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NCERT Solutions · Class 12 Chemistry · NCERT Chemistry Part I and II Part II · Chapter 7

Chapter 7: Alcohols, Phenols and Ethers (Organic Chemistry)

Answers to all 11 intext questions and 33 exercises of Chapter 7, Alcohols, Phenols and Ethers (NCERT Class 12 Chemistry Part II, 2026-27): classification and IUPAC names, preparation from alkenes, carbonyls and Grignard reagents, boiling point and solubility, acidity of phenols, dehydration and its mechanism, Reimer–Tiemann and Kolbe reactions, Williamson synthesis and its limits, cleavage of ethers by HI, and reactions of anisole. All 44 questions are answered, with the key answer highlighted.

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Acid-catalysed hydration of alkenes follows Markovnikov's rule; hydroboration–oxidation gives the anti-Markovnikov alcohol. Grignard reagents with HCHO give 1° alcohols, with other aldehydes 2°, with ketones 3°. Reactivity with HCl/ZnCl₂ (Lucas test): 3° > 2° > 1°. Phenol is more acidic than ethanol because the phenoxide ion is resonance-stabilised; electron-withdrawing groups (–NO₂) at ortho/para increase acidity. Williamson synthesis: alkoxide (or phenoxide) + primary alkyl halide.

Intext questions

7.1

Classify the following as primary, secondary and tertiary alcohols: (i)

(CH X_{3}) X_{3} C - CH X_{2} OH

(ii)

H X_{2} C = CH - CH X_{2} OH

(iii)

CH X_{3} - CH X_{2} - CH X_{2} - OH

(iv)

C X_{6} H X_{5} - CH (OH) - CH X_{3}

(v)

C X_{6} H X_{5} - CH X_{2} - CH (OH) - CH X_{3}

(vi)

C X_{6} H X_{5} - CH = CH - C (OH) (CH X_{3}) X_{2}

Solution

Primary: (i), (ii), (iii) (the –OH carbon is attached to one other carbon)
Secondary: (iv), (v)
Tertiary: (vi)

Primary: (i), (ii), (iii). Secondary: (iv), (v). Tertiary: (vi).

7.2

Identify allylic alcohols in the above examples.

Solution

An allylic alcohol has the –OH on a carbon next to a C=C double bond: (ii) $H X_{2} C = CH - CH X_{2} OH$ and (vi) $C X_{6} H X_{5} CH = CH - C (OH) (CH X_{3}) X_{2}$ .

(ii) and (vi).

7.3

Name the following compounds according to IUPAC system. (i)

CH X_{3} CH X_{2} - CH (CH X_{2} Cl) - CH (CH X_{2} OH) - CH (CH X_{3}) CH X_{3}

(ii)

CH X_{3} - CH (CH X_{3}) - CH X_{2} - CH (OH) - CH (CH X_{2} OH) - CH X_{3}

(iii) cyclohexanol with Br at C-3 (iv)

H X_{2} C = CH - CH (OH) - CH X_{2} CH X_{2} CH X_{3}

(v)

CH X_{3} - C (CH X_{3}) = C (Br) - CH X_{2} OH

Solution

(i) 4-Chloro-3-ethyl-2-(1-methylethyl)butan-1-ol
(ii) 2,5-Dimethylhexane-1,3-diol
(iii) 3-Bromocyclohexan-1-ol
(iv) Hex-1-en-3-ol
(v) 2-Bromo-3-methylbut-2-en-1-ol

As listed above.

7.4

Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal? (i)

CH X_{3} - CH (CH X_{3}) - CH X_{2} OH

(ii) cyclohexylmethanol.

Solution

Methanal with a Grignard reagent RMgX gives the primary alcohol $RCH X_{2} OH$ :

(i) $(CH X_{3}) X_{2} CHMgBr + HCHO (CH X_{3}) X_{2} CHCH X_{2} OMgBr H X_{2} O (CH X_{3}) X_{2} CHCH X_{2} OH$
(ii) $C X_{6} H X_{11} MgBr + HCHO C X_{6} H X_{11} CH X_{2} OMgBr H X_{2} O C X_{6} H X_{11} CH X_{2} OH$ (cyclohexylmagnesium bromide)

(i) Isopropylmagnesium bromide + HCHO, then hydrolysis. (ii) Cyclohexylmagnesium bromide + HCHO, then hydrolysis.

