OH

CH N5 1

3-methylpentan-2-ol

3

4

2

3 2

3

3

+ a →

CH CH CH CH

CH ONa

CH

Sodium-(3-methyl)-pentoxi

3 2

3

3

de

Alcohols,�Phenols�and�Ethers 311

OH

Phenol

CHCl + NaOH3 aq.

ONa– +

CHCl2

Intermediate

NaOH

ONa– +

CHO

H+

OH

CHO

Salicylaldehyde

OH

NaOH

ONa

(i) CO2

(ii) H+

OH

COOH

2-hydroxybenzoicacid

(Salicylic acid)

Sodiumphenoxide

(ii)


(ii) C H OH + HBr C H Br + H O2 5 2 5 2

Bromoethane

→

(iii) CH CH CH CH

CH ONa

CH C H Br2 53 2

3

3

+ →

CH CH CH CH

CH OC H

CH

2 5

2-ethoxy-3-methylpentan

3 2

3

3

e

Questions 11. Which of the following is an appropriate set of reactantsfor the preparation of 1-methoxy-4-nitrobenzene and why?

(i) (ii)

This reaction involves preparation of ether by Williamson synthesis. Ittakes place by S 2N attack of an alkoxide ion on primary alkyl halide.

Solution.

(ii)

Because here the alkyl halide used is primary. Moreover inmethod (i) the C Br− bond has some double bond character.

Question 12. Predict the products of the following reactions :(i) CH CH CH O CH3 2 2 3 + →HBr

(ii)

(iii)

(iv) (CH ) C OC H3 3 2 5 HI →


Br

NO2

+ CH ONa3

ONa

NO2

+ CH Br3

ONa

NO2

+ CH Br3

OCH3

NO2

+ NaBr

1-methoxy-4-nitrobenzene

OC H2 5

+ HBr

OC H2 5Conc. H SO2 4

Conc. HNO3


(i) and (ii) R R X R R X′ + → ′ O H OH+ i.e., primary halide isformed.

(iii) Nitration (introduction of NO2 group) takes place at o- andp-positions.

(iv) Tertiary iodide and primary alcohol are formed.

Solution. (i) CH CH CH O CH HBr

Methoxy propane

3 2 2 3 + →

CH CH CH OH CH Br3

Propanol3 2 2 +

(ii)

(iii)

(iv) (CH ) C OC H3 3 2 5

butyl ethyl ether

HI →−t

(CH ) C I C H OH3 3 2 5

butyl iodide Ethanolt − +

ExercisesQuestion 1. Write IUPAC names of the following compounds.

(i) CH CH CH C CH

CH OH CH

CH

3 3

3 3

3

(ii) H C CH

OH

CH CH CH

OH C H

CH3

2 5

2 2 CH3


OC H2 5

+ HBr

OH

+ C H Br2 5

PhenolEthoxy benzene

OC H2 5 OC H2 5

2-ethoxy-nitrobenzene

Ethoxy benzene

Conc. H SO2 4

Conc. HNO3NO2

1

2

+

OC H2 5

O N2

3

4

2

1

4.-ethoxy-nitrobenzene

Exercises

(i) and (ii) R R X R R X′ + → ′ O H OH+ i.e., primary halide isformed.

(iii) Nitration (introduction of NO2 group) takes place at o- andp-positions.

(iv) Tertiary iodide and primary alcohol are formed.

Solution. (i) CH CH CH O CH HBr

Methoxy propane

3 2 2 3 + →

CH CH CH OH CH Br3

Propanol3 2 2 +

(ii)

(iii)

(iv) (CH ) C OC H3 3 2 5

butyl ethyl ether

HI →−t

(CH ) C I C H OH3 3 2 5

butyl iodide Ethanolt − +

ExercisesQuestion 1. Write IUPAC names of the following compounds.

(i) CH CH CH C CH

CH OH CH

CH

3 3

3 3

3

(ii) H C CH

OH

CH CH CH

OH C H

CH3

2 5

2 2 CH3


OC H2 5

+ HBr

OH

+ C H Br2 5

PhenolEthoxy benzene

OC H2 5 OC H2 5

2-ethoxy-nitrobenzene

Ethoxy benzene

Conc. H SO2 4

Conc. HNO3NO2

1

2

+

OC H2 5

O N2

3

4

2

1

4.-ethoxy-nitrobenzene


(iii) CH CH CH

OH OH

CH3 3

(iv) HO CH

OH

CH CH OH

2 2

(v) (vi)

(vii) (viii)

(ix) CH O CH CH

CH

CH3 2

3

3

(x) C H O C H6 5 2 5 (xi) C H O C H6 5 7 15 −( )n

(xii) CH CH O CH

CH

CH CH3 2

3

2 3

Prefix according to the carbon chain; for alcohol (—OH) use suffix -ol; tell

the position of substitutents and functional group. For ether ( ) O ,

prefix alkoxy is used. For —OH group attached directly to benzene ring,

suffix phenol is used.

