NCERT Solutions for Class 12 Chemistry Chapter 7: Alcohols, Phenols and Ethers
These Class 12 Chemistry Chapter 7 solutions cover Alcohols, Phenols and Ethers from the NCERT textbook (session 2026–27). You get every NCERT Exercises question and every Intext question reproduced verbatim and solved step by step — IUPAC names, structures, mechanisms (hydration, dehydration, ether cleavage), preparation routes and named reactions (Kolbe’s, Reimer–Tiemann, Williamson) — in clean, exam-ready CBSE style.
An alcohol has one or more –OH groups attached to an sp3 carbon of an aliphatic system; a phenol has –OH attached to an sp2 carbon of an aromatic ring; an ether has an oxygen bonded to two alkyl/aryl groups (R–O–R′). The chapter covers their classification, IUPAC nomenclature, structures, methods of preparation, physical properties (boiling point and solubility trends linked to hydrogen bonding), and the chemical reactions of each class — including the acidity of alcohols and phenols, esterification, dehydration, oxidation, electrophilic aromatic substitution of phenols, named reactions (Kolbe’s and Reimer–Tiemann), Williamson ether synthesis and the cleavage of ethers by hydrogen halides.
Key Concepts & Definitions
Classification: alcohols/phenols are mono-, di-, tri- or polyhydric by the number of –OH groups; monohydric alcohols are 1°, 2° or 3° by the carbon bearing –OH. Ethers are simple (symmetrical) or mixed (unsymmetrical).
Acidity order: phenols are far more acidic than alcohols (phenoxide ion is resonance-stabilised). Among alcohols, acidity falls 1° > 2° > 3°. Electron-withdrawing groups (–NO2) at o-/p- positions raise phenol’s acidity; electron-releasing groups (–CH3) lower it.
Boiling points: alcohols > ethers ≈ alkanes of comparable mass, because alcohols form intermolecular hydrogen bonds that ethers and alkanes lack.
7.1 Classify the following as primary, secondary and tertiary alcohols:
(i) (CH3)3C–CH2OH (ii) H2C=CH–CH2OH (iii) CH3CH2CH2OH(iv) C6H5–CH(OH)–CH3 (v) C6H5–CH2–CH(OH)–CH3 (vi) (CH3CH−)CH–C(CH3)2OH
ANSWERPrimary (1°): (i), (ii), (iii) — the –CH2OH carbon is attached to only one other carbon.Secondary (2°): (iv) and (v) — the C bearing –OH is attached to two carbons.Tertiary (3°): (vi) — the C bearing –OH is attached to three carbons.
7.2 Identify allylic alcohols in the above examples.
ANSWERAllylic alcohols have –OH on an sp3 carbon next to a C=C double bond. These are (ii) (prop-2-en-1-ol) and (vi), which has the –OH-bearing carbon adjacent to the double bond.
7.3 Name the following compounds according to IUPAC system.
(i)–(v) (structures given in the textbook).
7.4 Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.
ANSWERMethanal (HCHO) reacting with any Grignard reagent R–MgX gives a primary alcohol R–CH2OH after hydrolysis. General route:R–MgX + HCHO → R–CH2–OMgX H2O/H+→ R–CH2OH + Mg(OH)X.To get the required alcohol R–CH2OH, choose the Grignard reagent R–MgX whose R is the group attached to the –CH2OH (e.g. for 2-methylpropan-1-ol take isopropyl magnesium bromide and react with HCHO).
7.5 Write structures of the products of the following reactions:
(i) propene + H2O/H+ (ii) reaction giving a phenyl acetate ester (iii) 2-methylbut-1-ene + H2O/H+.
ANSWER (per NCERT key)(i) Markovnikov hydration of propene → propan-2-ol, CH3–CH(OH)–CH3.(ii) The product is phenyl ethanoate, C6H5–CH2–C(=O)–OCH3 as per the NCERT key (the ester formed at the –OH centre).(iii) Markovnikov hydration → CH3CH2C(CH3)(OH)CH3, i.e. 2-methylbutan-2-ol (–OH on the more substituted carbon).
