I've read in a book that the main factor for determining SN2 reaction rate is steric hindrance. The lesser it is, the faster the reaction.

So consider this question:

$\ce{KI}$ in acetone undergoes SN2 reaction with each of $\ce{P,Q,R}$ and $\ce{S}$. The rates of reaction vary as:

enter image description here

I do not understand how $\ce{S}$ is the fastest. In my opinion it should be the slowest as it is very sterically hindered. What is going on here?

Source: Joint Entrance Exam (JEE) 2013 India


3 Answers 3


There are a number of factors that can influence the rate of an $\mathrm{S_{N}2}$ reaction. Solvent, leaving group stability, attacking group nucleophilicity, steric factors and electronic factors.

In the series of compounds you've presented, all of these parameters are held constant except for the steric and electronic factors. Considering only steric factors for a moment we would order the compounds as P>S~R>Q. But now let's consider electronic factors.

Considering the allyl system in compound R, the transition state looks something like this

enter image description here

image source

The p orbitals in the adjacent double bond overlap with the p orbital at the $\mathrm{S_{N}2}$ reaction center (remember that $\mathrm{S_{N}2}$ transition states are roughly $\ce{sp^2}$ hybridized) and stabilizes the transition state through resonance delocalization. Note that the transition state is electron rich, there is a full unit of negative charge in the transition state.

When a carbonyl is placed next to the $\mathrm{S_{N}2}$ reaction center, we replace the $\ce{C=C}$ double bond in the above drawing with a $\ce{C=O}$ double bond. Again there is overlap and resonance stabilization of the transition state. But in the carbonyl case the stabilization is even greater than with the $\ce{C=C}$ double bond because the carbonyl carbon is positively polarized and inductively stabilizes the electron rich (remember that negative charge) transition state even more.

enter image description here

image source

While we would expect the allyl case (R) to be faster than the ethyl case, in your series of compounds apparently it is not faster than the methyl case.

(Note: often the allyl system will react a bit faster than the methyl analogue, but the two rates are usually close)

However, the additional inductive stabilization from the carbonyl allows it to react faster than the methyl compound so our final reactivity order becomes S>P>R>Q.

  • 3
    $\begingroup$ You beat me by a few seconds :-) . I wanted to add that the iodine electrons are diffuse and hence the partial charge of the attacked carbon does not matter that much. It's only about the transition state stabilization. I wouldn't say that the allylic stabilization is in general less important than the reduction of steric hindrance when going from methyl to primary substituted carbon. Or am I wrong in thinking that? +1 $\endgroup$
    – Jori
    Mar 21, 2015 at 18:17
  • $\begingroup$ Hey , thanks for the answer. Okay I'm now confused. Rajesh's answer says S is the best because of inductive effect which actually gives that carbon a slight positive charge. You're saying that because of orbital overlap it gets a slight negative charge. What is correct? $\endgroup$ Mar 21, 2015 at 18:18
  • $\begingroup$ @Jori No you're right Jori, often the ally compound will react faster than the methyl analogue. $\endgroup$
    – ron
    Mar 21, 2015 at 18:18
  • $\begingroup$ Or am I misunderstanding and you're actually saying that transition state is stabilized because its extra negative charge is 'taken away' from it thus stabilizing it? $\endgroup$ Mar 21, 2015 at 18:23
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    $\begingroup$ Aah I see. By the way I just checked Jerry march's Advanced organic chemistry and it gives rates as following - Methyl 30 Isobutyl 0.03 Ethyl 1 Neopentyl 10^-5 Propyl 0.4 Allyl 40 Butyl 0.4 Benzyl 120 Isopropyl 0.025 So it means Allyl is faster than methyl. What do you think? Do you know of any other resource we can check? $\endgroup$ Mar 21, 2015 at 18:41

The reaction rate of SN2 depends primarily on the leaving group involved. Since in your question, $\ce{-Cl}$ is common in all 4 options we move on to the next important criteria.

The partial positive charge on carbon (or the electrophilicity of the carbon) is considered more important than the steric effect that you mentioned. Look at Wikipedia's first line.

The nucleophile attacks the more electrophilic carbon. This can be easily compared by the Inductive effect of the substituents.

This gives us the correct answer as option B.

  • $\begingroup$ Nice , I'm glad that you told me about that factor because it is not in any book. $\endgroup$ Mar 24, 2015 at 16:16

If you'll look at the transition state in SN2 mechanism, you'll find that there's a partial negative charge on carbon. In option S, The leaving group is attached to a 1 degree carbon and there will be a -M by the carbonyl group making the negative charge most stable. In P, we have a 1 degree carbon with minimum hindrance. In R, we have a 1 degree carbon and on which the negative charge can be stabilized by resonance. In Q, we have a 2 degree carbon means steric hindrance, further the +I effect of CH3 will make the negative charge on carbon even worse.

So correct answer is B.


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