MCQ
$C{H_3}C{H_2}C \equiv N\xrightarrow{X}C{H_3}C{H_2}CHO$ The compound $X$ is
- ✓$SnCl_2/HCl/H_2O$, boil
- B$H_2/Pd -BaSO_4$
- C$LiAlH_4$/ethe
- D$NaBH_4/ ether / H_3O^+$
✓
Answer
Correct option: A.
$SnCl_2/HCl/H_2O$, boil
a
Completing the given reaction
Completing the given reaction
$C{{H}_{3}}-C{{H}_{2}}-C\equiv N$ $\xrightarrow{{SnC{l_2}\,/\,HCl}}$ $C{H_3}C{H_2}CH = NH$
$\xrightarrow[{(Ether)}]{{{H_2}O\,,\,boil}}$ $C{H_3}C{H_2}CHO\, + \,N{H_4}Cl$
It is stephen's reduction.
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