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The process I have seen in class to activate a carbonyl function is acetalisation.

One of the steps involved is the following :

http://i.imgur.com/WA0W7bC.png

So I was wondering, is that a $\ce{S_{N}1}$ or a $\ce{S_{N}2}$ ?

(PS : drawing molecules with paint is awful, anyone knows a good free molecules drawer ?)

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2 Answers 2

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It is not an $\ce{S_{N}1}$ or an $\ce{S_{N}2}$ reaction. It is commonly referred to as an "addition-elimination" reaction. You've drawn the key piece of evidence in your diagram - the tetrahedral intermediate. Not an $\ce{S_{N}1}$ reaction because there is no carbonium ion formed in this reaction. This is evidenced by many facts such as the lack of carbonium ion type rearrangements when substituents are present alpha to the carbonyl. Nor is it an $\ce{S_{N}2}$ reaction, the attacking group does not start to bond and the leaving group does not start to leave and break its bond simultaneously. Instead there is a real tetrahedral intermediate that exists in equilibrium with the reactant as well as the product. You "add" a nucleophile to the protonated carbonyl compound to form the tetrahedral intermediate, then you "eliminate" one the substituents in the tetrahedral intermediate and either go back to reactant or forward to product.

enter image description here

Here's another, more specific drawing for the case at hand.

enter image description here

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    $\begingroup$ See the picture I've added to my post. $\endgroup$
    – ron
    Jun 12, 2014 at 18:51
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    $\begingroup$ 1) You had the word "carbonyl" in your title, so I assumed you were starting from a carbonyl compound - so I drew one in my diagram, 2) I replaced R2 with X, my drawing is a little more general, If R2 is an alkyl group, then it can't leave and we would just have an equilibrium between the ketal (the tetrahedral intermediate) and the starting ketone, but it is still an example of an "addition-elimination" reaction, 3) I use ChemSketch freeware for my drawings $\endgroup$
    – ron
    Jun 12, 2014 at 19:05
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    $\begingroup$ Since you're starting with a ketone, the reaction is called ketalization. The final product, the ketal, is in equilibrium with the starting ketone. For most ketones the equilibrium lies far to the ketone side (not much ketal present). Still, for the reasons I presented above, it is an addition-elimination reaction that proceeds through a tetrahedral intermediate. $\endgroup$
    – ron
    Jun 12, 2014 at 19:47
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    $\begingroup$ Yes, you are correct, the term "ketal" is now out of favor. I've added a second drawing does it help? $\endgroup$
    – ron
    Jun 12, 2014 at 20:43
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    $\begingroup$ Good catch, error in my drawing; fixed it. Does it make sense now? $\endgroup$
    – ron
    Jun 12, 2014 at 20:55
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This is an SN1-like process, the intermediate is an alyoxonium ion (R1R2C=O+R), which, having a double-bond resonance form is a somewhat better intermediate than a free carbocation, as in the tradition SN1.

There is too much steric hinderance for this to be a SN2 reaction with a reasonable reaction rate. That's a definite no-go.

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  • $\begingroup$ The carbocation is (R1)(R2)C-OR with a lacunary central C, how is that equivalent to a (R1)(R2)C=OR structure ? $\endgroup$
    – Hippalectryon
    Jun 12, 2014 at 16:49
  • $\begingroup$ Resonances forms involving donating a lone pair from the oxygen to the carbocation forms a double bond $\endgroup$
    – Lighthart
    Jun 13, 2014 at 22:17
  • $\begingroup$ I got it now with @ron 's drawings, thank you :) $\endgroup$
    – Hippalectryon
    Jun 13, 2014 at 22:24

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