Methyl 4-formylbenzoate, $p - OHC - C X_{6} H X_{4} - COOCH X_{3}$ , is treated in sequence with (i) one equivalent of ethylene glycol, dry $HCl$ (aldehyde selectively protected); (ii) $LiAlH X_{4}$ , dry ether; (iii) dilute $H X_{3} O X^{+}$ . The final product is

Answer: C

The aldehyde forms a cyclic acetal, which is inert to hydride.

LiAlH X_{4}

reduces the ester to

- CH X_{2} OH

. Aqueous acid then removes the acetal and regenerates the aldehyde, giving

p - OHC - C X_{6} H X_{4} - CH X_{2} OH

. Without protection both groups would be reduced to the diol.