In one of my organic chemistry lessons, it is possible to produce 2-methylbut-2-ene and 2-methylbut-1-ene from 2-bromo-2-methylbutane using a base, using the mechanism E2. The percentage of the two products 2-methylbut-2-ene and 2-methylbut-1-ene depends on the base. Using NaOEt, we obtain 69% of 2-methylbut-2-ene and 31% of 2-methylbut-1-ene. Using t-BuOK, we obtain 28% of 2-methylbut-2-ene and 73% of 2-methylbut-1-ene. The reason mentioned for these percentages is that NaOEt is more nucleophilic and t-BuOK is more basic than NaOEt. But it doesn't really answer my question. Why using these different bases favor one product or the other one? (See picture attached, and yes it's in french) enter image description here

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    $\begingroup$ Consider the steric hindrance affecting the two bases. tBuO- is considerably more hindered than EtO-. To produce 2-methylbut-2-ene requires the removal of a more hindered proton than 2-methylbut-1-ene. $\endgroup$
    – Waylander
    Mar 31 at 21:17
  • $\begingroup$ This issue has been addressed previously on ChemSE. $\endgroup$
    – user55119
    Apr 1 at 16:25

1 Answer 1


The reason for this is because t-BuO is more hindered than EtO-. So as you might have studied during deprotonation, if a bulky base is removing the Hydrogen cation, it will not be stable if the proton is removed from the 2nd carbon, therefore it is removed from the last carbon.

enter image description here

You can also check this article


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