So, give the name of alcohol as substituent with position (in alphabetical

order) + prefix of carbon chain + primary suffix (ane, ene, yne) + ‘ol’ with

position and give the name of ether as alkoxy alkane (where alkoxy

smaller R group and alkane is main chain)

Solution. (i) CH CH CH

CH OH

C

CH

CH

CH

2, 2, 4-trimethyl

5

3

4 3

3 3

2

3

1

3

pentan-3-ol


CH3

OH

CH3

OHCH3

OH

CH3

CH3

OH

CH3


(ii) H C CH

OH

CH CH CH

OH C H

CH CH3

2 5

5-ethyl h

1 2 3

2

4 5 6

2

7

3

eptane-2, 4-diol

(iii) CH CH CH

OH OH

CH

Butane-2, 3-diol

1

3

2 3 4

3

(iv) HO CH CH

OH

CH OH

Propane-1, 2, 3-triol

1

2

2 3

2

(v) (vi)

(vii) (viii)

(ix) CH O CH CH

CH

CH

1-methoxy-2-methyl propane

3

1

2

2

3

3

3

(x) C H O C H6 5 2 5

Ethoxybenzene

(xi) C H O C H6 5 7 15

1-phenoxy heptane

−( )n

(xii) CH CH O CH

CH

CH CH3

1

2-ethoxybutane

2

2

3

3

2

4

3

Question 2. Write structures of the compounds whose IUPAC namesare as follows.

(i) 2-methylbutan-2-ol (ii) 1-phenylpropan-2-ol

(iii) 3, 5-dimethylhexane-1,�3, 5-triol (iv) 2, 3-diethylphenol

(v) 1-ethoxypropane (vi) 2-ethoxy-3-methylpentane


2,5-dimethyl phenol

CH3

CH3

OH

1

23

4

56

2,6-dimethyl phenol

CH3

CH3

OH1

23

4

5

6

CH3

OH1

2-methylphenol

2

CH3

OH

12

3

4

4-methyl phenol


(vii) Cyclohexylmethanol (viii) 3-cyclohexylpentan-3-ol

(ix) Cyclopent-3-en-1-ol (x) 3-chloromethylpentan-1-ol

(i) Draw the main carbon chain according to the root word of carbonchain and then attach the other groups according to the positionsmentioned.

(ii) Suffix ‘al’ represents −OH group and prefix ‘alkoxy’ repersents ‘RO’group.

Solution. (i) CH C

OH

CH

CH CH

2-methylbutan-2-ol

1

3

2

3

3

2

4

3

(ii) C H CH CH

OH

CH6 5

1-phenylpropan-2-ol

1

2

2 3

3

(iii) CH

OH

CH C

CH

OH

CH C

OH

CH

CH

3, 5-dime

1

2

2

2

3

3

4

2

5

3

6

3

thylhexane-1, 3-5-triol

(iv) (v) C H O CH CH CH2 5

1-ethoxypropane

1

2

2

2

3

3

(vi) CH CH CH

OC H CH

CH CH

2 5

2-ethoxy-3-methyl

3

1 2 3

3

2

4

3

5

pentane

(vii) (viii)


OH

C H2 5

C H2 5

12

3

2,3-diethyl phenol

CH OH2

Cyclohexyl methanol

CH —CH —C—CH —CH3 2 2 3

OH

3-cyclohexylpentan-3-ol

1 2 3 4 5


(ix)

(x) HO CH CH CH

CH Cl

CH CH

2

3-chloromethylpent

1

2

2

2

3 4

2

5

3

an-1-ol

Question 3. (i) Draw the structures of all isomeric alcohols of molecularformula C H O5 12 and give their IUPAC names.

(ii) Classify the isomers of alcohols in question 3. (i) as primary,secondary and tertiary alcohols.

(i) Make different isomers by varying the number of C-atoms incontinuous chain, the position of CH3 and —OH groups.

(ii) Primary ( )1° alchols have CH OH2 ; group Secondary ( )2° haveCH OH ; group and Tertiary ( )3° have C OH .

Solution. (i) Molecular formula C H O5 12 represents eight isomericalkanols. These are :

(a) CH CH CH CH CH OH5 4 3 2 1

Pentan-1-ol3 2 2 2 2

(b) CH CH

CH

CH CH OH

3-methyl butan-1-ol

4

3

3

3

2

2

1

2

(c) CH CH CH

CH

CH OH

2-methyl butan-1-ol

4

3

3

2

2

3

1

2

(d) CH C

CH

CH

CH OH3 1

2, 2-dimethyl propan-1-ol

3

3

2

3

2

(e) CH CH CH CH

OH

CH

Pentan-2-ol

5

3

4

2

3

2

2 1

3


1

2

3

OH

Cyclopent-3-en-1-ol


(f) CH CH CH

CH OH

CH4 3 2 1

3-methyl butan-2-ol

3

3

3

(g) CH C

OH

CH

CH CH

2-methyl butan-2-ol

1

3

3

2 3

2

4

3

(h) CH CH CH

OH

CH CH

Pentan-3-ol

5

3

4

2

3 2

2

1

3

(ii) Primary : (a), (b), (c) and (d); Secondary (e), (f), (h) Tertiary : (g).

