Major end product of the following sequence of reaction is:
$C H_{3} C H_{2} C H_{2} C O N H_{2} C a {(O H)}_{2}, C l_{2} - ------- \to Δ X H N O_{2} - --- \to Z$

C H_{3} C H_{2} C H_{2} N H_{2}

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C H_{3} C H_{2} C H_{2} O H

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C H_{3} C H_{2} C H_{2} C O O H

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C H_{3} C H_{2} C O O H

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Solution

The correct option is B $C H_{3} C H_{2} C H_{2} O H$
$C H_{3} C H_{2} C H_{2} C O N H_{2} C a {(O H)}_{2}, C l_{2} - ------- \to Δ C H_{3} C H_{2} C H_{2} N H_{2} X H N O_{2} - --- \to C H_{3} C H_{2} C H_{2} O H Z$
The first step is the Hoffmann's bromamide reaction in which butanamide gives n-propyl amine which contains one carbon atom less.
In the second reaction, nitrous acid reacts with n-propyl amine to form n propyl alcohol.

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C H_{3} C H_{2} C O O H

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- ----- \to A l c . K O H

C H_{3} C H_{2} C O O H C l_{2}, a l c o h o l i c K O H - ----------- \to r e d P

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