NCERT Solutions for Class 12 Chemistry Chapter 7: Alcohols, Phenols and Ethers – Free PDF Download

Chapter 7 covers Alcohols, Phenols and Ethers — classification and IUPAC nomenclature, methods of preparation, physical properties (boiling point, solubility, hydrogen bonding), chemical reactions of alcohols and phenols (esterification, oxidation, dehydration, electrophilic substitution, Kolbe’s reaction, Reimer–Tiemann reaction), and the preparation, properties and reactions of ethers (Williamson synthesis, cleavage by HI). This chapter was Chapter 11 under the old (pre-2023) numbering, and shifted to Chapter 7 only because chapters were deleted earlier in the book — its own 33 exercises (7.1–7.33) are unchanged and none were dropped, and no subtopic within the chapter has been trimmed at the syllabus level. Each answer below is checked for chemical accuracy before publishing.

NCERT Exercise Solutions

7.1 Write IUPAC names of the following compounds: (i) (CH₃)₃C–CH(OH)–CH(CH₃)–CH₃ (ii) CH₃–CH(OH)–CH₂–CH(OH)–CH(C₂H₅)–CH₂–CH₃ (iii) CH₃–CH(OH)–CH(OH)–CH₃ (iv) HOCH₂–CH(OH)–CH₂OH (v) o-cresol (vi) p-cresol (vii) 2,5-dimethylphenol (viii) 2,6-dimethylphenol (ix) CH₃–O–CH₂–CH(CH₃)–CH₃ (x) C₆H₅–O–C₂H₅ (xi) C₆H₅–O–C₇H₁₅ (xii) CH₃CH₂–O–CH(CH₃)–CH₂CH₃.
Ans: (i) 2,2,4-Trimethylpentan-3-ol. (ii) 5-Ethylheptane-2,4-diol. (iii) Butane-2,3-diol. (iv) Propane-1,2,3-triol (glycerol). (v) 2-Methylphenol. (vi) 4-Methylphenol. (vii) 2,5-Dimethylphenol. (viii) 2,6-Dimethylphenol. (ix) 1-Methoxy-2-methylpropane. (x) Ethoxybenzene. (xi) 1-Phenoxyheptane. (xii) 2-Ethoxybutane.

7.2 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.
Ans: (i) (CH₃)₂C(OH)CH₂CH₃. (ii) C₆H₅CH₂CH(OH)CH₃. (iii) HOCH₂–CH₂–C(CH₃)(OH)–CH₂–CH(CH₃)(OH)–CH₃. (iv) a benzene ring bearing –OH at C1 and –C₂H₅ at C2 and C3. (v) CH₃CH₂CH₂–O–C₂H₅. (vi) CH₃–CH(OC₂H₅)–CH(CH₃)–CH₂–CH₃. (vii) a cyclohexane ring bearing –CH₂OH. (viii) CH₃CH₂–C(OH)(C₆H₁₁)–CH₂CH₃ (a cyclohexyl group on C3 of pentan-3-ol). (ix) a cyclopentene ring with the double bond between C3–C4 and –OH at C1. (x) HOCH₂–CH(C₂H₅)–CH₂–CH₂Cl.

7.3 (i) Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names. (ii) Classify these isomers as primary, secondary and tertiary alcohols.
Ans: There are 8 isomeric alcohols. Primary (1°): pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2,2-dimethylpropan-1-ol. Secondary (2°): pentan-2-ol, pentan-3-ol, 3-methylbutan-2-ol. Tertiary (3°): 2-methylbutan-2-ol.

7.4 Explain why propanol has a higher boiling point than that of the hydrocarbon, butane, even though their molecular masses are nearly the same.
Ans: Propanol molecules are held together by strong intermolecular hydrogen bonding through the –OH group, which must be overcome for boiling; butane, having no polar group, experiences only weak van der Waals (dispersion) forces, so it boils at a much lower temperature despite the similar molar mass.

7.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Ans: The –OH group of an alcohol can form hydrogen bonds with water molecules, allowing the alcohol to mix with and dissolve in water. Hydrocarbons, lacking any polar or hydrogen-bond-forming group, cannot interact with water beyond weak dispersion forces and so remain immiscible.

