Question
The organic product formed in the reaction
\[{{C}_{6}}{{H}_{5}}COOH\xrightarrow[II\,{{H}_{3}}{{O}^{+}}]{I\,LiAl{{H}_{4}}}\]
A.
\[{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH\]
B.
\[{{C}_{6}}{{H}_{5}}COOH\,\And C{{H}_{4}}\]
C.
\[{{C}_{6}}{{H}_{5}}C{{H}_{3}}\,\And \,C{{H}_{3}}OH\]
D.
\[{{C}_{6}}{{H}_{5}}C{{H}_{3}}\,\And \,C{{H}_{4}}\]
Answer :
\[{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH\]
Solution :
TIPS/FORMULAE :
$$LiAl{H_4}$$ is a reducing agent, it reduces $$ - COOH$$ group to $$ - C{H_2}OH$$ group.
\[{{C}_{6}}{{H}_{5}}COOH\xrightarrow{LiAl{{H}_{4}}}{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH\]