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Question 46

The major product formed in the following reaction is

CO$$_2$$H + CO$$_2$$Et compound treated with (i) LiBH$$_4$$/EtOH (ii) H$$_3$$O$$^+$$

Solution

The starting material contains two key carbonyl-based functional groups attached to the benzene ring via aliphatic chains:

A carboxylic acid group ($$-\text{CO}_2\text{H}$$)
An ester group ($$-\text{CO}_2\text{Et}$$)

Lithium borohydride ($$\text{LiBH}_4$$) is a selective reducing agent. It is more reactive than $$\text{NaBH}_4$$ but milder than $$\text{LiAlH}_4$$.

It selectively reduces esters to primary alcohols ($$-\text{CH}_2\text{OH}$$).

Reduction: $$\text{LiBH}_4$$ in $$\text{EtOH}$$ selectively attacks the ester carbonyl of the $$-\text{CH}_2\text{CO}_2\text{Et}$$ group, reducing it down to a primary alcohol. The $$-\text{CH}_2\text{CH}_2\text{CO}_2\text{H}$$ group remains completely untouched.

Acidic Workup: Treatment with $$\text{H}_3\text{O}^+$$ protonates the alkoxide intermediate to yield the stable alcohol and ensures the carboxylic acid group remains in its protonated form ($$-\text{CO}_2\text{H}$$).

Correct Option B

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