JEE Chemistry 2023 Question with Solution

Given: The substrate contains both an ester group $\left(\text{CO}_2\text{Et}\right)$ and a carboxylic acid group $\left(\text{CO}_2\text{H}\right)$. The reagents are $\text{LiBH}_4/\text{EtOH}$ followed by $\text{H}_3\text{O}^+$.

Find: The major product formed in the reaction.

Reaction mechanism: Lithium borohydride $\left(\text{LiBH}_4\right)$ is a selective reducing agent that reduces esters $\left(\text{CO}_2\text{Et}\right)$ to alcohols while leaving carboxylic acids $\left(\text{CO}_2\text{H}\right)$ unchanged.

Step-by-step process:

1. The ester group $\left(\text{CO}_2\text{Et}\right)$ is reduced by $\text{LiBH}_4$ in ethanol $\left(\text{EtOH}\right)$ to form a primary alcohol $\left(-\text{CH}_2\text{CH}_2\text{OH}\right)$.
2. The carboxylic acid group $\left(\text{CO}_2\text{H}\right)$ remains unchanged during the reaction.
3. Protonation with $\text{H}_3\text{O}^+$ ensures the stability of the final product.

The final product contains an unchanged carboxylic acid group and a newly formed primary alcohol group, corresponding to option $\left(4\right)$.

Therefore, the correct option is C.