Question

Reduction of acetyl chloride with ${H_2}$ in presence of $Pd$ and $BaS{O_4}$ gives:
(A)$C{H_3}COC{H_3}$
(B)${C_2}{H_5}OH$
(C)$C{H_3}COOH$
(D)$C{H_3}CHO$

Answer 1

Hint: To solve the question we need to decide what kind of reaction is taking place and the name of the reaction. Here we observe that there is addition of hydrogen taking place so it is an Example of Hydrogenation reaction. This reaction is called the Rosenmund reaction. It is named after German chemist Karl Wilhelm Rosenmund.

Complete answer:
Rosenmund Reaction- Some of the basic points that need to be remembered here are-
(1)It is a catalytic hydrogenation reaction.
(2)This reaction selectively reduces Acyl chloride into Aldehyde.
(3)The Palladium $(Pd)$ catalyst which is used must be poisoned in order to avoid the further reduction of Aldehyde into primary Alcohol (to deactivate it). This primary alcohol can react with Acyl chlorides to form esters. The common poisons used to reduce palladium activity are thioquinanthrene and thiourea.
(4) Barium Sulphate $(BaS{O_4})$ is used to reduce the activity of palladium catalyst because it has low surface area.
(5) The combination of Palladium and Barium Sulphate is called Lindlar’s Catalyst.
In the question we are given Acetyl Chloride which is $C{H_3}COCl$ and the reaction occurs as under-
$C{H_3}COCl\xrightarrow{{{H_2},Pd/BaS{O_4}}}C{H_3}CHO + HCl$
Therefore there is a production of Acetaldehyde and Hydrochloric Acid in this process.
So option (D) is correct.

Note:
The Rosenmund reduction can be used to produce several Alkyl and Aryl Aldehydes. It can also be used in the production of saturated fatty acids. However the reaction cannot be used in the production of Formaldehyde because Formyl chloride is highly unstable at room temperature.