Reduction of acetyl chloride with $H_{2}$ in presence of Pd and $B a S O_{4}$ gives :

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

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C_{2} H_{5} O H

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C H_{2} C O O H

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C H_{3} C H O

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Solution

The correct option is D $C H_{3} C H O$
$C H_{3} - O | | C - C l + H_{2} P d / B a S O_{4} - ----- \to C H_{3} - O | | C - H a c e t a l d e h y d e + H C l$

This reaction is known as Rosenmund reaction.

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