In the given reaction CH_3-CH_2-C(=O)-CH_2-CO | vedclass

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(B) The starting material contains both a ketone and an ester group.$LiAlH_4$ is a strong reducing agent that reduces both groups.To selectively reduc

Vedclass · Accepted Answer

$HO-CH_2-CH_2-OH + H^{+}$

Vedclass · Answer

(B) The starting material contains both a ketone and an ester group.$LiAlH_4$ is a strong reducing agent that reduces both groups.To selectively reduce the ester group to an alcohol while keeping the ketone group intact,the ketone must first be protected as a cyclic ketal using ethylene glycol $(HO-CH_2-CH_2-OH)$ in the presence of an acid catalyst $(H^{+})$.The subsequent acidic workup $((ii) H_2O/H^{+})$ after reduction removes the protecting group and restores the ketone functionality.

In the given reaction $CH_3-CH_2-C(=O)-CH_2-COOC_2H_5$ $\xrightarrow{[X]} (A)$ $\xrightarrow{(i) LiAlH_4, (ii) H_2O/H^{+}} CH_3-CH_2-C(=O)-CH_2-CH_2OH + C_2H_5OH$,$[X]$ will be:

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