Single correct option
Match the organic transformations in List-I with the statements in List-II and choose the correct option.
List-I
P. Benzamide treated with and aqueous , then warmed
Q. Benzaldehyde treated with concentrated aqueous
R. Acetophenone treated with excess and (i.e. ), then warmed
S. Benzenediazonium chloride treated with (and )
List-II
1. A yellow crystalline solid is formed as a by-product
2. The organic product is a nitrile
3. The organic product gives a positive carbylamine test
4. A hydride ion is transferred from one molecule to the carbonyl carbon of another
5. A dichlorocarbene-type reactive intermediate is involved
List-I
P. Benzamide treated with and aqueous , then warmed
Q. Benzaldehyde treated with concentrated aqueous
R. Acetophenone treated with excess and (i.e. ), then warmed
S. Benzenediazonium chloride treated with (and )
List-II
1. A yellow crystalline solid is formed as a by-product
2. The organic product is a nitrile
3. The organic product gives a positive carbylamine test
4. A hydride ion is transferred from one molecule to the carbonyl carbon of another
5. A dichlorocarbene-type reactive intermediate is involved
- A. P→3, Q→4, R→1, S→2
- B. P→3, Q→1, R→4, S→2
- C. P→2, Q→4, R→1, S→3
- D. P→4, Q→3, R→2, S→1
Show answer and solution
Answer: (A) P→3, Q→4, R→1, S→2
P is the Hofmann bromamide degradation: . The product aniline is a primary aromatic amine and therefore gives the carbylamine (isocyanide) test, so P→3. Q is the Cannizzaro reaction; benzaldehyde has no -hydrogen, and a hydride ion is transferred from the tetrahedral alkoxide intermediate to a second aldehyde molecule, giving benzyl alcohol and sodium benzoate: Q→4. R is the haloform reaction of a methyl ketone; gives sodium benzoate and yellow iodoform (): R→1. S is the Sandmeyer reaction giving benzonitrile, : S→2. Item 5 (a carbene) belongs to the Reimer-Tiemann reaction, which is not listed. The correct code is P→3, Q→4, R→1, S→2.
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