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NCERT solutions for Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers

NCERT solutions for Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers

#CBSE Class 12th
Shivani PooniaUpdated on 05 Sep 2025, 11:45 AM IST

Have you ever wondered the things like soft drinks, perfumes, antiseptics, and disinfectants that we use in our daily lives come from? So, all of these come from oxygen-containing organic compounds like alcohol, such as ethanol used in soft drinks, beverages, and sanitizers, while Phenols are used in antiseptics and disinfectants, and Ether is used in the manufacturing of perfumes. These alcohol phenol and ether ncert solutions discusses three important classes of compounds that feature oxygen-containing organic compounds which are widely used in the medical industry and in daily life.

This Story also Contains

  1. NCERT Solutions for Class 12 Chemistry Chapter 7 Download PDF
  2. NCERT Solutions for Class 12 Chemistry Chapter 7 (Intext Questions Exercise)
  3. NCERT Solutions for Class 12 Chemistry Chapter 7 (Exercise Questions with Answers)
  4. Class 12 Chemistry NCERT Chapter 7: Higher Order Thinking Skills (HOTS) Questions
  5. Approach to Solve Questions of Class 12 Chemistry Chapter 7
  6. Topics and Subtopics Covered in the NCERT Textbook
  7. What Extra Should Students Study Beyond the NCERT for JEE?
  8. NCERT Solutions Class 12 Chemistry Chapter-Wise
  9. NCERT Solutions for Class 12 Subject-wise
  10. NCERT Books and NCERT Syllabus
NCERT solutions for Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers
Alcohols, Phenols and Ethers

The NCERT Solution for Class 12 Chemistry explains various concepts that we observe in our daily lives. When the OH group replaces the hydrogen atom in aliphatic hydrocarbons, Alcohols are formed same goes for phenols when the OH group replaces the hydrogen of aromatic hydrocarbons, then Phenols are formed while Ethers are formed when alkoxy (R-O) or aryloxy (Ar-O) group substitutes the hydrogen atom in hydrocarbons. These NCERT Solutions provide a structured approach to understanding the classification, properties, preparation, and reactions. Students will learn about important concepts like nomenclature, structure, and uses of organic compounds containing oxygen. This article includes some higher-order thinking skills (HOTS) questions and approach to solve questions.

Also Read:

NCERT Solutions for Class 12 Chemistry Chapter 7 Download PDF

Students can download the NCERT Solutions for Class 12 Chemistry Chapter 7 PDF for free. These solutions are designed to help you understand the fundamental concepts and solve textbook questions with ease.

Download PDF

NCERT Solutions for Class 12 Chemistry Chapter 7 (Intext Questions Exercise)

Detailed alcohol phenol and ether ncert solutions for intext questions are given below. Download the free PDF and prepare for your exams with confidence.

Question 7.1 (1) Classify the following as primary, secondary and tertiary alcohols:

int1

Answer :

To classify we look at the OH bonded carbon.

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

Question 7.1 (2) Classify the following as primary, secondary and tertiary alcohols:

int2

Answer :

To classify we look at the OH bonded carbon.

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

Question 7.1 (3) Classify the following as primary, secondary and tertiary alcohols:

$CH_3 -CH_2- CH_2 -OH$

Answer :

To classify we look at the OH bonded carbon.

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

Question 7.1 (4) Classify the following as primary, secondary and tertiary alcohols:

int4

Answer :

To classify we look at the OH bonded carbon.

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

Question 7.1 (5) Classify the following as primary, secondary and tertiary alcohols:

int5

Answer :

To classify we look at the OH bonded carbon.

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

Question 7.1 (6) Classify the following as primary, secondary and tertiary alcohols :

1649963634534

Answer:

To classify we look at the OH bonded carbon.

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

Question 7.2 Identify allylic alcohols in the above examples.

Answer :

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

Question 7.3 (1) Name the following compounds according to IUPAC system.

Answer :

3-Chloromethyl-2-isopropylpentan-1-ol

Question 7.3 (2) Name the following compounds according to IUPAC system.

Answer :

2, 5-Dimethylhexane-1, 3-diol

Question 7.3 (3) Name the following compounds according to IUPAC system.

