KI in acetone undergoes $\mathrm{S_N2}$ reaction with (1) and (2). Compare the rates of the reactions.

1: methyl chloride; 2: phenacyl chloride

Like many kinds of these questions there are many factors involved. One is the steric effect, and, since steric effects plays a central role in inversion of reactant and there is a very bulky group attached to (2), I think (1) should be more reactive. Also, (2) will serve as a better $\mathrm{S_N1}$ substrate as the carbocation generated will be resonance stabilised.

And so using these two reasons I thought (1) is more reactive than (2), but, according to the answer, (2) is almost 500 times more reactive than (1), and I am not able to see the reason for this mammoth difference.

  • 2
    $\begingroup$ That carbocation is not resonance stabilised. $\endgroup$ Jan 3, 2017 at 18:29
  • $\begingroup$ Yeah, the carbocation is not resonance stabilised. Try drawing resonating structures (if you can). If you face it anyways, you will end up with a positive charge on Oxygen. Moreover you already know that Halogen exchange reaction goes with SN2, so the carbocation formation is pointless (no SN1). $\endgroup$ Jan 3, 2017 at 19:20
  • $\begingroup$ If you are given this question in an exam it's really easy to identify which is more reactive by hybridization. As in (1) hydrogen is already stable that refers to its not much reactive. we already know something which is stable is less reactive and vice versa then looking at the second compound is not stable hence would be more reactive than one. $\endgroup$
    – user84460
    Oct 6, 2019 at 23:07

1 Answer 1


First, let me point out that a rate difference of 500 is really not that large. There are solvolysis reactions with relative rate differences on the order of $\mathrm{10^{10}}$ or greater (1).

The transition state for an $\mathrm{S_{N}2}$ reaction involves hypercoordinate (or hypervalent) bonding. The transition state is, more or less, a trigonal bipyramid structure and the bonding at the central carbon atom ($\ce{Nu-C-X}$) involves a 3 center-4 electron bond (see the above link and links therein for more detail on this concept). With 5 electron pairs being shared by this central carbon, anything that removes electron density from the central carbon atom will prove beneficial and accelerate the reaction.

SN2 Reaction Pathway

$\mathrm{S_{N}2}$ Reaction Pathway

In your molecule #2, a benzoyl group is attached to the central carbon involved in the $\mathrm{S_{N}2}$ process. You are correct that such a group is larger than a hydrogen and should decrease the reaction rate due to steric effects. However the size of the benzoyl group is not exceptionally large and the steric effect will likely be small. More importantly, the benzoyl group will also exert an electronic effect. The central carbon in an $\mathrm{S_{N}2}$ reaction is $\mathrm{sp^2}$ hybridized [e.g. leading to a trigonal bipyramid structure as mentioned above] and has a $\mathrm{p}$ orbital. This $\mathrm{p}$ orbital can interact with the adjacent carbonyl group through resonance. This resonance effect will tend to remove (delocalize) electron density from the central carbon atom. Also, the carbonyl group will inductively remove electron density from the central carbon atom. These electronic effects will remove electron density from the central carbon atom thereby accelerating the reaction as observed.

(1) For example, the relative rates of solvolysis of 7-tosyloxynorbornane and anti-7-tosyloxynorbornene in acetic acid are $\mathrm{1:10^{11}}$

  • $\begingroup$ I Think I have got you (almost) ...I have just one problem with your statement "However the size of the benzoyl group is not exceptionally large and the steric effect will likely be small" . $\endgroup$
    – Freelancer
    Jan 4, 2017 at 3:15
  • $\begingroup$ As I think Steric effect will not be small after all there is the benzene ring and a carbonyl group both hugely obstructing the attack of Nucleophile... $\endgroup$
    – Freelancer
    Jan 4, 2017 at 3:17
  • $\begingroup$ But then whenever I come across these questions I always feel like we are using the creationist method(i.stack.imgur.com/bS6S0.gif) $\endgroup$
    – Freelancer
    Jan 4, 2017 at 3:27
  • 1
    $\begingroup$ Funny and it's easy to see how folks might view things the way the cartoon suggests, but to actually publish a paper a lot of experiments are done to rule out \ rule-in various hypotheses. As to the size of the $\ce{-CO-Ph}$ group, we can look at something called A-values which are used as a measure of the size of various substituents. The A-value for a methyl group is 1.7, it is 3 for a larger phenyl group. I can't find a value for $\ce{-COPh}$, but $\ce{-COMe}$ $\endgroup$
    – ron
    Jan 4, 2017 at 3:41
  • 1
    $\begingroup$ is only 1.17, $\ce{-COCl}$ is 1.25. All of these values are smaller than the A-value for a methyl group and I doubt that $\ce{-COPh}$ would be much different from $\ce{-COMe}$ or $\ce{-COCl}$. So I would expect $\ce{COPh}$ to be sterically rather "small". $\endgroup$
    – ron
    Jan 4, 2017 at 3:45

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