7.6 Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl–ZnCl2 (b) HBr and (c) SOCl2.
(i) Butan-1-ol (ii) 2-Methylbutan-2-ol
ANSWER(i) Butan-1-ol (CH3CH2CH2CH2OH):(a) with HCl/ZnCl2 → 1-chlorobutane, CH3CH2CH2CH2Cl (slow, no turbidity at room temp).(b) with HBr → 1-bromobutane, CH3CH2CH2CH2Br.(c) with SOCl2 → 1-chlorobutane + SO2 + HCl.(ii) 2-Methylbutan-2-ol (3°):(a) with HCl/ZnCl2 → 2-chloro-2-methylbutane (immediate turbidity, 3° reacts fastest).(b) with HBr → 2-bromo-2-methylbutane.(c) with SOCl2 → 2-chloro-2-methylbutane + SO2 + HCl.
7.7 Predict the major product of acid catalysed dehydration of (i) 1-methylcyclohexanol and (ii) butan-1-ol.
ANSWER (per NCERT key)(i) 1-Methylcyclohexanol → 1-methylcyclohexene (the more substituted, Saytzeff alkene).(ii) Butan-1-ol → a mixture of but-1-ene and but-2-ene; but-2-ene is the major product, formed via rearrangement of the 1° carbocation to the more stable 2° carbocation.
7.8 Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
ANSWERIn o- and p-nitrophenoxide ions, the negative charge on oxygen is delocalised over the ring and, crucially, onto the oxygen atoms of the –NO2 group. The –NO2 group (electron-withdrawing, −M effect) provides extra resonance structures in which the negative charge sits on the nitro oxygens at the ortho/para positions.This additional delocalisation makes the o-/p-nitrophenoxide ion much more stable than phenoxide, so the parent nitrophenols release H+ more readily and are stronger acids than phenol. (m-Nitrophenol shows no such extra delocalisation of the negative charge onto –NO2, so it is less acidic than the o-/p- isomers.)
7.9 Write the equations involved in the following reactions:
(i) Reimer–Tiemann reaction (ii) Kolbe’s reaction
7.10 Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
ANSWERConvert ethanol to ethyl bromide: C2H5OH + HBr → C2H5Br.Convert 3-methylpentan-2-ol to its sodium alkoxide: CH3CH2CH(CH3)CH(CH3)OH + Na → CH3CH2CH(CH3)CH(CH3)ONa.Williamson coupling (the alkoxide attacks the primary ethyl bromide, SN2): alkoxide + C2H5Br → CH3CH2CH(CH3)CH(CH3)–O–C2H5, i.e. 2-ethoxy-3-methylpentane + NaBr. (Using the alkoxide of ethanol with the secondary halide would favour elimination.)
7.11 Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?
(i) p-nitro alkyl halide + sodium methoxide (ii) sodium p-nitrophenoxide + methyl iodide
ANSWERSet (ii) — sodium p-nitrophenoxide + methyl iodide (CH3I) — is appropriate.Williamson synthesis works only when the alkyl halide is reactive towards SN2. An aryl halide does not undergo SN2, so set (i) (aryl halide + alkoxide) fails. In set (ii) the methyl iodide is a primary (methyl) halide that readily undergoes SN2 with the phenoxide ion, giving 1-methoxy-4-nitrobenzene.
