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

NCERT Solutions for Class 12 Chemistry Chapter 11 – Free PDF Download

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. These NCERT Solutions for Class 12 Chemistry help the students with some exercises, worksheets and assignments in practising important structures and formulas from aldehydes ketones and carboxylic acids topic. Referring to these solutions will strengthen the conceptual knowledge among students.

Moreover, the NCERT Solutions for Class 12 Chemistry presents you with accurate answers to textbook questions together with the extra questions and prepared by subject experts at BYJU’S according to the latest term – I CBSE Syllabus 2021-22. These solutions will also impact your understanding of this complex topic and aid you in preparing your own notes. Download the NCERT Solutions for Class 12 Chemistry Chapter 11 from the below-mentioned link to kickstart your preparations.

Download NCERT Solutions Class 12 Chemistry Chapter 11 PDF:-Download Here

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Class 12 Chemistry NCERT Solutions Chapter 11 Alcohols, Phenols and Ethers – Important Questions

Q 11.1 :

Write IUPAC names of the following compounds:

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.1-i

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.1-ii

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 :

Write structures of the compounds whose IUPAC names are 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

(vii) Cyclohexylmethanol

(viii) 3 – Cyclohexylpentan – 3 – ol

(ix) Cyclopent – 3 – en – 1 – ol

(x) 4-Chloro-3-ethylbutan-1-ol.

Solution :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.2-1

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.2-2

 

Q 11.3 :

(i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question.

Solution :

(i) The structures & IUPAC names of all isomeric alcohols with a molecular formula of C5H12O are shown below:

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

Pentan – 1 – ol (1 °)

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.3

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:

Explain why propanol has higher boiling point than that of the hydrocarbon, butane.

Solution:

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.4

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:

Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Solution :

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.5As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.

 

Q 11.6:

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

Solution :

The hydroboration – oxidation reaction is the reaction where 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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.6

 

Q 11.7:

Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.

Solution :NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.7

 

Q 11.8:

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

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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.8

 

Q 11.9:

Give the equations of reactions for the preparation of phenol from cumene.

Solution :

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.9-1

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.9-2

 

Q 11.10:

Write the mechanism of hydration of ethene to yield ethanol.

Solution :

There are three steps that 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 an electrophilic attack of H3 O + :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.10 Step1

Step 2:

Nucleophilic attack of water on carbocation :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.10 Step2

Step 3:

Deprotonation to form ethanol :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.10 Step3

 

Q 11.11:

Write a chemical reaction for the preparation of phenol from chlorobenzene.

Solution :

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.11

Q 11.12:

You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.

Solution :NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.12

 

Q 11.13:

Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?

Solution :

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.13-1

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.13-2

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.13-3

 

Q 11.14:

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

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 H2.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.14-1

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.14-2

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.14-3

 

Q 11.15:

Explain why ortho nitrophenol is more acidic than ortho methoxyphenol.

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, a methoxy group is an electron – releasing group. Hence, it increases the electron density in the O – H bond and thus, losing proton is not possible easily. Therefore, ortho – nitrophenol is more acidic than ortho – methoxyphenol.

 

Q 11.16:

Explain how does the –OH group attached to a carbon of benzene ring activate 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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.16

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

 

Q 11.17:

Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in the presence of aqueous NaOH.

Solution :

(i)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.17-1

(ii)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.17-2

(iii)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.17-3

(iv)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.17-4

 

Q 11.18:

 Explain the following with an example.
(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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.18(i) Kolbes 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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.18(ii) Reimer Tiemann reaction

This reaction is known as the Reimer – Tiemann reaction.

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.18(ii)-1

(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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.18(iii) 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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.18(iii)-1

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:

Write the mechanism of acid 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 :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.19 Step1

Step 2 :

A carbocation is formed ( rate-determining step ) :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.19 Step2

Step 3 :

Ethane is formed by elimination of proton:

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.19 Step3

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:

How are the following conversions 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.

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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.20(i)

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.20(ii)

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.20(iii)

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.20(iv)

Q 11.21:

Name the reagents used in the following reactions:
(i) Oxidation of primary alcohol to a carboxylic acid.
(ii) Oxidation of 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.

Solution :

(i) NaBH4 or LiAlH4 to acidified KMnO4

(ii) Pyridinium chlorochromate (PCC)

(iii) Bromine water

(iv) Acidified potassium permanganate

(v) Concentrated H2SO4 or H3PO4

(vi) Catalytic hydrogenation or sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4)

 

Q 11.22:

Give a reason for the higher boiling point of ethanol in comparison to methoxymethane.

