NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers is an essential study material that is required for all students studying class 12 chemistry. The solutions for class 12 chemistry chapter 11 pdf help the students with some exercises, worksheets, and assignments in practicing important structures and formulas from aldehydes ketones and carboxylic acids topic. This chapter 11 presents you with proficient answers to textbook questions together with the extra questions and exemplar questions prepared by subject experts at BYJU’S. These solutions will also impact your understanding of this complex topic and aid you in preparing your own notes.

Alcohols and phenols are classified on the basis of the number of -OH groups present. Compounds containing one -OH group are known as monohydride alcohols and phenols. The terms dihydric, trihydric or polyhydric are used, when the compounds contain two, three or more -OH groups respectively.

Class 12 NCERT Solutions for Alcohols, Phenols and Ethers

After studying this chapter alcohol, phenols and ethers students can learn how to write the names of alcohols, phenols, and ethers according to the IUPAC system of nomenclature.

  • The reactions involved in the process of making alcohol from Alcohols, Phenols and Ethers. The reactions involved in the preparation of phenols from benzene sulphonic acids, haloarenes, cumene and from diazonium salts.
  • The reaction for the preparation of ethers from alcohols, alkyl halides and sodium alkoxides.
  • The physical properties of ethers, phenols and alcohols.

Subtopics of Class 12 Chemistry Chapter 11 – Alcohols, Phenols and Ethers

  1. Classification
  2. Nomenclature 
  3. Structures of Functional groups
  4. Alcohols and phenols
  5. Some Commercially Important alcohols
  6. Ethers

Class 12 Chemistry NCERT Solutions (Alcohols, Phenols and Ethers) – Important Questions

Q 11.1 :

Give IUPAC names of the compounds given below :



Solution :

(i) 2, 2, 4 -Trimethylpentan – 3 – ol

(ii) 5 – Ethylheptane – 2, 4 – diol

(iii) Butane – 2, 3 – diol

(iv) Propane – 1, 2, 3 – triol

(v) 2-Methylphenol

(vi) 4 – Methyl phenol

(vii) 2, 5 – Dimethylphenol

(viii) 2, 6 – Dimethylphenol

(ix) 1 – Methoxy – 2 – methyl propane

(x) Ethoxy benzene

(xi) 1 – Phenoxyheptane

(xii) 2 – Ethoxybutane


Q 11.2 :

Give the structures of the compounds whose IUPAC names are given below :

(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

(vii) Cyclohexylmethanol

(viii) 3 – Cyclohexylpentan – 3 – ol

(ix) Cyclopent – 3 – en – 1 – ol

(x) 3 – Chloromethylpentan – 1 – ol.

Solution :





Q 11.3 :

(i) Give the structures & IUPAC names of all isomeric alcohols which have a molecular formula as :

C 5 H 12 O

(ii) Categorize the isomers of alcohols given in the Q 11.3 ( i ) as primary, secondary & tertiary alcohols.

Solution :

(i) The structures & IUPAC names of all isomeric alcohols with a molecular formula of C 5 H 12 O are shown below:

(a) CH3-CH2-CH2-CH2-CH2-OH

Pentan – 1 – ol ( 1 ° )


2 – Methylbutan – 2 – ol ( 3 ° )

(ii) Primary alcohol : Pentan – 1 – ol ; 2 – Methylbutan – 1 – ol ;

3 – Methylbutan – 1 – ol ; 2, 2 – Dimethylpropan – 1 – ol

Secondary alcohol: Pentan – 2 – ol ; 3 – Methylbutan – 2 – ol ;

Pentan – 3 – ol

Tertiary alcohol : 2 – methylbutan – 2 – ol


Q 11.4:

Given reasons for the following statement:

Hydrocarbon, butane has a lower boiling point than propanol.


The presence of – OH group makes Propanol undergo intermolecular H-bonding. Butane, while on the other side does not have the same privilege


Hence, additional energy would be required to break the intermolecular hydrogen bonds.

This is the reason why hydrocarbon butane has a lower boiling point than propanol.


Q 11.5:

The solubility of Alcohols is comparatively more in water than that of the solubility of hydrocarbons of comparable molecular masses. Justify the above fact.

Solution :

Due to the presence of – OH group, alcohols form H – bonds with water.

