Question

Choose the correct answer among the alternatives given, One mole of Ethyl acetate on treatment with an excess of ${\mathrm{LiAlH}}_4$ in dry Ether and subsequent acidification produces:

• A. $1\mathrm{mol}$ of Acetic acid + $1\mathrm{mol}$ of Ethyl alcohol
• B. $1\mathrm{mol}$ of Ethyl alcohol + $1\mathrm{mol}$ of Methyl alcohol
• C. $2\mathrm{mol}$ of Ethyl alcohol
• D. $1\mathrm{mol}$ of 2-Butanol

Answer 1

The correct option is C

$2\mathrm{mol}$ of Ethyl alcohol

Explanation for Correct answer:

Option (C) $2\mathrm{mol}$ of Ethyl alcohol

• One mole of Ethyl acetate on treatment with an excess of ${\mathrm{LiAlH}}_4$ in dry Ether undergo reduction to produce $2\mathrm{mol}$ of Ethyl alcohol.

$\underset{\mathrm{Ethyl}\mathrm{acetate}}{{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5}\underset{\mathrm{dryether}}{\overset{{\mathrm{LiAlH}}_4/{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }}\underset{\mathrm{Ethyl}\mathrm{alcohol}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}}+\underset{\mathrm{Ethyl}\mathrm{alcohol}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}}$

Therefore, Ethyl Acetate on treatment with ${\mathrm{LiAlH}}_4$ gives 2 moles of Ethyl alcohol so, option (C) $2\mathrm{mol}$ of Ethyl alcohol.