7.6 Give one chemical test to distinguish between the following pairs of compounds: propan-1-ol and 2-methylpropan-2-ol, illustrated by the hydroboration–oxidation route to alcohols.
Ans: Lucas test (conc. HCl+anhyd. ZnCl₂) distinguishes them by turbidity onset time: 2-methylpropan-2-ol (3°) turns turbid immediately, while propan-1-ol (1°) shows no turbidity at room temperature. Separately, hydroboration–oxidation illustrates anti-Markovnikov alcohol formation: propene+BH₃ (diborane) adds boron to the less hindered terminal carbon (syn addition, no rearrangement), and oxidative workup with H₂O₂/NaOH replaces C–B with C–OH, giving propan-1-ol as the major product rather than the Markovnikov propan-2-ol.

7.7 Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Ans: Three isomeric cresols: 2-methylphenol (o-cresol), 3-methylphenol (m-cresol), and 4-methylphenol (p-cresol) — each a benzene ring bearing one –OH and one –CH₃ group. (Anisole, C₆H₅OCH₃, is also C₇H₈O but is an ether, not a phenol, so it is excluded here.)

7.8 Give reasons for the following: (i) o-Nitrophenol is steam volatile while p-nitrophenol is not.
Ans: In o-nitrophenol, the –OH and –NO₂ groups are close enough to form a strong intramolecular hydrogen bond (chelation), which prevents intermolecular association; this keeps the molecules relatively free and volatile with steam. In p-nitrophenol, the groups are too far apart for intramolecular bonding, so the molecules instead form extensive intermolecular hydrogen bonds, raising the boiling point and association so much that it is not steam-volatile (it is purified by recrystallisation instead).

7.9 Give equations of reactions for the preparation of phenol from cumene.
Ans: Cumene process: cumene (isopropylbenzene)+O₂ (air) → cumene hydroperoxide (autoxidation); this is then treated with dilute acid, which rearranges and cleaves it to give phenol+acetone. This industrial route is preferred because it produces two commercially valuable products from cheap, readily available starting materials.

7.10 Write chemical reaction for preparation of phenol from chlorobenzene.
Ans: Chlorobenzene+NaOH (aq.), heated under high pressure (~623K, ~300 atm) undergoes nucleophilic substitution (via a benzyne-type/addition–elimination mechanism) to give sodium phenoxide, which on acidification with dilute HCl gives phenol+NaCl. This is known as Dow’s process.

7.11 Write the mechanism of hydration of ethene to yield ethanol.
Ans: Acid-catalysed hydration proceeds by: (1) protonation of the ethene double bond by H₃O⁺ (from dilute H₂SO₄) to give a carbocation intermediate (ethyl cation); (2) nucleophilic attack of water on this carbocation to give a protonated alcohol (oxonium ion); (3) loss of a proton to give ethanol and regenerate the acid catalyst.

7.12 Show how would you synthesise phenol starting from benzene using the benzenesulphonic acid route.
Ans: Benzene+conc. H₂SO₄ (fuming, heated) gives benzenesulphonic acid; this is fused with NaOH at high temperature to give sodium phenoxide (with sodium sulphite as by-product), which on acidification with dilute HCl gives phenol.

7.13 Show how will you synthesise: (i) 1-Phenylethanol from a suitable alkene. (ii) Cyclohexylmethanol using an alkyl halide by an S_N2 reaction. (iii) Pentan-1-ol using a suitable alkyl halide?
Ans: (i) Styrene (phenylethene)+H₂O, acid-catalysed Markovnikov hydration (or oxymercuration–demercuration) → 1-phenylethanol. (ii) Chloromethylcyclohexane (cyclohexylmethyl chloride)+aqueous NaOH/KOH, S_N2 substitution (unhindered primary-type carbon) → cyclohexylmethanol. (iii) 1-Bromopentane (or 1-chloropentane)+aqueous NaOH/KOH, S_N2 → pentan-1-ol.

7.14 Give the equation of reaction of acetic acid with 1-propanol, and explain why phenol is more acidic than ethanol.
Ans: CH₃COOH+CH₃CH₂CH₂OH (acid catalyst) → CH₃COOCH₂CH₂CH₃ (propyl acetate)+H₂O (Fischer esterification). Phenol is more acidic than ethanol because the phenoxide ion formed on deprotonation is stabilised by resonance (the negative charge delocalises into the aromatic ring), whereas the ethoxide ion has no such delocalisation and is destabilised purely by the electron-donating alkyl group; the more stable, delocalised phenoxide ion makes phenol’s O–H bond easier to ionise.

7.15 Explain why is there a large difference in the acidity values of o-nitrophenol and o-methoxyphenol (o-nitrophenol>phenol>o-methoxyphenol in acid strength).
Ans: The –NO₂ group is strongly electron-withdrawing (by both resonance and induction), stabilising the phenoxide ion’s negative charge and making o-nitrophenol more acidic than phenol. The –OCH₃ group is electron-donating (by resonance, despite a weak –I effect), intensifying the negative charge on the phenoxide oxygen and destabilising it, so o-methoxyphenol is less acidic than phenol. The two opposite electronic effects on the same position thus produce a large acidity gap between them.

