Since the molecule takes $2$ moles of $B{r_2}$. Therefore it is alkyne. Also it gives white ppt with Tollen’s reagent therefore acidic $H$ is present. Hence it is 1-Butyne.
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$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH = C{H_2}\,\,\,\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}\xrightarrow[{(ii)\,NaB{H_4}}]{{(i)\,Hg{{(OAc)}_2};{H_2}O}}$
$(i)\,\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus - C{H_3}} \\
| \\
{\,\,\,\,\,\,OC{H_3}}
\end{array}$
$(ii)\,\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus - C{H_3}} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iii)\begin{array}{*{20}{c}}
{C{H_3} - \mathop C\limits^ \oplus {H_2} - NH} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iv)\,C{H_3} - \mathop C\limits^ \oplus {H_2}$
