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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.

Curly arrow diagrams

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Sometimes, especially in introductory courses the instructor will try to keep things "focused" in order to promote learning. Still, it's unfortunate that the instructor couldn't respond in a more positive and stimulating way to your question.

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).

enter image description here

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.

enter image description here

$\ce{S_{N}1}$ 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).

enter image description here

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What are the curved line segments between the central carbon and the oxygen atom in your first diagram? –  Jay_One Jul 20 at 21:53
    
Good question, I should've explained it. The curved lines above and below the C-O sigma bond are meant to represent the pi electrons. The arrow below the bottom curved line represents the electron flow of these pi electrons away from carbon and onto oxygen. –  ron Jul 20 at 22:00
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Ron's explanation is very good. One more reason why these reactions are not explored in introductory courses is that they're generally not applied in synthesis of more complex compounds, which is one of the main goals of organic chemistry. –  jerepierre Jul 21 at 15:17

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