In a nucleophilic substitution reaction , which one of the following undergoes complete inversion of configuration ?

(A) $\ce{C_6H_5CCH_3C_6H_5Br}$
(B) $\ce{C_6H_5CHCH_3Br}$
(C) $\ce{C_6H_5CHC_6H_5Br}$
(D) $\ce{C_6H_5CH_2Br}$

Clearly, the question was asking which of the following compounds underwent $\mathrm{S_N2}$ most easily, i.e the case in which the nucleophilic approach was least sterically hindered.

I eliminated (A) and (C) immediately. I then went with option (B). We had pushed the phenyl group far away from the carbon to which the bromine was attached. It would thus not sterically hinder the nucleophilic.

However, the correct answer was (D) . The explanation: The PhEt group is bulkier than the PhMe group, and hence causes more steric interference.

So , what is the fallacy in my argument? Does pushing bulky groups away have any effect on steric interference? If so, then how far away should we push them? If not, then why not? I guess my key doubt is, under what circumstances would approach of the nucleophile be sterically compromised?

  • $\begingroup$ I think it has to do with the fact that benzylic halogen undergo a better SN2 reaction due to the stabilisation of the SN2 transition state. Hence is more reactive than normal primary halides.Try referring here mendelset.com/category/keywords/sn2-mechanism.. $\endgroup$ – booma vijay Apr 15 at 13:06
  • $\begingroup$ Please don't add spaces before punctuation marks... also, do A and C have two phenyl groups on the central carbon? More common to write that as $\ce{(C6H5)2C(CH3)Br}$... $\endgroup$ – orthocresol Apr 15 at 13:07
  • $\begingroup$ @orthocresol Correct, they do. I just copied the question, as it was, from my book. $\endgroup$ – Aspirant Apr 15 at 13:18
  • $\begingroup$ @boomavijay Would it not prefer $S_N1$ ? $\endgroup$ – Aspirant Apr 15 at 13:35
  • 1
    $\begingroup$ @Zenix Ah, good. Thanks. I misread the methyl as another phenyl... $\endgroup$ – Zhe Apr 15 at 17:44

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