7.12 Predict the products of the following reactions:
(i) CH3–CH2–CH2–O–CH3 + HBr →(ii) [anisole derivative] + HI → (iii) [aryl alkyl ether] reaction (iv) (CH3)3C–OC2H5 + HI →
ANSWER (per NCERT key)(i) The lower (smaller) alkyl group forms the alkyl bromide (SN2 on the less hindered C): products are CH3CH2CH2OH + CH3Br.(ii) For an aryl alkyl ether (anisole-type), the alkyl–O bond cleaves to give phenol + alkyl iodide (the aryl–O bond is stronger and does not break).(iii) Same principle — cleavage at the weaker (alkyl) C–O bond gives the phenol and the corresponding alkyl iodide.(iv) (CH3)3C–OC2H5 + HI → (CH3)3C–I + C2H5OH. Here cleavage follows SN1 because the 3° carbocation (CH3)3C+ is stable, so the tertiary group becomes the iodide.
NCERT Exercises — Solutions
7.1 Write IUPAC names of the following compounds (i)–(xii).
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.
ANSWER (condensed structures)(i) CH3–C(CH3)(OH)–CH2–CH3(ii) C6H5–CH2–CH(OH)–CH3(iii) HOCH2–CH2–C(CH3)(OH)–CH2–C(CH3)2–OH(iv) Benzene ring with –OH at C1 and –C2H5 groups at C2 and C3.(v) CH3CH2CH2–O–CH2CH3(vi) CH3–CH(OC2H5)–CH(CH3)–CH2–CH3(vii) C6H11–CH2OH (cyclohexyl ring bearing a –CH2OH group)(viii) cyclohexyl–C(OH)(C2H5)2 (C3 of pentane bears –OH and a cyclohexyl group)(ix) cyclopentene ring with –OH at C1 and the double bond between C3–C4(x) ClCH2–CH(C2H5)–CH2–CH2OH
7.3 (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names. (ii) Classify the isomers as primary, secondary and tertiary alcohols.
7.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
ANSWERPropanol molecules contain an –OH group and form strong intermolecular hydrogen bonds with one another. Butane (a hydrocarbon) has only weak van der Waals (London) forces.Breaking the extensive hydrogen-bond network in propanol needs much more energy, so propanol boils at a higher temperature than butane even though their molar masses are comparable.
7.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
ANSWERThe –OH group of an alcohol can form hydrogen bonds with water molecules, releasing energy that helps the alcohol dissolve. Hydrocarbons have no such group and cannot hydrogen-bond with water.Hence lower alcohols are appreciably soluble (lower ones miscible in all proportions), whereas hydrocarbons of comparable mass are essentially insoluble. (Solubility falls as the hydrophobic alkyl chain grows.)
7.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
ANSWERIn hydroboration–oxidation, an alkene first adds diborane (BH3)2 to give a trialkylborane; boron attaches to the sp2 carbon carrying more hydrogen atoms. The trialkylborane is then oxidised by H2O2 in aqueous NaOH to give an alcohol. The net result is addition of water against Markovnikov’s rule, in excellent yield.Example: 3CH3CH=CH2 + (BH3)2 → (CH3CH2CH2)3B H2O2/OH−→ CH3CH2CH2OH (propan-1-ol, the anti-Markovnikov product).
7.7 Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
ANSWERThe three monohydric (cresol) phenols are the o-, m- and p- methylphenols:(i) o-CH3–C6H4–OH — 2-Methylphenol (o-cresol)(ii) m-CH3–C6H4–OH — 3-Methylphenol (m-cresol)(iii) p-CH3–C6H4–OH — 4-Methylphenol (p-cresol)(Benzyl alcohol, C6H5CH2OH, and anisole are isomers of C7H8O but are not phenols, so they are excluded.)
7.8 While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
ANSWERo-Nitrophenol is steam volatile.It has intramolecular hydrogen bonding (chelation between –OH and the neighbouring –NO2), so its molecules are not associated and it is volatile. p-Nitrophenol has intermolecular hydrogen bonding, which associates the molecules, raises its boiling point and makes it non-volatile in steam.
7.9 Give the equations of reactions for the preparation of phenol from cumene.
7.10 Write chemical reaction for the preparation of phenol from chlorobenzene.
ANSWERChlorobenzene is fused with NaOH at 623 K and 320 atm to give sodium phenoxide:C6H5Cl + 2NaOH 623 K, 320 atm→ C6H5ONa + NaCl + H2O.Acidification of sodium phenoxide gives phenol: C6H5ONa + HCl → C6H5OH + NaCl.
