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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

Edited By Shivani Poonia | Updated on Mar 27, 2025 03:43 PM IST | #CBSE Class 12th
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CBSE Class 12th  Exam Date : 29 Mar' 2025 - 29 Mar' 2025

Chemistry is something that gives us the reason behind things we observe around us whether it is the fizz of soft drinks or the working of medicines. Alcohols, Phenols, and Ethers are three important classes of compounds that feature oxygen-containing organic compounds that are widely used in the medical industry and daily life. 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.

This Story also Contains
  1. NCERT Solutions for Class 12 Chemistry Alcohols, Phenols and Ethers ( Intext Questions Ex 7.1 to 7.12)
  2. Chemistry Chapter 7 NCERT Solutions Insights
  3. Topics NCERT Class 12 Chemistry Alcohols, Phenols and Ethers
  4. NCERT Solutions Class 12 Chemistry Chapter Wise
  5. NCERT Solutions for Class 12 Subject-wise
  6. Also Check NCERT Books and NCERT Syllabus here:

Class 12 Chemistry chapter 7 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 hydrogen atom in hydrocarbons. NCERT Chapter 7 Class 12 Chemistry Solutions provides 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. The chapter also covers essential reaction mechanisms like dehydration of alcohol, electrophilic substitution in phenols, and Williamson’s synthesis of ethers.

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NCERT Solutions of class 12 chemistry Alcohols, Phenols, and Ether is one of the essential sources of preparation for organic chemistry whether students are preparing for class 12 CBSE boards or other competitive exams like NEET, JEE, BITSAT, etc. Solutions are prepared by our subject experts in a very comprehensive way that helps students understand the concepts used in solving questions in a very simple and easy way. Scroll down to learn more details about the NCERT solutions for Class 12 Chemistry Chapter 7.

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NCERT Solutions for Class 12 Chemistry Alcohols, Phenols and Ethers ( Intext Questions Ex 7.1 to 7.12)

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:

CH3CH2CH2OH

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.

1649963673264

Answer :

3-Chloromethyl-2-isopropylpentan-1-ol


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

1649963717497

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.

1649963863346

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


1649964024243

Answer :

1649964043498

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

1649964371846

Answer :

Product of the given reaction is-

1649964168123

(a) HCLZnCl2 with Butan-1-ol

Answer :

Primary alcohols do no react with Lucas’ reagent.

Hence no reaction.


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

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?

1649965008877

Answer :

(ii) is appropriate because CH3Br is nucleophile whereas CH3ONa is also a strong base. So elimination will be dominating in (i).

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.H2SO4+conc.HNO3

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 Alcohols, Phenols and Ethers - Exercise Questions

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(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(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 7.3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O
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 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 (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 7.7 Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O .

Answer :

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

Phenol: C6H5OH

C7H8O(C6H5OH)(CH2)

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.H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents .

Answer :

When benzene reacts with conc.H2SO4 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 SN2 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.

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 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 KMnO4 solution.

Answer :

Oxidation of propane-1-ol with alkaline KMnO4 solution gives propanoic acid.

CH3CH2CH2OHCH3CH2COOHpropane1olpropanoicacid


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

(ii) Bromine in CS2 with phenol.

Answer :

Bromine in CS2 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 S N 2 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: C2H5OCH3 and C2H5OC6H5


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 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 Benzyl alcohol.

Answer :

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

12th_chemistry_11_1_8_1466307644_14

Question 7.20(iii) How are the following conversions carried out?
(iii) Ethyl magnesium chloride 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 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 (KMnO4) or acidified K2Cr2O7


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 KMnO4 (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 LiAlH4 or NaBH4


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 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 the2-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 SN2 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 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 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 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(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 7.33 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

Chemistry Chapter 7 NCERT Solutions Insights

Alcohols Phenols and Ethers Class 12 NCERT solutions have answers to a total of 12 topic-wise questions and 33 questions in the exercise. In the CBSE board 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 7 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 are the basic compound for the formation of fragrances.

Topics NCERT Class 12 Chemistry Alcohols, Phenols and Ethers

7.1 Classification

7.2 Nomenclature

7.3 Structures of Functional Groups

7.4 Alcohols and Phenols

7.5 Some Commercially Important Alcohols

7.6 Ether

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Frequently Asked Questions (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 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 7 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 7 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 Alcohols, Phenols & Ethers 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 ?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.
  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
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How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
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Get Results: In just two days, the results are made public.
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Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

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Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

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Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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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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