Find the product of the following reaction

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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) ?

  • 4
    $\begingroup$ It is similar to dow's process, which takes place via Sn Benzyne mechanism $\endgroup$ – Jayadithya 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:

enter image description here

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


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