A) \[{{C}_{6}}{{H}_{5}}COOH+C{{H}_{3}}OH\]
B) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH+C{{H}_{3}}OH\]
C) \[{{C}_{6}}{{H}_{5}}CHO+C{{H}_{3}}COOH\]
D) All of the above products
Correct Answer: B
Solution :
The reduction of\[{{C}_{6}}{{H}_{5}}COOC{{H}_{3}}\]with\[LiAl{{H}_{4}}\] produces two different alcohols. \[{{C}_{6}}{{H}_{5}}COOC{{H}_{3}}\xrightarrow[{}]{LiAl{{H}_{4}}}\underset{benzyl\text{ }alcohol}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH}}\,\]\[+\underset{methyl\text{ }alcohol}{\mathop{C{{H}_{3}}OH}}\,\]
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