MCQ
$\,$
- A
- ✓
- C
- D
✓
Answer
Correct option: B.
b
$Ph - CH = CH - C = N\xrightarrow{{C{H_3}MgBr/{H_3}{O^ + }}}$ $\begin{array}{*{20}{c}}
{Ph - CH - CH - C - NH} \\
{|\,} \\
{\,\,\,\,\,\,C{H_3}\,}
\end{array}$ $\xrightarrow{{NaB{H_4},\Delta }}\begin{array}{*{20}{c}}
{Ph - CH - C{H_2} - C{H_2} - N{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$Ph - CH = CH - C = N\xrightarrow{{C{H_3}MgBr/{H_3}{O^ + }}}$ $\begin{array}{*{20}{c}}
{Ph - CH - CH - C - NH} \\
{|\,} \\
{\,\,\,\,\,\,C{H_3}\,}
\end{array}$ $\xrightarrow{{NaB{H_4},\Delta }}\begin{array}{*{20}{c}}
{Ph - CH - C{H_2} - C{H_2} - N{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
In the above reaction, the reactant undergoes $1,4$ Michael addition first then the formed intermediate undergoes reduction in the presence of $NaBH _{4}$ to form the desired product.
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