Question 4. Explain why propanol has higher boiling point than that of

the hydrocarbon, butane?

Boiling point is directly proportional to the intermolecular forces existingin a compound.

Solution. Although butane and propanol have comparable molecularmasses (58 and 60 respectively) but, in propanol, polar —OH group ispresent, due to which strong intermolecular hydrogen bonding existsbetween its molecules. Whereas in butane weak van der Waals’ forcesof attraction are the only forces between the molecules. Therefore,propanol has higher boiling point (391 K) as compared to that of butane(309 K).

…… ……

……

……

+ − + − + −

H O

C H

H O

C H

H O

C H3 7 3 7 3 7

δ δ δ δ δ δ

Question 5. Alcohols are comparatively more soluble in water than

hydrocarbons of comparable molecular masses. Explain this fact.

Consider polar nature of water as well as alcohol; presence of hydrogenbonding.

Solution. Water and alcohols both are polar in nature. When an alcohol

is dissolved in water, it forms hydrogen bonds with water by breaking

the H-bonding already existing between the water molecules.

Hydrocarbons are non-polar in nature and do not form H-bonds withwater molecules. Therefore, alcohols are readily soluble in waterwhereas hydrocarbons are not.

…… ……

……

……+

− + − + −

R

R

O

H

H O

H

H O

Hydrogen bonding

δ

δ δ δ δ δ

among alcohol and water



Question 6. What is meant by hydroboration-oxidation reaction?Illustrate it with an example.

Addition of diborane to alkenes followed by oxidation.

Solution. The addition of diborane to alkenes to form trialkyl boranesfollowed by their oxidation by alkaline H O2 2 to form alcohols is knownas hydroboration-oxidation reaction. For example

CH CH==CH H BH CH CH CH

H

Propene Dibroane3 2 2 2 3 2 + →

( )

BH2

↓ CH CH==CH3 2

↓ ← −

3 2 2

2

3 22 2 3

H O OH

H O

CH CH==CH(CH CH CH3

,) B

(CH CH CH BH3 2 2 2)

3CH CH CH OH OH3

Propan-1-ol Boric acid

+2 2 3B( )

Alcohol formed by this reaction appears as it has been formed byaddition of water in a way contrary to the Markownikoff’s rule.

Question 7. Give the structures and IUPAC names of monohydricphenols of molecular formula, C H O7 8 .

Ortho (1, 2); para (1, 4); meta (1, 3) Draw these isomers according to thementioned positions of —OH and CH3 groups.

Solution. Following three isomers are formed fromC H O7 8 .

Question 8. While separating a mixture of ortho and para nitrophenols bysteam distillation, name the isomer which will be steam volatile. Give reason.

Isomer, which is steam volatile is less strongly bonded. The other one isstrongly bonded by H-bonding.

Solution. Ortho-nitrophenol is steam volatile. The reason is chelationdue to intramolecular H-bonding. So, it can be separated by steamdistillation from p-nitrophenol. Para-nitrophenol is not steam volatiledue to presence of intermolecular H-bonding.


OH

CH31

2

2-methyl phenolor

- cresol (I)o

OH

CH3

12

3-methyl phenolor

cresol (II)m

3

OH

CH3

12

3

4

4-methyl phenolor

-cresol (III)p


Question 9. Give the equations of reactions for the preparation ofphenol from cumene.

Oxidation of cumene into cumene hydroperoxide, then acidic hydrolysis.

Solution.

Note This method is employed for the preparation of phenol on

commercial scale, as the by product is also an important reagent.

Question 10. Write chemical reaction for the preparation of phenol fromchlorobenzene.

Reaction of chlorobenzene with sodium hydroxide followed byacidification of the product formed.


–

N+OO

OH

Intramolecular H-bonding

o-nitrophenol

H—O N—O+ –

H—O N—O+ –

O OIntermolecularH-bonding-nitrophenolp

H C—CH3

CH3

O2

H C—C—O—O—H3

CH3

H+

H O2

Cumeneor

(Isopropyl benzene)

Cumenehydro-peroxide

OH

+ CH —C—CH3 3

O

Phenol Propanone(Acetone)


Solution.

Question 11. Write the mechanism of hydration of ethene to yieldethanol.

The steps involved in this process are1. Protonation of alkene.2. Nucleophilic attack of water on carbocation.3. Deprotonation to form an alcohol.

Solution. The mechanism involves 3 steps :

Step I : Protonation of alkene to form carbocation by electrophile

attack of H O3+ .

H O + H H O2+

3+→

Step II: Nucleophilic attack of water on carbocation formed.

Step III : Deprotonation to form corresponding alcohol.

