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The major products $P$ and $Q$ formed in the following reactions of benzonitrile are
Options:
- A
- B
- C
- D
Solution:
1898 Upvotes
Verified Answer
The correct answer is:
When $\mathrm{C}_d \mathrm{H}_5 \mathrm{CN}$ reacts with $\mathrm{Na} / \mathrm{E}$ toH, $\mathrm{C}_e \mathrm{H}_5 \mathrm{CN}$
undergo for reduction and $-\mathrm{C}=\mathrm{N}$, changes to $-\mathrm{CH}_2-\mathrm{NH}_2$.
Thus, the product $(P)$ is
The reaction occurs as follows :
When $\mathrm{C}_s \mathrm{H}_5 \mathrm{CN}$ reacts with (i) $\mathrm{HCl}$ (conc. cold) and (ii) $\mathrm{Br}_2 / \mathrm{KOH}$, it gives corresponding primary amine with one $\mathrm{C}$-atom less, as in parent formula (using Hoffmann-Bromamide degradation reaction). Thus, the final product is
The whole reaction is as follows:
Hence, option (c) is the correct answer.
undergo for reduction and $-\mathrm{C}=\mathrm{N}$, changes to $-\mathrm{CH}_2-\mathrm{NH}_2$.
Thus, the product $(P)$ is
The reaction occurs as follows :
When $\mathrm{C}_s \mathrm{H}_5 \mathrm{CN}$ reacts with (i) $\mathrm{HCl}$ (conc. cold) and (ii) $\mathrm{Br}_2 / \mathrm{KOH}$, it gives corresponding primary amine with one $\mathrm{C}$-atom less, as in parent formula (using Hoffmann-Bromamide degradation reaction). Thus, the final product is
The whole reaction is as follows:
Hence, option (c) is the correct answer.
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