An alcohol on oxidation gives $C H_{3} C O O H$ and $C H_{3} C H_{2} C O O H$ . The alcohol is :

C H_{3} C H_{2} O H

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C H_{3} C H (O H) C H_{2} C H_{3}

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(C H_{3})_{2} C (O H) C H_{2} C H_{3}

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C H_{3} C H (O H) C H_{2} C H_{2} C H_{3}

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Solution

The correct option is C $C H_{3} C H (O H) C H_{2} C H_{2} C H_{3}$
Since the secondary alcohol on oxidation gives two different acids, acetic acid and propanoic acid, the alcohol is secondary alcohol and will contain 5 C atoms.

$C H_{3} C H (O H) C H_{2} C H_{2} C H_{3} o x i d a t i o n, e x t r e m e c o n d i t i o n - ---------------- \to C H_{3} C O O H + C H_{3} C H_{2} C O O H$

Hence, option D is correct.

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