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

NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols and Ethers - This article covers solutions of NCERT class 12 chemistry chapter 11 Alcohols, Phenols and Ethers. You are going to study the chemistry of three classes of compounds which are alcohols, phenols and ethers and also this chapter will discuss the reactions involved in the preparation of alcohols, phenols and ethersAlcohols are formed when a hydrogen atom in an aliphatic hydrocarbon is replaced by -OH group and Phenols are formed when a hydrogen atom in an aromatic hydrocarbon is replaced by –OH group while Ethers are formed by the substitution of a H-atom in a hydrocarbon by an alkoxy(R-O) or by an aryloxy(Ar-O) group. CBSE NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols and Ethers has answers to a total of 12 topic wise questions and 33 questions in the exercise. In CBSE boards exam, the weightage of this chapter is 4 marks hence it is recommended to solve all the exercises of the book to get good marks. You will find all the solutions of NCERT class 12 chemistry chapter 11 Alcohols, Phenols and Ethers here free of cost.

The NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols and Ethers are prepared by chemistry subject experts. These NCERT solutions help students in their preparation of the CBSE Board exam and in the competitive exams like JEE Mains, NEET etc. NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols and Ethers start with the IUPAC nomenclature of alcohols, phenols and ethers followed by the topics:- reactions involved in the preparation of alcohols from aldehydes, ketones, alkenes and carboxylic acids and discuss reactions involved in the preparation of phenols from benzene sulphonic acids, haloarenes,cumene and diazonium salts and also discuss reactions involved in the preparation of ethers from alcohols and alkyl halides. Scroll down to get solutions of NCERT class 12 chemistry Chapter 11 Alcohols, Phenols and Ethers.

Alcohols are the basic compound for the formation of detergents, phenols are the basic compound for the formation of antiseptics and ethers is the basic compound for the formation of fragrances.

## Topics and Sub-topics of NCERT Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers-

11.1 Classification

11.2 Nomenclature

11.3 Structures of Functional Groups

11.4 Alcohols and Phenols

11.5 Some Commercially Important Alcohols

11.6 Ethers

## Solutions to In-Text Questions Ex 11.1  to 11.12

Question

To classify we look at the OH bonded carbon.

Here, only 1 carbon is attached to it, hence it is primary alcohol.

Question

To classify we look at the OH bonded carbon.

Here, only 1 carbon is attached to it, hence it is primary alcohol.

Question

$CH_3 -CH_2- CH_2 -OH$

To classify we look at the OH bonded carbon.

Here, only 1 carbon is attached to it, hence it is primary alcohol.

Question

To classify we look at the OH bonded carbon.

Here, 2 carbons are attached to it, hence it is secondary alcohol.

Question

To classify we look at the OH bonded carbon.

Here, 2 carbons are attached to it, hence it is secondary alcohol.

Question

To classify we look at the OH bonded carbon.

Here, 3 carbons are attached to it, hence it is tertiary alcohol.

Question

The alcohols (ii) and (vi) are allylic alcohols. Because -C=C-C-OH is the skeleton of allylic alcohol.

Question

3-Chloromethyl-2-isopropylpentan-1-ol

Question

2, 5-Dimethylhexane-1, 3-diol

Question

3-Bromocyclohexanol

Question

Hex-1-en-3-ol

Question

2-Bromo-3-methylbut-2-en-1-ol

The reaction of a suitable Grignard reagent on the methanal is mentioned below:

Question

Question

Product of the given reaction is-

Question

Product of the given reaction is

(a)$HCL- ZnCl_2$   with Butan-1-ol

Primary alcohols do no react with Lucas’ reagent.

Hence no reaction.

Reaction of  $HCl - ZnCl_2$  with 2-Methylbutan-2-ol

Reaction of HBr with Butan-1-ol

Reaction of HBr with 2-Methylbutan-2-ol

Reaction of $SOCl_2$  with Butan-1-ol

Reaction of $SoCl_2$  with 2-Methylbutan-2-ol

Dehydration of 1-methylcyclohexanol

1-Methylcyclohexene is the major product.

Question

Dehydration of  butan-1-ol

But-2-ene is the major product.