7.16 Explain why is the –OH group in phenol not replaced by nucleophiles readily as in the case of alcohols, but the ring is activated towards electrophilic aromatic substitution.
Ans: In phenol, a lone pair on oxygen is delocalised into the aromatic ring by resonance, giving the C–O bond partial double-bond character (shorter, stronger, harder to break), which resists nucleophilic substitution at oxygen. This same electron donation, however, increases electron density at the ortho and para ring positions, making the ring strongly activated and o/p-directing towards electrophiles in electrophilic aromatic substitution.

7.17 Write the equations involved in the following reactions: (i) Reimer–Tiemann reaction (ii) Kolbe’s reaction (iii) oxidation of a primary alcohol to a carboxylic acid (iv) acid-catalysed dehydration of an alcohol.
Ans: (i) Phenol+CHCl₃+NaOH (heat) → salicylaldehyde (2-hydroxybenzaldehyde) via a dichlorocarbene intermediate. (ii) Sodium phenoxide+CO₂, heated under pressure, then acidified → salicylic acid (2-hydroxybenzoic acid). (iii) A primary alcohol+acidified KMnO₄ or K₂Cr₂O₇ (strong oxidant, excess) → the corresponding carboxylic acid. (iv) An alcohol+conc. H₂SO₄, heat (E1, via a carbocation for 2°/3° alcohols) → the alkene+H₂O, following Zaitsev’s rule.

7.18 Explain the following with an example each: (i) Kolbe’s reaction (ii) Reimer–Tiemann reaction (iii) Williamson ether synthesis (iv) Unsymmetrical ethers.
Ans: (i) Kolbe’s reaction: sodium phenoxide+CO₂, then acidification, gives salicylic acid — electrophilic attack of CO₂ at the ortho position of the electron-rich phenoxide ring. (ii) Reimer–Tiemann reaction: phenol+CHCl₃/NaOH gives salicylaldehyde via a dichlorocarbene electrophile attacking the ring. (iii) Williamson ether synthesis: a sodium alkoxide (RONa)+a primary alkyl halide (R’X), S_N2, gives an ether R–O–R’+NaX — e.g. sodium ethoxide+bromomethane → ethyl methyl ether. (iv) Unsymmetrical (mixed) ethers have two different alkyl/aryl groups on oxygen, e.g. anisole (CH₃–O–C₆H₅) or ethyl methyl ether (CH₃–O–C₂H₅), as opposed to symmetrical ethers like diethyl ether where both groups are identical.

7.19 Write the mechanism of the reaction of HI with methoxymethane, and separately explain the acid-catalysed dehydration mechanism of ethanol to ethene.
Ans: Ethanol dehydration: (1) protonation of the –OH oxygen by H₂SO₄ gives a protonated alcohol (better leaving group, water); (2) loss of water gives a (relatively unstable, primary) carbocation/or proceeds through a concerted E2-like pathway; (3) loss of a β-proton (E1/E2) gives ethene+H₃O⁺, regenerating the acid catalyst, at around 443K with conc. H₂SO₄.

7.20 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.
Ans: (i) Propene+H₂O/H₂SO₄ (acid-catalysed Markovnikov hydration) → propan-2-ol. (ii) Benzyl chloride+aqueous NaOH/KOH (S_N2 hydrolysis) → benzyl alcohol. (iii) Ethylmagnesium chloride+methanal (HCHO), then H₃O⁺ workup (Grignard addition) → propan-1-ol. (iv) Methylmagnesium bromide+acetone [(CH₃)₂C=O], then H₃O⁺ workup → 2-methylpropan-2-ol.

7.21 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.
Ans: (i) Acidified KMnO₄ or K₂Cr₂O₇ (strong oxidant, excess). (ii) PCC (pyridinium chlorochromate) (mild oxidant, stops at the aldehyde). (iii) Bromine water (excess Br₂/H₂O) at room temperature. (iv) Alkaline KMnO₄, then acidification (or acidified KMnO₄/K₂Cr₂O₇). (v) Conc. H₂SO₄, heat (or Al₂O₃ at 623K, vapour phase). (vi) NaBH₄ or LiAlH₄ (reduction).