1649963822470

Answer :

3-Bromocyclohexanol

Question 7.3 (4) Name the following compounds according to IUPAC system.

Answer :

Hex-1-en-3-ol

Question 7.3 (5) Name the following compounds according to IUPAC system.

1649963909613

Answer :

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

Question 7.4 (1) Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?

1649963974883

Answer :

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

1649963954035

Question 7.4 (2) Show how are the following alcohols prepared by the reaction of a suitable
Grignard reagent on methanal ?

1649964024243

Answer :

1649964043498

Question 7.5 (1) Write structures of the products of the following reactions:

Answer :

1649964112055

Question 7.5 (2) Write structures of the products of the following reactions:

1649964371846

Answer :

The product of the given reaction is-

1649964168123

Question 7.5 (3) Write structures of the products of the following reactions:

Answer : 1649964224157

Product of the given reaction is

1649964252232

Question 7.6 (1) Give structures of the products you would expect when each of the following alcohol reacts with

(a) $HCL- ZnCl_2$ with Butan-1-ol

Answer :

Primary alcohols do no react with Lucas’ reagent.

Hence no reaction.

Question 7.6 (2) Give structures of the products you would expect when each of the following alcohol reacts with $HCl - ZnCl_2$ with 2-Methylbutan-2-ol

Answer :

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

1649964290479

Question7.6 (3) Give structures of the products you would expect when each of the following alcohol reacts with HBr with Butan-1-ol

Answer :

Reaction of HBr with Butan-1-ol

1649964332727

Question 7.6 (4) Give structures of the products you would expect when each of the following alcohol reacts with HBr with 2-Methylbutan-2-ol

Answer :

Reaction of HBr with 2-Methylbutan-2-ol

1649964449101

Question 7.6 (5) Give structures of the products you would expect when each of the following alcohol reacts with $SOCl_2$ with Butan-1-ol

Answer :

Reaction of $SOCl_2$ with Butan-1-ol

1649964487196

Question 7.6 (6) Give structures of the products you would expect when each of the following alcohol reacts with $SoCl_2$ with 2-Methylbutan-2-ol

Answer :

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

1649964525510

Question 7.7(1) Predict the major product of acid catalysed dehydration of 1-methylcyclohexanol

Answer :

Dehydration of 1-methylcyclohexanol

1649964563324

1-Methylcyclohexene is the major product.

Question 7.7 (2) Predict the major product of acid catalysed dehydration of butan-1-ol

Answer :

Dehydration of butan-1-ol

1649964596061

But-2-ene is the major product.

Question 7.8 Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

Answer :

Resonance structure of ortho-nitrophenol

1649964631687

Resonance structure of para-nitrophenol

1649964652729

Question 7.9 (1) Write the equations involved in the following reactions: Reimer - Tiemann reaction

Answer :

Reimer - Tiemann reaction

1649964703424

Question 7.9 (2) Write the equations involved in the following reactions: Kolbe’s reaction

Answer :

Kolbe’s reaction

1649964746071

Question 7.10 Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Answer :

1.Reaction of ethanol with hydrogen bromide.

1649964791076

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

1649964898068

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

1649964914852

Question 7.11 Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?

Answer :

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

Question 7.12 (1) Predict the products of the following reactions:

$CH_3 -CH_2 -CH_2 -O - CH_3 +HBr$

Answer :

Reaction is

1649965070178

Question 7.12 (2) Predict the products of the following reactions:

1649965128401

Answer :

The reaction between ethoxybenzene and HBr is

1649965110827

Question 7.12 (3) Predict the products of the following reactions:

1649965169971

Answer :

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

1649965188865

Question 7.12 (4) Predict the products of the following reactions:

1649965226084

Answer :

Reaction between ter - butyl ethyl ether and HI

1649965242940

NCERT Solutions for Class 12 Chemistry Chapter 7 (Exercise Questions with Answers)

Given below the clear and simple solutions to all exercise questions of Class 12 Chemistry Chapter 7. Download and study these answers to prepare well for exams.

