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$C H_{2} B r - C H_{2} B r \to_{Δ}^{Z n d u s t} A$ $\overset{H_{2} / N i}{\to} B \to_{O_{2}}^{{(C H_{3} C O O)}_{2} M n} C .$ Here ‘C’ is

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$C H_{3} N O_{2}$

$C H_{3} C O O H$

$C H_{3} - O - N = O$

$C_{2} H_{5} - O - N = O$

answer is B.

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Detailed Solution

${A is CH}_{2} {= CH}_{2}$

${B is CH}_{3} - {CH}_{3}$

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