7.5

Write structures of the products of the following reactions: (i)

CH X_{3} - CH = CH X_{2} H X_{2} O / H X^{+}

(ii) 2-oxocyclohexyl–

CH X_{2} COOCH X_{3}

(methyl 2-(2-oxocyclohexyl)acetate)

NaBH X_{4}

(iii)

CH X_{3} CH X_{2} CH (CH X_{3}) CHO NaBH X_{4}

Solution

(i) $CH X_{3} CH (OH) CH X_{3}$ , propan-2-ol (Markovnikov addition of water)
(ii) The ring ketone is reduced to a secondary alcohol; $NaBH X_{4}$ does not reduce the ester: methyl 2-(2-hydroxycyclohexyl)acetate
(iii) $CH X_{3} CH X_{2} CH (CH X_{3}) CH X_{2} OH$ , 2-methylbutan-1-ol

(i) propan-2-ol; (ii) the hydroxy ester (only the C=O of the ketone is reduced); (iii) 2-methylbutan-1-ol.

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)

SOCl X_{2}

. (i) Butan-1-ol (ii) 2-Methylbutan-2-ol.

Solution

(i) Butan-1-ol: (a) $CH X_{3} CH X_{2} CH X_{2} CH X_{2} Cl$ (very slow at room temperature; needs heating), (b) $CH X_{3} CH X_{2} CH X_{2} CH X_{2} Br$ , (c) $CH X_{3} CH X_{2} CH X_{2} CH X_{2} Cl$ .

(ii) 2-Methylbutan-2-ol: (a) $(CH X_{3}) X_{2} CClCH X_{2} CH X_{3}$ (fast, turbidity at once), (b) $(CH X_{3}) X_{2} CBrCH X_{2} CH X_{3}$ , (c) $(CH X_{3}) X_{2} CClCH X_{2} CH X_{3}$ .

(i) 1-chlorobutane, 1-bromobutane, 1-chlorobutane. (ii) 2-chloro-2-methylbutane, 2-bromo-2-methylbutane, 2-chloro-2-methylbutane.

7.7

Predict the major product of acid catalysed dehydration of (i) 1-methylcyclohexanol and (ii) butan-1-ol.

Solution

(i) 1-Methylcyclohexene (the more substituted alkene, by Saytzeff's rule).
(ii) But-2-ene, $CH X_{3} CH = CHCH X_{3}$ : the primary carbocation formed first rearranges by a hydride shift to the secondary carbocation, which loses $H X^{+}$ to give the more substituted alkene (some but-1-ene also forms).

(i) 1-Methylcyclohexene. (ii) But-2-ene (major).

7.8

Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

Solution

In o- and p-nitrophenoxide, the negative charge on the phenoxide oxygen is delocalised round the ring and onto the oxygens of the nitro group:

the charge moves from O⁻ to the ortho and para carbons of the ring;
when the –NO₂ is at the ortho or para position, one more structure puts the charge on the nitro group: the ring carbon bonded to N becomes C=N, and N carries $= O$ and $- O X^{-}$ ( $O = C X_{6} H X_{4} = N X^{+} (O -) - O X^{-}$ type quinonoid form).

This extra delocalisation (−R effect of $NO X_{2}$ ) stabilises the phenoxide ion more than in phenol, so o- and p-nitrophenols are stronger acids.

The phenoxide charge spreads onto the nitro oxygens through a quinonoid structure, giving extra stabilisation.

7.9

Write the equations involved in the following reactions: (i) Reimer–Tiemann reaction (ii) Kolbe's reaction.

Solution

(i) Reimer–Tiemann: phenol with chloroform and aqueous NaOH gives salicylaldehyde (an –CHO group enters ortho to –OH):

$C X_{6} H X_{5} OH + CHCl X_{3} + 3 NaOH 340 K o - HOC X_{6} H X_{4} CHO + 3 NaCl + 2 H X_{2} O$

(through the dichloromethyl intermediate $o - NaOC X_{6} H X_{4} CHCl X_{2}$ , hydrolysed and then acidified)

(ii) Kolbe's reaction: sodium phenoxide with $CO X_{2}$ under pressure gives salicylic acid:

$C X_{6} H X_{5} ONa + CO X_{2} 400 K, 4 - 7 atm o - HOC X_{6} H X_{4} COONa H X^{+} o - HOC X_{6} H X_{4} COOH$

(i) Phenol + CHCl₃ + NaOH → salicylaldehyde. (ii) Sodium phenoxide + CO₂ (400 K, 4–7 atm), then H⁺ → salicylic acid.