7.11 Write the mechanism of hydration of ethene to yield ethanol.
ANSWERStep 1 — protonation: the π electrons of ethene attack H+ (from H3O+) to form a carbocation: CH2=CH2 + H+ → CH3–CH2+.Step 2 — nucleophilic attack of water: CH3–CH2+ + H2O → CH3–CH2–O+H2 (protonated alcohol / oxonium ion).Step 3 — deprotonation: CH3–CH2–O+H2 → CH3CH2OH + H+. The H+ is regenerated, confirming its catalytic role.
7.12 You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
7.13 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?
ANSWER(i) Markovnikov hydration of styrene: C6H5–CH=CH2 + H2O/H+ → C6H5–CH(OH)–CH3 (1-phenylethanol; –OH on the benzylic carbon).(ii) SN2 of chloromethylcyclohexane with aqueous KOH: C6H11–CH2Cl + KOH(aq) → C6H11–CH2OH (cyclohexylmethanol) + KCl.(iii) SN2 of 1-chloropentane with aqueous KOH: CH3(CH2)4Cl + KOH(aq) → CH3(CH2)4OH (pentan-1-ol) + KCl.
7.14 Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
ANSWER(i) With sodium metal: 2C6H5OH + 2Na → 2C6H5ONa + H2↑.(ii) With aqueous NaOH: C6H5OH + NaOH → C6H5ONa + H2O.Comparison: Phenol (pKa ≈ 10) is far more acidic than ethanol (pKa ≈ 15.9) — about a million times. The phenoxide ion is resonance-stabilised (negative charge delocalised over the ring), whereas in the ethoxide ion the charge is localised on oxygen and the electron-releasing alkyl group destabilises it. Phenol reacts with NaOH; ethanol does not.
7.15 Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
ANSWERIn o-nitrophenol the –NO2 group is electron-withdrawing (−I and −M). It pulls electron density away from the O–H bond and stabilises the resulting phenoxide ion by delocalising the negative charge, so the proton is released easily — the compound is strongly acidic.In o-methoxyphenol the –OCH3 group is electron-releasing (+M). It increases electron density on oxygen and destabilises the phenoxide ion, making proton loss harder. Hence o-nitrophenol is more acidic than o-methoxyphenol.
7.16 Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
ANSWERThe oxygen of the –OH group has lone pairs that are delocalised into the benzene ring (+M / resonance effect). This increases the electron density of the ring, especially at the ortho and para positions, making it more attractive to electrophiles.Thus phenol is more reactive than benzene towards electrophilic aromatic substitution, and the incoming group is directed mainly to the ortho and para positions.
7.17 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.
ANSWER(i) CH3CH2CH2OH alk. KMnO4→ CH3CH2COOH (propanoic acid) + H2O.(ii) Phenol + Br2 (in CS2, low temp) → mainly p-bromophenol (with some o-bromophenol) + HBr.(iii) Phenol + dil. HNO3 (298 K) → mixture of o-nitrophenol and p-nitrophenol + H2O.(iv) Phenol + CHCl3 + 3NaOH (Reimer–Tiemann), then hydrolysis/H+ → salicylaldehyde (o-hydroxybenzaldehyde).
7.18 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: phenoxide (from phenol + NaOH) undergoes electrophilic substitution with CO2 to give salicylic acid. C6H5ONa + CO2 → (after H+) o-HO–C6H4–COOH.(ii) Reimer–Tiemann reaction: phenol + CHCl3/NaOH introduces a –CHO group at the ortho position, giving salicylaldehyde after hydrolysis.(iii) Williamson ether synthesis: an alkyl halide reacts with a sodium alkoxide/aryloxide (SN2) to form an ether. e.g. C2H5ONa + CH3Br → C2H5OCH3 + NaBr.(iv) Unsymmetrical (mixed) ether: an ether whose two groups attached to oxygen are different, e.g. C2H5–O–CH3 (ethyl methyl ether).