Question 12. You are given benzene, conc. H SO2 4 and NaOH. Write theequations for the preparation of phenol using these reagents.

(i) Sulphonation of benzene.

(ii) Reaction withNaOHand water.


C C + H O H

H

+� C C

H

++ H O2

C � C

H

C

H

++ H O2 C O

H

H+

C

H

C O

H

+H + H O2 C

H

C

OH

+ H O3+

Cl

+ NaOH623 K

300 atm

O Na– +

HCl

OH

Chloro-benzene

Sodiumphenoxide

Phenol


Solution.

Question 13. Show how will you synthesise :(i) 1-phenylethanol form�a�suitable alkene.

(ii) Cyclohexylmethanol using�an�alkyl�halide�by�an S 2N reaction.

(iii) pentan-1-ol using�a�suitable�alkyl�halide?

(i) Acidic hydration of ethenyl benzene.

(ii) Alkaline hydrolysis of cyclohexyl methyl bromide.

(iii) Alkaline hydrolysis of 1-bromopentane.

Solution. (i) C H HC == H + H O6 5 2 2

Ethenylbenzene(Styrene)

Marko →C

wnikoff'srule

Dil.H SO2 4

C H CH

OH

CH6 5 3

1-phenyl ethanol

1 2

(iii) CH CH CH CH CH Br + KOH3 2 2 2 21-bromopentane

5 4 3 2 1

( )aq→ CH CH CH CH CH OH

5

3 2

3

2 2 2

Pentan-1-ol

4 2 1

+ KBr


–NaHSO , NaOH (Fuse)3

ONaOH

H+

Sodium phenoxidePhenol

CH Br2

+ NaOH( )aq

CH OH2

+ NaBr

Cyclohexylmethyl bromide

Cyclohexylmethanol

(ii)

H SO (Conc.)2 4

Heat

SO H3

NaOH

–H O2

SO Na3– +

Benzene Benzenesulphonic

acid

Sodium benzenesulphonate


Question 14. Give two reactions that show the acidic nature of phenol.Compare acidity of phenol with that of ethanol.

Reaction of phenol with Na and NaOH; comparing of acidity of phenoland ethanol.

Solution. Following reactions show the acidic nature of phenol :

(i) Reaction with sodium H2 gas is produced.

(ii) Reaction with sodium hydroxide Forms sodium salt and water.

Comparison of acidic character of phenol and ethanol

Phenol is more acidic than ethanol because after losing a proton

( )H+ , phenol forms phenoxide ion which is stabilised by resonance

whereas ethoxide ion is not.

Question 15. Explain why is ortho nitrophenol more acidic than orthomethoxyphenol?

(i) NO2- electron withdrawing group. OCH3-electron releasing group.

(ii) Presence of electron withdrawing group increases the stability ofphenoxide ion while presence of electron releasing group decreasesits stability.

(iii) Higher is the stability of phenoxide ion formed, more is the acidiccharacter.


IV V

OO

OH

+ NaOH

ONa

+ H O2

Phenol Sodium phenoxide

OH

2 + 2Na

ONa

2 + H2

Phenol Sodium phenoxide

I

O

II

O

III

O


Solution. Nitro ( )NO2 group is electron withdrawing whereas methoxy

(OCH3) group is electron releasing in nature. o-nitrophenol produces

H+ ions easily but methoxyphenol does not. This is becauseo-nitrophenoxide ion is stabilised due to resonance. This is not true witho-methoxyphenoxide ion. The two negative charges repel each otherthereby destabilising it.

Note Also give resonating structures of o-nitrophenoxide ion for Q. 8

intext questions.

Question 16. Explain how does the —OH group attached to a carbon ofbenzene ring activate it towards electrophilic substitution?

Resonance effect due to —OH group.

Solution. In the presence of attacking electrophile, —OH group exerts+R effect on the benzene ring. So, electron density in the ring increasesparticularly at the ortho and para positions. When an electrophileattacks, substitution takes place readily at these (o pand −) positions.


O—H O—H O—H

O—H O—H

O

O

CH3O

O

CH3

O—H

OCH3

O

OCH3

+ H+

O—H

NO2

O

NO2

+ H+

o-nitro-phenol


Question 17. Give equation of the following reactions.(i) Oxidation�of propan-1-ol with�alkaline KMnO4 solution.

(ii) Bromine�in CS2 with�phenol.

(iii) DiluteHNO3 with�phenol.

(iv) Treating�phenol�with�chloroform�in�presence�of�aqueous NaOH.

(i) Alcohol oxidises to carboxylic acid.

(ii) o-and p- bromophenol are formed.

(iii) Nitration (introduction of NO2 group) at o- and p- positions.

(v) Reimer-Tiemann reaction involves introduction of —CHO group atortho position, salicylaldehyde is formed.