7.10

Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Solution

The alkyl halide must be primary (a secondary one would eliminate), so ethanol is made into bromoethane and the secondary alcohol into its alkoxide:

$C X_{2} H X_{5} OH + HBr C X_{2} H X_{5} Br + H X_{2} O$
$2 CH X_{3} CH (OH) CH (CH X_{3}) CH X_{2} CH X_{3} + 2 Na 2 CH X_{3} CH (ONa) CH (CH X_{3}) CH X_{2} CH X_{3} + H X_{2}$
$CH X_{3} CH (ONa) CH (CH X_{3}) CH X_{2} CH X_{3} + C X_{2} H X_{5} Br CH X_{3} CH (OC X_{2} H X_{5}) CH (CH X_{3}) CH X_{2} CH X_{3} + NaBr$

Convert ethanol to C₂H₅Br and 3-methylpentan-2-ol to its sodium alkoxide; then SN2 gives 2-ethoxy-3-methylpentane.

7.11

Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why? (i) 1-bromo-4-nitrobenzene +

CH X_{3} ONa

(ii) sodium 4-nitrophenoxide +

CH X_{3} Br

Solution

(ii) is appropriate. Sodium 4-nitrophenoxide is the nucleophile and attacks methyl bromide by SN2, which works very well at a methyl carbon.

In (i), an SN2 reaction at the aryl carbon is not possible: the C–Br bond of an aryl halide has partial double-bond character and the ring blocks back-side attack; also the strongly basic methoxide would cause side reactions.

(ii): phenoxide + CH₃Br reacts by SN2 at an unhindered methyl carbon; aryl halides do not undergo SN2.

Exercises

7.1

Write IUPAC names of the following compounds: (i)

CH X_{3} - CH (CH X_{3}) - CH (OH) - C (CH X_{3}) X_{2} - CH X_{3}

(ii)

H X_{3} C - CH (OH) - CH X_{2} - CH (OH) - CH (C X_{2} H X_{5}) - CH X_{2} - CH X_{3}

(iii)

CH X_{3} - CH (OH) - CH (OH) - CH X_{3}

(iv)

HO - CH X_{2} - CH (OH) - CH X_{2} - OH

(v) phenol with

CH X_{3}

ortho to OH (vi) phenol with

CH X_{3}

para to OH (vii) phenol with

CH X_{3}

groups at C-2 and C-5 (viii) phenol with

CH X_{3}

groups at C-2 and C-6 (ix)

CH X_{3} - O - CH X_{2} - CH (CH X_{3}) - CH X_{3}

(x)

C X_{6} H X_{5} - O - C X_{2} H X_{5}

(xi)

C X_{6} H X_{5} - O - C X_{7} H X_{15}

(n-) (xii)

CH X_{3} CH X_{2} - O - CH (CH X_{3}) - CH X_{2} - CH X_{3}

Solution

(i) 2,2,4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butane-2,3-diol
(iv) Propane-1,2,3-triol
(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

As listed above.

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.

Solution

(i) $(CH X_{3}) X_{2} C (OH) CH X_{2} CH X_{3}$
(ii) $C X_{6} H X_{5} CH X_{2} CH (OH) CH X_{3}$
(iii) $HOCH X_{2} CH X_{2} C (OH) (CH X_{3}) CH X_{2} C (OH) (CH X_{3}) CH X_{3}$
(iv) a benzene ring with OH on C-1 and $C X_{2} H X_{5}$ on C-2 and C-3
(v) $CH X_{3} CH X_{2} CH X_{2} - O - CH X_{2} CH X_{3}$
(vi) $CH X_{3} CH (OC X_{2} H X_{5}) CH (CH X_{3}) CH X_{2} CH X_{3}$
(vii) $C X_{6} H X_{11} CH X_{2} OH$ (cyclohexane ring bearing $- CH X_{2} OH$ )
(viii) $CH X_{3} CH X_{2} C (OH) (C X_{6} H X_{11}) CH X_{2} CH X_{3}$
(ix) a five-membered ring with OH on C-1 and a C=C between C-3 and C-4
(x) $HOCH X_{2} CH X_{2} CH (C X_{2} H X_{5}) CH X_{2} Cl$

As written above.

7.3

(i) Draw the structures of all isomeric alcohols of molecular formula

C X_{5} H X_{12} O

and give their IUPAC names. (ii) Classify the isomers of alcohols in question 7.3 (i) as primary, secondary and tertiary alcohols.