7.19 Write the mechanism of acid dehydration of ethanol to yield ethene.
ANSWERStep 1 (protonation): CH3CH2OH + H+ → CH3CH2–O+H2 (protonated alcohol; good leaving group).Step 2 (formation of carbocation, slow/rate-determining): CH3CH2–O+H2 → CH3–CH2+ + H2O.Step 3 (loss of proton): CH3–CH2+ → CH2=CH2 + H+. The acid (H+) is regenerated; removing ethene drives the equilibrium forward. (Conc. H2SO4, 443 K.)
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.
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.
ANSWER(i) Acidified KMnO4 (or acidified K2Cr2O7).(ii) Pyridinium chlorochromate (PCC) [or CrO3 in anhydrous medium].(iii) Bromine water (Br2 in water).(iv) Acidified KMnO4 (strong oxidising agent).(v) Conc. H2SO4 (or 85% H3PO4) with heat.(vi) NaBH4 (or LiAlH4) — reduction of the ketone.
7.22 Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
ANSWEREthanol (CH3CH2OH) has an –OH group, so its molecules associate through intermolecular hydrogen bonding. Methoxymethane (CH3OCH3) has no O–H bond and cannot hydrogen-bond with itself (only weak dipole–dipole forces).Because more energy is needed to break the hydrogen bonds in ethanol, it has a much higher boiling point than its isomer methoxymethane, despite identical molar mass.
7.23 Give IUPAC names of the following ethers (i)–(vi).
7.24 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
ANSWER(i) 1-Propoxypropane: sodium propoxide + 1-bromopropane. CH3CH2CH2ONa + CH3CH2CH2Br → CH3CH2CH2–O–CH2CH2CH3 + NaBr.(ii) Ethoxybenzene: sodium phenoxide + ethyl bromide (phenol must be the aryloxide). C6H5ONa + C2H5Br → C6H5–O–C2H5 + NaBr.(iii) 2-Methoxy-2-methylpropane: use sodium tert-butoxide + methyl iodide (the methyl halide must be primary; a 3° halide would eliminate). (CH3)3C–ONa + CH3I → (CH3)3C–O–CH3 + NaI.(iv) 1-Methoxyethane: sodium ethoxide + methyl iodide. C2H5ONa + CH3I → C2H5–O–CH3 + NaI.
7.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
ANSWERThe reaction is an SN2 attack of an alkoxide on an alkyl halide, so it works best when the alkyl halide is primary. With secondary and especially tertiary halides, the alkoxide (a strong base) causes elimination instead of substitution.Example: CH3ONa + (CH3)3C–Br gives only 2-methylpropene (elimination), not tert-butyl methyl ether. To make a mixed 3° ether, the 3° group must come from the alkoxide and the primary group from the halide: (CH3)3C–ONa + CH3Br → (CH3)3C–O–CH3. Also, aryl halides do not react (no SN2 at aryl C).
7.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
ANSWERHeating propan-1-ol with conc. H2SO4 at about 413 K gives 1-propoxypropane (dipropyl ether) by bimolecular (SN2) dehydration:2 CH3CH2CH2OH conc. H2SO4, 413 K→ CH3CH2CH2–O–CH2CH2CH3 + H2O.Mechanism: Step 1 — protonation of one alcohol: C3H7OH + H+ → C3H7–O+H2. Step 2 — a second (neutral) alcohol molecule acts as nucleophile and attacks this protonated alcohol (SN2), displacing water: C3H7OH + C3H7O+H2 → C3H7–O+(H)–C3H7 + H2O. Step 3 — loss of H+ gives the ether C3H7–O–C3H7.