Question 7.1 (1) Write IUPAC names of the following compounds:

1649965274378

Answer :

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

Question 7.1 (2) Write IUPAC names of the following compounds:

1649965319080

Answer :

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

Question 7.1 (3) Write IUPAC names of the following compounds:

1649965354978

Answer :

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

Question 7.1 (4) Write IUPAC names of the following compounds:

1649965421081

Answer :

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

Question 7.1 (5) Write IUPAC names of the following compounds:

1649965465500

Answer :

IUPAC name of the given compound is 2-Methylphenol

Question 7.1 (6) Write IUPAC names of the following compounds:

1649965524003

Answer :

IUPAC name of the given compound is 4-Methylphenol

Note : Also called p-cresol

Question 7.1 (7) Write IUPAC names of the following compounds:

1649965563584

Answer :

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

Question 7.1 (8) Write IUPAC names of the following compounds:

1594976548377

Answer :

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

Question 7.1 (9) Write IUPAC names of the following compounds:

1649965645747

Answer :

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

Question 7.1 (10) Write IUPAC names of the following compounds:

1649965690874

Answer :

IUPAC name of the given compound is Ethoxybenzene

Question 7.1 (11) Write IUPAC names of the following compounds:

1649965735782

Answer :

IUPAC name of the given compound is 1-Phenoxyheptane

Question 7.1 (12) Write IUPAC names of the following compounds:

1594976564404

Answer :

IUPAC name of the given compound is 2-Ethoxybutane

Question 7.2 Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol

Answer :

Structure of 2-Methylbutan-2-ol

1649965777770

Question 7.2 Write structures of the compounds whose IUPAC names are as follows: (ii) 1-Phenylpropan-2-ol

Answer :

Structure of 1-Phenylpropan-2-ol

1649965829440

Question 7.2(iii) Write structures of the compounds whose IUPAC names are as follows: (iii) 3,5-Dimethylhexane –1, 3, 5-triol

Answer :

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

1649965957466

Question 7.2(iv) Write structures of the compounds whose IUPAC names are as follows: (iv) 2,3 – Diethylphenol

Answer :

structure of 2,3 – Diethylphenol

1649965990130

Question 7.2(v) Write structures of the compounds whose IUPAC names are as follows:
(v) 1 – Ethoxypropane

Answer :

structure of 1 – Ethoxypropane

1649966018689

Question 7.2(vi) Write structures of the compounds whose IUPAC names are as follows:
(vi) 2-Ethoxy-3-methylpentane

Answer :

structure of 2-Ethoxy-3-methylpentane

1649966071934

Question 7.2(vii) Write structures of the compounds whose IUPAC names are as follows:
(vii) Cyclohexylmethanol

Answer :

Structure of Cyclohexylmethanol

1649966113638

Question 7.2(viii) Write structures of the compounds whose IUPAC names are as follows:
(viii) 3-Cyclohexylpentan-3-ol

Answer :

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

1649966143832

Question 7.2(ix) Write structures of the compounds whose IUPAC names are as follows:
(ix) Cyclopent-3-en-1-ol

Answer :

structure of Cyclopent-3-en-1-ol

1649966178429

Question 7.2(x) Write structures of the compounds whose IUPAC names are as follows:
(x) 4-Chloro-3-ethylbutan-1-ol.

Answer :

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

1649966222407

Question 7.3 (i) Draw the structures of all isomeric alcohols of molecular formula $C_5 H_{12}O$
and give their IUPAC names.

Answer :

The structures of all isomeric alcohols of C 5 H 12 O are given below:

1649966270037
Pentan-1-ol
Pentan-2-ol
1519019424935836
Pentan-3-ol
3-Methylbutan-1-ol
2,2-Dimethylpropan-1-ol
3-Methylbutan-2-ol
2-Methylbutan-2-ol

Question 7.3 (ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols.

Answer :

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

Question 7.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

Answer :

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.

Question 7.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Answer :

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.

Question 7.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

Answer :

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.

Question 7.7 Give the structures and IUPAC names of monohydric phenols of molecular formula, $C_7 H_8 O$ .

Answer :

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

Phenol: $C_6 H_5 OH$

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

1649966313305

Question 7.8 While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Answer :

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.

Question 7.9 Give the equations of reactions for the preparation of phenol from cumene.

Answer :

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.