Solution

Structure	IUPAC name	Type
CH₃CH₂CH₂CH₂CH₂OH	Pentan-1-ol	1°
CH₃CH(OH)CH₂CH₂CH₃	Pentan-2-ol	2°
CH₃CH₂CH(OH)CH₂CH₃	Pentan-3-ol	2°
(CH₃)₂CHCH₂CH₂OH	3-Methylbutan-1-ol	1°
CH₃CH₂CH(CH₃)CH₂OH	2-Methylbutan-1-ol	1°
(CH₃)₂C(OH)CH₂CH₃	2-Methylbutan-2-ol	3°
(CH₃)₂CHCH(OH)CH₃	3-Methylbutan-2-ol	2°
(CH₃)₃CCH₂OH	2,2-Dimethylpropan-1-ol	1°

Eight isomers: four primary, three secondary, one tertiary (2-methylbutan-2-ol).

7.4

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

Solution

Propanol and butane have similar molecular masses (60 and 58), but propanol molecules are held together by intermolecular hydrogen bonds through their –OH groups. Extra energy is needed to break these, so propanol boils much higher. Butane has only weak van der Waals forces.

Propanol molecules are hydrogen-bonded; butane molecules are not.

7.5

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

Solution

The –OH group of an alcohol forms hydrogen bonds with water molecules, so alcohols dissolve. Hydrocarbons are non-polar and cannot form hydrogen bonds with water, so they do not dissolve.

Alcohols form hydrogen bonds with water; hydrocarbons cannot.

7.6

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

Solution

An alkene adds diborane ( $B X_{2} H X_{6}$ ) to form a trialkylborane, which is then oxidised by alkaline hydrogen peroxide to an alcohol. The –OH ends up on the less substituted carbon (overall anti-Markovnikov addition of water).

$3 CH X_{3} CH = CH X_{2} + \frac{1}{2} B X_{2} H X_{6} (CH X_{3} CH X_{2} CH X_{2}) X_{3} B H X_{2} O X_{2}, OH X^{-} 3 CH X_{3} CH X_{2} CH X_{2} OH + B (OH) X_{3}$

Propene gives propan-1-ol (acid-catalysed hydration would give propan-2-ol).

Addition of B₂H₆ to an alkene then oxidation with H₂O₂/OH⁻; e.g. propene → propan-1-ol (anti-Markovnikov).

7.7

Give the structures and IUPAC names of monohydric phenols of molecular formula

C X_{7} H X_{8} O

Solution

$CH X_{3} C X_{6} H X_{4} OH$ (cresols):

2-Methylphenol (o-cresol)
3-Methylphenol (m-cresol)
4-Methylphenol (p-cresol)

2-, 3- and 4-methylphenol.

7.8

While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Solution

o-Nitrophenol is steam volatile. It forms an intramolecular hydrogen bond (between –OH and the neighbouring –NO₂), so its molecules are not held to each other. p-Nitrophenol forms intermolecular hydrogen bonds, which hold molecules together and make it much less volatile.

o-Nitrophenol, because of intramolecular H-bonding (p-isomer is associated by intermolecular H-bonds).

7.9

Give the equations of reactions for the preparation of phenol from cumene.

Solution

Cumene is oxidised by air to cumene hydroperoxide: $C X_{6} H X_{5} CH (CH X_{3}) X_{2} + O X_{2} C X_{6} H X_{5} C (CH X_{3}) X_{2} OOH$
The hydroperoxide is treated with dilute acid: $C X_{6} H X_{5} C (CH X_{3}) X_{2} OOH H X^{+} H X_{2} O C X_{6} H X_{5} OH + CH X_{3} COCH X_{3}$

Acetone is obtained as a by-product.

Cumene + O₂ → cumene hydroperoxide; with dilute acid → phenol + acetone.

7.10

Write chemical reaction for the preparation of phenol from chlorobenzene.

Solution

$C X_{6} H X_{5} Cl + NaOH 623 K 300 atm C X_{6} H X_{5} ONa HCl C X_{6} H X_{5} OH$

(Dow's process: fusion with NaOH under high temperature and pressure, then acidification.)

Chlorobenzene + NaOH at 623 K, 300 atm → sodium phenoxide; acidify → phenol.

7.11

Write the mechanism of hydration of ethene to yield ethanol.

Solution

Step 1: protonation of the alkene by $H X_{3} O X^{+}$ to form a carbocation: $CH X_{2} = CH X_{2} + H X_{3} O X^{+} CH X_{3} CH X_{2} X^{+} + H X_{2} O$
Step 2: nucleophilic attack of water on the carbocation: $CH X_{3} CH X_{2} X^{+} + H X_{2} O CH X_{3} CH X_{2} OH X_{2} X^{+}$
Step 3: loss of a proton: $CH X_{3} CH X_{2} OH X_{2} X^{+} + H X_{2} O CH X_{3} CH X_{2} OH + H X_{3} O X^{+}$

Protonation → ethyl carbocation; attack by water → protonated ethanol; deprotonation → ethanol.

