I'm wondering if the hydrogen stolen during an $E_1$ reaction has to be antiperiplanar/anticoplanar like the hydrogen in an $E_2$ reaction.

Intuitively I'd say no, because the carbocation is flat so there's less steric hindrance than an E2 reaction, but I want to check.

As an example, we could use 3-methyl-2-bromobutane. Once that carbocation forms and shifts, does it matter which hydrogen is stolen?


2 Answers 2


In an E2 reaction, there is a strong preference for the hydrogen to be coplanar with the leaving group and pointed in an anti-direction to the leaving group. In an E1 reaction, the hydrogen does not have to be anti-periplanar to the leaving group because in the E1 reaction the deprotonation event happens in a separate mechanistic step, after the leaving group has left. So there's no need to define the relative orientation of the hydrogen and leaving group in the E1 reaction.

However, for the E2 reaction, remember that the anti-periplanar arrangement is preferred because it places the C-H bonding orbital anti to the C-LG antibonding orbital. In the transition state, those are the orbitals that reorganize to create the C-C pi bond.

With respect to the E1 reaction, there is no C-LG antibonding orbital, but there is an empty p-orbital on the carbocation. The C-H bond should be parallel to the p-orbital to create a similar (but strictly speaking, not anti-periplanar) alignment of orbitals.

enter image description here


It actually does not matter whether or not it is antiperiplanar, because once the carbocation is formed, there is no sense of antiperiplanar. The two carbon atoms are in the same plane now. For a planar carbocation, all arrangements are equivalent(i.e. no sense of stereoisomerism). It does not matter which side the base attacks on. Also, since the bond angle in an $sp^2$ system is greater than that of an $sp^3$ system, the steric resistance to the base is less, which is the point you mentioned.

However, steric factors are not much significant in eliminations where protons are involved, because only a proton is required to be abstracted, and the incoming nucleophile(acting as a base ) is not required to attack or attach to a site.

  • $\begingroup$ Three or less carbon atoms are always in the same plane, therefore your last statement is quite confusing. $\endgroup$ Dec 9, 2014 at 8:51
  • $\begingroup$ Yes, you're right. I should not have written $\alpha-\beta$ $\endgroup$
    – Shubham
    Dec 9, 2014 at 9:01

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.