1649966356497

Question 7.10 Write chemical reaction for the preparation of phenol from chlorobenzene.

Answer :

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

1649966463015

Question 7.11 Write the mechanism of hydration of ethene to yield ethanol.

Answer :

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.

1649966496746

Step 2. Nucleophilic attack of water on carbocation .

1649966514431

Step 3. Deprotonation to form an alcohol.

1649966531637

Question 7.12 You are given benzene, $conc. H_2SO_4$ and NaOH. Write the equations for the preparation of phenol using these reagents .

Answer :

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

1649966572842

Question 7.13 Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.

Answer :

Styrene on acid catalysed hydration gives 1-phenylethanol.

1649966607271

Question 7.13 (2) Show how will you synthesise:(ii) cyclohexylmethanol using an alkyl halide by an $SN^{2}$ reaction.

Answer :

On adding NaOH to chloromethylcyclohexane, cyclohexy methanol is formed.

1649966644075

Question 7.13 Show how will you synthesise:
(iii) pentan-1-ol using a suitable alkyl halide?

Answer :

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

1649966679228

Question 7.14 Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Answer :

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.

Question 7.15 Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?

Answer :

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.

Question 7.16 Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Answer :

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.

Question 7.17(i) Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline $KMnO_4$ solution.

Answer :

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

$\\CH_{3}CH_{2}CH_{2}OH \rightarrow CH_{3}CH_{2}COOH\\$

Question 7.17(ii) Give equations of the following reactions:

(ii) Bromine in $CS_2$ with phenol.

Answer :

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

1649966712433

Question 7.17(iii) Give equations of the following reactions:

(iii) Dilute HNO3 with phenol.

Answer :

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

1649966745746

Question 7.17(iv) Give equations of the following reactions:
(iv) Treating phenol with chloroform in presence of aqueous NaOH.

Answer :

Treating phenol with chloroform in presence of aqueous NaOH.

1649966793128

This reaction is known as the Reimer-Tiemann reaction.

Question 7.18(i) Explain the following with an example.
(i) Kolbe’s reaction.

Answer :

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.

1649966822710

Question 7.18(ii) Explain the following with an example.
(ii) Reimer-Tiemann reaction.

Answer :

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

1649966859823

Question 7.18(iii) Explain the following with an example. (iii) Williamson ether synthesis.

Answer :

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

1649966930561

Question 7.18(iv) Explain the following with an example.(iv) Unsymmetrical ether.

Answer :

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

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

Question 7.19 . Write the mechanism of acid dehydration of ethanol to yield ethene.

Answer :

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.

1649966962919

Question 7.20(i) How are the following conversions carried out?
(i) Propene $\rightarrow$ Propan-2-ol.

Answer :

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

1649966998750

Question 7.20(ii) How are the following conversions carried out?
(ii) Benzyl chloride $\rightarrow$ Benzyl alcohol.

Answer :

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

Question 7.20(iii) How are the following conversions carried out?
(iii) Ethyl magnesium chloride $\rightarrow$ Propan-1-ol.

Answer :

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

1649967048713

Question 7.20(iv) How are the following conversions carried out?
(iv) Methyl magnesium bromide $\rightarrow$ 2-Methylpropan-2-ol.

Answer :

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

1649967086833

Question 7.21(i) Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.

Answer :

Acidic/neutral/alkaline potassium permanganate $(KMnO_4)$ or acidified $K_2Cr_2O_7$

Question 7.21(ii) Name the reagents used in the following reactions:
(ii) Oxidation of a primary alcohol to aldehyde.

Answer :

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

Question 7.21(iii) Name the reagents used in the following reactions:
(iii) Bromination of phenol to 2,4,6-tribromophenol.

Answer :

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

Question 7.21(iv) Name the reagents used in the following reactions:

(iv) Benzyl alcohol to benzoic acid.

Answer :

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

Question 7.21(v) Name the reagents used in the following reactions:
(v) Dehydration of propan-2-ol to propene.

Answer :

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

Question 7.21(vi) Name the reagents used in the following reactions:
(vi) Butan-2-one to butan-2-ol.

