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I have a question regarding the reactivity of alkyl halide for E2. I know that the general rule is tertiary alkyl halide reacts the fastest with E2, and primary alkyl halide reacts the slowest.

I was told that the reasoning behind this trend is due to stability of product based on the number of substituents the alkene have. However, I wanted to know if this trend still holds if both a primary alkyl halide and a tertiary alkyl halide results in the same number of substituents.

For example, consider the two pairs below: enter image description here

Here, the first one is primary alkyl halide, while the second one is tertiary alkyl halide. Therefore, I could decide automatically that the latter will have a higher reactivity toward E2. However, if I draw the E2 product for both alkyl halide, I get the same alkene.

enter image description here The book this question is from states that the tertiary alkyl halide will have a higher E2 reactivity, but provides no further justification.

Am I missing something important here? I cannot seem to understand why the 2nd compound has a higher E2 reactivity? I am guessing that maybe it could be due to the fact that it is more difficult for the OH to get in proximity of the beta hydrogen of the first compound. However, I wanted to see if there is a more clear justification for why the tertiary alkyl halide hear has a higher E2 reactivity.

Could it be that a primary alkyl halide is more susceptible to an SN2 reaction? However, does that mean it loses reactivity toward an E2?

It would be great if I could get an explanation for this! Thank you :)

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  • $\begingroup$ What you are missing is that the tBu cation is more stable and more likely to form $\endgroup$
    – Waylander
    Commented Nov 26, 2023 at 7:53
  • $\begingroup$ Wouldn't that be more relevant for E1 reaction? E2 is concerted and therefore I thought the formation of cation wasn't a contributing factor. $\endgroup$
    – Jaehyun Ahn
    Commented Nov 26, 2023 at 14:19
  • $\begingroup$ Compounds don't get to chose the mechanism. I think it would be closer to E1 for t-butyl no matter how you tried to manipulate it. $\endgroup$
    – Mithoron
    Commented Nov 26, 2023 at 22:36
  • $\begingroup$ Given that the base is a strong base, shouldn't it be E2? $\endgroup$
    – Jaehyun Ahn
    Commented Nov 27, 2023 at 0:20

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