NCERT Solutions for Class 12 Chemistry Chapter 7 Alcohols, Phenols Phenols and Ethers
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Page 1
NCERT
SOLUTIONS
CLASS - 12th
aglase .co
SOLUTIONS
CLASS - 12th
aglase .co
Page 2
Class : 12th
Subject : Chemistry
Chapter : 11
Chapter Name : Alcohols, Phenols Phenols and Ethers
Q11.1 Classify the following as primary, secondary and tertiary alcohols:
Answer. Primary alcohol (i), (ii), (iii)
Secondary alcohol (iv), (v)
Tertiary alcohol (vi)
Page : 325 , Block Name : Intext Questions
Q11.2 Identify allylic alcohols in the above question.
Answer. The alcohols given in (ii) and (vi) are allylic alcohols.
Page : 325 , Block Name : Intext Questions
Q11.3 Name the following compounds according to IUPAC system.
Subject : Chemistry
Chapter : 11
Chapter Name : Alcohols, Phenols Phenols and Ethers
Q11.1 Classify the following as primary, secondary and tertiary alcohols:
Answer. Primary alcohol (i), (ii), (iii)
Secondary alcohol (iv), (v)
Tertiary alcohol (vi)
Page : 325 , Block Name : Intext Questions
Q11.2 Identify allylic alcohols in the above question.
Answer. The alcohols given in (ii) and (vi) are allylic alcohols.
Page : 325 , Block Name : Intext Questions
Q11.3 Name the following compounds according to IUPAC system.
Page 3
Answer. (i) 3-Chloromethyl-2-isopropylpentan-1-ol
(ii) 2, 5-Dimethylhexane-1, 3-diol
(iii) 3-Bromocyclohexanol
(iv) Hex-1-en-3-ol
(v) 2-Bromo-3-methylbut-2-en-1-ol
Page : 328 , Block Name : Intext Questions
Q11.4 Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal?
(ii) 2, 5-Dimethylhexane-1, 3-diol
(iii) 3-Bromocyclohexanol
(iv) Hex-1-en-3-ol
(v) 2-Bromo-3-methylbut-2-en-1-ol
Page : 328 , Block Name : Intext Questions
Q11.4 Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal?
Page 4
Page : 328 , Block Name : Intext Questions
Q11.5 Write structures of the products of the following reactions:
+
H2 O/H
(i) CH 3 − CH = CH2 →
(ii)
(iii)
Answer. Missing
Page : 328 , Block Name : Intext Questions
Q11.6 Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl –
ZnCl (b) HBr and (c) SOCl .
2 2
(i) Butan-1-ol
(ii) 2-Methylbutan-2-ol
Q11.5 Write structures of the products of the following reactions:
+
H2 O/H
(i) CH 3 − CH = CH2 →
(ii)
(iii)
Answer. Missing
Page : 328 , Block Name : Intext Questions
Q11.6 Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl –
ZnCl (b) HBr and (c) SOCl .
2 2
(i) Butan-1-ol
(ii) 2-Methylbutan-2-ol
Page 5
Page : 343 , Block Name : Intext Questions
Q11.7 Predict the major product of acid catalysed dehydration of
(i) 1-methylcyclohexanol and (ii) butan-1-ol
Q11.7 Predict the major product of acid catalysed dehydration of
(i) 1-methylcyclohexanol and (ii) butan-1-ol
Page 6
Page : 343 , Block Name : Intext Questions
Q11.8 Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the
corresponding phenoxide ions.
Q11.8 Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the
corresponding phenoxide ions.
Page 7
Resonance structures of m-nitrophenoxide ion. It can be observed that the presence of nitro groups increases the
stability of phenoxide Ion.
Page : 343 , Block Name : Intext Questions
Q11.9 Write the equations involved in the following reactions:
(i) Reimer - Tiemann reaction
(ii) Kolbe’s reaction
stability of phenoxide Ion.