Answer :

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

Question 7.22 Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

Answer :

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 7.23(i) Give IUPAC names of the following ethers:

1649967122741

Answer :

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

Question 7.23(ii) Give IUPAC names of the following ethers:

1649967185868

Answer :

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

Question 7.23(iii) Give IUPAC names of the following ethers:

1649967223743

Answer :

IUPAC names of the given ether is 4-Nitroanisole

Question 7.23(iv) Give IUPAC names of the following ethers:

1649967263588

Answer :

IUPAC names of the given ether is 1-Methoxypropane

Question 7.23(v) Give IUPAC names of the following ethers:

1649967298676

Answer :

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

Question 7.23(vi) Give IUPAC names of the following ethers:

1649967335363

Answer :

IUPAC names of the given ether is Ethoxybenzene

Question 7.24(i) Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane

Answer :

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

1649967375627

Question 7.24(ii) Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(ii) Ethoxybenzene

Answer :

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

1649967410030 with NaBr as a side product.

Question 7.24(iii) Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(iii) 2-Methoxy-2-methylpropane

Answer :

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

1649967458991

Question 7.24(iv) Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis: (iv) 1-Methoxyethane

Answer :

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

1649967498923

Question 7.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.

Answer :

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.

Question 7.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.

Answer :

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

Mechanism of this reaction:

Formation of protonated alcohol.

1649967562663

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

1649967580906

Formation of ethene by the elimination of a proton.

1649967599892

Question 7.27 Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Answer :

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.

Question 7.28(i) Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane

Answer :

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

1649967642248

Question 7.28(ii) Write the equation of the reaction of hydrogen iodide with: (ii) methoxybenzene

Answer :

Methoxybenzene reacts with HI to give phenol and iodomethane.

1649967678156

Question 7.28(iii) Write the equation of the reaction of hydrogen iodide with: (iii) benzyl ethyl ether.

Answer :

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

1649967720536

Question 7.29(i) Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution

Answer :

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.

Question 7.29(ii) Explain the fact that in aryl alkyl ethers (ii) it directs the incoming substituents to ortho and para positions in benzene ring.

Answer :

1649967762793

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.

7.30. Write the mechanism of the reaction of HI with methoxymethane.

Answer :

Following is the mechanism:

1. Protonation of methoxymethane

1649967808094

2. Nucleophilic attack of $I^-$

1649967828420

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.

1649967848758

Question 7.31(i) Write equations of the following reactions:
(i) Friedel-Crafts reaction – alkylation of anisole.

Answer :

Fridel Craft reaction(Alkylation):

1649967890235

Question 7.31(ii) Write equations of the following reactions: (ii) Nitration of anisole.

Answer :

Nitration of anisole:

1649967923121

Question 7.31(iii) Write equations of the following reactions: (iii) Bromination of anisole in ethanoic acid medium.

Answer :

Bromination of anisole in ethanoic acid medium:

1649967952984

Question 7.31(iv) Write equations of the following reactions: (iv) Friedel-Craft’s acetylation of anisole.

Answer :

Friedel-Craft’s acetylation of anisole:

1649967984977

Question 7.32 Show how would you synthesise the following alcohols from appropriate alkenes?

1594976641206

Answer :

(i) 1649968028846

(ii) 1649968047019

(iii) 1649968066505

(iv) 1649968087729

Question 7.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes
place:

1594976469500
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.

Answer :

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

1649968132938

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Class 12 Chemistry NCERT Chapter 7: Higher Order Thinking Skills (HOTS) Questions

Selected Higher Order Thinking Skills questions from Chapter 7 to strengthen your concepts are given below. These NCERT Solutions for Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers will help you practise advanced problems and improve exam readiness.

Question 1. Which one of the following, with HBr, will give a phenol?

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Answer:

Hence, the correct answer is option (2).

Question 2. Select the incorrect statement about Phenol.

(1) Phenol is a white crystalline solid.

(2) Phenol is soluble in hot water

(3) Phenol does not have antifungal and antibacterial properties.

(4) Phenol is soluble in NaOH.

Answer:

Properties of phenols:

  1. Phenols are either colourless liquids or white crystalline solids.
  2. Phenols are soluble in hot water.
  3. The boiling points of phenols are much higher than the corresponding aromatic hydrocarbons and haloarenes.
  4. Phenols have antifungal and antibacterial properties. Thus used as disinfectants and antiseptics.
  5. Phenols are poisonous in nature but act as antiseptics and disinfectants.
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Hence, the correct answer is option (3).