5.1As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.


Q 11.6:

Explain with a suitable example the hydroboration – oxidation reaction?

Solution :

The hydroboration – oxidation reaction is the reaction borane is added in order for the oxidation to take place. For example, propan – 1 – ol is formed by making propene undergo the hydroboration – oxidation reaction. In the above reaction, the reaction between propene and diborane ( BH 3 )2 takes place in order to form trialkyl borane which acts an additional product. This additional product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.



Q 11.7:

Give the structures & IUPAC names of monohydric phenols of molecular formula,

C 7 H 8 O.

Solution :7.1


Q 11.8:

While separating a mixture of para & ortho nitro phenols by steam distillation, mention the isomer which will be steam volatile. Reason your solution.

Solution :

Intramolecular H – bonding is present in o – nitrophenol & p – nitrophenol. In p – nitrophenol, the molecules are strongly associated due to the existence of intermolecular bonding.

therefore, o – nitrophenol is steam volatile.




Q 11.9:

Explain the reactions for the preparation of phenol from cumene with the equation.

Solution :

To synthesize phenol, cumene is initially oxidized in the presence of air of cumene hydro peroxide.


Followed by, treating cumene hydroxide with dilute acid to prepare phenol & acetone as byproducts.



Q 11.10:



Write the mechanism of hydration of ethene to yield ethanol.

Solution :

There are three steps which are involved in the mechanism of hydration of ethene to form ethanol. these steps are as follows:

Step 1 :

Protonation of ethene to form carbocation by electrophilic attack of H3 O + :


Step 2:

Nucleophilic attack of water on carbocation :


Step 3:

Deprotonation to form ethanol :



Q 11.11:

Explain reaction for the preparation of phenol from chlorobenzene with the equation.

Solution :

Chlorobenzene is combined with NaOH ( at 623 K & 320 atm pressure ) to prepare sodium phenoxide, resulting in phenol on acidification.


Q 11.12:

Give the equation for the following reaction:

preparation of phenol using the following reagents à benzene, conc. H 2 SO 4 & NaOH. Write the equations for the

Solution : 12.1


Q 11.13:

Give the procedure to synthesize the following :

(i) 1 – phenylethanol from a suitable alkene.

(ii) cyclohexylmethanol using an alkyl halide by an SN 2 reaction.

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

Solution :

(i) By acid – catalyzed hydration of ethylbenzene ( styrene ), 1 – phenylethanol can be synthesized.


(ii) When chloromethyl cyclohexane is treated with sodium hydroxide, cyclohexyl methanol is obtained.


(iii) When 1 – chloropentane is treated with NaOH, pentan – 1 – ol is produced.



Q 11.14:

Show that phenol is of acidic nature by giving two reactions. Also , Compare acidity of ethanol with that of phenol.

Solution :

The acidic nature of phenol can be proven with the two reactions given below :

(i) Phenol reacts with sodium to give sodium phenoxide, liberating H 2.


(ii)Phenol reacts with sodium hydroxide to give sodium phenoxide & water as by – products.


The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance & gets stabilized whereas ethoxide ion does not.



Q 11.15:

Give reasons for the following statement: ortho-nitrophenol is more acidic than ortho methoxy phenol.

Solution :

The nitro – group is an electron-withdrawing group. The existence of this group in the ortho position decreases the electron density in the O – H bond. Consequently, it is easier to give away a proton. Furthermore, the o -nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Therefore, ortho nitrophenol is a stronger acid. In contrast, methoxy group is an electron – releasing group. Hence, it increases the electron density in the O – H bond & thus, losing proton is not possible easily. Therefore, ortho – nitrophenol is more acidic than ortho – methoxyphenol.


Q 11.16:

Give detailed explanation for the following reaction: – OH group attached to a carbon of benzene ring activates it towards electrophilic substitution

Solution :

The density of the electron increases in the benzene ring as the – OH group acts as an electron donating group. This is clearly shown in the resonance structure of phenol given here


As a result, the benzene ring is activated towards electrophilic substitution.


Q 11.17:

Write appropriate equations of the following reactions:

(i) Oxidation of propan – 1 – ol with alkaline KMnO 4 solution.

(ii) Bromine in C S 2 with phenol.

(iii) Dilute HNO 3 with phenol.

