Reduction of acetyl chloride with \(\mathrm{H}_{2}\) in presence of \(\mathrm{Pd}\) and…

Reduction of acetyl chloride with \(\mathrm{H}_{2}\) in presence of \(\mathrm{Pd}\) and \(\mathrm{BaSO}_{4}\) gives:
  1. \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\)
  2. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)
  3. \(\mathrm{CH}_{2} \mathrm{COOH}\)
  4. \(\mathrm{CH}_{3} \mathrm{CHO}\)

Solution

\(\mathrm{C H}_{3}-\stackrel{\huge \mathrm{O} \atop ||}{\mathrm{C}}-\mathrm{C l}+\mathrm{H}_{2}\stackrel{\mathrm{P d} / \mathrm{BaSO}_{4}}{\longrightarrow} \underset{\text{acetaldehyde}}{\mathrm{C} \mathrm{H}_{3}-\stackrel{\huge \mathrm{O} \atop ||}{\mathrm{C}}-\mathrm{H}+}\mathrm{HCl}\)
This reaction is known as Rosenmund reaction.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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