In the following sequence of reactions the maximum number of atoms present in molecule ‘ $C$ ’ in one plane is_______
$\mathrm{A}\stackrel{\mathrm{R}\mathrm{e}\mathrm{d}\mathrm{h}\mathrm{o}\mathrm{t}}{\underset{\mathrm{C}\mathrm{u}\mathrm{t}\mathrm{u}\mathrm{b}\mathrm{e}}{\to }}\mathrm{B}\stackrel{\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{l}\left(1.\mathrm{e}\mathrm{q}\right)}{\underset{\mathrm{A}\mathrm{n}\mathrm{h}\mathrm{y}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{s}\mathrm{A}\mathrm{l}\mathrm{C}{\mathrm{l}}_3}{\to }}\mathrm{C}$
(A is a lowest molecular weight alkyne)

Solution

Number of atoms in one plane = $13$

Hence, the correct answer is 13.