6
$\begingroup$

For E2 reactions, why is a strong base like $\ce{NaOH}$ or $\ce{RONa}$ needed? Whereas for E1, even a weak base like $\ce{H2O}$ could be used.

Wikipedia states:

E2 typically uses a strong base, it needs a chemical strong enough to pull off a weakly acidic hydrogen.

However it does not explain why. The lone pair of the base directly attacks the hydrogen, regardless of E1 or E2.

$\endgroup$
11
$\begingroup$

E2 is a concerted mechanism. The alpha proton in an E2 reaction substrate is also only weakly acidic because the bond between the alpha proton and the alpha carbon is relatively strong; there is only some inductive withdrawal of electron density from the C-H bond by the halogen.

Therefore, we need a strong base to rip the proton away from the alpha carbon.

enter image description here

E1 however involves two steps. In E1 the leaving group leaves, and then the base attacks.

When the leaving group leaves, a carbon lacking an octet is formed. This carbon is highly electronegative and will withdraw electron density. This makes any alpha proton more acidic as hydrogen is less electronegative than plain old carbon - much less a carbon deficient of an octet in a covalent framework.

In other words, electron withdrawal by the carbocation weakens the bond holding on to the alpha proton, and a weaker base suffices for taking away the proton.

Note that I don't particularly like the below picture because it doesn't show anything abstracting the proton. Still the general idea is correct; electron density from the alpha C-H bond flows toward the hypovalent carbon. Heck, we can draw valid, hyperconjugative resonance structures for this baby.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ I'm fairly certain you understand what is going on here, but the drawing used in the E1 part of your answer is incorrect. The C-H bond containing the hydrogen to be abstracted must be aligned with the carbocation's p-orbital during the abstraction step. $\endgroup$ – ron Oct 2 '14 at 21:03
  • $\begingroup$ @ron you mean the anti co planar requirement? Isn't that for e2 only? $\endgroup$ – Dissenter Oct 2 '14 at 21:16
  • 2
    $\begingroup$ Yes and no. Yes, the anti co-planar arrangement is a requirement for the E2. No, in the E1 case the C-H bond must be co-planar with the carbocation p-orbital (syn or anti have no meaning here). Otherwise the p-orbital developing from the C-H bond will not be aligned with the carbocation p-orbital and a pi bond will not result. The lack of partial pi bond formation in the hydrogen abstraction TS would dramatically increase the activation energy for this step. $\endgroup$ – ron Oct 2 '14 at 21:24
  • $\begingroup$ Okay, will fix (after my orgo exam tonight). $\endgroup$ – Dissenter Oct 2 '14 at 21:37
1
$\begingroup$

Also, if you tried using a weak base with, for example, a primary substrate then you would see $\rm S_N2$ and $E_2$ (however this is not practical due to the slowness of the reaction).

You may recall rate of $$\rm E_2=k[Substrate][Base]$$ while $E_1$ does not have a rate dependence on the base.

$\endgroup$

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.