identify 'A' and 'B' in the following reaction:
$C H_{3} - C H = C H_{2} H B r - -- \to^{'} A^{'} a l c K O H - ---- \to^{'} B^{'}$

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Solution

$C H_{3} - C H = C H_{2} p r o p e n e H B r - -- \to C H_{3} - C H_{2} - C H_{2} - B r 1 - p r o p y l b r o m i d e a l c . K O H - ---- \to C H_{3} - C H = C H_{2} p r o p e n e + K B r + H_{2} O$
When propene reacts with $H B r$ it undergoes hydrohalogenation reaction to form $1 - b r o m o p r o p a n e$ ( $A$ ) as a product.
When compound $(A)$ reacts with $a l c . K O H$ , it undergoes $β$ -elimination reaction to form $p r o p e n e$ ( $B$ ) as a product.

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