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$C H_{3} C O O H (a q) \overset{K O H (a q)}{\to} A \to_{}^{E l e c t r o l y s i s} B$
$B \to_{{(C H_{3} C O O)}_{2} M n}^{O_{2}} C \to_{Δ}^{N a O H + C a O} D$
The compound D is

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$C H_{3} - C H_{2} - C H_{3}$

$C H_{3 -} C O O H$

$C H_{4}$

$C H_{3} - C H_{3}$

answer is C.

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Detailed Solution

A is $C H_{3} C O O K (a q)$

B is $H_{3} C - C H_{3}$

C is $C H_{3} - C O O H$

D is $C H_{4}$

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