7.27 Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
ANSWERAcid dehydration of 2° and 3° alcohols proceeds through stable carbocations, where elimination competes strongly over substitution. The alkene (formed by loss of a proton) is the favoured product instead of the ether.So heating a 2° or 3° alcohol with acid gives mainly the corresponding alkene, not the ether. Acidic bimolecular dehydration is therefore suitable only for unhindered primary alcohols.
7.28 Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane (ii) methoxybenzene (iii) benzyl ethyl ether.
ANSWER(i) CH3CH2CH2–O–CH2CH2CH3 + HI → CH3CH2CH2I + CH3CH2CH2OH (excess HI converts the alcohol further to the iodide).(ii) C6H5–O–CH3 + HI → C6H5OH (phenol) + CH3I. The alkyl–O bond breaks; the aryl–O bond is stronger, so no iodobenzene forms.(iii) C6H5CH2–O–C2H5 + HI → C6H5CH2I (benzyl iodide) + C2H5OH. (Benzyl is more reactive, forming a stabilised benzylic cation; here the alkyl–O bonds are both at sp3 carbons.)
7.29 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.
ANSWER(i) Activation: the lone pairs on the oxygen of the –OR group are delocalised into the ring (+M effect), raising the ring’s electron density. The ring therefore reacts faster with electrophiles than benzene does.(ii) o/p-direction: the resonance structures that result from this delocalisation place the extra electron density specifically at the ortho and para carbons, so the electrophile attacks there. (The para product usually dominates because the ortho position suffers steric hindrance from the –OR group.)
7.30 Write the mechanism of the reaction of HI with methoxymethane.
ANSWERStep 1 (protonation): CH3–O–CH3 + HI → CH3–O+(H)–CH3 + I− (an oxonium ion; HI is acidic enough to protonate the ether oxygen).Step 2 (SN2 attack): the good nucleophile I− attacks the less hindered methyl carbon and displaces a neutral alcohol: I− + CH3–O+(H)–CH3 → CH3I + CH3OH.Step 3 (with excess HI): the methanol formed reacts further — CH3OH + HI → CH3I + H2O — so the overall product with excess HI is 2 CH3I.
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.
7.32 Show how would you synthesise the following alcohols from appropriate alkenes?
(i) 1-methylcyclohexan-1-ol (3°) (ii) cyclohexylmethanol (iii) cyclohexan-1-ol (with –OH on the ring) (iv) a 1-substituted alcohol.
ANSWER(i) 1-Methylcyclohexan-1-ol: Markovnikov hydration of 1-methylcyclohexene — the –OH adds to the more substituted (3°) carbon. (Use H2O/H+.)(ii) Cyclohexylmethanol: hydroboration–oxidation of methylenecyclohexane (the exocyclic alkene C6H10=CH2) with (BH3)2, then H2O2/OH− — anti-Markovnikov, –OH on the terminal CH2.(iii) For an alcohol whose –OH is on the less substituted carbon of an alkene, use hydroboration–oxidation (anti-Markovnikov). For –OH on the more substituted carbon, use acid hydration (Markovnikov).(iv) Choose the route by the position of –OH: Markovnikov (H2O/H+) places –OH on the more substituted carbon; hydroboration–oxidation places it on the less substituted carbon. Pick the alkene whose double bond, after the chosen addition, gives the target alcohol.
7.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: (rearranged product is 2-bromo-2-methylbutane). 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 the 3rd carbon atom.)
ANSWERStep 1 (protonation): the –OH of 3-methylbutan-2-ol [(CH3)2CH–CH(OH)–CH3] is protonated by HBr to give a good leaving group (–O+H2).Step 2 (loss of water): water leaves to form a secondary carbocation at C2: (CH3)2CH–C+H–CH3.Step 3 (1,2-hydride shift): a hydride ion (H−) migrates from C3 to C2, converting the 2° cation into a more stable tertiary carbocation: (CH3)2C+–CH2–CH3.Step 4 (nucleophilic capture): Br− attacks the 3° carbocation to give the rearranged product 2-bromo-2-methylbutane, (CH3)2C(Br)–CH2–CH3.