7.12

You are given benzene, conc.

H X_{2} SO X_{4}

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

Solution

$C X_{6} H X_{6} + H X_{2} SO X_{4} (conc \cdot) C X_{6} H X_{5} SO X_{3} H + H X_{2} O$ (sulphonation)
$C X_{6} H X_{5} SO X_{3} H + NaOH C X_{6} H X_{5} SO X_{3} Na + H X_{2} O$
$C X_{6} H X_{5} SO X_{3} Na + 2 NaOH fuse C X_{6} H X_{5} ONa + Na X_{2} SO X_{3} + H X_{2} O$
$2 C X_{6} H X_{5} ONa + H X_{2} SO X_{4} 2 C X_{6} H X_{5} OH + Na X_{2} SO X_{4}$ (acidification)

Sulphonate benzene, fuse the sodium salt with NaOH to sodium phenoxide, then acidify to phenol.

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?

Solution

(i) Acid-catalysed hydration of styrene (Markovnikov): $C X_{6} H X_{5} CH = CH X_{2} + H X_{2} O H X^{+} C X_{6} H X_{5} CH (OH) CH X_{3}$
(ii) $C X_{6} H X_{11} CH X_{2} Br + NaOH (aq) C X_{6} H X_{11} CH X_{2} OH + NaBr$ (SN2 at a primary carbon)
(iii) $CH X_{3} (CH X_{2}) X_{4} Cl + NaOH (aq) CH X_{3} (CH X_{2}) X_{4} OH + NaCl$

(i) Hydration of styrene. (ii) Cyclohexylmethyl bromide + aq. NaOH. (iii) 1-Chloropentane + aq. NaOH.

7.14

Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Solution

With sodium, phenol gives hydrogen: $2 C X_{6} H X_{5} OH + 2 Na 2 C X_{6} H X_{5} ONa + H X_{2}$
With sodium hydroxide, it forms a salt: $C X_{6} H X_{5} OH + NaOH C X_{6} H X_{5} ONa + H X_{2} O$ (ethanol does not react with NaOH)

Phenol ( $p K_{a} \approx 10$ ) is far more acidic than ethanol ( $p K_{a} \approx 15.9$ ). The phenoxide ion is stabilised by resonance (the negative charge spreads over the ring), whereas the ethoxide ion is destabilised by the electron-releasing ethyl group.

Phenol reacts with Na and with NaOH; it is more acidic than ethanol because the phenoxide ion is resonance-stabilised.

7.15

Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

Solution

The –NO₂ group withdraws electrons (−I and −R effects). It stabilises the o-nitrophenoxide ion by delocalising its negative charge, so the proton is lost more easily. The –OCH₃ group donates electrons (+R effect), which destabilises the o-methoxyphenoxide ion, so o-methoxyphenol is a weaker acid.

–NO₂ withdraws electrons and stabilises the phenoxide; –OCH₃ donates electrons and destabilises it.

7.16

Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Solution

The lone pair on the oxygen of –OH is delocalised into the ring by resonance (+R effect). This increases the electron density in the ring, especially at the ortho and para positions, so electrophiles attack the ring more readily, at those positions.

Resonance donation of the O lone pair raises electron density at the ortho and para positions.

7.17

Give equations of the following reactions: (i) Oxidation of propan-1-ol with alkaline

KMnO X_{4}

solution. (ii) Bromine in

CS X_{2}

with phenol. (iii) Dilute

HNO X_{3}

with phenol. (iv) Treating phenol with chloroform in presence of aqueous NaOH.

Solution

(i) $CH X_{3} CH X_{2} CH X_{2} OH alk \cdot KMnO X_{4} CH X_{3} CH X_{2} COOH$ (propanoic acid)
(ii) $C X_{6} H X_{5} OH + Br X_{2} CS X_{2} 273 K o - BrC X_{6} H X_{4} OH + p - BrC X_{6} H X_{4} OH$ (mainly para; monobromination in a non-polar solvent)
(iii) $C X_{6} H X_{5} OH dil \cdot HNO X_{3} 298 K o - O X_{2} NC X_{6} H X_{4} OH + p - O X_{2} NC X_{6} H X_{4} OH$
(iv) $C X_{6} H X_{5} OH + CHCl X_{3} aq \cdot NaOH then H X^{+} o - HOC X_{6} H X_{4} CHO$ (salicylaldehyde, Reimer–Tiemann)

(i) propanoic acid; (ii) o- and p-bromophenol; (iii) o- and p-nitrophenol; (iv) salicylaldehyde.

