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

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

Edited By Sumit Saini | Updated on Aug 16, 2022 01:18 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers - This article covers NCERT Class 12 Chemistry solutions chapter 11. 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 ethers.

Alcohols 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 an H-atom in a hydrocarbon by an alkoxy(R-O) or by an aryloxy(Ar-O) group. 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 for the CBSE Board exam and in competitive exams like JEE Mains, NEET etc.

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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. Alcohols Phenols and Ethers Class 12 also discuss reactions involved in the preparation of ethers from alcohols and alkyl halides. Scroll down to know more details about the NCERT solutions for Class 12 Chemistry Chapter 11 PDF download.

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

Solutions to In-Text Questions Ex 11.1 to 11.12

Question 11. 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 11.1 (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 11.1 (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 11.1 (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 11.1 (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 11.1 (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 11.2 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 11.3 (1) Name the following compounds according to IUPAC system.

1649963673264

Answer :

3-Chloromethyl-2-isopropylpentan-1-ol


Question 11.3 (2) Name the following compounds according to IUPAC system.

1649963717497

Answer :

2, 5-Dimethylhexane-1, 3-diol


Question 11.3 (3) Name the following compounds according to IUPAC system.

1649963822470

Answer :

3-Bromocyclohexanol


Question 11.3 (4) Name the following compounds according to IUPAC system.

1649963863346

Answer :

Hex-1-en-3-ol


Question 11.3 (5) Name the following compounds according to IUPAC system.

1649963909613

Answer :

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


Question 11.4 (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 11.5 (2) Write structures of the products of the following reactions: efwefew

1649964371846

Answer :

Product of the given reaction is-

1649964168123


Question 11.5 (3) Write structures of the products of the following reactions: sdgsfwe

Answer : 1649964224157

Product of the given reaction is

1649964252232


Question 11.6 (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 11.7(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 11.7 (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 11.8 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 11.10 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 11.11 Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?

1649965008877

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 11.12 (1) Predict the products of the following reactions:

CH_3 -CH_2 -CH_2 -O - CH_3 +HBr

Answer :

Reaction is

1649965070178


Question 11.12 (2) Predict the products of the following reactions:

1649965128401

Answer :

The reaction between ethoxybenzene and HBr is

1649965110827


Question 11.12 (3) Predict the products of the following reactions:

1649965169971

Answer :

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

1649965188865

Question 11.12 (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 11 Alcohols, Phenols and Ethers - Exercise Questions

Question 11.1 (1) Write IUPAC names of the following compounds:

1649965274378

Answer :

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


Question 11.1 (2) Write IUPAC names of the following compounds:

1649965319080

Answer :

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


Question 11.1 (3) Write IUPAC names of the following compounds:

1649965354978

Answer :

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


Question 11.1 (4) Write IUPAC names of the following compounds:

1649965421081

Answer :

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


Question 11.1 (5) Write IUPAC names of the following compounds:

1649965465500

Answer :

IUPAC name of the given compound is 2-Methylphenol


Question 11.1 (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 11.1 (7) Write IUPAC names of the following compounds:

1649965563584

Answer :

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

Question 11.1 (8) Write IUPAC names of the following compounds:

15949765483771594976545954

Answer :

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


Question 11.1 (9) Write IUPAC names of the following compounds:

1649965645747

Answer :

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


Question 11.1 (10) Write IUPAC names of the following compounds:

1649965690874

Answer :

IUPAC name of the given compound is Ethoxybenzene


Question 11.1 (11) Write IUPAC names of the following compounds:

1649965735782

Answer :

IUPAC name of the given compound is 1-Phenoxyheptane


Question 11.1 (12) Write IUPAC names of the following compounds:

15949765644041594976562819

Answer :

IUPAC name of the given compound is 2-Ethoxybutane


Question 11.2(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 11.2(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 11. 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
1519019423390745
Pentan-2-ol
1519019424935836
Pentan-3-ol
1519019421852179
3-Methylbutan-1-ol
1519019421074148
3-Methylbutan-1-ol
1519019422610673
2,2-Dimethylpropan-1-ol
1519019424169279
3-Methylbutan-2-ol
1519019425724622
2-Methylbutan-2-ol


Question 11. 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 11.4 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 11.5 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 11.6 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 (BH 3 ) 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 11.7 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 11.8 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 11.9 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 11.10 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 11.11 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 11.12 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 11.13 Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.

Answer :

Styrene on acid catalysed hydration gives 1-phenylethanol.

1649966607271


Question 11.13 (2) Show how will you synthesise:
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.

