We have been learning the E1cB mechanism in organic chemistry, but we have not been delving into the motivation as to why this mechanism occurs instead of others. Consider the following mechanism:

E1cB Mechanism

I am confused as to why we push the electrons to the left, creating an anion on the oxygen, instead of simply performing an E2 reaction on the right hand side. What is the functional reason for drawing this intermediate? Why does it arise?

  • 1
    $\begingroup$ In terms of the difference between the 3 main types of elimination (E) reactions, this earlier answer may be helpful. $\endgroup$
    – ron
    Apr 9 '18 at 16:48

-OH is a poor leaving group and so E1 and E2 are not favourable. In the case of E1, the formation of a primary or secondary carbocation is not stable either.

For E1cb, the intermediate enolate is more stable and since the removal of the -OH is not in the rate determining step, it is more favourable than E1.


In the two mechanisms, two bonds in the substrate are broken. In the E1 of t-butyl bromide, the C-Br bond breaks first to leave a tertiary carbocation. Secondly, a C-H bond breaks to form isobutylene. In the E1cB one bond is also broken before the other one. In your example, a C-H bond breaks to form a carbanion, then the C-OH bond is broken afterward. In the E1 example, the C-Br bond is more labile. In the E1cB, it is the C-H bond.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.