Page : 343 , Block Name : Intext Questions
Q11.9 Write the equations involved in the following reactions:
(i) Reimer - Tiemann reaction
(ii) Kolbe’s reaction
Page 8
Page : 343 , Block Name : Intext Questions
Q11.10 Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-
methylpentan-2-ol.
Answer. In Williamson synthesis, an alkyl halide reacts with an alkoxide ion. Also, it is an SN2 reaction. In the
reaction, alkyl halides should be primary having the least steric hindrance. Hence, an alkyl halide is obtained from
ethanol and alkoxide ion from 3- methylpentan-2-ol.
Page : 350 , Block Name : Intext Questions
Q11.11 Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene
and why?
Q11.10 Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-
methylpentan-2-ol.
Answer. In Williamson synthesis, an alkyl halide reacts with an alkoxide ion. Also, it is an SN2 reaction. In the
reaction, alkyl halides should be primary having the least steric hindrance. Hence, an alkyl halide is obtained from
ethanol and alkoxide ion from 3- methylpentan-2-ol.
Page : 350 , Block Name : Intext Questions
Q11.11 Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene
and why?
Page 9
Answer. Set (ii) is an appropriate set of reactants for the preparation of I-methoxy-4- nitrobenzene.
In set (i), sodium methoxide (CH30Na) is a strong nucleophile as well as a strong base.
Hence, an elimination reaction predominates over a substitution reaction.
Page : 350 , Block Name : Intext Questions
Q11.12 Predict the products of the following reactions:
In set (i), sodium methoxide (CH30Na) is a strong nucleophile as well as a strong base.
Hence, an elimination reaction predominates over a substitution reaction.
Page : 350 , Block Name : Intext Questions
Q11.12 Predict the products of the following reactions:
Page 10
Page : 350 , Block Name : Intext Questions
Q11.1 Write IUPAC names of the following compounds:
Q11.1 Write IUPAC names of the following compounds:
Page 11
Answer. (i) 2, 2, 4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2, 4-diol
(iii) Butane-2, 3-diol
(iv) Propane-I, 2, 3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2, 5-Dimethylphenol
(viii) 2, 6-Dimethylphenol
(ix) I-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) I-Phenylheptane
(xii) 2-Ethoxybutane
Page : 352 , Block Name : Exercise
Q11.2 Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol
(ii) 1-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane –1, 3, 5-triol
(iv) 2,3 – Diethylphenol
(v) 1 – Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(ii) 5-Ethylheptane-2, 4-diol
(iii) Butane-2, 3-diol
(iv) Propane-I, 2, 3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2, 5-Dimethylphenol
(viii) 2, 6-Dimethylphenol
(ix) I-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) I-Phenylheptane
(xii) 2-Ethoxybutane
Page : 352 , Block Name : Exercise
Q11.2 Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol
(ii) 1-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane –1, 3, 5-triol
(iv) 2,3 – Diethylphenol
(v) 1 – Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
Page 12
(viii) 3-Cyclohexylpentan-3-ol
(ix) Cyclopent-3-en-1-ol
(x) 4-Chloro-3-ethylbutan-1-ol
(ix) Cyclopent-3-en-1-ol
(x) 4-Chloro-3-ethylbutan-1-ol
Page 13
Page : 352 , Block Name : Exercise
Q11.3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols.
Q11.3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols.
Page 14
Page : 352 , Block Name :Exercise
Q11.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer. Propanol undergoes intermolecular H-bonding because of the presence of —OH group.
On the other hand, butane does not
Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point
than hydrocarbon butane.
Page : 352 , Block Name :Exercise
Q11.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Explain this fact.
Answer. Alcohols form H-bonds with water due to the presence of hydrocarbons cannot form H-bonds with water. —
OH group. However,
As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Page : 352 , Block Name : Exercise
Q11.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Answer. The addition Of borane followed by Oxidation is known as the hydroboration-oxidation reaction.
For example, propan-l-ol is produced by the hydroboration-oxidation reaction
of propene. In this reaction, propene reacts with diborane (BH3)2 to form trialkyl borane as an addition product.
