Question

In the reaction CH$_3$COCl + H$_2$ \xrightarrow{Pd-BaSO$_4$ .......... the product is –}

• A. Ketone
• B. Aldehyde
• C. Acid
• D. Alcohol

Hint:

Rosenmund reduction is a selective method for preparing aldehydes from acyl chlorides. Pd–BaSO$_4$ ensures partial reduction only up to aldehyde, not alcohol.

Answer 1

The Correct Option is B

Step 1: Recognize the reaction.
The given reaction is the Rosenmund reduction. In this reaction, an acid chloride (R–COCl) is reduced to an aldehyde (R–CHO).

Step 2: Role of catalyst.
- The catalyst is Pd (palladium) deposited on BaSO$_4$.
- BaSO$_4$ acts as a poison to palladium, preventing complete reduction to alcohol.
- Thus, the reduction stops at the aldehyde stage.

Step 3: Apply to given case.
CH$_3$COCl $\xrightarrow{H_2/Pd-BaSO_4}$ CH$_3$CHO
So, acetyl chloride is reduced to acetaldehyde.
Step 4: Final Answer.
The product is an aldehyde.
\[ \boxed{\text{Aldehyde (CH$_3$CHO)}} \]