Solution. (i) CH CH CH OH C3 2 24

Propan-1-ol (Oxidation)

KMnO /OH→

−H CH CHO3 2Propanal

→−

(Oxidation)

KMnO /OH

Propanoic acid

43 2CH CH COOH

Question 18. Explain the following with an example.(i) Kolbe's reaction.

(ii) Reimer-Tiemann reaction.

(iii) Williamson�ether�synthesis.

(iv) Unsymmetrical�ether.


OH

Br /CS2 2

OH

Br

OH

Br

+ + HBr

Phenol o-bromo-phenol

p-bromo-phenol

(ii)

OH

HNO (dil.)3

OH

NO2

OH

NO2

+ + H O2

Phenol o-nitrophenol p-nitrophenol

(iii)

OH

CHCl /NaOH ( )3 aq

ONa

CHO

Phenol

343 K(iv)

H O/H2+

OH

CHO

Salicyl-aldehyde


(i) (Reaction of phenol withNaOHand CO2), Salicylic acid is formed.

(ii) Reaction of phenol with CHCl + NaOH3 aq. → salicylaldehyde.

(iii) Alkyl halide + sod. alkoxide → Ether + Sod. halide.

(iv) Ethers containing two different alkyl/aryl groups.

Solution. (i) Kolbe's reaction When phenol reacts with sodiumhydroxide, reactive phenoxide ion is formed. So, electrophilicaromatic substitution takes place withCO2 under a pressure of 4-7atm followed by acidification and ortho hydroxybenzoic acid(Salicylic acid) is formed. This is called Kolbe's reaction.

(ii) Reimer-Tiemann reaction When phenol is treated withchloroform in the presence of sodium hydroxide, —CHO group isintroduced at ortho position of benzene ring. This is calledReimer-Tiemann reaction.

(iii) Williamson ether synthesis Ethers (symmertical andunsymmertrical) are prepared by this method. It involves thereaction of an alkyl halide with sodium alkoxide.

R X R R R X ′ → ′+ ONa O + Na

e.g., 1. (CH ) C ONa + CH Br3 3 3 → CH O—C CH + NaBr3 3( )3

2. C H OH + NaOH C H ONa C H O6 5 6 5 6 5→ → RX

R

(iv) Unsymmetrical ether An ether in which alkyl or aryl groupsattached to the oxygen atom are different, is known asunsymmetrical ether. For example, ethyl methyl ether(C H OCH )2 5 3 , methyl phenyl ether (CH O C H3 6 5 ) etc.


OH

NaOH

ONa

(i) CO2

(ii) H+

OH

COOH

2-hydroxybenzoicacid

(Salicylic acid)

12

Phenol Sodiumphenoxide

OH

Phenol

CHCl + NaOH3 aq.

ONa– +

CHCl2

Intermediate

NaOH

ONa– +

CHOH+

OH

CHO

Salicylaldehyde


Question 19. Write the mechanism of acid catalysed dehydration ofethanol to yield ethene.

Step involved in acid catalysed dehydration are

(i) Protonation of alcohol. (ii) Formation of carbocation.

(iii) Elimination of a proton.

Solution. Mechanism of acid catalysed dehydration of ethanolStep I : Protonation of ethanol

H

H

H

H

H

H

H

H

H

• •

• •+C C O H + H H C C

Ethanol

Fast� O H+

Protonated ethanol(Ethyl oxonium ion)

• •

H

Step II : Formation of carbocation

H

H

H

H

H H

H

H

H

H

O

• •

+ +C C O H H C C + H

Slow

2�

Step III : Elimination of a proton

H

H

H

H

H

H

HC

H

H

== ++C C C H

Ethene

+�

Note The acid used in Step I is released in Step III. To drive the

equilibrium to the right, ethene is removed as it is formed.

Question 20. How are the following conversion carried out?(i) Propene → propan-2-ol.

(ii) Benzyl�chloride → Benzyl�alcohol.

(iii) Ethyl�magnesium�chloride → Propan-1-ol.

(iv) Methyl�magnesium�bromide → 2-methylpropan-2-ol.

(i) Acidic hydration of propene.

(ii) Alkaline hydrolysis of benzyl chloride.

(iii) Reaction with methanal in dry ether followed by acid hydrolysis.

(iv) Reaction with propanone in dry ether followed by acid hydrolysis .

Solution. (i) CH CH== CH + conc.H SO3 2 2 4Propene

→

CH CH

OSO H

CH3

3

Isopropyl hydrogen s

3

ulphate

→

+Boil

H O

Propan-2-ol

21

3

3

3 2 4CH CH

OH

CH H SO2



(ii)

(iii) CH CH MgCl + H C3 2

Ethyl magnesiumchloride

Methanal

H

O

→

ether

Dry

Adduct

H C H

H CCH

OMgCl

2 3

→

+H O HH

CH CH

OH

H2

2 33

/1

2

C

Propan-1-ol

(iv) CH Mg + H C

CH

O

3 3

3

Methyl magnesiumbromide

Propano

Br C

ne (Acetone)

→ H C C CH

CH

OMgBr

Aduct

3

3

3

→

+H O HH C

23

1 2

3

3

3

/C

CH

OH

CH

2 - methyl propan - 2 - ol

Question 21. Name the reagents used in the following reactions.(i) Oxidation�of�a�primary�alcohol�to�carboxylic�acid.

