The final product B in the following reaction is :
$C H_{3} C H_{2} C H = C H_{2} N B S, h v - ------ \to (C_{4} H_{3} C O_{2})_{2} A C H_{3} S N a - ---- \to C H_{2} O H B$

methyl I-methylallysulphide

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allyl methyl sulphide

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n-butyl methyl sulphide

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diallyl sulphide

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Solution

The correct option is A methyl I-methylallysulphide
$C H_{3} C H_{2} C H = C H_{2}$
$↓$
( Allylic addition of $B r$ )
$B r$
|
$C H_{3} - C H - C H = C H_{2} (A)$
$↓$ $C H_{3} S N a,$ $C H_{3} O H, (S N 2)$

$C H_{3} - C H - C H = C H_{2}$
|
$S C H_{3}$
So, (A) is correct answer.

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