Extra Practice Questions
Short Answer Type Questions
Q1. Why is phenol a stronger acid than ethanol?
ANSWEROn losing H+, phenol gives a phenoxide ion whose negative charge is delocalised over the aromatic ring (resonance), making it stable. In ethoxide the charge is localised on oxygen and the +I effect of the ethyl group destabilises it. Hence phenol ionises far more easily and is the stronger acid.
Q2. What is the Lucas test and what does it distinguish?
ANSWERLucas reagent is conc. HCl + anhydrous ZnCl2. It distinguishes 1°, 2° and 3° alcohols by the rate of turbidity (formation of insoluble alkyl chloride): 3° gives turbidity immediately, 2° in about 5 minutes, and 1° only on heating.
Q3. Why does o-nitrophenol have a lower boiling point than p-nitrophenol?
ANSWERo-Nitrophenol has intramolecular H-bonding (chelation), so its molecules do not associate, giving a lower boiling point and steam volatility. p-Nitrophenol has intermolecular H-bonding, which associates molecules and raises its boiling point.
Q4. Write the product when phenol is treated with bromine water.
ANSWERPhenol + 3Br2 (bromine water) → 2,4,6-tribromophenol (white precipitate) + 3HBr. The strongly activating –OH group allows polybromination even without a catalyst.
Q5. Why are alcohols weaker acids than water?
ANSWERThe electron-releasing alkyl group in an alcohol increases electron density on oxygen, destabilising the alkoxide ion and reducing the polarity of the O–H bond. So water is a better proton donor (stronger acid) than an alcohol, and alkoxide ions are stronger bases than OH−.
Long Answer Type Questions
Q1. Describe the mechanism of acid-catalysed dehydration of an alcohol to an alkene, using ethanol, and state why tertiary alcohols dehydrate most easily.
ANSWERThe reaction is E1. (1) The –OH is protonated by the acid to form –O+H2, a good leaving group. (2) Water leaves in the slow, rate-determining step to give a carbocation (for ethanol, CH3CH2+). (3) A β-proton is lost to form the alkene (CH2=CH2) and regenerate H+. Because the rate-determining step is carbocation formation, the order of ease of dehydration follows carbocation stability: 3° > 2° > 1°. Tertiary alcohols give the most stable (3°) carbocations, so they dehydrate under the mildest conditions and fastest. The major alkene follows Saytzeff’s rule (more substituted, more stable).
Q2. Compare the methods of preparation and the reactivity of ethers towards cleavage by HX, explaining which bond breaks in alkyl aryl ethers.
ANSWEREthers are prepared by (a) acidic dehydration of (primary) alcohols and (b) Williamson synthesis (alkoxide + 1° alkyl halide, SN2). Ethers are the least reactive functional group but are cleaved by hot, concentrated HX (order HI > HBr > HCl). The reaction starts by protonating the ether oxygen; the halide then attacks the less hindered alkyl carbon (SN2) or, if a 3° carbocation can form, by SN1. In alkyl aryl ethers (e.g. anisole) the alkyl–oxygen bond breaks, giving phenol + alkyl halide, because the aryl–O bond has partial double-bond character (oxygen lone pair conjugated with the ring) and the sp2 aryl carbon cannot undergo nucleophilic substitution. With excess HX the alcohol product can be converted further to a second alkyl halide.
Q3. Explain, with the help of resonance and inductive effects, how substituents change the acidity of phenols.
ANSWERPhenol’s acidity depends on how stable the phenoxide ion is. Electron-withdrawing groups (–NO2, –CHO, –X) remove electron density (−I/−M) and delocalise the negative charge of the phenoxide, stabilising it and increasing acidity; the effect is strongest at the ortho and para positions where direct resonance delocalisation onto the group is possible (e.g. p-nitrophenol pKa 7.1 vs phenol 10.0). Electron-releasing groups (–CH3, –OCH3) push electron density into the ring (+I/+M), destabilise the phenoxide and decrease acidity (cresols are less acidic than phenol). Thus 2,4,6-trinitrophenol (picric acid) is a very strong acid, while p-cresol is weaker than phenol.
