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 Clayden's Organic Chemistry (2nd ed.), 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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2 Answers 2


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 therefore 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).

  • $\begingroup$ What are the curved line segments between the central carbon and the oxygen atom in your first diagram? $\endgroup$ Commented Jul 20, 2014 at 21:53
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    $\begingroup$ 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. $\endgroup$
    – ron
    Commented Jul 20, 2014 at 22:00
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    $\begingroup$ 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. $\endgroup$
    – jerepierre
    Commented Jul 21, 2014 at 15:17
  • $\begingroup$ Wouldn’t an $\mathrm{S_N2'}$ mechanism be even better to show $\mathrm{S_N2}$ at $\mathrm{sp^2}$ carbons? $\endgroup$
    – Jan
    Commented Nov 20, 2015 at 13:56
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    $\begingroup$ The other answer seems to be challenging your stand. Anything to say to what the user Schrodinger mentioned? $\endgroup$ Commented Dec 20, 2017 at 3:03

Not a bad question, but a poor answer. The person who mentioned $\mathrm{S_N2}$' is on a much better track.

Nucleophilic attack on a carbonyl or 1,4-Michael acceptor is not an $\mathrm{S_N2}$ reaction. It is also very misleading. It's like saying a chicken and a human are the same thing because they both have two legs.

$\mathrm{S_N2}$ reactions involve nucleophilic attack on the sigma-antibonding orbital. In those two examples, (carbonyl & 1,4-Michael acceptor), the nucleophile is attacking the $\pi$-antibonding orbital. Moreover, there is no inversion of configuration, and it is not a concerted bond-breaking/bond-making reaction. So no, it is in no way an $\mathrm{S_N2}$ reaction.

Back to the original question: in the $\mathrm{S_N2}$ case, the lowest energy unoccupied molecular orbital is the $\ce{C-C}$ $\pi$-antibonding orbital. Thus, the nucleophile will overlap with that orbital. The $\ce{C-X}$ $\sigma$-antibonding orbital, which is what the nucleophile would attack in the mechanism you're drawing, is higher in energy, and the resulting transition state would be higher in energy compared to the reaction in which the $\ce{C-C}$ $\pi$-antibonding orbital is attacked. Thus, the reaction does not proceed in that manner.


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