A) \[SnC{{l}_{2}}/HCl/{{H}_{2}}O,boil\]
B) \[{{H}_{2}}/Pd-BaS{{O}_{4}}\]
C) \[LiAI{{H}_{4}}/ether\]
D) \[NaB{{H}_{4}}/ether/{{H}_{3}}{{O}^{+}}\]
Correct Answer: A
Solution :
[a] It is Stephen's reaction. \[C{{H}_{3}}C{{H}_{2}}C\equiv N\xrightarrow{SnC{{l}_{2}}/HCl}C{{H}_{3}}C{{H}_{2}}CH\equiv NH.HCl\] \[\xrightarrow{{{H}_{2}}O}C{{H}_{3}}C{{H}_{2}}CHO+N{{H}_{4}}Cl\]
You need to login to perform this action.
You will be redirected in
3 sec