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

Solution :


(ii) 17.2




Q 11.18:

Give a detailed explain for each of the following.

(i) Kolbe’s reaction.

(ii) Reimer – Tiemann reaction.

(iii) Williamson ether synthesis.

(iv) Unsymmetrical ether.

Solution :

(i) Kolbe’s reaction:

Sodium phenoxide is formed when phenol is treated with sodium hydroxide. Ortho – hydroxybenzoic acid as the main product when sodium phenoxide is treated with carbon dioxide, followed by acidification, it undergoes electrophilic substitution. This reaction is known as Kolbe’s reaction.


(ii) Reimer – Tiemann reaction:

A -CHO group is introduced at the ortho position of the benzene ring when phenol is treated with chloroform ( CHCl 3 ) in the presence of sodium hydroxide.


This reaction is known as the Reimer – Tiemann reaction.

Salicylaldehyde is produced when the intermediate is hydrolyzed in the presence of alkalis.


(iii) Williamson ether synthesis:

A chemical method to synthesize symmetrical & unsymmetrical ethers by making alkyl halides to react with sodium alkoxides is called Williamson ether synthesis.


The above reaction includes Sn 2 attack of the alkoxide ion on the alkyl halide. In the case of primary alkyl halides, better results are obtained.


Only If the alkyl halide is tertiary or secondary, in that case, elimination competes over substitution.

(iv) Unsymmetrical ether:

When an oxygen atom has two groups on two of its side, it is called Unsymmetrical ether. ( i.e., the number of carbon atoms is unequal ). For eg: ethyl methyl ether ( CH 3 – O − CH 2 CH 3 ).


Q 11.19:

Give a detailed explanation of acid – catalyzed dehydration of ethanol to yield ethene.

Solution :

The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps :

Step 1 :

Formation of ethyl oxonium by protonation of ethanol :


Step 2 :

Carbocation is formed ( rate determining step ) :


Step 3 :

Ethane is formed by elimination of proton:


The acid is released in Step 3 which was being consumed in Step 1. It is removed to shift the equilibrium in a forward direction, after ethene is formed.


Q 11.20:

Explain the procedure for following conversions.

(a) Benzyl chloride à Benzyl alcohol

(b) Propene à Propan – 2 – ol

(c) Methyl magnesium bromide à 2 – Methylpropan – 2 – ol.

(d) Ethyl magnesium chloride à Propan – 1 – ol.

Solution :

(i) when propene is reacted with water in the presence of an acid as a catalyst , as a result we obtain propan – 2 – ol.


(ii) If then benzyl alcohol is obtained when benzyl chloride is reacted with NaOH which is then followed by acidification .


(iii) An adduct is the obtained when ethyl magnesium chloride is reacted with methanal , as a result propan – 1 – ol on hydrolysis is obtained.


(iv) On treating methyl magnesium bromide with propane, an adduct is obtained which results in 2 – methylpropane – 2 – ol on hydrolysis.


Q 11.21:

List out the the reagents used in the reactions given below :

(i) Butan – 2 – one to butan – 2 – ol.

(ii) Dehydration of propan – 2 – ol to propene.

(iii) Benzyl alcohol to benzoic acid

(iv) Bromination of phenol to 2 , 4 , 6 – tribromophenol.

(v) Oxidation of a primary alcohol to aldehyde.

(vi) Oxidation of a primary alcohol to carboxylic acid.

Solution :

(i) NaBH4 or LiAlH4

(ii) 85 % phosphoric acid

(iii) Acidified potassium permanganate

(iv) Bromine water

(v) Pyridinium chlorochromate ( PCC )

(vi) Acidified potassium permanganate


Q 11.22:

State reasons why ethanol has a higher boiling point when compared to methoxymethane.

