Ethyl alcohol on oxidation with acidified $K_{2} C r_{2} O_{7}$ gives

C H_{3} C O C H_{3}

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H C O O H

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C H_{3} C O O H

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H C H O

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Solution

The correct option is C $C H_{3} C O O H$
Ethyl alcohol on oxidation with acidified $K_{2} C r_{2} O_{7}$ gives $C H_{3} C O O H$ . Acidified $K_{2} C r_{2} O_{7}$ and $K M n O_{4}$ converts primary alcohol to carboxylic acid. Hence the correct answer is option (c).

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K_{2} C r_{2} O_{7}

Ethyl alcohol on oxidation with acidified potassium permanganate gives:

K_{2} C r_{2} O_{7}

K M n O_{4}

K_{2} C r_{2} O_{7}

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