Why is -OCH3 more strongly activating than -CH3 in electrophilic aromatic substitution?

Question

Facts I know: 1) More the electron density in benzene ring, the faster the reaction. 2) Lone pair on -OCH3 group undergoes resonance and makes ortho-para positions electron rich, and so does -CH3 by hyperconjugation. 3) -OCH3 group shows -I (electron withdrawing) and -CH3 shows +I (electron releasing).

Now, -OCH3 and -CH3 both undergo resonance and hyperconjugation respectively and both make ortho/para positions equally electron rich.

So the factor that should come in play is I effect. As the -CH3 group releases electron into benzene ring by +I effect, and -OCH3 withdraws electron by -I, what should be more activating is the -CH3 group.

Where am i wrong?

Answer 1

Since we are talking about the rates of these reactions, we really want to examine the structures of the transition states in the rate determining step in these reactions, rather than the starting materials. For electrophilic aromatic substitution, there is a high energy carbocation, the Wheland intermediate, formed when the benzene derivative attacks the electrophile. Formation of this intermediate is normally the rate determining step. Since this carbocation is high in energy, the transition state leading to it is going to be similar in structure, according to the Hammond Postulate. So, looking at the influence of varying substituents on the stability of these Wheland intermediates is a good approach to justifying differences in reaction rates (not to mention selectivities).

If we take the resonance theory approach, we can draw many resonance structures for eg. the para-substitution intermediate, putting the positive charge ortho and para to the "E" substituent we just added. Looking at the methyl-substituted benzene substrate (toluene), resonance theory would allow us to draw an additional "no bond" resonance structure, where we break the C-H bond and have a proton associated with a neutral methylidene cyclohexadiene. This is the resonance theory approach to hyperconjugation. This resonance contributor can justify why toluene is more reactive than benzene. However, how important is the contribution of a resonance structure where we have one less σ-bond and a positive charge on a tiny hydrogen atom?

Looking at anisole (methoxybenzene) the same way, resonance theory would lead us to draw a resonance structure where the non-bonding (lone) pair of electrons on oxygen is used to make an additional π-bond and the oxygen ends up with the formal positive charge. It may not feel nice to put a formal positive charge on the oxygen, but this resonance contributor has a lot going for it - we have gained an additional π-bond relative to the other resonance structures and all non-hydrogen atoms satisfy the octet rule. The "loss" of the lone pair is no problem - as a non-bonding pair of electrons they were already relatively high in energy. For these reasons, this is an excellent contributor to the overall structure of this intermediate - probably the most important.

Taken together, and using resonance theory, the above can help justify the differences in reactivity. Another approach would be to use molecular orbital theory. Hyperconjugation would be viewed as overlap of the C-H σ-orbital with the adjacent carbocation π-system. The efficiency of this overlap is poor because there is a big energy difference between a low energy C-H σ-bond and π-system of the carbocation. In contrast the energy difference between the oxygen non-bonding (lone) pair and the π-system is relatively small and overlap is therefore efficient.