From USNCO 2004, Q58:

The reaction between which pair of reactants occurs the fastest for $[\ce{OH-}] = \pu{0.010 M}$?

(A) $\ce{CH3CH2CH2CH2Cl + OH-}$

(B) $\ce{(CH3)3CCl + OH-}$

(C) $\ce{CH3CH2CH2CH2Br + OH-}$

(D) $\ce{(CH3)3CBr + OH-}$

The following is my thought process:

  • This reaction occurs through one of the SN2, SN1, E1 or E2 pathways.
  • All pathways prefer better leaving groups. $\ce{Br-}$ is a better leaving group than $\ce{Cl-}$, hence (A) and (B) can be eliminated.
  • The nucleophile is strongly basic, so will favor E1 and E2 reactions.

However, I am unable to choose between E1 and E2, and option (D) or option (C), respectively.

Also, if there is any error in my reasoning, could it be pointed out?


2 Answers 2


in this question, I think, we are not talking about priority of SN2 or E2, but according to the Clayden, (D) would go through E2 and (C) would choose SN2.

But we know that reactivity of tertiary halides are much more than primary ones. So, I think (D) is the right choice, either.

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(D), a tertiary halide, undergoes E2 much faster than (C), the primary halide.

  • $\begingroup$ How do I choose between E1 and E2 mechanisms? Does the low concentration of hydroxide rule out E2? $\endgroup$ Commented Jan 14, 2019 at 9:17
  • $\begingroup$ I don't think you can rule out a mechanism entirely, there's always a certain proportion of product that comes from the other mechanism. $\endgroup$ Commented Jan 14, 2019 at 9:22
  • $\begingroup$ As far as I understand, the reactivity of tertiary halides is greater than primary halides in both E1 and E2 reactions. $\endgroup$ Commented Jan 14, 2019 at 9:23

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