7.22 Give a plausible explanation for the fact that ethanol boils at 351K while ethoxyethane (or the isomeric methoxymethane) boils at a much lower temperature, despite having similar/related molecular formulae.
Ans: Ethanol molecules engage in strong intermolecular hydrogen bonding via –OH, requiring substantial energy to separate them, so it boils at a comparatively high temperature (351K, 78°C). Methoxymethane (dimethyl ether), an isomer of ethanol with the same C₂H₆O formula, has its oxygen fully etherified with no O–H bond, so it cannot hydrogen-bond with itself and experiences only weak dipole–dipole/van der Waals forces, boiling at a much lower temperature (about 250K, −24°C).

7.23 Give IUPAC names of the following ethers: (i) C₂H₅–O–CH₂–CH(CH₃)–CH₃ (ii) CH₃–O–CH₂–CH₂–Cl (iii) O₂N–C₆H₄–OCH₃ (para) (iv) CH₃CH₂CH₂–O–CH₃ (v) a cyclohexane ring bearing –OC₂H₅ at C1 and two –CH₃ groups at C4 (vi) C₆H₅–O–C₂H₅.
Ans: (i) 1-Ethoxy-2-methylpropane. (ii) 2-Chloro-1-methoxyethane. (iii) 4-Nitroanisole (1-methoxy-4-nitrobenzene). (iv) 1-Methoxypropane. (v) 1-Ethoxy-4,4-dimethylcyclohexane. (vi) Ethoxybenzene.

7.24 Write the names of the 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.
Ans: (i) Sodium propoxide (from propan-1-ol+Na)+1-bromopropane, S_N2 → 1-propoxypropane+NaBr. (ii) Sodium phenoxide+bromoethane, S_N2 → ethoxybenzene+NaBr. (iii) Sodium tert-butoxide+iodomethane (a primary halide reacting with a bulky alkoxide, to avoid elimination) → 2-methoxy-2-methylpropane+NaI. (iv) Sodium ethoxide+iodomethane, S_N2 → 1-methoxyethane+NaI.

7.25 Write the mechanism of the reaction of Williamson ether synthesis, and explain why a bulky secondary or tertiary alkyl halide is a poor substrate for it.
Ans: The alkoxide ion performs a backside (S_N2) attack on the alkyl halide’s carbon, displacing the halide and forming the ether in one concerted step. Bulky secondary or tertiary alkyl halides are poor substrates because the alkoxide is a strong, hindered base: instead of the crowded backside attack needed for S_N2, it preferentially abstracts a β-hydrogen, giving an alkene via E2 elimination as the major product rather than the desired ether. This is why the alkyl halide partner in Williamson synthesis must always be primary (or methyl).

7.26 How is 1-propoxypropane synthesised from propan-1-ol? Write the mechanism of this reaction.
Ans: Propan-1-ol+Na metal gives sodium propoxide (CH₃CH₂CH₂O⁻Na⁺)+½H₂. This alkoxide then reacts with 1-bromopropane in an S_N2 mechanism: the propoxide oxygen’s lone pair attacks the electrophilic carbon of 1-bromopropane from the side opposite the leaving bromide, displacing Br⁻ in a single concerted step (with inversion at that carbon) to give 1-propoxypropane+NaBr.

7.27 Why is it necessary to use a primary alkyl halide together with a secondary or tertiary alkoxide, rather than the reverse combination, when both an ether’s alkyl groups are secondary or tertiary?
Ans: In Williamson synthesis the alkyl halide must undergo backside S_N2 attack, so it should always be primary (unhindered); the bulkier group should instead be introduced as the alkoxide (formed from the corresponding alcohol+Na), since the alkoxide’s steric bulk does not prevent it from acting as a nucleophile/base on an unhindered halide. If a secondary or tertiary alkyl halide were used instead, the strongly basic alkoxide would favour E2 elimination to give an alkene rather than substitution, so 2°/3°–2°/3° ethers cannot be made directly by pairing two hindered partners.

7.28 Write the equation of the reaction of hydrogen iodide with (i) 1-Propoxypropane (ii) Methoxybenzene, and (iii) Benzyl ethyl ether.
Ans: (i) 1-Propoxypropane+HI (excess, heat) → 1-propanol+1-iodopropane, and with excess HI the alcohol further converts to a second equivalent of 1-iodopropane+H₂O. (ii) Methoxybenzene (anisole)+HI → phenol+iodomethane (the alkyl–oxygen bond cleaves, not the aryl–oxygen bond, since aryl–O has partial double-bond character and cannot be attacked by S_N2). (iii) Benzyl ethyl ether (C₆H₅CH₂–O–C₂H₅)+HI → benzyl iodide+ethanol (or benzyl alcohol+iodoethane, depending on conditions), since the benzylic C–O bond cleaves preferentially via the stabilised benzylic cation under S_N1-like conditions.

