Vinylic halides do not show nucleophilic substitution at all, because of their stability due to resonance. But elimination to form alkynes is possible. No nucleophile at all can displace the halide ion in a nucleophilic substitution, but a strong base suffices for elimination.

Why is this?

  • 1
    $\begingroup$ I wonder if that is really the case. Source please? I could imagine nucleophilic substitution would more difficult due to the increased bond strength, but because of resonance structures it is impossible? Idk. $\endgroup$
    – Jori
    May 17, 2015 at 14:33
  • $\begingroup$ But a halogen attached to a carbonyl atom, also resonance stabilised, can be removed by nucleophilic attack. $\endgroup$
    – Charles
    May 18, 2015 at 10:09

1 Answer 1


In the $\ce{sp^2}$ transition state for an $\mathrm{S_{N}}2$ reaction

  • the attacking nucleophile, the central carbon and the leaving group prefer a linear arrangement
  • the 3 remaining groups around carbon are arranged with 120° bond angles

enter image description here

(image source)

Now, let's look at ethylene (as a model for a vinyl halide) and let's use the bent bond description of ethylene because it makes it easy to visualize what's happening (the bent bond model has been shown to be equivalent to the pi bond model in all aspects, see below for the same explanation phrased in terms of the pi bond model).

enter image description here

(image source)

Replace one of the ethylene hydrogens with halide and imagine what happens as we approach the transition state. The bond angle in the two-membered ring wants to be small, but as we approach the transition state it needs to open up wider to 120°! This would create a lot of strain. It is this large strain energy that causes the $\mathrm{S_{N}}2$ reaction at $\ce{sp^2}$ hybdridized vinylic carbons to be a very high energy pathway and therefore less likely to occur.

On the other hand, the elimination reaction at a vinylic carbon is quite facile. Here as the base removes a proton we build up negative charge on an $\ce{sp^2}$ hybridized carbon. An $\ce{sp^2}$ orbital has more s-character than an $\ce{sp^3}$ orbital; the more s-character in an orbital the lower the energy of an electron in that orbital. Therefore, negative charge is more stable in an $\ce{sp^2}$ orbital than a $\ce{sp^3}$ orbital. Consequently it is easier to remove a proton when the negative charge will reside on an $\ce{sp^2}$ hybridized carbon compared to an $\ce{sp^3}$ hybridized carbon (and that's why it is even easier [lower energy] to remove a proton from an $\ce{sp}$ hybridized carbon).

Explanation Based on Pi Bond Model

As expected for equivalent models, the bent bond explanation can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state. It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons are in higher energy unhybridized p orbitals and this is not offset by the energy lowering experienced by the electrons as some bonds rehybridize from $\ce{sp^2}$ to $\ce{sp}$) thereby increasing the energy required to reach the transition state.


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