JEE Chemistry Alcohols Phenols and Ethers MCQs Set 08

Answer: (c)

Explanation:
1. Cyclopentylmethanol is protonated by H₃PO₄ and loses water to give a primary carbocation.
2. A ring expansion (1,2-alkyl shift) converts the primary carbocation into the more stable secondary cyclohexyl cation.
3. Loss of a proton gives cyclohexene, option (c).

Teacher's Note:
1. Ring expansion relieves strain and gives a more stable cation.
2. The product has six carbons in the ring.

Answer: (d) 2-Methylbutan-2-ol

Explanation:
1. Lucas reagent reacts through a carbocation (SN₁), so tertiary alcohols react immediately.
2. 2-Methylbutan-2-ol is tertiary; 2-pentanol is secondary and reacts slowly; the other two are primary.

Teacher's Note:
1. Lucas test: 3° at once, 2° in about 5 min, 1° only on heating.
2. Faster carbocation formation means faster reaction.

Answer: (c) CH₃CONH₂ → (LiAlH₄)

Explanation:
1. (a) Primary amines react with nitrous acid to give alcohols (with N₂).
2. (b) Grignard reagent adds to acetaldehyde to give propan-2-ol.
3. (c) LiAlH₄ reduces an amide to an amine (CH₃CH₂NH₂), not an alcohol.
4. (d) Aqueous KOH substitutes Cl by OH to give ethanol.

Teacher's Note:
1. Amides are reduced to amines.
2. Check each option for the product.

Answer: (c) NaHCO₃

Explanation:
1. Cumene is oxidised by air to cumene hydroperoxide (P), which on acid treatment gives phenol (Q) and acetone.
2. Phenol reacts with Na (liberates H₂), with NaOH (forms sodium phenoxide) and with Zn dust on heating (gives benzene).
3. Phenol is a weaker acid than carbonic acid, so it does not react with NaHCO₃.

Teacher's Note:
1. Phenol is weaker than H₂CO₃.
2. Phenol + Zn dust gives benzene.

Answer: (b) a > c > d > b

Explanation:
1. Pentan-1-ol forms intermolecular hydrogen bonds, so it has the highest boiling point (138 C).
2. Pentanal is polar with dipole–dipole attraction (103 C).
3. Ethoxyethane is only weakly polar (35 C).
4. n-Butane has only weak van der Waals forces (about 0 C).
5. Order: a > c > d > b.

Teacher's Note:
1. H-bond > dipole–dipole > weak polarity > non-polar for similar mass.
2. Use actual boiling points to check.

Answer: (a) CH₃CH₂CH(CH₃)CH₂OH + CH₃CH₂I

Explanation:
1. With HI, the iodide attacks by SN₂ at the less hindered alkyl group of the ether.
2. The ethyl group is less hindered than the 2-methylbutyl group, which is branched next to the carbon attached to oxygen.
3. So ethyl iodide and 2-methylbutan-1-ol are formed.

Teacher's Note:
1. In SN₂ cleavage the less hindered carbon is attacked.
2. Branching near the reacting carbon slows SN₂.

Answer: (b) C₆H₅CH₂I + C₆H₅OH

Explanation:
1. The protonated ether is cleaved at the alkyl (benzyl) side, because the C(aryl)–O bond is not broken by SN₂.
2. The benzyl cation or SN₂ attack by I⁻ gives benzyl iodide, and the leaving group is phenol.

Teacher's Note:
1. Aryl–O bond stays intact.
2. Phenol never turns into iodobenzene with HI.

Answer: (b)

Explanation:
1. Aniline reacts with nitrous acid at low temperature to give benzenediazonium chloride (A).
2. Warming the diazonium salt with water replaces N₂+ by OH, giving phenol (B), with release of N₂ and HCl.

Teacher's Note:
1. Diazonium salts hydrolyse to phenols on warming.
2. The –NH₂ group is replaced; no second substituent is introduced.

Answer: (a) CH₃–O–C(CH₃)₃

Explanation:
1. This is Williamson synthesis, an SN₂ reaction.
2. The alkyl halide is methyl bromide (no hindrance) and the alkoxide is the bulky sodium tert-butoxide, so substitution occurs and not elimination.
3. The product is methyl tert-butyl ether, CH₃OC(CH₃)₃.

Teacher's Note:
1. Use the primary or methyl halide and the bulky alkoxide in Williamson synthesis.
2. A tertiary halide would give an alkene.

Answer: (c) (CH₃)₂CCl–CH(CH₃)₂

Explanation:
1. The secondary alcohol forms a secondary carbocation (CH₃)₃C–CH(+)–CH₃.
2. A methyl shift converts it into the more stable tertiary cation (CH₃)₂C(+)–CH(CH₃)₂.
3. Chloride attacks this cation to give 2-chloro-2,3-dimethylbutane, (CH₃)₂CCl–CH(CH₃)₂.

Teacher's Note:
1. Carbocations rearrange to become more stable.
2. Neopentyl-type systems usually undergo a methyl shift.

Answer: (b) 1-Bromopropane

Explanation:
1. Alcoholic KOH eliminates HBr from 2-bromopropane to give propene (A).
2. HBr in the presence of peroxide adds in the anti-Markovnikov way (Kharasch effect): Br goes to the terminal carbon.
3. B is 1-bromopropane, CH₃CH₂CH₂Br.

