Class 12 Chemistry

Chapter 7 Solutions — Alcohols, Phenols and Ethers

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Step-by-step NCERT solutions for Alcohols, Phenols and Ethers (Chapter 7, CBSE Class 12 Chemistry) — every question and answer worked out in full, not just the final result. You can also read the Alcohols, Phenols and Ethers textbook chapter.

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All 33 questions in Alcohols, Phenols and Ethers are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

1.
Write IUPAC names of the following compounds:
- (i) CH₃-CH(CH₃)-CH(OH)-C(CH₃)₂-CH₃
- (ii) H₃C-CH(OH)-CH₂-CH(OH)-CH(C₂H₅)-CH₂-CH₃
- (iii) CH₃-CH(OH)-CH(OH)-CH₃
- (iv) HO-CH₂-CH(OH)-CH₂-OH
- (v) benzene ring: OH at C₁, CH₃ at C₂
- (vi) benzene ring: CH₃ at C₁, OH at C₄ (para)
- (vii) benzene ring: OH at C₁, CH₃ at C₂ and C₅
- (viii) benzene ring: OH at C₁, CH₃ at C₂ and C₆
- (ix) CH₃-O-CH₂-CH(CH₃)-CH₃
- (x) C₆H₅-O-C₂H₅
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
- (viii) 3-Cyclohexylpentan-3-ol
- (ix) Cyclopent-3-en-1-ol
- (x) 4-Chloro-3-ethylbutan-1-ol
Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names. (ii) Classify the isomers of alcohols in question 7.3(i) as primary, secondary and tertiary alcohols.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Give the structures and IUPAC names of monohydric phenols of molecular formula C₇H₈O.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Give the equations of reactions for the preparation of phenol from cumene.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Write the mechanism of hydration of ethene to yield ethanol.
You are given benzene, conc. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
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?
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
Explain how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
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.
18.
Explain the following with an example.
- (i) Kolbe's reaction.
- (ii) Reimer-Tiemann reaction.
- (iii) Williamson ether synthesis.
- (iv) Unsymmetrical ether.
Write the mechanism of acid dehydration of ethanol to yield ethene.
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.
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.
- (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.
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
23.
Give IUPAC names of the following ethers (structures from figure):
- (i) C₂H₅-O-CH₂-CH(CH₃)₂
- (ii) CH₃-O-CH₂-CH₂-Cl
- (iii) O₂N-C₆H₄-O-CH₃ (para)
- (iv) CH₃-CH₂-CH₂-O-CH₃
- (v) cyclohexane ring: gem-dimethyl at C₁, OC₂H₅ at C₄
- (vi) benzene ring with -OC₂H₅
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
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
28.
Write the equation of the reaction of hydrogen iodide with:
- (i) 1-propoxypropane
- (ii) methoxybenzene
- (iii) benzyl ethyl ether.
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.
Write the mechanism of the reaction of HI with methoxymethane.
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.
32.
Show how would you synthesise the following alcohols from appropriate alkenes? (Structures from figure:
- (i) 1-methylcyclohexan-1-ol;
- (ii) 4-methylheptan-4-ol;
- (iii) pentan-2-ol;
- (iv) 2-cyclohexylbutan-2-ol)
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.)

Sample

Worked examples

3 questions from Alcohols, Phenols and Ethers, solved step by step.

Exercises

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

Propanol (C₃H₇OH, MW = 60) and butane (C₄H₁₀, MW = 58) have comparable molecular masses.

Propanol has a much higher boiling point (370 K) than butane (273 K) because:

- Propanol molecules form strong intermolecular hydrogen bonds through the -OH group (O is highly electronegative; O-H...O hydrogen bonds have energy ~20-40 kJ/mol).

- These hydrogen bonds associate propanol molecules strongly, requiring more energy (higher temperature) to break them during vaporisation.

Butane is a non-polar hydrocarbon. The only intermolecular forces are weak van der Waals (London dispersion) forces. These are much weaker than hydrogen bonds, so butane evaporates at a much lower temperature.

Answer: Propanol has a higher boiling point than butane because propanol molecules are associated by strong intermolecular hydrogen bonds through -OH groups, which are absent in butane (non-polar hydrocarbon with only weak van der Waals forces).

Exercises

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

Solubility in water depends on the ability of a solute to interact with water molecules.

Alcohols contain the -OH group, which can form hydrogen bonds with water molecules:

R-OH...O(H)-H and H-O-H...O(H)-R

This hydrogen bonding between alcohol and water makes alcohols soluble in water.

Hydrocarbons are non-polar molecules. They cannot form hydrogen bonds with water and have no significant dipole-dipole interactions. They interact with water only through very weak London dispersion forces. Since 'like dissolves like', non-polar hydrocarbons are sparingly soluble or insoluble in the polar solvent, water.

Hence, lower alcohols (methanol, ethanol, propan-1-ol) are completely miscible with water, while hydrocarbons of comparable molecular mass are essentially insoluble.

Answer: Alcohols form hydrogen bonds with water through their -OH group, making them soluble. Hydrocarbons are non-polar and cannot hydrogen-bond with water, so they are sparingly soluble. The -OH group of alcohols makes them much more water-soluble than hydrocarbons of similar molecular mass.

Exercises

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

Hydroboration-oxidation is a two-step method for converting an alkene to an alcohol with anti-Markovnikov regioselectivity.

Step 1 — Hydroboration: Diborane (B₂H₆) adds across the double bond. Boron attaches to the less substituted (more H-bearing) carbon, and H attaches to the more substituted carbon. This is a syn addition (both B and H add to the same face).

Step 2 — Oxidation: The organoborane is treated with H₂O₂ in the presence of NaOH, replacing the C-B bond with C-OH with retention of configuration.

Net result: The -OH group appears on the carbon that would have received H in Markovnikov addition (the less substituted carbon of the alkene) — anti-Markovnikov product.

Example: Propene → propan-1-ol

CH₃-CH=CH₂ + B₂H₆ → CH₃-CH₂-CH₂-B (trialkylborane)

CH₃-CH₂-CH₂-B + H₂O₂/NaOH → CH₃-CH₂-CH₂-OH (propan-1-ol) + B(OH)₃

Compare: acid-catalysed hydration of propene would give propan-2-ol (Markovnikov product).

Hydroboration-oxidation gives the anti-Markovnikov alcohol in excellent yield.

Answer: Hydroboration-oxidation converts an alkene to an anti-Markovnikov alcohol: Step 1 — B₂H₆ adds (B to less-substituted C); Step 2 — H₂O₂/NaOH replaces C-B with C-OH. Example: CH₃-CH=CH₂ → CH₃CH₂CH₂OH (propan-1-ol).

All 33 questions in Alcohols, Phenols and Ethers are solved step by step in the Solutions PDF above.

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Chapter 7 Solutions — Alcohols, Phenols and Ethers

Overview

What these solutions cover

Exercises

Worked examples

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

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

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

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