7.18

Explain the following with an example. (i) Kolbe's reaction. (ii) Reimer–Tiemann reaction. (iii) Williamson ether synthesis. (iv) Unsymmetrical ether.

Solution

(i) Kolbe's reaction: sodium phenoxide reacts with $CO X_{2}$ (400 K, 4–7 atm) and the product is acidified to give salicylic acid: $C X_{6} H X_{5} ONa + CO X_{2} o - HOC X_{6} H X_{4} COONa H X^{+} o - HOC X_{6} H X_{4} COOH$
(ii) Reimer–Tiemann reaction: phenol with chloroform and aqueous NaOH gives salicylaldehyde (–CHO enters ortho): $C X_{6} H X_{5} OH + CHCl X_{3} + NaOH o - HOC X_{6} H X_{4} CHO$
(iii) Williamson synthesis: an alkyl halide reacts with a sodium alkoxide or phenoxide by SN2: $CH X_{3} CH X_{2} Br + C X_{6} H X_{5} ONa C X_{6} H X_{5} OCH X_{2} CH X_{3} + NaBr$
(iv) Unsymmetrical ether: an ether with two different groups on oxygen, e.g. $CH X_{3} OC X_{2} H X_{5}$ (methoxyethane), $C X_{6} H X_{5} OCH X_{3}$ (anisole)

As explained with the examples above.

7.19

Write the mechanism of acid dehydration of ethanol to yield ethene.

Solution

With conc. $H X_{2} SO X_{4}$ at 443 K:

Step 1: protonation of the –OH: $CH X_{3} CH X_{2} OH + H X^{+} CH X_{3} CH X_{2} OH X_{2} X^{+}$
Step 2 (slow): loss of water gives a carbocation: $CH X_{3} CH X_{2} OH X_{2} X^{+} CH X_{3} CH X_{2} X^{+} + H X_{2} O$
Step 3: loss of a β-hydrogen as $H X^{+}$ : $CH X_{3} CH X_{2} X^{+} CH X_{2} = CH X_{2} + H X^{+}$

Protonation, loss of water to form CH₃CH₂⁺, then loss of H⁺ to give ethene.

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.

Solution

(i) $CH X_{3} CH = CH X_{2} + H X_{2} O H X^{+} CH X_{3} CH (OH) CH X_{3}$
(ii) $C X_{6} H X_{5} CH X_{2} Cl + NaOH (aq) C X_{6} H X_{5} CH X_{2} OH + NaCl$
(iii) $C X_{2} H X_{5} MgCl + HCHO C X_{2} H X_{5} CH X_{2} OMgCl H X_{2} O CH X_{3} CH X_{2} CH X_{2} OH$
(iv) $CH X_{3} MgBr + CH X_{3} COCH X_{3} (CH X_{3}) X_{3} COMgBr H X_{2} O (CH X_{3}) X_{3} COH$

(i) Acid-catalysed hydration. (ii) Aqueous NaOH. (iii) With methanal, then hydrolysis. (iv) With acetone, then hydrolysis.

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.

Solution

(i) Acidified $KMnO X_{4}$ (or acidified $K X_{2} Cr X_{2} O X_{7}$ )
(ii) Pyridinium chlorochromate (PCC) in $CH X_{2} Cl X_{2}$ (or heated Cu at 573 K)
(iii) Bromine water
(iv) Acidified (or alkaline, then acidified) $KMnO X_{4}$
(v) Conc. $H X_{2} SO X_{4}$ (or 85% $H X_{3} PO X_{4}$ ) at about 440 K
(vi) $NaBH X_{4}$ or $LiAlH X_{4}$ (or $H X_{2}$ /Ni)

(i) acidified KMnO₄; (ii) PCC; (iii) bromine water; (iv) KMnO₄; (v) conc. H₂SO₄ at 440 K; (vi) NaBH₄ or LiAlH₄.

7.22

Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

Solution

Ethanol molecules are joined by intermolecular hydrogen bonds (–O–H···O). Methoxymethane has no H on oxygen and cannot hydrogen-bond with itself. So ethanol (351 K) boils much higher than methoxymethane (248 K), although both are $C X_{2} H X_{6} O$ .

Ethanol forms intermolecular hydrogen bonds; methoxymethane cannot.

