A) \[{{H}_{2}}/Ni\]
B) \[NaB{{H}_{4}}\]
C) \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}/{{H}^{+}}\]
D) Both A and B
Correct Answer: B
Solution :
\[NaB{{H}_{4}}\] and \[LiAl{{H}_{4}}\] attacks only carbonyl group and reduce it into alcohol group. They do not attack on double bond. \[\underset{\text{cinnamic aldehyde}}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CHCHO}}\,\xrightarrow{NaB{{H}_{4}}}\]\[\underset{\text{cinnamic alcohol}}{\mathop{{{C}_{6}}{{H}_{5}}-CH=CH.C{{H}_{2}}OH}}\,\]
You need to login to perform this action.
You will be redirected in
3 sec