Resonance structure of ortho-nitrophenol

Resonance structure of para-nitrophenol

Reimer - Tiemann reaction

Kolbe’s reaction

1.Reaction of ethanol with hydrogen bromide.

2. Reaction of 3-methylpentan-2-ol with sodium

3. Reaction of product formed in 1st and reaction with the product formed in the 2nd reaction.

(ii) is appropriate because $\dpi{80} CH_3Br$  is nucleophile whereas$\dpi{80} CH_3ONa$ is also a strong base. So elimination will be dominating in (i).

Question

Reaction is

Question

The reaction between ethoxybenzene and HBr is

Question

Reaction between ethoxybenzene and $\dpi{80} Conc. \:H_{2}SO_{4} \:+\:conc. HNO_{3}$

Question

Reaction between ter-butyl ethyl ether and HI

## NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers - Exercise Questions

Question

IUPAC name of the given compound is 2,2,4-Trimethylpentan-3-ol

Question

IUPAC name of the given compound is 5-Ethylheptane-2,4-diol

Question

IUPAC name of the given compound is Butane-2,3-diol

Question

IUPAC name of the given compound is  Propane-1,2,3-triol

Question

IUPAC name of the given compound is 2-Methylphenol

Question

IUPAC name of the given compound is 4-Methylphenol

Note : Also called p-cresol

Question

IUPAC name of the given compound is 2,5-Dimethylphenol

Question

IUPAC name of the given compound is 2,6-Dimethylphenol

Question

IUPAC name of the given compound is 1-Methoxy-2-methylpropane

Question

IUPAC name of the given compound is Ethoxybenzene

Question

IUPAC name of the given compound is 1-Phenoxyheptane

Question

IUPAC name of the given compound is 2-Ethoxybutane

Structure of  2-Methylbutan-2-ol

Structure of 1-Phenylpropan-2-ol

structure of 3,5-Dimethylhexane –1, 3, 5-triol

structure  of 2,3 – Diethylphenol

structure  of 1 – Ethoxypropane

structure of 2-Ethoxy-3-methylpentane

Structure of Cyclohexylmethanol

The structure of 3-Cyclohexylpentan-3-ol is as follows:

structure of Cyclopent-3-en-1-ol

structure of 4-Chloro-3-ethylbutan-1-ol

The structures of all isomeric alcohols of C5H12O are given below:

 Pentan-1-ol Pentan-2-ol Pentan-3-ol 3-Methylbutan-1-ol 3-Methylbutan-1-ol 2,2-Dimethylpropan-1-ol 3-Methylbutan-2-ol 2-Methylbutan-2-ol

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

Propanol forms intermolecular H-bonds because of the presence of -OH group while butane cannot. To break these bonds, extra energy will be required. This causes a higher boiling point for propanol as compared to butane.

Alcohols form hydrogen bonds with water due to the presence of –OH group whereas hydrocarbons cannot. Due to this inter molecular hydrogen bonding, alcohols are more soluble in water.

Hydroboration-oxidation reaction also called HBO reaction is the addition of borane followed by oxidation to produce alcohol.

Eg: Hydroboration-oxidation reaction of propene. In this reaction, propene reacts with diborane (BH3)2 to form trialkyl borane. This addition product is oxidized by hydrogen peroxide in the presence of aqueous sodium hydroxide to form propan-1-ol.

The structures and IUPAC names of monohydric phenols of molecular formula, $C_7 H_8 O$.

Phenol: $C_6 H_5 OH$

$C_7 H_8 O \rightarrow (C_6 H_5 OH)(C H_2)$

Due to inter-molecular H bonding in para-nitrophenol, it gets tightly bonded with water. But ortho nitrophenol has intra-molecular H bonding and hence is steam volatile.

Cumene (isopropylbenzene) is oxidised in the presence of air to form cumene hydroperoxide. On treating with dilute acid it is converted to phenol and acetone.

Chlorobenzene, when fused with NaOH, produces sodium phenoxide which on acidification produces Phenol.

Question

Ethanol is yielded from ethene by acid catalysed hydration.

