- ✓$BeC{l_2}$
- B$NaCl$
- C$C{H_4}$
- D${N_2}$
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$\begin{array}{*{20}{l}}
{(u){H_2}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} CHC{H_3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (v){\mkern 1mu} C{H_2}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} CHCl{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (w){\mkern 1mu} C{H_3}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} H_2^ + {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (x){\mkern 1mu} H - C \equiv C - H} \\
{(y){\mkern 1mu} C{H_3}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} N{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (z){\mkern 1mu} {{(C{H_3})}_2}C\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{N} N{H_2}}
\end{array}$
The product formed is