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Molecular Formula

A compound A ...

Question

A compound A having molecular formula $C_{4} H_{9} B r$ on reaction with alcoholic $K O H$ gives a compound B. Bromination of B gives compound C. compound C on treatment with soda lime gives a gaseous compound D. The gas D when passed through ammonical silver nitrate solution forms white precipitate. Identify A, B, C and D and write the reactions involved.

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Solution

$B r - C H_{2} - C H_{2} - C H_{2} - C H_{3} a l c . - -- \to K O H C H_{2} = C H - C H_{2} - C H_{3} B r_{2} - -- \to C C l_{4} B r - C H_{2} - C H | - C H_{2} C H_{3}$
$A$ $B$ $B r$ $C$
$↓ (S o d a l i m e) (N a O H + C a O)$
$C H_{3} C H_{2} - C \equiv C^{-} A g^{+} ↓ [A g (N H_{3})_{2}]^{+} \leftarrow -------------- - (a m m o n i c a l s i l v e r n i t r a t e) H - C \equiv C - C H_{2} - C H_{3} t e r m i n a l a l k y n e$
Alkyl bromide (A) on reaction with alcoholic $K O H$ gives a alkene B which on bromination gives dibromobutane (C). C on treatment with soda lime gives a terminal alkyne D which gives white ppt with ammonical silver nitrate solution where alkynes should be terminal.

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