Teacher's Note:
1. Peroxide effect applies to HBr only.
2. Anti-Markovnikov means Br goes to the carbon with more hydrogens.

Answer: (b)

Explanation:
1. The acetylide ion CH₃–C≡C⁻ is a strong nucleophile and opens the epoxide at the less hindered CH₂ carbon, giving the alkoxide CH₃CH(O⁻)CH₂–C≡C–CH₃.
2. The alkoxide is a strong nucleophile and is methylated by CH₃I (Williamson step) to give the methyl ether.
3. The product is CH₃CH(OCH₃)CH₂C≡C–CH₃, option (b); the triple bond is not changed.

Teacher's Note:
1. Epoxide opening is followed by O-alkylation.
2. The C≡C remains in the product.

Answer: (c) (i) < (ii) < (iii)

Explanation:
1. Reaction with sodium involves breaking the O–H bond, so the more acidic alcohol reacts faster.
2. Alkyl groups donate electrons and reduce acidity, so acidity (and reactivity with Na) is primary > secondary > tertiary.
3. Increasing reactivity: (i) < (ii) < (iii).

Teacher's Note:
1. Reactivity with Na: 1° > 2° > 3°.
2. Reactivity towards HX (cleaving C–O) follows the opposite order.

Answer: (a) I < II < III < IV

Explanation:
1. Compound I is phenol; the C–O bond has double bond character, so it does not undergo acid dehydration at all.
2. Compound II is cyclohexanol, a plain secondary alcohol.
3. Compound III is allylic, so it gives a resonance-stabilised cation and dehydrates more easily than II.
4. Compound IV gives a pentadienyl-type (conjugated) cation, and loss of water or a proton gives benzene, so it is the easiest.
5. Order: I < II < III < IV.

Teacher's Note:
1. More stable carbocation means easier dehydration.
2. Conjugation stabilises the cation.

Answer: (d) CH₃CH=CH₂, CH₃CH₂CH₂Br, CH₃CH₂CH₂OCH₃

Explanation:
1. Alcoholic KOH gives propene (X).
2. HBr with peroxide gives 1-bromopropane (Y) by anti-Markovnikov addition.
3. Sodium methoxide and 1-bromopropane react by Williamson synthesis to give methyl propyl ether, CH₃CH₂CH₂OCH₃ (Z).

Teacher's Note:
1. The OR group of the alkoxide enters the molecule.
2. CH₃ONa supplies the methoxy group.

Answer: (d) C₆H₅OH and CH₃COCH₃

Explanation:
1. Cumene is oxidised by air to cumene hydroperoxide.
2. Acid-catalysed rearrangement and hydrolysis cleave it into phenol and acetone.
3. This is the cumene process, the main industrial source of phenol.

Teacher's Note:
1. Remember: cumene process gives phenol + acetone.
2. The C₆H₅ group migrates to oxygen.

Answer: (c) (A)→(ii), (B)→(iii), (C)→(iv), (D)→(i)

Explanation:
1. (A) Williamson synthesis is the reaction of an alkoxide with an alkyl halide, (ii).
2. (B) R–O–R' with different groups is an unsymmetrical (mixed) ether, (iii).
3. (C) p-Nitrophenol molecules are joined by intermolecular hydrogen bonds (higher boiling point than the ortho isomer), (iv).
4. (D) Acetylation of phenol with acetyl chloride and pyridine gives phenyl acetate, (i).

Teacher's Note:
1. Match each term to its best description.
2. Ortho-nitrophenol has intramolecular H-bonding; para has intermolecular.

Answer: (c) (i) I₂/NaOH; (ii) H+

Explanation:
1. A methyl ketone is converted to a carboxylic acid with one carbon less by the haloform reaction (I₂/NaOH, then H+).
2. The ring C=C is untouched by I₂/NaOH. KMnO₄ would oxidise the double bond, Tollens' reagent oxidises only aldehydes and benzoyl peroxide is a radical initiator.

Teacher's Note:
1. Methyl ketone plus I₂/NaOH gives the carboxylate and iodoform.
2. KMnO₄ would also cleave C=C.

Answer: (a) CH₃CH(OH)CH₂CHO

Explanation:
1. The aldehyde group is more reactive than the ketone group, so one equivalent of glycol protects the aldehyde as a cyclic acetal (A).
2. LiAlH₄ reduces the free ketone to a secondary alcohol; the acetal is stable to LiAlH₄.
3. H₃O+ removes the acetal and restores the CHO group, giving CH₃CH(OH)CH₂CHO.

Teacher's Note:
1. Protect the aldehyde first, then reduce the ketone.
2. Acetals are stable to bases and hydrides but hydrolysed by aqueous acid.

Answer: (b)

Explanation:
1. The alcohol is protonated and loses water to give a primary carbocation next to the five-membered ring.
2. Ring expansion (1,2-shift) gives a six-membered cyclohexadienyl cation, which is stabilised by delocalisation.
3. Loss of a proton gives the aromatic ring, benzene, which is very stable.

Teacher's Note:
1. Aromatic stability drives the reaction.
2. Ring expansion: five-membered ring plus the CH₂ becomes a six-membered ring.