Question

Which of the following reagents is used in the reaction shown below?
Benzoyl chloride → Benzaldehyde

• A. DIBAI-H
• B. SnCl\(_2\)/HCl
• C. H\(_2\)/Pd – BaSO\(_4\)
• D. Dimethyl cadmium

Hint:

Rosenmund reduction: R–COCl + H\(_2\) \(\xrightarrow{\text{Pd/BaSO}_4}\) R–CHO + HCl. The BaSO\(_4\) and a poison prevent further reduction to R–CH\(_2\)OH.

Answer 1

The Correct Option is C

Step 1: Understand the question.
We must change benzoyl chloride ($C_6H_5COCl$) into benzaldehyde ($C_6H_5CHO$). This means reducing an acid chloride down to an aldehyde and stopping there.
Step 2: Note the danger of over-reduction.
A strong reducer would push past the aldehyde and make an alcohol. So we need a gentle, controlled method.
Step 3: Recall the Rosenmund reduction.
The Rosenmund reduction uses hydrogen gas with palladium spread on barium sulphate. The catalyst is weakened on purpose so it stops at the aldehyde.
Step 4: Write the change.
\[ C_6H_5COCl \xrightarrow{H_2,\ Pd\text{-}BaSO_4} C_6H_5CHO \]
Step 5: Rule out the others.
$SnCl_2/HCl$ is for nitriles or nitro groups. Dimethyl cadmium turns acid chlorides into ketones, not aldehydes. So those do not fit.
Step 6: Choose the answer.
The reagent is $H_2/Pd\text{-}BaSO_4$, which is option 3. \[ \boxed{H_2/Pd\text{-}BaSO_4} \]