MCQs & Assertion–Reason
1. The correct order of acidity is:
(a) ethanol > water > phenol (b) phenol > water > ethanol (c) water > phenol > ethanol (d) phenol > ethanol > water
2. Anti-Markovnikov addition of water to an alkene is achieved by:
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: Phenol is more acidic than ethanol.
Reason: The phenoxide ion is stabilised by resonance, whereas the negative charge in the ethoxide ion is localised on oxygen.
A-R 2. Assertion: o-Nitrophenol is steam volatile.
Reason: o-Nitrophenol has intramolecular hydrogen bonding, so its molecules are not associated.
A-R 3. Assertion: Tertiary alcohols are easily oxidised to ketones.
Reason: The carbon bearing the –OH group in a tertiary alcohol has no hydrogen atom.
A-R 4. Assertion: In the reaction of anisole with HI, iodobenzene is not formed.
Reason: The aryl–oxygen bond is stronger and the sp2 aryl carbon does not undergo nucleophilic substitution.
A-R 5. Assertion: Williamson synthesis of tert-butyl methyl ether fails when (CH3)3C–Br is treated with sodium methoxide.
Reason: With a tertiary halide the alkoxide acts as a base, so elimination dominates over substitution.
Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).
Common Mistakes & Exam Tips
Watch out for these
Do not give the Markovnikov product for hydroboration–oxidation — it is anti-Markovnikov.
For alkyl aryl ethers + HX, the alkyl–O bond breaks (phenol + alkyl halide); never write iodobenzene.
Tertiary alcohols do not get oxidised to ketones under normal conditions — the –OH carbon has no H.
In Williamson synthesis, never pair a 3° halide with an alkoxide (gives alkene). Put the bulky group on the alkoxide, the primary group on the halide.
Remember the dehydration of butan-1-ol gives mainly but-2-ene (carbocation rearrangement), not but-1-ene.
Use correct reagents: PCC/CrO3 stops at the aldehyde; KMnO4 goes all the way to the acid.
How to score full marks in this chapter
Practise IUPAC naming until it is automatic (number from the end nearest –OH; lowest locants). Learn the three mechanisms cold — acid hydration, acid dehydration (E1) and ether cleavage by HI — and write each in three clear steps with the rate-determining step marked. Memorise the acidity order (carboxylic acid > phenol > water > alcohol) and justify it with resonance. For conversions, state the reagent and conditions (temperature, catalyst) explicitly — many marks are for reagents. Keep the named reactions (Kolbe’s, Reimer–Tiemann, Williamson) ready with one balanced equation each.
Frequently Asked Questions
What does Class 12 Chemistry Chapter 7 cover?
Chapter 7, Alcohols, Phenols and Ethers, covers the classification, IUPAC nomenclature, structures, preparation, physical properties and chemical reactions of these three classes of compounds — including acidity trends, esterification, dehydration, oxidation, electrophilic substitution of phenols, named reactions (Kolbe’s, Reimer–Tiemann), Williamson ether synthesis and ether cleavage by HX.
Why is phenol more acidic than ethanol?
Because the phenoxide ion formed on ionisation is stabilised by resonance (the negative charge is delocalised over the aromatic ring), while in the ethoxide ion the charge stays localised on oxygen and is destabilised by the +I effect of the ethyl group. So phenol releases H+ much more readily (about a million times more acidic than ethanol).
Are these NCERT Solutions for Class 12 Chemistry Chapter 7 free?
Yes. All solutions are free and follow the official NCERT Chemistry textbook for session 2026–27, with every Intext and Exercises question solved step by step.