The mechanism:

Step 1. Protonation of alkene to form carbocation by electrophilic attack of hydronium ion.

Step 2. Nucleophilic attack of water on carbocation.

Step 3. Deprotonation to form an alcohol.

When benzene reacts with $\dpi{80} conc. \:H_2SO_4$  and heat it gives benzene sulphonic acid ,and after sulphonic this acid  with NaOH then it gives phenol

Styrene on acid catalysed hydration gives 1-phenylethanol.

On adding NaOH to chloromethylcyclohexane, cyclohexy methanol is formed.

when 1-chloropentane reacts with NaOH it gives pantan-1-ol

1. Phenol reacts with sodium to give sodium phenoxide, liberating hydrogen gas.

2. Phenol reacts with sodium hydroxide to give sodium phenoxide and water.

Phenol is more acidic than ethanol. This is because phenol after losing a proton becomes phenoxide ion which is highly stable due to resonance whereas ethoxide ion does not.

Ortho-nitrophenol is more acidic than ortho-methoxyphenol. The presence of the nitro group, which is an electron withdrawing group, at the ortho position decreases the electron density in the O-H bond. Also, the o-nitrophenoxide ion formed after the loss of protons is stabile due to resonance. Hence, ortho nitrophenol is a stronger acid. Whereas the methoxy group is an electron-releasing group. Thus, it increases the electron density in the O-H bond.

The -OH group is an electron-donating group (EDG). Thus, it increases the electron density in the benzene ring in the resonance structure of phenol. As a result, the benzene ring is activated towards electrophilic substitution.

Oxidation of propane-1-ol with alkaline $KMnO_4$ solution gives propanoic acid.

$\\CH_3CH_2CH_2OH \rightarrow CH_3CH_2COOH\\propane-1-ol\:\:\:\:\:\:\:\:propanoic \:acid$

Question

(ii) Bromine in$CS_2$ with phenol.

Bromine in $CS_2$  with phenol produces a mixture of o-bromo phenol and p-bromo phenol is formed.

Question

(iii) Dilute HNO3 with phenol.

When dilute HNO3 reacts with phenol it gives o-bromo phenol and p-bromo phenol

Treating phenol with chloroform in presence of aqueous NaOH.

This reaction is known as the Reimer-Tiemann reaction.

Question

Kolbe’s reaction:  Phenol with carbon dioxide under pressure followed by treating the product with sulphuric acid produces Ortho-hydroxybenzoic acid (salicylic acid). Phenoxide ion generated is more reactive than phenol towards electrophilic aromatic substitution. Hence, it undergoes electrophilic substitution with carbon dioxide, a weak electrophile.

On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at the ortho position of the benzene ring. This reaction is known as Reimer - Tiemann reaction

Question

Williamson ether synthesis is a reaction forming ether from a primary alkyl halide via SN2 reaction.

Question

If the alkyl or aryl groups attached to the oxygen atom are different, then it is mixed or unsymmetrical ether.

Eg: $C_2H_5-O-CH_3\ and\ C_2H_5-O-C_6H_5$

Alcohols undergo dehydration (removal of a molecule of water) to form alkenes on treating with a protic acid. Ethanol undergoes dehydration by heating it with concentrated sulphuric acid at 443 K.

Acid catalysed hydration of propene produces propan-2-ol.

Benzyl chloride treated with NaOH followed by acidification produces benzyl alcohol.

Ethyl magnesium chloride treated with formaldehyde followed by hydrolysis produces propan-1-ol.

Methyl magnesium bromide treated with propane, gives 2-methylpropane-2-ol on hydrolysis.

Acidic/neutral/alkaline potassium permanganate $\dpi{100} (KMnO_4)$ or acidified $\dpi{100} K_2Cr_2O_7$

The reagent used for oxidation of primary alcohol to aldehyde is Pyridinium chlorochromate (PCC).

Reagents used in the bromination of phenol to 2,4,6-tribromophenol is Bromine water

Question

(iv) Benzyl alcohol to benzoic acid.

Reagent used in the benzyl alcohol to benzoic acid isAcidified $KMnO_4$  (potassium permanganate)

Reagents used in the dehydration of propan-2-ol to propene is Concentrated Phosphoric acid.

