Single correct option
Match each reaction in List-I with the description of its major organic product in List-II.
List-I
(P) 2,3-Dimethylbutane-2,3-diol heated with dilute
(Q) Benzamide treated with and aqueous , then warmed
(R) Benzaldehyde and formaldehyde treated with concentrated (organic product other than formate)
(S) Benzil () boiled with aqueous followed by acidification
List-II
(1) A cyclic amide
(2) A primary aromatic amine
(3) A ketone containing a quaternary carbon atom
(4) An -hydroxy carboxylic acid
(5) A primary alcohol in which the carbinol carbon is attached to a phenyl ring
List-I
(P) 2,3-Dimethylbutane-2,3-diol heated with dilute
(Q) Benzamide treated with and aqueous , then warmed
(R) Benzaldehyde and formaldehyde treated with concentrated (organic product other than formate)
(S) Benzil () boiled with aqueous followed by acidification
List-II
(1) A cyclic amide
(2) A primary aromatic amine
(3) A ketone containing a quaternary carbon atom
(4) An -hydroxy carboxylic acid
(5) A primary alcohol in which the carbinol carbon is attached to a phenyl ring
- A. P→3, Q→2, R→5, S→4
- B. P→3, Q→2, R→4, S→5
- C. P→1, Q→2, R→5, S→4
- D. P→3, Q→1, R→5, S→2
Show answer and solution
Answer: (A) P→3, Q→2, R→5, S→4
P: the pinacol rearrangement. Protonation and loss of water gives a tertiary cation, then a methyl shift gives pinacolone, , a ketone with a quaternary carbon (3). Q: the Hofmann bromamide degradation converts into with loss of the carbonyl carbon as carbonate, a primary aromatic amine (2). R: crossed Cannizzaro reaction. Formaldehyde is the better hydride donor and is oxidised to formate, while benzaldehyde is reduced to benzyl alcohol, , a primary alcohol with a benzylic carbinol carbon (5). S: benzilic acid rearrangement of a non-enolisable 1,2-diketone gives after acidification, an -hydroxy acid (4). Cyclic amide (1) belongs to a Beckmann product and is not used. The correct code is P→3, Q→2, R→5, S→4.
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