In 2-bromo cyclohexanone,why bromine prefers to occupy axial rather than equatorial position?

Question

I read that the substituents generally tend to occupy equatorial position rather than axial because diaxial interactions make the molecule unstable. But why is 2-bromocyclohexanone an exceptional case? I also read that it’s because of the dipolar interactions. What actually are these dipolar interactions and how do they stabilise 2-bromocyclohexanone with bromine being on axial position. How do they outweigh the effect of diaxial interactions.

Answer 1

I think there must have been some misunderstanding. This stabilization is working for all 2‑halocyclohexanones (but does not always cause the axial conformer to predominate), where bromo- one is not a special case, but has been used just as an example.

There is a combination of various effects, but the dipolar one is supposed to cause a destabilization by dipolar repulsion in the equatorial-X conformer. See the difference of partial charges distribution:

Measured or calculated molar fractions (%) of the axial conformers in different phases/solvents (taken from Yoshinaga, Fabiana, et al. "Conformational analysis of 2-halocyclohexanones: an NMR, theoretical and solvation study." Journal of the Chemical Society, Perkin Transactions 2 9 (2002): 1494-1498. doi:10.1039/B204635K):

$$ \begin{array}{c*}\hline & \text{vapor} & \text{liquid} & \ce{CCl4} & \ce{CHCl3} & \ce{MeCN} & \text{DMSO} \\ \hline \ce{F} & 64 & 5 & 28 & 13 & 2 & 2 \\ \ce{Cl} & 86 & 26 & 62 & 42 & 15 & 13 \\ \ce{Br} & 92 & 53 & 81 & 66 & 36 & 33 \\ \ce{I} & 96 & 87 & 92 & 85 & 65 & 63\\ \end{array} $$

Overall stabilization effects in axial-X conformation of 2‑halocyclohexanones clearly increases in the order of F < Cl < Br < I.

However, the ‘gauche effect’ participates, which is stabilizing the equatorial conformation in largest extent for F, also for Cl, has almost zero effect for Br, and has a negative effect for I.