Reagent used in the butan-2-one to butan-2-ol is $LiAlH_4\ or\ NaBH_4$

Ethanol undergoes intermolecular hydrogen bonding due to the presence of -OH group. Therefore, extra energy is required to break those hydrogen bonds. Whereas methoxymethane does not make H-bonds and hence ethanol has a higher boiling point than methoxymethane.

Question

IUPAC names of the given ether is 1-Ethoxy-2-methylpropane

Question

IUPAC names of the given ether is 2-Chloro-1-methoxyethane

Question

IUPAC names of the given ether is 4-Nitroanisole

Question

IUPAC names of the given ether is 1-Methoxypropane

Question

IUPAC names of the given ether is 4-Ethoxy-1, 1-dimethylcyclohexane

Question

IUPAC names of the given ether is Ethoxybenzene

Names of reagents and equations for the preparation of the 1-Propoxypropane ether by Williamson’s synthesis:-

Names of reagents and equations for the preparation of the Ethoxybenzene ether by Williamson’s synthesis:-

with NaBr as side product.

Names of reagents and equations for the preparation of the2-Methoxy-2-methylpropane ether by Williamson’s synthesis:-

Names of reagents and equations for the preparation of the1-Methoxyethane ether by Williamson’s synthesis:-

Williamson synthesis involves $S_N2$ attack by alkoxide ion on a primary alkyl halide. But if secondary or tertiary alkyl halides are taken then alkenes would be produced because elimination would take place. This is because alkoxides are nucleophiles as well as strong bases.

Propan-1-ol on dehydration using protic acids such as sulphuric acid gives 1-propoxypropane.

Mechanism of this reaction:

Formation of protonated alcohol.

Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction.

Formation of ethene by the elimination of a proton.

The formation of ethers by dehydration of a primary alcohol is an SN2 reaction. In case of secondary or tertiary alcohols, the alkyl group is hindered and hence elimination dominates substitution. Therefore alkenes are formed in place of ethers.

1-propoxypropane reacts with HI to give propan-1-ol and 1-iodopropane as the products.

Methoxybenzene reacts with HI to give phenol and iodomethane.

Benzyl ethyl ether reacts with HI to give benzyl iodide and ethanol.

Due to the +R effect of the alkoxy group, it increases the electron density of the benzene ring pushing electrons into the ring making the benzene ring activated towards electrophilic substitution reactions.

The above resonating structures shows that the electron density increases more at the ortho and para positions as compared to the meta positions. Hence, the alkoxy group directs the incoming substituents to ortho and para positions in the benzene ring.

Following is the mechanism:

1. Protonation of methoxymethane

2. Nucleophilic attack of $I^-$

3. If HI is in excess, then methanol formed in step 2 reacts with another HI molecule and gets converted to methyl iodide at a high temperature.

Fridel Craft reaction(Alkylation):

Question

Nitration of anisole:

Bromination of anisole in ethanoic acid medium:

Friedel-Craft’s acetylation of anisole:

(i)

(ii)

(iii)

(iv)

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.

Mechanism for the  reaction  3-methylbutan-2-ol is treated with HBr

## NCERT Solutions Class 12 Chemistry

 Chapter 1 CBSE NCERT solutions for class 12 chapter 1 The Solid State Chapter 2 NCERT solutions for class 12 chemistry chapter 2 Solutions Chapter 3 Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry Chapter 4 CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics Chapter 5 Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry Chapter 6 NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements Chapter 7 CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements Chapter 8 Solutions of NCERT class 12 chemistry chapter 8 The d and f block elements Chapter 9 NCERT solutions for class 12 chemistry chapter 9 Coordination compounds Chapter 10 Solutions of NCERT class 12 chemistry chapter 10 Haloalkanes and Haloarenes Chapter 11 CBSE NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols and Ethers Chapter 12 Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids Chapter 13 NCERT solutions for class 12 chemistry chapter 13 Amines Chapter 14 CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules Chapter 15 Solutions of NCERT class 12 chemistry chapter 15 Polymers Chapter 16 NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

## NCERT Solutions for Class 12 Subject wise

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