Solution :

Ethanol experiences intermolecular H – bonding because of the presence of – OH group, which results in the 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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.22

 

Q 11.23:

Give IUPAC names of the following ethers:

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.23

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 and 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 :

(i)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.24(i)

(ii)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.24(ii)

(iii)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.24(iii)

(iv)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.24(iv)

 

Q 11.25 :

Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers

Solution :

The reaction of Williamson synthesis includes SN2 attack of an alkoxide ion on a primary alkyl halide.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.25-1

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 the 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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.25-2

 

Q 11.26 :

How is 1-propoxypropane synthesised from propan-1-ol? Write the 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.

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.26-1

The mechanism of this reaction involves the following three steps :

Step 1 : Protonation

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.26 Step1

Step 2 : Nucleophilic attack

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.26 Step2

Step 3 : Deprotonation

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.26 Step3

 

Q 11.27:

Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

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 the case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination of dominates substitution. Hence, in place of ethers, alkenes are formed.

 

Q 11.28:

Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether

Solution :

(i)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.28(i)

(ii)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.28(ii)

(iii)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.28(iii)

 

Q 11.29:

Explain the fact that in aryl alkyl ethers

(i) the alkoxy group activates the benzene ring towards electrophilic substitution and

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

Solution :

(i)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.29(i)-1

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 :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.29(i)-2

 

Therefore, a 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 and ortho positions compared to meta position. Consequently, the substituent which is incoming is directed to the para and ortho positions in the benzene ring.

 

Q 11.30:

Write the mechanism of the reaction of HI with methoxymethane.

Solution :

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

Step 1 : Protonation of methoxymethane :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.30 Step1

Step 2 : Nucleophilic attack of I :

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.30 Step2

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.NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.30 Step3

 

Q 11.31:

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-Craft’s acetylation of anisole.

Solution :

(i)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.31(i)

(ii)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.31(ii)

(iii)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.31(iii)

(iv)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.31(iv)

 

Q11.32:

Show how would you synthesise the following alcohols from appropriate alkenes?

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.32

Solution :

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.32-1

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

ol.NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.32(iii)-1

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

(iv)NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.32(iv)

 

Q 11.33:

When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.33

Give a mechanism for this reaction.
(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.

Solution :

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

Step 1 : Protonation

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.33 Step1

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.33 Step2

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

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.33 Step3

Step 4 : Nucleophilic attack

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers Q 11.33 Step4

The Central Board of Secondary Education (CBSE) is one of the most preferred educational boards in the country. The schools affiliated with CBSE are advised to follow the NCERT curriculum as well as the NCERT books. The NCERT Solutions for Class 12 are prepared by the subject matter experts of BYJU’S so that students can easily understand the concepts of the chapter Alcohols, Phenols and Ethers.

NCERT Solutions for Class 12 Chemistry Chapter 11 deals with alcohols and phenols and their classifications based on 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. Develop a strong foundation on these topics by solving the exemplar questions of this chapter.

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

Chapter 11 Alcohols, Phenols and Ethers of Class 12 Chemistry is categorized under the term – I CBSE Syllabus for 2021-22. After studying this chapter about alcohol, phenols and ethers using the NCERT Solutions, 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

Why Opt for BYJU’S?

BYJU’S provides the latest NCERT Solutions, updates, notifications, sample papers and previous year questions. All the resources provided by us are the result of the experience and expertise of the subjects teachers at BYJU’S. These study resources will help you pass your exams with good grades.

Frequently Asked Questions on NCERT Solutions for Class 12 Chemistry Chapter 11

How can I learn Chapter 11 of NCERT Solutions for Class 12 Chemistry faster?

Before starting with the chapter, students must first get a clear idea about the concepts related to IUPAC names and structures. By referring to the NCERT Solutions for Class 12 Chemistry Chapter 11, students can understand the classification of compounds in a systematic way. In this chapter, students will learn about the classification of mono, di, tri or polyhydric alcohols as per the latest term – I CBSE Syllabus. Preparation of alcohols, phenols and ethers are also explained in a simple language for a better understanding among students.

What concepts will I learn in Chapter 11 of NCERT Solutions for Class 12 Chemistry?

In Chapter 11 of NCERT Solutions for Class 12 Chemistry, students will learn about important concepts like –
1. Classification
2. Nomenclature
3. Structures of Functional groups
4. Alcohols and phenols
5. Some Commercially Important alcohols
6. Ethers

Does the NCERT Solutions for Class 12 Chemistry Chapter 11 help students with their first term exam preparation?

The marks obtained in the Class 12 term – I exams will help students to build their career paths. Before preparing for the first term exam, students must first understand the term-wise syllabus and marks weightage for the concepts as per the latest update on the term – I CBSE guidelines. Students having doubts while answering the questions from the textbook can make use of the NCERT Solutions designed by the faculty at BYJU’S. These solutions are prepared by keeping in mind the understanding abilities of Class 12 students. It mainly improves the logical and analytical thinking skills among students which are important for the first term exam.

 

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