7.23

Give IUPAC names of the following ethers: (i)

C X_{2} H X_{5} OCH X_{2} - CH (CH X_{3}) - CH X_{3}

(ii)

CH X_{3} OCH X_{2} CH X_{2} Cl

(iii)

O X_{2} N - C X_{6} H X_{4} - OCH X_{3}

(p) (iv)

CH X_{3} CH X_{2} CH X_{2} OCH X_{3}

(v) cyclohexane ring with

OC X_{2} H X_{5}

on C-1 and two

CH X_{3}

on C-4 (vi)

C X_{6} H X_{5} OC X_{2} H X_{5}

Solution

(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane
(iii) 1-Methoxy-4-nitrobenzene (4-nitroanisole)
(iv) 1-Methoxypropane
(v) 1-Ethoxy-4,4-dimethylcyclohexane
(vi) Ethoxybenzene

As listed above.

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.

Solution

(i) $CH X_{3} CH X_{2} CH X_{2} ONa + CH X_{3} CH X_{2} CH X_{2} Br (CH X_{3} CH X_{2} CH X_{2}) X_{2} O + NaBr$
(ii) $C X_{6} H X_{5} ONa + C X_{2} H X_{5} Br C X_{6} H X_{5} OC X_{2} H X_{5} + NaBr$
(iii) $(CH X_{3}) X_{3} CONa + CH X_{3} Br (CH X_{3}) X_{3} COCH X_{3} + NaBr$ (the tertiary part must be the alkoxide; tert-butyl bromide with methoxide would give 2-methylpropene)
(iv) $CH X_{3} CH X_{2} ONa + CH X_{3} Br CH X_{3} CH X_{2} OCH X_{3} + NaBr$ (or $CH X_{3} ONa + C X_{2} H X_{5} Br$ )

Use the alkoxide/phenoxide with a primary (or methyl) halide, as in the equations above.

7.25

Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.

Solution

Tertiary (and often secondary) alkyl halides give alkenes, not ethers, because the alkoxide acts as a base: $(CH X_{3}) X_{3} CBr + CH X_{3} ONa (CH X_{3}) X_{2} C = CH X_{2} + CH X_{3} OH + NaBr$ . So $(CH X_{3}) X_{3} COCH X_{3}$ must be made from $(CH X_{3}) X_{3} CONa + CH X_{3} Br$ .
Aryl halides do not undergo SN2, so aryl alkyl ethers cannot be made from $C X_{6} H X_{5} Br + RONa$ ; the phenoxide and an alkyl halide must be used instead.

Tertiary halides undergo elimination with alkoxides, and aryl halides do not react by SN2.

7.26

How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.

Solution

By heating propan-1-ol with conc. $H X_{2} SO X_{4}$ at about 413 K (excess alcohol):

$2 CH X_{3} CH X_{2} CH X_{2} OH H X_{2} SO X_{4} 413 K CH X_{3} CH X_{2} CH X_{2} OCH X_{2} CH X_{2} CH X_{3} + H X_{2} O$

Mechanism (SN2):

Protonation: $CH X_{3} CH X_{2} CH X_{2} OH + H X^{+} CH X_{3} CH X_{2} CH X_{2} OH X_{2} X^{+}$
A second alcohol molecule attacks the protonated alcohol and water leaves: $CH X_{3} CH X_{2} CH X_{2} OH + CH X_{3} CH X_{2} CH X_{2} OH X_{2} X^{+} CH X_{3} CH X_{2} CH X_{2} O X^{+} (H) CH X_{2} CH X_{2} CH X_{3} + H X_{2} O$
Loss of $H X^{+}$ gives the ether.

Heat propan-1-ol with conc. H₂SO₄ at 413 K; protonation, SN2 attack by a second alcohol molecule, then deprotonation.

7.27

Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Solution

Secondary and tertiary alcohols easily form stable carbocations, which lose a proton to give alkenes (elimination) instead of being attacked by another alcohol molecule. Also, the bulky groups hinder the SN2 attack needed to form an ether.

They readily form carbocations that eliminate to alkenes rather than giving ethers.

7.28

Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether.

Solution

(i) $CH X_{3} CH X_{2} CH X_{2} OCH X_{2} CH X_{2} CH X_{3} + HI CH X_{3} CH X_{2} CH X_{2} I + CH X_{3} CH X_{2} CH X_{2} OH$ (with excess HI, both give $CH X_{3} CH X_{2} CH X_{2} I$ )
(ii) $C X_{6} H X_{5} OCH X_{3} + HI C X_{6} H X_{5} OH + CH X_{3} I$ (the O–aryl bond is not broken)
(iii) $C X_{6} H X_{5} CH X_{2} OC X_{2} H X_{5} + HI C X_{6} H X_{5} CH X_{2} I + C X_{2} H X_{5} OH$ (the stable benzyl carbocation forms, SN1)

(i) 1-iodopropane + propan-1-ol; (ii) phenol + iodomethane; (iii) benzyl iodide + ethanol.

