Chlorobenzene in which the carbon bearing chlorine (C-1) is enriched in ${}_{}^{}{}_{}^{14}\mathrm{C}$ is treated with $\mathrm{KNH}{}_2$ in liquid ammonia. Neglecting isotope effects, what fraction of the aniline formed carries the $\mathrm{NH}{}_2$ group on the labelled carbon?

Loss of $\mathrm{HCl}$ gives a benzyne whose triple bond joins the labelled C-1 and the neighbouring unlabelled C-2. The amide ion adds to either carbon with equal probability because the two ends are equivalent apart from the isotope. Half of the aniline therefore has $\mathrm{NH}{}_2$ on the labelled carbon and half on the adjacent carbon.