Alcohols: classification and dehydration: NEET MCQs with answers

Answer: (D) 2-phenylpropan-2-ol

In 2-phenylpropan-2-ol [C₆H₅-C(CH₃)₂(OH)], the hydroxyl group is attached to a benzylic carbon that is bonded to three other carbons (one phenyl group and two methyl groups), making it a tertiary (3°) benzylic alcohol. 1-Phenylethan-1-ol is a secondary benzylic alcohol, whereas phenylmethanol is primary.

Answer: (B) C₆H₅-CH(OH)-CH₃

In benzylic alcohols, the -OH group is attached to an sp³-hybridised benzylic carbon adjacent to an aromatic ring. In 1-phenylethanol (C₆H₅-CH(OH)-CH₃), the benzylic carbon is attached to one hydrogen, one phenyl group, and one methyl group, making it a secondary benzylic alcohol.

Answer: (A) 2-Methylbut-2-ene

Protonation of neopentyl alcohol and departure of water is accompanied by a 1,2-methyl shift to form the stable tertiary 2-methylbutan-2-yl carbocation. Elimination of a β-proton from C3 then gives 2-methylbut-2-ene as the most stable, most substituted (Zaitsev) major product.

Answer: (D) 2.24 L

Each −OH gives ½ H₂. methanol has 1 −OH, so moles H₂ = 0.2 × 1/2 = 0.1; volume = 2.24 L.

Answer: (D) 2-Methylbut-3-en-2-ol

In 2-methylbut-3-en-2-ol [CH₂=CH-C(CH₃)₂(OH)], the -OH group is attached to an sp³ hybridised carbon atom adjacent to a carbon-carbon double bond (allylic carbon), and this carbon is bonded to three carbon groups, making it a tertiary (3°) allylic alcohol.

Answer: (B) (III) > (IV) > (I) > (II)

Acid-catalysed dehydration proceeds via carbocation intermediates. The rate of dehydration is directly proportional to carbocation stability. The carbocation formed from (III) is benzylic (C₆H₅-CH⁺-CH₃), which is stabilized by resonance with the benzene ring and is more stable than the 3° carbocation from (IV), which in turn is more stable than the 2° carbocation from (I), and the 1° carbocation from (II) is least stable. Hence, the order is (III) > (IV) > (I) > (II).

Answer: (C) 2.24 L

Each −OH gives ½ H₂. ethane-1,2-diol has 2 −OH, so moles H₂ = 0.1 × 2/2 = 0.1; volume = 2.24 L.

Answer: (B) 3.36 L

Each −OH gives ½ H₂. glycerol (propane-1,2,3-triol) has 3 −OH, so moles H₂ = 0.1 × 3/2 = 0.15; volume = 3.36 L.

Answer: (A) 2-Phenylbut-2-ene

Protonation and loss of water from 3-phenylbutan-2-ol yields a secondary carbocation at C2: CH₃-CH(Ph)-CH(+)-CH₃. A 1,2-hydride shift from C3 gives a highly stable benzylic-tertiary carbocation: CH₃-C(+)(Ph)-CH₂-CH₃. Subsequent elimination of a proton yields 2-phenylbut-2-ene, which is stabilized by both hyperconjugation and conjugation with the phenyl ring.

Answer: (A) 3

For C₆H₁₄O, tertiary alcohols have an -OH group on a carbon attached to three alkyl groups. The possible alkyl combinations on the carbinol carbon are: 2-methylpentan-2-ol, 3-methylpentan-3-ol, and 2,3-dimethylbutan-2-ol. Thus, exactly 3 constitutional tertiary alcohol isomers exist.

Answer: (D) 1.12 L

Each −OH gives ½ H₂. methanol has 1 −OH, so moles H₂ = 0.1 × 1/2 = 0.05; volume = 1.12 L.

Answer: (A) 11.2 L

Each −OH gives ½ H₂. methanol has 1 −OH, so moles H₂ = 1 × 1/2 = 0.5; volume = 11.2 L.