[tezha]

Dear organic chemists,

A textbook I use states that both compounds on the left have faster rates for Sn2 reactions and that it is due resonance stabilisation of the transition state. Can someone please explain exactly how the resonance happens in this case? because all I could predict to happen is conjugation of the p orbital on the beta carbon with the p orbital of the transition state, but not exactly resonance, because as far as I understand, the carboun would have to expand it's octet? I'm really confused with this one. here are the compounds:

[opsomath]

This is an interesting question and one which I have had trouble explaining to my students in the past. It is definitely experimentally true that benzylic, allylic, and alpha-carbonyl halides react faster in SN2 reactions than their non-conjugated cousins. My initial inclination would be to explain this in terms of orbitals. To do the SN2, the reaction has to go through a 5-membered transition state where there is essentially a p-orbital with the nucleophile coming on one side and the leaving group leaving on the other. This p-orbital will overlap well with the p-orbitals on the conjugated group, so the transition state gets stabilized.

It's interesting to me that the carbonyl group has the same effect as the C=C bond. Generally these two groups have opposite effects. Perhaps the effects are for different reasons? I can make a pretty decent argument that the antibonding orbital of the carbonyl group overlaps with the antibonding orbital of the alkyl halide, lowering the energy of the LUMO and thus the barrier to nucleophile attack by the Hammond postulate. What do y'all think?

[fledarmus]

There is also a steric effect, especially when the pi systems are lined up so they will overlap with the making/breaking bond. The flat double bond or aromatic ring system doesn't occupy as much space as an sp3 system would, even with just hydrogen substituents.

[opsomath]

Ok, this just occurred to me...do you think there could be some neighboring-group participation with the carbonyl group? Depending on what conformation you draw of the alpha-halo carbonyl, you can picture the leaving group interacting with the electron-deficient carbonyl carbon as it leaves. It would be interesting to look up SN2 reaction rates of, say, 2-bromocyclohexanone vs. 3-bromo-2-hexanone, where in the cyclic compound the conformation is locked in and the carbonyl carbon cannot directly interact with the leaving group as it leaves.

[tezha]

Ok, this just occurred to me...do you think there could be some neighboring-group participation with the carbonyl group? Depending on what conformation you draw of the alpha-halo carbonyl, you can picture the leaving group interacting with the electron-deficient carbonyl carbon as it leaves. It would be interesting to look up SN2 reaction rates of, say, 2-bromocyclohexanone vs. 3-bromo-2-hexanone, where in the cyclic compound the conformation is locked in and the carbonyl carbon cannot directly interact with the leaving group as it leaves.

So your'e saying that the electron deficient carbonyl carbon is somehow attracting the leaving group? I haven't stumbled upon such intramolecular interaction in the only organic chem textbook i'm using. Anyway, although both your answers were really detailed and made me understand these phenomena better, I still don't understand where exactly there is resonance? Resonance is no the same thing as conjugation, but the textbook stated the process as resonance stabilization. Can I say that the authors used wrong terminology?

[Dan]

I think in the case of the carbonyl, the rate increase is normally explained by either:

1. An additional bonding interaction between the C-O pi* and the lone pair of the incoming nucleophile in the transition state.

2. Overlap of the C-O pi* and the C-Cl sigma* to create a more electrophilic LUMO.

Both are essentially the same argument, just from different directions.

This is all in Clayden et al somewhere.

[opsomath]

The carbonyl interacting with the leaving group thing is highly speculative. Ignore it for your purposes.

Basically, the transition state of the SN2 reaction is "conjugated" with the pi-orbitals of the carbonyl/vinyl group next to it. This conjugation reduces its energy, and therefore lowers the barrier to the reaction resulting in a faster reaction. Maybe drawing the five-membered transition state, complete with the p-orbitals on the neighboring group, would help?

Personally, I always just thought of it like this: having a double bond adjacent speeds up all substitution reactions by weakening the carbon-leaving group bond. The one exception is that carbonyls don't speed up SN1 reactions, because they are electron-withdrawing and you need electron-donating groups for carbocation stability.