A) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\]
B) \[{{C}_{6}}{{H}_{5}}CH=CHC{{H}_{2}}OH\]
C) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{2}}CHO\]
D) \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}CHOHC{{H}_{3}}\]
Correct Answer: B
Solution :
\[NaB{{H}_{4}}\]being a mild reducing agent does not affect\[C=C\]but cause reduction of\[CHO\]group. \[{{C}_{6}}{{H}_{5}}CH=CHCHO\xrightarrow{NAB{{H}_{4}}}\]\[{{C}_{6}}{{H}_{5}}CH=CHC{{H}_{2}}OH\]
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