Solution :

Ethanol experiences intermolecular H – bonding because of the presence of – OH group, which results in association of molecules. additional energy is necessary to break these hydrogen bonds. Conversely, methoxymethane does not experience H – bonding. Therefore, ethanol has a higher boiling point when compared to methoxymethane






Q 11.23:

Name of the ethers given below :


Solution :

(i) 1 – Ethoxy – 2 – methylpropane

(ii) 2 – Chloro – 1 – methoxyethane

(iii) 4 – Nitroanisole

(iv) 1 – Methoxypropane

(v) 1 – Ethoxy – 4 , 4 – dimethylcyclohexane

(vi) Ethoxybenzene


Q 11.24 :

Write the names of reagents & equations for the preparation of the following ethers by

Williamson’s synthesis :

(i) 1 – Propoxypropane

(ii) Ethoxybenzene

(iii) 2 – Methoxy – 2 – methylpropane

(iv) 1 – Methoxyethane

Solution :


(ii) 24.2




Q 11.25 :

Explain with examples the drawbacks of Williamson synthesis for preparing specific types of ethers.

Solution :

The reaction of Williamson synthesis includes SN 2 attack of an alkoxide ion on a primary alkyl halide.


while, tertiary alkyl halides or secondary alkyl halides were to be substituted in place of primary alkyl halides, in that case elimination would contend over substitution. This results in formation of alkenes. This happens because alkoxides are nucleophiles as well as strong bases. Therefore, they react with alkyl halides, which results in an elimination reaction.



Q 11.26 :

How is 1 – propoxy propane synthesized from propan – 1 – ol ? Write mechanism of this reaction.

Solution :

1 – propoxypropane can be synthesized from propan – 1 – ol by dehydration.

Propan – 1 – ol undergoes dehydration in the presence of protic acids ( such as H 2 SO 4 , H 3 PO 4 ) to give 1 – propoxypropane.


The mechanism of this reaction involves the following three steps :

Step 1 : Protonation


Step 2 : Nucleophilic attack


Step 3 : Deprotonation



Q 11.27:

Reason the following statement : Acid dehydration of tertiary or secondary alcohols is not a suitable technique for synthesis of ethers.

Solution :

The synthesis of ethers with dehydration of alcohol is a bimolecular reaction ( SN 2 ) which involves the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of ethers, alkenes are formed.


Q 11.28:

Give equations for the reaction between the following compound & hydrogen iodide :

  1. a) Benzyl ethyl ether
  2. b) Methoxybenzene
  3. c) 1 – propoxypropane

Solution :





Q 11.29:

Give explanation for the following statements

(i) In case of aryl alkyl ethers , benzene ring is activated by the alkoxy group towards electrophilic substitution

(ii) aryl alkyl ethers direct the arriving substituent to para & ortho positions in benzene ring.

Solution :


The electron density in the benzene ring boosts up because of the existence of + R effect of the alkoxy group In aryl alkyl ethers which is given in the resonance configuration below :



Therefore, benzene ring is activated by the alkoxy group towards electrophilic substitution.

(ii) The above resonance configuration , we infer that the electron density is higher at the para & ortho positions compared to meta position. consequently, the substituent which is incoming is directed to the para & ortho positions in the benzene ring.


Q 11.30:

Give the steps involved in the reaction of HI with methoxymethane.

Solution :

The steps of the reaction of methoxymethane with HI are given below :

Step 1 : Protonation of methoxymethane :


Step 2 : Nucleophilic attack of I :


Step 3 : The methanol synthesized in the above step is made to react with another HI molecule which then converts to methyl iodide. This is done at a high temperature & HI is present in excess.30.3


Q 11.31:

Give equations with structures for the reactions given below :

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

(ii) Nitration of anisole.

(iii) Bromination of anisole in ethanoic acid medium.

(iv) Friedel – Craft’s acetylation of anisole.

Solution :


(ii) 31.2





Explain how the following alcohols can be synthesized from appropriate alkenes.


Solution :

The alcohols given above can be synthesized by applying Markovnikov’s rule of acid – catalyzed hydration of appropriate alkenes.


Acid-catalyzed hydration of pent-2-ene also produces pentan-2-ol but along with pentan3-


Thus, the first reaction is preferred over the second one to get pentan-2-ol.



Q 11.33:

As, 3 – methylbutan – 2 – ol is treated with HBr, the reaction below takes place :


Explain the steps involved in the above reaction.

( Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbonation by a hydride ion shift from 3 rd carbon atom )

Solution :

The steps of the reaction given above involve the following procedure :

Step 1 : Protonation


Step 2 : Forming 2 ° carbonation by eliminating the water molecule


Step 3 : Re – arranging by the shifting hydride – ion


Step 4 : Nucleophilic attack


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