This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
Q11.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer. Propanol undergoes intermolecular H-bonding because of the presence of —OH group.
On the other hand, butane does not
Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point
than hydrocarbon butane.
Page : 352 , Block Name :Exercise
Q11.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Explain this fact.
Answer. Alcohols form H-bonds with water due to the presence of hydrocarbons cannot form H-bonds with water. —
OH group. However,
As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Page : 352 , Block Name : Exercise
Q11.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Answer. The addition Of borane followed by Oxidation is known as the hydroboration-oxidation reaction.
For example, propan-l-ol is produced by the hydroboration-oxidation reaction
of propene. In this reaction, propene reacts with diborane (BH3)2 to form trialkyl borane as an addition product.
This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
Page 15
Page : 352 , Block Name : Exercise
Q11.7 Give the structures and IUPAC names of monohydric phenols of molecular formula,C H O7 a
Page : 352 , Block Name : Exercise
Q11.8 While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will
be steam volatile. Give reason.
Answer. Intramolecular H-bonding is present in 0-nitrophenol and p-nitrophenol. In p- nitrophenol, the molecules
are strongly associated due to the presence Of intermolecular bonding. Hence, 0-nitrophenol is steam volatile.
Page : 352 , Block Name : Exercise
Q11.9 Give the equations of reactions for the preparation of phenol from cumene.
Answer. To prepare phenol, cumene is rst oxidized in the presence of air of cumene hydroperoxide.
Q11.7 Give the structures and IUPAC names of monohydric phenols of molecular formula,C H O7 a
Page : 352 , Block Name : Exercise
Q11.8 While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will
be steam volatile. Give reason.
Answer. Intramolecular H-bonding is present in 0-nitrophenol and p-nitrophenol. In p- nitrophenol, the molecules
are strongly associated due to the presence Of intermolecular bonding. Hence, 0-nitrophenol is steam volatile.
Page : 352 , Block Name : Exercise
Q11.9 Give the equations of reactions for the preparation of phenol from cumene.
Answer. To prepare phenol, cumene is rst oxidized in the presence of air of cumene hydroperoxide.
Page 16
Then, cumene hydroxide is treated with dilute acid to prepare phenol and acetone as by- products.
Page : 352 , Block Name : Exercise
Q11.10 Write chemical reaction for the preparation of phenol from chlorobenzene
Answer. Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which
gives phenol on acidi cation.
Page : 352 , Block Name : Exercise
Q11.11 Write the mechanism of hydration of ethene to yield ethanol.
Page : 352 , Block Name : Exercise
Q11.10 Write chemical reaction for the preparation of phenol from chlorobenzene
Answer. Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which
gives phenol on acidi cation.
Page : 352 , Block Name : Exercise
Q11.11 Write the mechanism of hydration of ethene to yield ethanol.
Page 17
Page : 352 , Block Name : Exercise
Q11.12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using
these reagents.
Page : 353 , Block Name : Exercise
Q11.13 Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?
Q11.12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using
these reagents.
Page : 353 , Block Name : Exercise
Q11.13 Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?
Page 18
Page : 353 , Block Name : Exercise
Q11.14 Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Q11.14 Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Page 19
Page : 353 , Block Name : Exercise
Q11.15 Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?
The nitro-group is an electron-withdrawing group. The presence Of this group in the ortho position decreases the
electron density in the bond. As a result, it is easier to lose a proton. Also, the 0-nitrophenoxide ion formed after
the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid.
On the other hand, methoxy group is an electron-releasing group. Thus, it increases the electron density in the
bond and hence, the proton cannot be given out easily.
For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.
Page : 353 , Block Name : Exercise
Q11.16 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. Thus, it increases the electron density in the benzene ring
Q11.15 Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?
The nitro-group is an electron-withdrawing group. The presence Of this group in the ortho position decreases the
electron density in the bond. As a result, it is easier to lose a proton. Also, the 0-nitrophenoxide ion formed after
the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid.
