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What is the exact order of reactivity of alkyl halides in an E2 reaction?

Organic Chemistry by Paula Yurikanis Bruice says that the order is $3^\circ>2^\circ>1^\circ$. And I think this order is correct because the transition state of the tertiary alkyl halide will be better stabilized by more hyperconjugating structures as compared to the secondary and primary halides.

But my teacher made me write that the order is $3^\circ<2^\circ<1^\circ$ and I even saw the this order in a book.

So I'm kind of confused about the correct order in E2 mechanism. Any help will be appreciated.

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  • $\begingroup$ I think you are right about the order. $\endgroup$
    – Weijun Zhou
    Jan 28 '18 at 17:53
  • $\begingroup$ @WeijunZhou $3^\circ>2^\circ>1^\circ$ is that it?? sure?? $\endgroup$
    – Carrick
    Jan 28 '18 at 18:20
  • $\begingroup$ Sure, as you said it is related to the stability of the transition state. $\endgroup$
    – Weijun Zhou
    Jan 28 '18 at 18:26
  • $\begingroup$ @WeijunZhou know what buddy, I saw a dream last night where I saw this order for E2 mechanism: 3∘>2∘>1∘. Now I have a firm belief that this order is correct. Just like Kekule's structure of benzene which he predicted based on the dream he saw where a snake was whirling around a campfire lol. $\endgroup$
    – Carrick
    Jan 30 '18 at 15:20
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I believe you might be mixing up E1 and E2 elimination reactions. In the first case the reaction first proceeds by formation of a carbocation from the alkyl halide and then a proton extractio. In this case, the order of reactivity is indeed $3^∘>2^∘>1^∘$ as such is the order of carbocation stability - in order the E1 reaction to proceed the carbocation must be formed.

But the situation is completely different in an E2 elimination. In this case the mechanism is the abstraction of the proton that is oriented anti to the halide with the simultaneous formation of the double bond and halide leaving the molecule as anion. That means that in this case the stability of carbocation plays no role and the significant factor are the sterical considerations. If the carbon atoms around the halide make it hard for the base to get to the anti proton then the reaction is less likely to proceed via E2 mechanism.

But bear in mind that usually both E1 and E2 elimination can take place and because of that tertiary halides would usually undergo elimination more readily than primary halides - just not via E2 mechanism.

More details on the E2 mechanism, with pictures: Master Organic Chemistry

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  • $\begingroup$ Nobody is talking about carbocation. We are talking about transition state. $\endgroup$
    – Weijun Zhou
    Jan 28 '18 at 19:13
  • $\begingroup$ @Jan Rzymkowski I am aware of E1, E2 & E1cb and i'm really not mixing them up. According to you, the order should be $3^\circ<2^\circ<1^\circ$ right? but what about the stable transition state that can be attained when a tertiary halide undergoes elimination? $\endgroup$
    – Carrick
    Jan 28 '18 at 19:25
  • $\begingroup$ If you have a stable transition state with a carbocation then you have an E1 reaction, not E2 reaction. As I said, tertiary halide is generally more reactive than primary, because of that possible carbocation - but it's not more reactive in E2 mechanism. $\endgroup$
    – Jan Rzymkowski
    Jan 28 '18 at 20:07
  • $\begingroup$ Here you have a transition state for E2 mechanism: iverson.cm.utexas.edu/courses/310M/ReactMoviesFl05%20/… If your transition state is a carbocation, then the reaction proceeds via E1 mechanism, not E2 mechanism. $\endgroup$
    – Jan Rzymkowski
    Jan 28 '18 at 20:09
  • $\begingroup$ Not a transition state with a carbocation. In fact a carbocation cannot be called a transition state. It is a reaction intermediate, and please stop talking about E1 here. We are not confused about what is E1 and what is E2. For your reference here is one pdf: home.iitk.ac.in/~madhavr/CHM102/Lec13.pdf. Check page 6. Here is another: ocw.uci.edu/upload/files/51b_chapter8_f2014.pdf. Maybe the problem is that we have different understanding of the word "reactivity". $\endgroup$
    – Weijun Zhou
    Jan 28 '18 at 20:18

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