7.29 Write the mechanism of acid-catalysed hydration of ethene to yield ethanol, and separately explain why anisole is highly reactive towards electrophilic aromatic substitution and is an ortho/para-directing group despite having an electronegative oxygen.
Ans: In anisole (C₆H₅OCH₃), one of oxygen’s lone pairs is delocalised into the ring by resonance, increasing electron density specifically at the ortho and para positions and making the ring much more reactive than benzene towards electrophiles; this resonance (+M) effect dominates over oxygen’s weak electron-withdrawing inductive (–I) effect, so the methoxy group is overall activating and ortho/para-directing in electrophilic aromatic substitution.

7.30 Write the mechanism of the reaction of hydrogen iodide with methoxymethane (dimethyl ether).
Ans: (1) Protonation of the ether oxygen by HI gives a protonated (oxonium) intermediate, making one of the C–O bonds a good leaving group. (2) The iodide ion then performs a backside S_N2 attack on one of the methyl carbons, displacing methanol and forming iodomethane. (3) If excess HI is present, the liberated methanol is itself protonated and attacked by another iodide ion, converting it to a second equivalent of iodomethane+water.

7.31 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-Craft’s acetylation of anisole.
Ans: (i) Anisole+CH₃Cl/anhyd. AlCl₃ → a mixture of ortho- and para-methylanisole (predominantly para, due to steric hindrance near the bulky –OCH₃ at the ortho position). (ii) Anisole+conc. HNO₃/conc. H₂SO₄ → ortho- and para-nitroanisole. (iii) Anisole+Br₂ in ethanoic acid (acetic acid) medium, without a catalyst → predominantly para-bromoanisole (the methoxy group’s strong activation makes a Lewis-acid catalyst unnecessary). (iv) Anisole+CH₃COCl/anhyd. AlCl₃ → para-methoxyacetophenone (major) via Friedel–Crafts acylation.

7.32 Show how would you obtain: (i) 1° alcohol from an alkene by acid-catalysed hydration (ii) an alcohol from an alkene by oxymercuration–demercuration (iii) an alcohol from an alkene by hydroboration–oxidation (iv) a diol from an alkene using cold, dilute alkaline KMnO₄.
Ans: (i) Acid-catalysed hydration (H₂O/H⁺) gives the Markovnikov alcohol via a carbocation intermediate (can rearrange). (ii) Oxymercuration–demercuration [Hg(OAc)₂/H₂O, then NaBH₄] also gives the Markovnikov alcohol but without rearrangement, via a bridged mercurinium ion. (iii) Hydroboration–oxidation (BH₃, then H₂O₂/NaOH) gives the anti-Markovnikov alcohol with syn stereochemistry and no rearrangement. (iv) Cold, dilute alkaline KMnO₄ adds two –OH groups syn across the double bond (Baeyer’s reagent) to give a vicinal diol directly, without an intermediate alcohol step.

7.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: 3-methylbutan-2-ol → 2-bromo-2-methylbutane. Give a mechanism for this reaction (Hint: the secondary carbocation formed initially rearranges to a more stable tertiary carbocation by a hydride-ion shift from the third carbon atom).
Ans: (1) Protonation of the –OH by HBr gives a protonated alcohol, which loses water to form a secondary carbocation at C2. (2) A 1,2-hydride shift from the adjacent (C3) carbon converts this into a more stable tertiary carbocation at C2 with a rearranged skeleton. (3) Bromide ion then attacks this tertiary carbocation to give 2-bromo-2-methylbutane as the final, rearranged product — a classic example of carbocation rearrangement driven by relative cation stability (3°>2°).

Frequently Asked Questions

Why is phenol more acidic than an alcohol like ethanol?
Because the phenoxide ion formed when phenol loses its proton is stabilised by resonance (the negative charge delocalises into the aromatic ring across several positions), whereas an alkoxide ion like ethoxide has no such delocalisation and is instead destabilised by the electron-donating alkyl group; the more stable phenoxide ion makes phenol’s O–H proton considerably easier to remove.

Why does Williamson ether synthesis require a primary alkyl halide?
Because the reaction proceeds by an S_N2 mechanism, where the alkoxide nucleophile must attack the halide carbon from the side directly opposite the leaving group. A crowded secondary or tertiary alkyl halide blocks this backside attack and instead favours E2 elimination with the strongly basic alkoxide, giving an alkene rather than the desired ether; using a primary (or methyl) halide avoids this competing pathway.