On the other hand, methoxy group is an electron-releasing group. Thus, it increases the electron density in the
bond and hence, the proton cannot be given out easily.
For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.
Page : 353 , Block Name : Exercise
Q11.16 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. Thus, it increases the electron density in the benzene ring
Page 20
as shown in the given resonance structure of phenol.
Page : 353 , Block Name : Exercise
Q11.17 Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Page : 353 , Block Name : Exercise
Q11.17 Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Page 21
Page : 353 , Block Name : Exercise
Q11.18 Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tiemann reaction.
(iii) Williamson ether synthesis.
(iv) Unsymmetrical ether
Answer. (i) Kolbe's reaction:
When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when
treated with carbon dioxide, followed by acidi cation, undergoes electrophilic substitution to give ortho-
hydroxybenzoic acid as the main product. This reaction is known as Kolbe's reaction.
(ii) Reimer—Tiemann reaction:
When phenol is treated with chloroform (CHCh) in the presence of sodium hydroxide, a —CHO group is introduced
at the ortho position of the benzene ring.
Q11.18 Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tiemann reaction.
(iii) Williamson ether synthesis.
(iv) Unsymmetrical ether
Answer. (i) Kolbe's reaction:
When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when
treated with carbon dioxide, followed by acidi cation, undergoes electrophilic substitution to give ortho-
hydroxybenzoic acid as the main product. This reaction is known as Kolbe's reaction.
(ii) Reimer—Tiemann reaction:
When phenol is treated with chloroform (CHCh) in the presence of sodium hydroxide, a —CHO group is introduced
at the ortho position of the benzene ring.
Page 22
Page : 353 , Block Name : Exercise
Q11.19 Write the mechanism of acid dehydration of ethanol to yield ethene
Q11.19 Write the mechanism of acid dehydration of ethanol to yield ethene
Page 23
Page : 353 , Block Name : Exercise
Q11.20 How are the following conversions carried out?
(i) Propene → Propan-2-ol.
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium chloride → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol
Page : 353 , Block Name : Exercise
Q11.21 Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
(ii) Oxidation of a primary alcohol to aldehyde.
Q11.20 How are the following conversions carried out?
(i) Propene → Propan-2-ol.
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium chloride → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol
Page : 353 , Block Name : Exercise
Q11.21 Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
(ii) Oxidation of a primary alcohol to aldehyde.
Page 24
(iii) Bromination of phenol to 2,4,6-tribromophenol.
(iv) Benzyl alcohol to benzoic acid.
(v) Dehydration of propan-2-ol to propene.
(vi) Butan-2-one to butan-2-ol.
Answer. (i) Acidi ed potassium permanganate
(ii) Pyridinium chlorochromate (PCC)
(iii) Bromine water
(iv) Acidi ed potassium permanganate
(v) 85% phosphoric acid
(vi) NaBH4 or LiAlH4
Page : 353 , Block Name : Exercise
Q11.22 Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer. Ethanol undergoes intermolecular H-bonding due to the presence of —OH group, resulting in the
association Of molecules. Extra energy is required to break these hydrogen bonds. On the other hand,
methoxymethane does not undergo H-bonding.
Hence, the boiling point of ethanol is higher than that of methoxymethane.
Page : 354 , Block Name : Exercise
Q11.23 Give IUPAC names of the following ethers
Answer. (i) I-Ethoxy-2-methylpropane
(ii) 2-chloro-l-methoxyethane
(iii) 4-Nitroanisole
(iv) I-Methoxypropane
(v) 1-Ethoxy-4, 4-dimethylcyclohexane
(vi) Ethoxybenzene
Page : 354 , Block Name : Exercise
Q11.24 Write the names of reagents and equations for the preparation of the following ethers by Williamson’s
synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
(iv) Benzyl alcohol to benzoic acid.
(v) Dehydration of propan-2-ol to propene.
(vi) Butan-2-one to butan-2-ol.
