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The following question came in Paper 2 of JEE Advanced 2014-

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Here we have $\ce{CH3-}$ as the nucleophile. There are 2 possible attacks:

  1. SN2 type attack on the chlorine group.
  2. Nucleophilic addition on the carbonyl.

The answer given is option D, which will happen by the 2nd pathway(nucleophilic addition).

Why is nucleophilic addition given a preference over SN2? I feel that nucleophilic addition should rather decrease the reaction rate because the carbon goes from sp2 to sp3, decreasing the bond angle and hence increasing the repulsion

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    $\begingroup$ Alkyl Grignards do not readily do SN2 nucleophilic attack without catalysis, certainly on on alkyl chlorides $\endgroup$
    – Waylander
    Commented Feb 10, 2019 at 8:38
  • $\begingroup$ Just wondering, could the coordination of the carbonyl oxygen by Mg in the Grignard reagent play a role? Because one might also ask why the Grignard reagent does not behave as a base and abstract the fairly acidic proton alpha to the carbonyl. $\endgroup$
    – user6376297
    Commented Feb 10, 2019 at 10:24
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    $\begingroup$ As to @Waylander's point, how could you possibly make CH3MgBr from CH3Br without getting ethane by an SN2 reaction? In addition, experience is worth a lot in answering this question. $\endgroup$
    – user55119
    Commented Feb 10, 2019 at 16:11
  • $\begingroup$ @user55119 in theory, even if it did react via SN2, you could add RBr to Mg such that there is never any excess RBr left in the reaction mixture. Analogous to how one would add carbonyl compound to LDA instead of LDA to carbonyl. I'm not saying it does react, I'm just saying that if it did there are ways of getting around it, so the argument isn't complete. $\endgroup$
    – orthocresol
    Commented Feb 10, 2019 at 16:26
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    $\begingroup$ Importantly "Here we have CH3− as the nucleophile" isn't true and is a source of OP's misunderstanding. @user6376297 That's part of real mechanism and methyl is still coordinated when this happens, only later it's intramolecularly migrated to carbonyl. $\endgroup$
    – Mithoron
    Commented Feb 10, 2019 at 19:28

1 Answer 1

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To begin this short discussion one has to differ between Sn2 and addition to the carbonyl group, as the former is driven by orbital interactions while the latter is driven by coulomb interactions. This can be explained with a closer look at the carbonyl function. due to the highly polarized π-bond between carbon and oxygen, carbon is rendered quite positive. It results that any nucleophile bearing a negative charge will be attracted to attack it. In addition the reaction of a carbanion and a ketone or aldehyde is irreversible! there is fundamentally no way in which the carbanion could ever be considered as a leaving group.

On the other hand Sn2 reactions appear on saturated carbon (in the best case methyl or primary) with at least one good leaving group (e.g. a halide or tosylate). Often the sigma bond connecting carbon and the leaving group is not even polarized, at least not significantly. Just consider the electronegativity of iodine, the BEST leaving group for SN2. That is why orbital interactions are more dominant in thess kind of reactions. It can be easily noticed that a potential nucleophile has to donate its electrons into a sigma antibonding orbital in order to react with the carbon electrophile. These kind of orbital can be considered as very high in energy as it is the counterpart of the sigma bonding orbital which is quite stable. Therefore the best nucleophiles for nucleophilic attack on saturated carbon is usually a second or third row nucleophile like sulphur or phosphorus (Wittig reaction).

Therefore for a carbanion the irreversible addition to the carbonyl group happens much faster, resulting in a alkoxid. The new formed alkoxid is in close proximity to the chlorine substituted carbon and therefore likely to form the five membered ring .

Any references that support everything written above can be found in Clayden, Organic chemistry second edition. Specifically the chapter nucleophilic substitution at saturated carbon will help.

Best regards, John

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    $\begingroup$ Carbonyl addition is certainly not driven by coulomb interactions. $\endgroup$
    – Zhe
    Commented Feb 11, 2019 at 20:22
  • $\begingroup$ but the attraction of the nucleophile and the electrophile can be considered electrostatic $\endgroup$
    – John
    Commented Feb 11, 2019 at 20:34
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    $\begingroup$ In that case, you might just say all chemistry is electrostatics. $\endgroup$
    – Zhe
    Commented Feb 11, 2019 at 20:40
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    $\begingroup$ My big issue here is really that both reactions are strongly dictated by orbital considerations. Otherwise, there wouldn't be such strong stereoelectronic considerations in each case (back aside attack, Bürgi-Dunitz angle). $\endgroup$
    – Zhe
    Commented Feb 11, 2019 at 20:45
  • $\begingroup$ But nevertheless there is a significant difference in the electrostatic interaction of a carbanion with a carbonyl or a chlorosubstituted saturated carbon. Which should render the attack on the ketone much faster, shouldn't it? $\endgroup$
    – John
    Commented Feb 11, 2019 at 20:56

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