Relative rates of some compounds towards SN2 is given below enter image description here enter image description here

The reason for reactivity of alpha carbonyl halides is given as enter image description here

It’s not clear what exactly is happening.


2 Answers 2


I have asked a question previously which concerns the same electrophilic system with the exception that nucleophilic attack happens on the carbonyl carbon and not on the α-carbon.

From the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on pp. 890-891:

Scheme from Clayden
The $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ orbitals add together to form a new, lower-energy molecular orbital, more susceptible to nucleophilic attack. But, if $\ce{X}$ is not a leaving group, attack on this orbital will result not in nucleophilic substitution but in addition to the carbonyl group. Again, this effect will operate only when the $\ce{C-X}$ and $\ce{C=O}$ bonds are perpendicular so that the orbitals align correctly.

This interaction when considered in the context of an attack on the carbonyl is the principle of the polar Felkin-Anh model used to describe a certain diastereoselectivity. However it can likewise be used to explain the lowering of the $\unicode{x3c3}^*(\ce{C-Br})$ orbital which is attacked in a nucleophilic substitution reaction. It may seem as though these statements are contradictory, but the orbital interaction will create a three-centre π-type double-antibonding orbital with only the contribution on oxygen being neglegible due to oxygen’s electronegativity being highest and thus its contribution to high-energy orbitals being lowest.

It will depend on your nucleophile what will actually happen. It can attack both on the carbonyl carbon and on the α-carbon. If the attack on the carbonyl carbon is reversible, only the attack on the α carbon leading to an observed $\mathrm{S_N2}$ process will be observed after the reaction — even if the carbonyl attack is faster. Only if the attack on the carbonyl is not reversible as is the case e.g. with Grignard reagents, a mixture of two products will be observed.

Naturally, if $\ce{X}$ is a very poor leaving group — e.g. $\ce{OTBS}$ — the attack on the carbonyl will dominate this way or that.

To sum up:

The reaction rate is increased because the $\unicode{x3c3}^*(\ce{C-X})$ orbital and the $\unicode{x3c0}^*(\ce{C=O})$ orbital linear combine to give a lower-energy orbital. Since this is the LUMO, the LUMO energy is lowered and nucleophilic attacks are facilitated.

  • $\begingroup$ Why does the presence of carbonyl on α-carbon of a substrate restrict SN1? $\endgroup$
    – Neha
    Jun 18, 2021 at 3:26
  • 3
    $\begingroup$ @Neha Draw mesomeric structures and the answer will become apparent. $\endgroup$
    – Jan
    Jun 18, 2021 at 11:05

Most of the time, a $\text{S}_\text{N}2$ reaction will involve a negatively charged nucleophile attacking an uncharged substrate (as is depicted in the second picture of the question). This negative charge can be stabilized effectively by the carbonyl group through conjugation. Note that in the example given in the first picture, the stabilization probably also involves the phenyl ring. Of course, the molecular structure must allow for alignment of the orbitals involved.

This mechanism is valid for most examples given in the table of the first picture bar the methoxy-methylchloride (MOM chloride, an old-fashioned protection agent for alcohols). In this case, I would think that the oxygen lowers the $\sigma^\ast$ orbital energy such that it may serve better as a point of attack.

  • $\begingroup$ Why does the presence of carbonyl on α-carbon of a substrate restrict SN1? $\endgroup$
    – Neha
    Jun 18, 2021 at 3:27
  • 1
    $\begingroup$ @Neha It makes the $\text{S}_\text{N}2$ so much faster that $\text{S}_\text{N}1$ does not stand a chance. $\endgroup$
    – TAR86
    Jun 18, 2021 at 17:29

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