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Q8.15(ii):
How will you bring about the following conversions in not more than two steps? (ii) Benzoic acid to Benzaldehyde

Solution :

Step 1: Conversion of Benzoic Acid to Benzoyl Chloride

To convert Benzoic acid ($C_6H_5COOH$) into an intermediate reactive species, we treat it with Thionyl chloride ($SOCl_2$). This reaction replaces the hydroxyl group of the carboxylic acid with a chlorine atom, forming an acid chloride.

Reaction:

$C_6H_5COOH + SOCl_2 \xrightarrow{\Delta} C_6H_5COCl + SO_2(g) + HCl(g)$

[Justification: Per the nucleophilic acyl substitution mechanism, Thionyl chloride is the preferred reagent as the by-products $SO_2$ and $HCl$ are gases, facilitating the forward equilibrium reaction.]

Step 2: Rosenmund Reduction to Benzaldehyde

The resulting Benzoyl chloride ($C_6H_5COCl$) is subjected to partial reduction using Hydrogen gas in the presence of a poisoned catalyst. This process is known as the Rosenmund reduction.

Reaction:

$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$

[Justification: Palladium ($Pd$) is supported on Barium sulphate ($BaSO_4$) to act as a "poisoned" catalyst. The $BaSO_4$ reduces the activity of the $Pd$ catalyst, preventing the further reduction of the generated aldehyde into a primary alcohol.]

Visual Representation of the Transformation

C₆H₅COOH SOCl₂ C₆H₅COCl H₂, Pd/BaSO₄ C₆H₅CHO

Summary of Reaction Conditions

Step Reagent Product
1 $SOCl_2$ Benzoyl Chloride
2 $H_2, Pd/BaSO_4$ Benzaldehyde

Final Solution: Benzoic acid is first converted to Benzoyl chloride using $SOCl_2$, followed by a Rosenmund reduction using $H_2$ in the presence of $Pd/BaSO_4$ to yield Benzaldehyde.


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