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Curtius and Hofmann Rearrangements

An aromatic c...

Question

An aromatic compound ' $A$ ' $(C_{7} H_{9} N)$ on reacting with $N a N O_{2} / H C l$ at $0 - 5^{0} C$ forms benzyl alcohol and nitrogen gas. The number of isomers possible for the compound ' $A$ ' is :

7

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6

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5

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3

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Solution

The correct option is C $5$

An aromatic compound 'A' ( $C_{7} H_{9} N$ ) on reacting with $N a N O_{2} / H C l$ at $0^{o} C$ forms benzyl alcohol and nitrogen gas. Hence, the aromatic compound 'A' is benzylamine.

$C_{6} H_{5} - C H_{2} - N H_{2} + H N O_{2} N a N O_{2} + H C l - ------- \to 0^{o} C C_{6} H_{5} - C H_{2} - O H + H_{2} O + N_{2} ↑$

The number of isomers possible for the compound 'A' is 5. These are as shown.
Option C is correct.

Suggest Corrections

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