Count the number of hydride ions ( $H X^{-}$ ) that must be delivered by $LiAlH X_{4}$ to convert one molecule of 4-cyanobenzoic acid, $NC - C X_{6} H X_{4} - COOH$ , completely to the aluminium alkoxide and amide stage of 4-(aminomethyl)phenylmethanol. Convention: the hydride that only removes the acidic O-H proton as $H X_{2}$ is counted; a carboxylate needs two further hydrides to reach the alkoxide; a nitrile needs two hydrides to reach $- CH X_{2} N X^{2 -}$ . The number of hydride equivalents is

Answer: A

Carboxylic acid: 1 (O-H deprotonation,

H X_{2}

evolved) + 2 (carboxylate to alkoxide) = 3. Nitrile: 2. Total = 3 + 2 = 5 hydrides.