# On baker nathan hyperconjugation

Recently I have been reading about the history of the Baker Nathan reaction.

In addition to the large rate increase for Me vs H, all the other alkyl groups showed decreases with respect to Me in a regular manner as would be predicted by reduced hyperconjugative ability.

By drawing resonance structures, it is easy to see the $\mathrm{Me < Et < {}^iPr <} {}^t\mathrm{Bu}$ predicted by the inductive effect:

Therefore the more electron donating (i.e. the stronger the I+ effect) the para group is , the more stabilized the transition structure is (black $\delta^+$ on the left diagram become smaller hence less electrostatic repulsion in the $\pi$ system, hence blue $\delta^+$ and the $\delta^-$ can be larger thus facilitating the reaction). Hence $\mathrm{Me < Et <} {}^i\mathrm{Pr <} {}^t\mathrm{Bu}$.

However, for hyperconjugation, the $\sigma$ bonds overlap with the $\pi$ system resulting in electrons to become more delocalised at the C-C bond between the alkyl group and the $\pi$ system. Therefore the more hyperconjugations present, once again the more stablised the transition structure hence a faster reaction.

Since both $\sigma_\mathrm{CC}$ and $\sigma_\mathrm{CH}$ bonds adjacent to the para carbon can potentially engage in hyperconjugation (and naively the $\sigma_\mathrm{CC}$, being larger in lobe size, should expect to overlap better with the $\pi$ system), does that mean the experimentally observed trend that $$\mathrm{Me > Et > {}^iPr >} {}^t\mathrm{Bu}$$

suggest hypderconjugation of $\sigma_\mathrm{CC}$ is much weaker than that of $\sigma_\mathrm{CH}$ hence justifying the quoted message?

If hyperconjugation of $\sigma_\mathrm{CC} < \sigma_\mathrm{CH}$ what is the underlying molecular orbital explanation?