Question

Consider the following reaction, and choose the correct option.

• A. Compound P is obtained by the hydrogenation of benzoyl chloride with Pd on $\text{BaSO}_4$.
• B. On treating compound P with saturated $\text{NaHCO}_3$ solution, brisk effervescence is observed.
• C. Compound P can be prepared by treating benzene with anhydrous $\text{AlCl}_3$ and $\text{CH}_3\text{COCl}$.
• D. On treatment with bromine water, compound P gives a white precipitate.

Hint:

To quickly master named reactions of carbonyl compounds: - Etard Reaction: $\text{Toluene} + \text{CrO}_2\text{Cl}_2 \rightarrow \text{Benzaldehyde}$ - Rosenmund Reduction: $\text{Acid Chloride} + \text{H}_2 / \text{Pd-BaSO}_4 \rightarrow \text{Aldehyde}$ Both are premier methods for synthesizing pure aromatic aldehydes without over-oxidation or over-reduction!

Answer 1

The Correct Option is A

Step 1: Read the reaction in the figure.
The starting material is toluene, \(C_6H_5CH_3\), and it is treated with chromyl chloride \(CrO_2Cl_2\) in \(CS_2\), then hydrolysed. This is the Etard reaction, so the product P is benzaldehyde \(C_6H_5CHO\). Let us keep that fact ready and test each option against it.
Step 2: Why P is benzaldehyde.
Chromyl chloride is a gentle oxidant. It oxidises the \(-CH_3\) group on the ring up to \(-CHO\) but stops there, so we do not over-oxidise to the acid. Hence P has an aldehyde group, written as \(C_6H_5CHO\).
Step 3: Check option 1 (Rosenmund route).
Hydrogenating benzoyl chloride \(C_6H_5COCl\) with \(H_2\) over \(Pd\) poisoned by \(BaSO_4\) is the Rosenmund reduction. The poison stops the reaction at the aldehyde: \[ C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl \] This gives benzaldehyde, exactly our P, so option 1 is correct.
Step 4: Check option 2 (NaHCO3 test).
Brisk effervescence with \(NaHCO_3\) needs a carboxylic acid \(-COOH\) to release \(CO_2\). Benzaldehyde has no acidic \(-COOH\) proton, so no gas comes out. Option 2 is wrong.
Step 5: Check option 3 (Friedel-Crafts).
Benzene with \(CH_3COCl\) and anhydrous \(AlCl_3\) gives acetophenone \(C_6H_5COCH_3\), a ketone, not benzaldehyde. Option 3 is wrong.
Step 6: Check option 4 and conclude.
A white precipitate with bromine water needs a strongly activating ring like phenol or aniline. The \(-CHO\) group is deactivating, so benzaldehyde gives no white precipitate. Option 4 is wrong. Only the Rosenmund statement survives.
\[ \boxed{\text{Compound P is obtained by hydrogenation of benzoyl chloride with } Pd/BaSO_4} \]