# Why do SN1 and SN2 reaction not occur at sp2 centres?

I was told in my organic chemistry course that $\text{S}_\text{N}1$ and $\text{S}_\text{N}2$ reactions did not occur at $\text{sp}^2$ centres. When I asked why, I was not given a satisfactory explanation. For $\text{S}_\text{N}2$ it was suggested that the reaction could not proceed with inversion of configuration, as this would disrupt the orbital overlap causing the pi-bond. I couldn't find any explanation in Organic Chemistry Second Edition by Clayden et al, or even a mention of it.

Could anyone please explain why? I have attached some curly arrow diagrams that seem reasonable enough to me, although I realise that being able to draw a reaction mechanism does not mean it is valid.

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These reactions do occur at $\ce{sp^2}$ hybridized carbon atoms, they are often just energetically more costly, and therefor somewhat less common. Consider when a nucleophile reacts with a carbonyl compound, the nucleophile attacks the carbonyl carbon atom in an $\ce{S_{N}2}$ manner. The electrons in the C-O pi bond can be considered as the leaving group and a tetrahedral intermediate is formed with a negative charge on oxygen. It is harder to do this with a carbon-carbon double bond (energetically more costly) because you would wind up with a negative charge on carbon (instead of oxygen), which is energetically less desirable (because of the relative electronegativities of carbon and oxygen).
If you look at the Michael addition reaction, the 1,4-addition of a nucleophile to the carbon-carbon double bond in an $\ce{\alpha-\beta}$ unsaturated carbonyl system, this could be viewed as an $\ce{S_{N}2}$ attack on a carbon-carbon double bond, but again, it is favored (lower in energy) because you create an intermediate with a negative charge on oxygen.
$\ce{S_{N}1}$ reactions at $\ce{sp^2}$ carbon are well documented. Solvolysis of vinyl halides in very acidic media is an example. The resultant vinylic carbocations are actually stable enough to be observed using nmr spectroscopy. The picture below helps explain why this reaction is so much more difficult (energetically more costly) than the more common solvolysis of an alkyl halide. In the solvolysis of the alkyl halide we produce a traditional carbocation with an empty p orbital. In the solvolysis of the vinyl halide we produce a carbocation with the positive charge residing in an $\ce{sp^2}$ orbital. Placing positive charge in an $\ce{sp^2}$ orbital is a higher energy situation compared to placing it in a p orbital (electrons prefer to be in orbitals with higher s density, it stabilizes them because the more s character in an orbital the lower its energy; conversely, in the absence of electrons, an orbital prefers to have high p character and mix the remaining s character into other bonding orbitals that do contain electrons in order to lower their energy).