Answer. (i) Acidi ed potassium permanganate
(ii) Pyridinium chlorochromate (PCC)
(iii) Bromine water
(iv) Acidi ed potassium permanganate
(v) 85% phosphoric acid
(vi) NaBH4 or LiAlH4
Page : 353 , Block Name : Exercise
Q11.22 Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer. Ethanol undergoes intermolecular H-bonding due to the presence of —OH group, resulting in the
association Of molecules. Extra energy is required to break these hydrogen bonds. On the other hand,
methoxymethane does not undergo H-bonding.
Hence, the boiling point of ethanol is higher than that of methoxymethane.
Page : 354 , Block Name : Exercise
Q11.23 Give IUPAC names of the following ethers
Answer. (i) I-Ethoxy-2-methylpropane
(ii) 2-chloro-l-methoxyethane
(iii) 4-Nitroanisole
(iv) I-Methoxypropane
(v) 1-Ethoxy-4, 4-dimethylcyclohexane
(vi) Ethoxybenzene
Page : 354 , Block Name : Exercise
Q11.24 Write the names of reagents and equations for the preparation of the following ethers by Williamson’s
synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
Page 25
Page : 354 , Block Name : Exercise
Q11.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of
ethers.
Answer. The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide.
But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then
elimination would compete over substitution. As a result, alkenes would be produced.
This is because alkoxides are nucleophiles as well as strong bases. Hence, they react
with alkyl halides, which results in an elimination reaction.
Page : 354 , Block Name : Exercise
Q11.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction
Q11.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of
ethers.
Answer. The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide.
But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then
elimination would compete over substitution. As a result, alkenes would be produced.
This is because alkoxides are nucleophiles as well as strong bases. Hence, they react
with alkyl halides, which results in an elimination reaction.
Page : 354 , Block Name : Exercise
Q11.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction
Page 26
Page : 354 , Block Name : Exercise
Q11.27 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 alcohol is a bimolecular reaction (SN2) involving the attack of an
alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of
secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution.
Hence, in place of ethers, alkenes are formed.
Page : 354 , Block Name : Exercise
Q11.28 Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether
Q11.27 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 alcohol is a bimolecular reaction (SN2) involving the attack of an
alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of
secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution.
Hence, in place of ethers, alkenes are formed.
Page : 354 , Block Name : Exercise
Q11.28 Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether
Page 27
Page : 354 , Block Name : Exercise
Q11.29 Explain the fact that in aryl alkyl ethers
(i) the alkoxy group activates the benzene ring towards electrophilic substitution and
(ii) it directs the incoming substituents to ortho and para positions in benzene ring.
(ii) It can also be observed from the resonance structures that the electron density increases more at the ortho and
para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para
positions in the benzene.
Page : 354 , Block Name : Exercise
Q11.30 Write the mechanism of the reaction of HI with methoxymethane.
Q11.29 Explain the fact that in aryl alkyl ethers
(i) the alkoxy group activates the benzene ring towards electrophilic substitution and
(ii) it directs the incoming substituents to ortho and para positions in benzene ring.
(ii) It can also be observed from the resonance structures that the electron density increases more at the ortho and
para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para
positions in the benzene.
Page : 354 , Block Name : Exercise
Q11.30 Write the mechanism of the reaction of HI with methoxymethane.
Page 28
Page : 354 , Block Name : Exercise
Q11.31 Write equations of the following reactions:
(i) Friedel-Crafts reaction – alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium.
(iv) Friedel-Crafts acetylation of anisole.
Q11.31 Write equations of the following reactions:
(i) Friedel-Crafts reaction – alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium.
(iv) Friedel-Crafts acetylation of anisole.
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Q11.32 Show how would you synthesise the following alcohols from appropriate alkenes?
Q11.32 Show how would you synthesise the following alcohols from appropriate alkenes?
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Q11.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
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.)
Q11.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
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.)
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Document Details
| Board / Org | NCERT |
| Exam | Class 12 |
| Type | Solution |
| Pages | 32 |
| Updated | 30 Apr 2026 |