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.

Solution

(i) The oxygen lone pair is delocalised into the ring (+R effect), increasing the electron density of the ring, so it reacts faster with electrophiles.

(ii) The resonance structures place the extra negative charge on the ortho and para carbons, so these positions have the highest electron density and the electrophile attacks there.

The +R effect of –OR increases ring electron density (activation), mostly at ortho and para positions (direction).

7.30

Write the mechanism of the reaction of HI with methoxymethane.

Solution

Step 1: protonation of the ether: $CH X_{3} - O - CH X_{3} + HI CH X_{3} - O X^{+} (H) - CH X_{3} + I X^{-}$
Step 2 (SN2): iodide attacks a methyl carbon and methanol leaves: $I X^{-} + CH X_{3} - O X^{+} (H) - CH X_{3} CH X_{3} I + CH X_{3} OH$
With excess HI, the methanol is also converted: $CH X_{3} OH + HI CH X_{3} I + H X_{2} O$

Protonation of the ether, then SN2 attack of I⁻ on CH₃ giving CH₃I and CH₃OH (excess HI turns CH₃OH into 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.

Solution

(i) $C X_{6} H X_{5} OCH X_{3} + CH X_{3} Cl anhyd \cdot AlCl X_{3} o - CH X_{3} C X_{6} H X_{4} OCH X_{3} + p - CH X_{3} C X_{6} H X_{4} OCH X_{3}$ (4-methoxytoluene major)
(ii) $C X_{6} H X_{5} OCH X_{3} conc \cdot HNO X_{3} / H X_{2} SO X_{4} o - O X_{2} NC X_{6} H X_{4} OCH X_{3} + p - O X_{2} NC X_{6} H X_{4} OCH X_{3}$ (4-nitroanisole major)
(iii) $C X_{6} H X_{5} OCH X_{3} + Br X_{2} CH X_{3} COOH p - BrC X_{6} H X_{4} OCH X_{3}$ (major) + a little 2-bromoanisole
(iv) $C X_{6} H X_{5} OCH X_{3} + CH X_{3} COCl anhyd \cdot AlCl X_{3} p - CH X_{3} COC X_{6} H X_{4} OCH X_{3}$ (4-methoxyacetophenone, major) + 2-methoxyacetophenone

Ortho and para products in each case, with para as the major product.

7.32

Show how would you synthesise the following alcohols from appropriate alkenes? (i) 1-methylcyclohexanol (ii) 4-methylheptan-4-ol (iii) pentan-2-ol (iv) 2-cyclohexylbutan-2-ol.

Solution

Acid-catalysed hydration ( $H X_{2} O / H X^{+}$ ) puts the –OH on the more substituted carbon (Markovnikov):

(i) 1-Methylcyclohexene + $H X_{2} O / H X^{+}$ → 1-methylcyclohexanol
(ii) 4-Methylhept-3-ene + $H X_{2} O / H X^{+}$ → 4-methylheptan-4-ol
(iii) Pent-1-ene + $H X_{2} O / H X^{+}$ → pentan-2-ol (pent-2-ene would give a mixture with pentan-3-ol)
(iv) 2-Cyclohexylbut-2-ene (or 2-cyclohexylbut-1-ene) + $H X_{2} O / H X^{+}$ → 2-cyclohexylbutan-2-ol

Hydration (H₂O/H⁺, Markovnikov) of the alkenes named above.

7.33

When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

CH X_{3} - CH (CH X_{3}) - CH (OH) - CH X_{3} HBr CH X_{3} - C (Br) (CH X_{3}) - CH X_{2} - CH X_{3}

. 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.)

Solution

Step I: protonation of the –OH: $(CH X_{3}) X_{2} CH - CH (OH) - CH X_{3} + H X^{+} (CH X_{3}) X_{2} CH - CH (OH X_{2} X^{+}) - CH X_{3}$
Step II: loss of water gives a secondary carbocation: $(CH X_{3}) X_{2} CH - CH X^{+} - CH X_{3}$
Step III: a 1,2-hydride shift from C-3 to C-2 gives the more stable tertiary carbocation: $(CH X_{3}) X_{2} C X^{+} - CH X_{2} - CH X_{3}$
Step IV: bromide attacks the tertiary carbocation: $(CH X_{3}) X_{2} C X^{+} CH X_{2} CH X_{3} + Br X^{-} (CH X_{3}) X_{2} CBrCH X_{2} CH X_{3}$ (2-bromo-2-methylbutane)

Protonation, loss of water (2° carbocation), 1,2-hydride shift to the 3° carbocation, then attack by Br⁻.

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