Question

The reaction of benzene with CO and HCl in the presence of anhydrous AlCl\(_3\) gives a compound X. X can also be obtained from which of the following reaction?

• A. Benzyl alcohol with CrO\(_3\)-H\(_2\)SO\(_4\)
• B. Toluene with KMnO\(_4\)|OH\(^-\)
• C. Benzoyl chloride with H\(_2\)-Pd/BaSO\(_4\)
• D. Benzonitrile with (i) LiAlH\(_4\) (ii) H\(_3\)O\(^+\)

Hint:

Remember these key named reactions for preparing benzaldehyde: - \textbf{Gattermann-Koch:} Benzene + CO/HCl + AlCl\(_3\)/CuCl. - \textbf{Gattermann:} Benzene + HCN/HCl + AlCl\(_3\). - \textbf{Rosenmund Reduction:} Benzoyl chloride + H\(_2\)/Pd-BaSO\(_4\). - \textbf{Stephen Reaction:} Benzonitrile + SnCl\(_2\)/HCl, then hydrolysis. - \textbf{Etard Reaction:} Toluene + CrO\(_2\)Cl\(_2\).

Answer 1

The Correct Option is C

Step 1: Identify compound X.
The reaction of benzene with a mixture of carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of a Lewis acid catalyst like anhydrous AlCl\(_3\) is known as the Gattermann-Koch reaction.
This reaction introduces a formyl group (-CHO) onto the benzene ring.
The product X is benzaldehyde (C\(_6\)H\(_5\)CHO).
Step 2: Analyze the reactions in the options to see which one also produces benzaldehyde.
(A) This option appears to have a typo. Oxidation of benzyl alcohol (C\(_6\)H\(_5\)CH\(_2\)OH) with a mild oxidizing agent like PCC would give benzaldehyde. Stronger agents like Jones reagent (CrO\(_3\)/H\(_2\)SO\(_4\)) would oxidize it further to benzoic acid. The image shows Benzyl alcohol with CrO3 - H2SO4, which would give benzoic acid. So this is incorrect.
(B) Toluene (C\(_6\)H\(_5\)CH\(_3\)) with alkaline KMnO\(_4\), followed by acidification, is a strong oxidation that converts the methyl group to a carboxylic acid group. The product is benzoic acid (C\(_6\)H\(_5\)COOH). So this is incorrect.
(C) Benzoyl chloride (C\(_6\)H\(_5\)COCl) is an acid chloride. Its reaction with H\(_2\) in the presence of a poisoned catalyst (Pd on BaSO\(_4\), sometimes with sulfur or quinoline) is the Rosenmund reduction.
This reaction selectively reduces an acid chloride to an aldehyde.
\( \text{C}_6\text{H}_5\text{COCl} + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4} \text{C}_6\text{H}_5\text{CHO} + \text{HCl} \).
This reaction produces benzaldehyde (X). So this is the correct option.
(D) Benzonitrile (C\(_6\)H\(_5\)CN) on reduction with a strong reducing agent like LiAlH\(_4\), followed by hydrolysis, gives a primary amine. The product would be benzylamine (C\(_6\)H\(_5\)CH\(_2\)NH\(_2\)). So this is incorrect. (If a milder reduction like Stephen reaction were used, it would give benzaldehyde).