# bromobenzene and NaOH reaction at high temperature and pressure

Find the product of the following reaction

Note that the carbon marked as a star is a special labelled carbon which may be considered as a $$\ce{C-14}$$ isotope

The given answer is (c)

Why would the product in option (b) form ? I know that it proceeds via SN2 mechanism but is there some sort of benzyne reaction where we would get the product in option (b) ?

• It is similar to dow's process, which takes place via Sn Benzyne mechanism Jun 13 at 8:55

The point is, as mentioned in the comment by @Jayadithya, that given the reaction conditions, the explicit «marking» a carbon atom as $$\ce{^{13}C}$$ (instead of the typical $$\ce{^{12}C}$$ / $$\ce{^{13}C}$$ according to the general isotopic distribution for this element) does not introduce a significant site selectivity for the reaction of the intermediate benzyne:
Thus, one anticipates an equal molar ratio of products formed along pathway a), and b). It is not the kinetic H-D isotopic effect you may observe e.g., with the $$S_N2$$ reactions