(ii) Oxidation�of�a�primary�alcohol�to�aldehyde.

(iii) Bromination of�phenol�to�2,4,6-tribromophenol.

(iv) Benzyl�alcohol�to�benzoic�acid.

(v) Dehydration�of propan -2-ol to�propene.

(vi) Butan-2-one to butan-2-ol.

(i) Oxidising agent (strong)

(ii) Weak oxidising agent

(iii) Br / H O2 2

(iv) Strong oxidising agent

(v) Dehydrating agent

(vi) Reducing agent.


CH Cl2

+ NaOH ( )aq

Hydrolysis

CH OH2

+ NaCl

Benzylchloride

Benzylalcohol


Solution. (i) Acidified K Cr O2 2 7 or neutral/acidic/alkaline KMnO4.

(ii) Pyridine chlorochromate (PCC) (C H N HCl CrO5 5+

3−) in CH Cl2 2.

(iii) Bromine water (Br /H O)2 2 .

(iv) Alkaline KMnO4 or acidified KMnO4

(v) 85%H PO3 4 at 440 K.

(vi) LiAlH4 or NaBH4.

Question 22. Give reason for the higher point of ethanol in comparisonto methoxymethane.

Presence of intermolecular H-bonding in ethanol, but not in ether.

Solution. Hydrogen bonding is present between the ethanol moleculeswhich is absent in the molecules of methoxymethane. So, a largeamount of energy is required to break these strong bonds in ethanol.

− − − − − −

− − −

− − −

− +

H O

C H

H O

C H

H O

C H2 5 2 5 2 5

δ δ

Question 23. Give IUPAC names of the following ethers.

(i) C H OCH CH

CH

CH2 5 2

3

3

(ii) CH OCH CH Cl3 2 2

(iii) O N C H OCH2 6 4 3 ( )p (iv) CH CH CH OCH3 2 2 3

(v) (vi)

Recall the rules of IUPAC nomenclature see solution 1 of excercise.

Solution. (i) C H O CH CH

CH

CH2 5

1-ethoxy-2-methylpropane

1

2

2

3

3

3

(ii) CH O CH CH Cl

2-chloro-1-methoxyethane3

1

2

2

2

(iii) O N C H OCH2 6 4 3 ( )p or

4-nitroanisole


CH3H C3

OC H2 5

OC H2 5

OCH3

NO2

12

3

4

4-nitroanisole


(iv) CH CH CH OCH3

3

2

2

1

2 3

1-methoxypropane

(v) (vi)

Question 24. Write the names of reagents and equations for thepreparation of the following ethers by Williamson’s synthesis.

(i) 1-propoxypropane

(ii) Ethoxybenzene

(iii) 2-methoxy-2-methylpropane

(iv) 1-methoxyethane

(i) R X R + ′ →O NaWilliamson's

synthesisR R X ′ +O Na

(ii) Alkyl halide should be primary.

Solution. (i) CH CH CH O Na + CH CH CH Br3 2 2+

3 2 2

Sodium propoxide 1-bromop

− ropane

Heat →

CH CH CH O CH CH CH NaBr3 2 2 2 2 3

1-propoxypropane

+

(ii)


CH3H C3

OC H2 5

1

2

34

1-ethoxy-4,4-dimethylcyclohexane

OC H2 5

Ethoxybenzene

O Na– +

Sodiumphenoxide

+ CH CH —Br3 2

Bromo-ethane

Heat

OCH CH2 3

Ethoxy-benzene

+ NaBr


(iii) CH C

CH

CH

ONa

butyl-alkoxide

3

3

3

− +

Sod. tert-

+ →−CH Br

Bromomethane

Heat3

CH C

CH

CH

OCH NaBr

2-methoxy-2-methylpropane

3

3

3

3

+

(iv) CH CH ONa + CH Br3 2+

3

Sodium ethoxide Bromomethane

Heat− → CH CH O CH3 2

1-methoxyethane3

+ NaBr

Question 25. Illustrate the examples the limitations of Williamson’ssynthesis for the preparation of certain types of ethers.

Tertiary alkyl halides readily give elimination product with strong baseslike alkoxide ions. The C—X bond of alkyl halide does not possess doublebond character i e. ., should not be in conjugation with multiple bond.

Solution. (i) In Williamson’s synthesis reaction for the preparation ofether, the alkyl halide used must be primary as tertiary halidesreadily undergo elimination with strong bases like C H ONa2 5 . Thisis be best understood by taking the example of preparation ofethyl tertiary butyl ether. For which the reactants used are ethylbromide and sodium tertiary butoxide, but not the tertiary butylchloride and sodium ethoxide.

