MCQ
When $C{H_3}C{H_2}CHC{l_2}$ is treated with $NaN{H_{2,}}$ the product formed is
- A$C{H_3} - CH = C{H_2}$
- B$C{H_3} - C \equiv CH$
- C$C{H_3}C{H_2}CH(N{H_2})(Cl)$
- ✓$C{H_3}C{H_2}C{(N{H_2})_2}$
✓
Answer
Correct option: D.
$C{H_3}C{H_2}C{(N{H_2})_2}$
d
It’s obvious.
It’s obvious.
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