Question
One mole of ethyl acetate on treatment with an excess of $$LiAl{H_4}$$ in dry ether and subsequent acidification produces
A.
$$1\,mol$$ acetic acid $$ +\, \,1\,mol$$ ethyl alcohol
B.
$$1\,mol$$ ethyl alcohol $$ + \,\,1\,mol$$ methyl alcohol
C.
$$2\,moles$$ of ethyl alcohol
D.
$$1\,mol$$ of $$2$$-butanol
Answer :
$$2\,moles$$ of ethyl alcohol
Solution :
\[\underset{\text{Ethyl}\,\,\text{acetate}}{\mathop{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}}}\,\xrightarrow[\text{dry}\,\text{ether}]{\frac{LiAl{{H}_{4}}}{{{H}_{3}}{{O}^{+}}}}\] \[C{{H}_{3}}C{{H}_{2}}OH+{{C}_{2}}{{H}_{5}}OH\]