Question 3. Given below are two statements:
Statement I:
The acidic strength of monosubstituted nitrophenol is higher than phenol because of the electron-withdrawing nitro group.
Statement II:
o-nitrophenol, m-nitrophenol and p-nitrophenol will have the same acidic strength as they have one nitro group attached to the phenolic ring

In light of the above statements, choose the most appropriate answer from the options given below:

Both Statement I and Statement II are incorrect.

Statement I is correct, but Statement II is incorrect.

Statement I is incorrect, but Statement II is correct.

Both Statement I and Statement II are correct.

Answer:

Electron-withdrawing group, O-Phenol increases its acidic strength

Statement I is correct

Now, the pKa values of nitro-substituted Phenols are given below:

Thus acidic strength of nitrophenols is

Para > Ortho > Meta

Hence, statement I is correct while statement II is incorrect

Hence, the correct answer is option (2).

CBSE Class 12th Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters

Approach to Solve Questions of Class 12 Chemistry Chapter 7

Alcohols, Phenols and Ethers NCERT Solutions is an essential study material that is required for all students preparing for Class 12 board exams as well as competitive exams. Here are some tips that students can follow to effectively solve questions:

1. Understand about nomenclature

Questions related to nomenclature are often asked in exams, hence knowing the IUPAC naming rules for alcohols, phenols, and ethers becomes important. For ethers, name the smaller alkyl group as a prefix, followed by the main chain

2. Learn about methods of preparation

Preparation reactions of Alcohols, Phenols, and Ethers from haloalkanes, alkenes, carbonyl compounds, etc. Learn about reagents and conditions, like the hydration of alkenes and Williamson synthesis for ethers

3. Physical Properties

Compare boiling points, solubility, and hydrogen bonding.

4. Reactivity Order and Mechanisms:

Questions related to reaction mechanisms play a very important role and are often asked in exams. Practice reaction mechanisms like dehydration of alcohols, oxidation, and acidic behaviour of phenols.

5. Practice Questions:

Students can refer to the solved examples in the textbook and then solve the in-text questions. Students can also access the NCERT Solutions for Class 12 Chemistry Chapter 7 PDF for quick revision of the concepts.

Topics and Subtopics Covered in the NCERT Textbook

All the topics and subtopics covered in the NCERT Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers are listed below:

7.1 Classification

7.1.1 Alcohols - Mono, Di, Tri, or Polyhydric Alcohols

7.1.2 Phenols- Mono, Di, and Trihydric Phenols

7.1.3 Ethers

7.2 Nomenclature

7.3 Structures of Functional Groups

7.4 Alcohols and Phenols

7.4.1 Preparation of Alcohols

7.4.2 Preparation of Phenols

7.4.3 Physical Properties

7.4.4 Chemical Reactions

7.5 Some Commercially Important Alcohols

7.6 Ethers

7.6.1 Preparation of Ethers

7.6.2 Physical Properties

7.6.3 Chemical Reactions

What Extra Should Students Study Beyond the NCERT for JEE?

Students need to focus on important concepts and reactions according to the table given below:

Grignard Reagent - 1✅✅
Grignard Reagent - 2✅✅
Reduction of Anhydrides and Esters✅❌
Reduction by LiAlH4 and NaBH4✅✅
Properties of Alcohols✅✅
Acylation and Oxidation of Alcohols✅✅
Haloform Reaction✅✅
Pinacol Pinacolone Rearrangement✅❌
Oxidative Cleavage with HIO4✅❌
Preparation of Phenol(I)✅✅
Preparation of Phenol(II)✅✅
Properties of Phenols✅✅
Reactions due to (-OH) group✅✅
Claisen's Rearrangement✅❌
Reaction with Conc. HNO3, Phthalic Anhydride✅✅
Reaction of Phenols with dil. HNO3✅❌
Williamson's Ether Synthesis✅✅

NCERT Solutions Class 12 Chemistry Chapter-Wise

NCERT chapter-wise solutions are given below:

