| \[{{C}_{6}}{{H}_{5}}CH=CHCHO\xrightarrow{x}{{C}_{6}}{{H}_{5}}CH=CHC{{H}_{2}}OH\] |
| In the above sequence X can be |
A) \[{{K}_{2}}C{{r}_{2}}{{O}_{5}}/{{H}^{+}}\]
B) \[NaB{{H}_{4}}\]
C) \[{{N}_{2}}/Ni\]
D) Both (b) and (c)
Correct Answer: B
Solution :
[b]| \[NaB{{H}_{4}}\] and \[LiAl{{H}_{4}}\] attacks only carbonyl group and reduce it into alcohol group. They do not attack on double bond. |
| \[\underset{Cinnamic\,\,aldehyde}{\mathop{{{C}_{6}}{{H}_{5}}CH=CHCHO}}\,\xrightarrow{NaB{{H}_{4}}}\underset{Cinnamic\,\,alcohol}{\mathop{{{C}_{6}}{{H}_{5}}CH=CH\cdot C{{H}_{2}}OH}}\,\] |
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