C H Br + Na O C

CH

CH

CH2 5+

Ethyl bromide

Sodium tertia

−

3

3

3

ry butoxide tertiary butyl ethyl

C H O C

CH

CH

CH2 5→

3

3

3

ether

+ NaBr

H C C

CH

CH

Br + Na O C H3 2 5

butyl bromide

Sodium e

3

3

Tert.

thoxide

2-methyl propene(Ether is

CH C==

CH

CH3

3

1

→

2

3

2

not formed)

+ +NaBr C H OH2 5

(ii) Aryl halides and vinyl halides cannot be used as substratesbecause they are less reactive in nucleophilic substitution.



Question 26. How is 1-propoxypropane synthesised from propan-1-ol?Write mechanism of this reaction.

(i) Williamson’s synthesis or

(ii) Dehydration of propan-1-ol.

Solution. Any one of the following two methods can be employed

(i) Williamson’s synthesis

(a) 3CH CH CH OH + PBr 3CH CH CH Br3 2 2 3 3 2 2

Propan-1-ol 1-bromop

→ropane

+ H PO3 3

(b) 2 CH CH CH OH + 2Na 2CH CH CH O Na3 2 2 3 2 2

Propan-1-ol Sodium p

→ −

ropoxide

++ H2

Reaction mechanism

CH CH CH ONa + CH CH CH Br3 2 2+

3 2 2

Sod. propoxide 1-bromopro

−

pane

Dry ether

Heat →

CH CH CH O CH CH CH NaBr3 2 2 2 2 3

1-propoxypropane

+

or

(ii) By dehydration of 1-propanol with conc. H SO2 4 at 413 K.

CH CH CH OH + H3 2 2+ 2 4

Propan-1-ol

H SO → CH CH CH O

H

H

3 2 2

Protonated 1-propanol

+

CH CH CH OH + CH CH CH O

H

H

3 2 2 3 2 2+

Propan-1-ol

K →• •

• •413

−H O,2

CH CH CH O

H

CH CH CH3 2 2

+

2 2 3

→−H+

,

CH CH CH O CH CH CH3 2 2 2 2 3

1-propoxypropane

Question 27. Preparation of ethers by acid dehydration of secondary ortertiary alcohols is not a suitable method. Give reason.

Due to steric hindrance, alkenes are formed by S 1N mechanism.

Solution. Ethers are formed as a result of acid dehydration by S 2N

mechanism (from primary alcohols). If secondary or tertiary alcohols are

used, due to steric hindrance, alkenes are formed and not ethers.



CH CH

CH

OH

Propan-2-ol

(2 alcohol)

Conc. H SO2 43

3

→

°

413 K Propene

(major)

CH CH==CH3 2

+

CH CH

CH

O CH

CH

CHDi-isopropyl ether

(minor)

3

3 3

3

CH C

CH

CH

OH

butyl alcohol

(3

3

3

3

°Tert.

)

Conc. H SO

413 K

2 42CH C==

CH

CH + H O →

3

3

2

Question 28. Write the equation of the reaction of hydrogen iodidewith : (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethylether.

Alcohols and alkyl or aryl iodides are formed. Generally primary halide(or lower halide) is formed in this reaction.

Solution. (i) CH CH CH O CH CH CH3 2 2 2 2 3

1-propoxypropane

HI

373 K →

CH CH CH OH CH CH CH I3 2 2 3 2 2

Propan-1-ol Iodopropane

+

Question 29. Explain the fact that in aryl alkyl ethers(i) the alkoxy group activates the benzene ring towards electrophilic

substitution.

(ii) it directs the incoming substituents to ortho and para positions in

benzene�ring.


OCH3

HI, 373 K

OH

Methoxybenzene Phenol

+ CH —I3

Iodomethane

(ii)

CH —O—C H2 2 5

HI, 373 K

CH I2

Benzyl ethyl ether Benzyl iodide

+ C H OH2 5

Ethanol

(iii)


Consider + R effect of the alkoxy group and more electron density at o-and p-positions.

Solution. In aryl alkyl ethers, +R effect of the alkoxy group ( –– )OR

increases the electron density in the benzene ring thereby activatingthe benzene ring towards electrophilic substitution reactions.

Electron density is more at o- and p-positions so o- and p-products aremainly formed during electrophilic substitution reactions.e.g.,

Question 30. Write the mechanism of the reaction of HI withmethoxymethane.

Steps involved in this process are(i) protonation (ii) formation of methanol and iodomethane.

Solution. When equimolar amounts of HI and methoxymethane are

taken, a mixture of methyl alcohol and iodomethane are formed.

Mechanism

Step I CH O CH H Ifast

Protonation3 3 + →

• •

• •CH O

H

CH I

Dimethyl oxonium ion

3 3

+• •

+ −

Step�II I CH O

H

CHSlow

S 2N+

→+

• •3 3 CH I + CH OHIodomethane Methanol

33

If HI is present in excess, CH OH3 formed in step II is furtherconverted into CH I3 .