NCERT solutions for class 12 Chemistry chapter 1 Solutions

NCERT Solutions for class 12 Chemistry chapter 2 Electrochemistry

NCERT solutions for class 12 Chemistry chapter 3 Chemical Kinetics

NCERT solutions for class 12 Chemistry chapter 4 The d and f block elements

NCERT solutions for class 12 Chemistry chapter 5 Coordination compounds

NCERT Solutions for class 12 Chemistry chapter 6 Haloalkanes and Haloarenes

NCERT solutions for class 12 Chemistry chapter 7 Alcohols, Phenols, and Ethers

NCERT Solutions for class 12 Chemistry chapter 8 Aldehydes, Ketones and Carboxylic Acids

NCERT solutions for class 12 Chemistry chapter 9 Amines

NCERT solutions for class 12 Chemistry chapter 10 Biomolecules

NCERT Solutions for Class 12 Subject-wise

NCERT Subject-wise solutions are given below:

NCERT Solutions for class 12 biology

NCERT solutions for class 12 maths

NCERT solutions for class 12 chemistry

NCERT Solutions for class 12 physics

NCERT Books and NCERT Syllabus

The NCERT books and syllabus links for class 12 are given below:

NCERT Books Class 12 Chemistry

NCERT Syllabus Class 12 Chemistry

NCERT Books Class 12

NCERT Syllabus Class 12

Frequently Asked Questions (FAQs)

Q: What are the main types of alcohols discussed in Class 12 Chemistry Chapter 7?
A:

Chapter 7,  discusses three types of alcohols primary, secondary, and tertiary alcohols. Primary alcohols have the hydroxyl group attached to a carbon atom that is bonded to only one other carbon. Secondary alcohols are connected to two other carbons, while tertiary alcohols have the hydroxyl group on a carbon that is connected to three other carbons.

Q: How are phenols different from alcohols?
A:

Phenols are compounds where the hydroxyl group is directly bonded to an aromatic benzene ring, whereas alcohols have the OH group attached to aliphatic carbon chains. This structural difference leads to varying physical and chemical properties.  

Q: What is the significance of the hydroxyl group in alcohols and phenols as discussed in NCERT Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers ?
A:

The hydroxyl group is crucial as it imparts unique properties to alcohols and phenols. It makes these compounds polar, allowing them to form hydrogen bonds with water, which contributes to their solubility in water.  

Q: What methods are commonly used for the preparation of alcohols?
A:

Common methods for preparing alcohols include fermentation, which converts sugars into ethanol; reduction of carbonyl compounds using reducing agents and hydration of alkenes in the presence of an acid.  

Q: What are the hazards of alcohols, phenols, and ethers?
A:

Hazards of alcohols, phenols, and ethers:

  • Alcohols can be harmful if ingested, inhaled, or absorbed through the skin.
  • Phenols can cause skin and eye irritation, and these are also harmful if inhaled. 
  • Ethers can cause dizziness and respiratory irritation. They can also form explosive peroxides if stored improperly.
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Can i improve my 12 result this year in PCM

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

Read Complete Answer
R
Rahul Mandal
8 Sep'25
how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

Read Complete Answer
G
gopi
4 Sep'25
Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

  • B.Tech in Computer Science Engineering (CSE) – most popular choice.

  • BCA (Bachelor of Computer Applications) – good for software and IT jobs.

  • B.Sc. Computer Science / IT – good for higher studies and research.

  • B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

Read Complete Answer
P
Prachi Kumari
4 Sep'25
I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

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G
gopi
3 Sep'25
12 arts previous years board paper 2025

To find Class 12 Arts board papers, go to the official website of your education board, then click on the Sample Papers, Previous Years Question Papers(PYQ) or Model Papers section, and select the Arts stream. You will find papers for the various academic year. You can then select the year of which you want to solve and do your practice. There are many other educational websites that post pyqs on their website you can also visit that.

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A
Anshika Gupta
10 Sep'25
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

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2.45×10−3 kg

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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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2,000 \; J - 5,000\; J

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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

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0.02

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3.125 × 10-2
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1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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decrease twice

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increase two fold
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less than 3

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more than 3 but less than 6
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