Most important points to remember when thinking of $\mathrm{S_N2}$ reactions are that they are nucleophilic reactions, a pentavalent transition state is formed, depends on steric hindrance, strength of the bonds, nucleophilic and substrate used i.e. what kind of leaving group is attached to it, etc. I know and understand these.

I also know that in the following conditions the rate of $\mathrm{S_N2}$ increases, but I don't seem to understand the reason behind it.

1) The presence of unsaturation on beta-$\ce{C}$ in primary halide increases $\mathrm{S_N2}$ reaction rate.

I really don't seem to understand this one, I know for the alpha $\ce{C}$ it would be difficult because the halide and the carbon will acquire partial double bond character which would make it very difficult to break.

2) The presence of carbonyl group at alpha carbon is most suitable for $\mathrm{S_N2}$ reaction.

My reason: it might be because of tautomerism, i.e., due to its conversion to enol form it will increase the reactivity.

3) Presence of hetero $\ce{O,N,S}$ atom at alpha carbon in primary halides increase $\mathrm{S_N2}$ reaction.

This too, I fail the understand.

  • 1
    $\begingroup$ 1) Please don’t use MathJax in titles due to searching issues. 2) Please use \ce{...} always and exclusively for chemical notations i.e. atoms and compounds. Note that $\mathrm{S_N2}$ reactions are better typeset using $\mathrm{S_N2}$ because the 2 should not be an index. 3) Please note that all punctuation marks except for opening brackets, opening quotation marks or mid-word apostrophes are followed by a space. (Groups of quotation marks — like at the end of this phrase — have a space at the end of the group.) That makes your post easier to read. $\endgroup$ – Jan Oct 27 '15 at 12:29

I know this question was from awhile ago, but I'll answer anyway.

a) The key to understanding this is the concept of the transition state. Below is an image taken from Khan Academy

Transition state

For an $\mathrm{S_N2}$ reaction to take place, there must be enough activation energy to reach the transition state. In this type of reaction, the transition state comprises an alpha carbon in a pseudo-$\mathrm{sp^2}$ state. The nucleophile and the leaving group are directly above and below the alpha carbon in this state, with atoms bound to the carbon in a horizontal plane.

The reason this reaction proceeds more quickly with an unsaturated beta-carbon is the pi-electrons from the unsaturated carbon stabilize the transition state. This lowers the activation energy of the reaction, making it more likely to happen with each molecular collision.

c) I'm assuming you're comparing $\ce{N, O}$ and $\ce{S to C}$. $\ce{N, O,}$ and $\ce{S}$ are more electronegative than carbon, which will impart a more positive-like characteristic on the alpha carbon, making it more susceptible to attack from the nucleophile.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.