Answer :

On adding NaOH to chloromethylcyclohexane, cyclohexy methanol is formed.

1649966644075


Question 11.13 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 11.14 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.

1519019443005476

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

1519019443784771

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 11.15 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 11.16 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 11.17(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_3CH_2CH_2OH \rightarrow CH_3CH_2COOH\\propane-1-ol\:\:\:\:\:\:\:\:propanoic \:acid


Question 11.17(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 11.17(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 11.17(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 11.18(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 11.18(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 11.18(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 S N 2 reaction.

1649966930561


Question 11.18(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 11.19 . 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 11.20(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 11.20(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.

12th_chemistry_11_1_8_1466307644_14


Question 11.20(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 11.20(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 11.21(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 11.21(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 11.21(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 11.21(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 11.21(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 11.21(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 11.22 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 11.23(i) Give IUPAC names of the following ethers:

1649967122741

Answer :

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


Question 11.23(ii) Give IUPAC names of the following ethers:

1649967185868

Answer :

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


Question 11.23(iii) Give IUPAC names of the following ethers:

1649967223743

Answer :

IUPAC names of the given ether is 4-Nitroanisole


Question 11.23(iv) Give IUPAC names of the following ethers:

1649967263588

Answer :

IUPAC names of the given ether is 1-Methoxypropane


Question 11.23(v) Give IUPAC names of the following ethers:

1649967298676

Answer :

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


Question 11.23(vi) Give IUPAC names of the following ethers:

1649967335363

Answer :

IUPAC names of the given ether is Ethoxybenzene


Question 11.24(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 11.24(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 side product.


Question 11.24(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 the2-Methoxy-2-methylpropane ether by Williamson’s synthesis:-

1649967458991


Question 11.24(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 11.25 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.

12th_chemistry_11_1_8_1466308233_67


Question 11.26 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 11.27 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 S N 2 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 11.28(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 11.28(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 11.28(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 11.29(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 11.29(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.


Question 11.30. 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 11.31(ii) Write equations of the following reactions: (ii) Nitration of anisole.

Answer :

Nitration of anisole:

1649967923121


Question 11.31(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 11.31(iv) Write equations of the following reactions: (iv) Friedel-Craft’s acetylation of anisole.

Answer :

Friedel-Craft’s acetylation of anisole:

1649967984977


Question 11.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes
place:

15949764695001594976466780
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

Chemistry Chapter 11 NCERT Solutions Insights

Alcohols Phenols and Ethers Class 12 NCERT solutions has answers to a total of 12 topic wise questions and 33 questions in the exercise. In the 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 NCERT solutions for Class 12 Chemistry chapter 11 Alcohols, Phenols and Ethers here free of cost.

By referring to the NCERT solutions for class 12, students can understand all the important concepts and practice questions well enough before their examination. 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

NCERT Solutions Class 12 Chemistry Chapter Wise

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As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

JEE Main Important Physics formulas

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

Benefits of NCERT Solutions for class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers

  • The answers presented in a comprehensive manner in the NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols and Ethers will help in understanding chapter easily.
  • It will be easy to revise because the detailed solutions will help you to remember the concepts and fetch you good marks.
  • Homework problems won't bother you anymore, all you need to do is check the detailed NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols and Ethers and you are good to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. Where can I find complete solutions of NCERT class 12 Chemistry?

For complete solutions refer to this website: https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry 

2. What are the important topics of class 12 chemistry chapter 11 alcohols, phenols and ethers ?

Important topics:

  • Chemical Reactions of Alcohols, Phenols & Ethers
  • Physical Properties of Alcohols, Phenols and Ethers
  • Preparation of Alcohols
  • Chemical Reactions of Ethers
  • Physical Properties of Ethers
  • Preparation of Ethers
  • Preparation of Phenols
  • Some Commercially Important Alcohols
  • Introduction and Classification of Alcohols, Phenols & Ethers
  • Nomenclature
3. What is the weightage of NCERT class 12 Chemistry chapter 11 in NEET?

This chapter holds weightage of 4 percent in NEET examination. Practice NEET previous year papers, NCERT question and NCERT exemplar questions for a good score.

4. What is the weightage of NCERT class 12 Chemistry chapter 11 in CBSE Board Exam ?

Questions worth 5-6 marks are asked from this chapter in CBSE Board exam. Follow the NCERT syllabus for good score in board exam.

5. What is the weightage of NCERT class 12 Chemistry chapter 11 in JEE Main?

This chapter holds the weightage of 4-5 Marks in JEE Main

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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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

Option 4)

0.67\; J

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 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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

Option 1)

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

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

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

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

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

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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