Why shifting of pi electron away from benzene ring?
Answers
Methyl Benzene
Ethyl Benzene
t-butyl benzene
I would say that t-butyl benzene would have the highest electron density as the alkyl group attached to it is bulkier than methyl and ethyl thus it will exert a greater +I effect pushing the electrons into the ring making it electron rich. Using same analogy ethyl benzene would have a greater electron density in the ring compared to methyl benzene.
However that is not the case. Something counter intuitive seems to be happening here. The order is methyl benzene >ethyl benzene >t-butyl benzene.
Well, tetrahydrocannabinol, you said it correct only the correct reason/explanation is hyperconjugation only. More the number of (alpha) hydrogen atoms on a C=C [isolated as in alkenes or in resonance as in benzene], more the possibility of hyperconjugation.
In benzene, this will shift the electron density from C-H bond of CH3 group towards the ring, making it electron rich.
Contrary to what could have been thought from +I effect of corresponding alkyl groups on the ring, hyperconjugation is more in CH3.
Since hyperconjugation involves complete shifting of electron from valence shell of atoms to the other, it is more "effective" that INDUCTIVE which just pushes electron little more towards more EN (more correctly, towards group with more -I) atom.