OCH3

Br2

OCH3

Br

OCH3

Br

+

Anisole o-bromo-anisole(minor)

p-bromo-anisole(major)

CH COOH3

OR OR OR

I II III�

�

OR OR

IV V

�


Step�III CH O H H Ifast

Protonation3 + →

• •

• •CH O

H

H I

Protonated methanol

3

+• •

+ −

Step�IV I CH O

H

HSlow

S 2N+

→+

• •3 CH I + H OIodomethane

23

Question 31. Write equations of the following reactions.

(i) Friedel Crafts�reaction–alkylation of anisole.

(ii) Nitration�of anisole.

(iii) Bromination of anisole in ethanoic acid�medium.

(iv) Friedel Crafts acetylation of anisole.

OCH3 group is o/p directing.

(i) Alkylation (i.e., introduction of R group ) at o- and p-positions.

(By anhyd. AlCl CS3 2− ).

(ii) Nitration (i.e., introduction of NO2 group) at o- and p-positions

(By H SO HNO2 4− 3 ).

(iii) Bromination (introduction of Br2 group) at o- and p-positions

(ByBr2 in CH COOH)3 .

(iv) Acetylation (introduction of RCO-group) at o- and p-positions

(By CH COCl3 -anhydrous AlCl3).

Solution. (i) Friedel Crafts reaction (Alkylation)

(ii) Nitration of anisole


OCH3

Anhyd. AlCl3

Anisole

CS2+ CH Cl3

Chloro-methane

OCH3

CH3

OCH3

CH3

+

2-methoxy toluene(Minor)

4-methoxy toluene(Major)

OCH3

H SO2 4

Anisole

HNO3

OCH3

2-nitroanisole(Minor)

NO2

OCH3

4-nitroanisole(Major)

NO2

+


(iii) Bromination of anisole

(iv) Friedel Crafts acetylation of anisole

Question 32. Show how would you synthesise the following alcoholsfrom appropriate alkenes?

(i) (ii)

(iii) (iv)

By hydration (addition of H O2 )of appropriate alkenes in accordance with.Markownikoff’s rule.


OCH3

Br in2

Anisole

Ethanoic acid

OCH3

o-bromoanisole (Minor)

Br

OCH3

p-bromoanisole (Major)

Br

+

OCH3

Anhyd. AlCl3

Anisole

+ CH COCl3

Ethanoylchloride

2-methoxyacetophenone

(Monor)

OCH3 OCH3

4-methoxy acetophenone(Major)

+

COCH3

COCH3

CH3

OH

OH

OH

OH


Solution.

Question 33. When 3-methylbutan-2-ol is treated with HBr, thefollowing reaction takes place

CH CH CH

CH OH

CH3

3

3

→– Br

HBrCH C

CH

Br

CH CH3

3

2 3

Give a mechanism for this reaction.

The secondary carbocation formed in the process rearranges to a morestable tertiary carbocation by a hydride ion shift from 3rd carbon atom.


4-methyl hept-3-ene

+ H–OHMarkownikoff's

rule

H+

(ii)

4-methyl heptan-4-ol

OH

+ H–OHH

+

OH

Pent-1-ene Penan-2-ol

(iii)

2-cyclohexylbut-2-ene

H+ H O2

–H+ OH

1

23

4

2-cyclohexylbutan-2-ol

(iv)

CH3

1-methylcyclohexene

H+

CH3

carbocationTert.

H O2(i)

CH3

O—H

H

CH3

1-methylcyclo-hexan-1-ol

OH–H+


Solution. Mechanism

CH CH CH

CH OH

CH

3-methylbutan-2-ol

HBr

–Br3

3

3

→−

CH CH

CH

CH CHH O2

3

3

3

→⊕

••

−

O

H

H

CH C

CH

H

CH CH

2 carbocation(Less stable)

H ion s3

3

3

⊕

°

− hift →

CH C

CH

CH CH3

3 carbocation(More stable)

Br

→⊕ −

°3

2 3 CH C

CH

Br

CH CH2

2-bromo-2-methylbutane

3

3

3

Selected NCERT Exemplar�Problems

Short�Answers�Type

Question 1. What is the structure and IUPAC name of glycerol?

Solution. CH CH CH2 2

Propane-1,2,3-triol

OH OH OH

Question 2. Write the IUPAC name of the following compounds.

(i) CH CH CH CH CH CH

CH OH C H OH

3

2

6 5 4 3 2 1

3

3 5

(ii)

Solution. (i) 3-ethyl -5- methylhexane-2, 4,-diol.(ii) 1-methoxy-3-nitrocyclohexane

Question 3. Write the IUPAC name of the compound given below.

CH CH C

CH

==C

CH OH

OH5 4 3

2

2

3 2

3

1

Solution. 3-methylpent-2-ene-1, 2-diol.


OCH3

NO2

1

23

ncert chemistry xii 11

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