Past exam of the chemistry B course of the University of Cambridge / 2021 / Paper 1 / B1 / a / iii / Solution

The methoxy groups activate the aromatic ring by electron donation through chemical resonance and direct electrophilic substitution ortho and para. Their para positions are already occupied by one another, so the first alkylation occurs at an available ortho position; the four initially vacant positions are equivalent by symmetry.

After placing the first tert-amyl group at position 2, position 5 is favored for the second substitution. It is para to that bulky group and ortho to the other methoxy group. Position 3 is crowded between methoxy and tert-amyl substituents, while position 6 is near the first tert-amyl group across the intervening methoxy-bearing carbon. The 2,5 arrangement distributes the bulky groups on opposite parts of the ring and preserves the equivalence seen in the spectrum.

Alkyl groups do not electronically deactivate the ring, so the stopping point needs a steric explanation. After two such substitutions the remaining positions are much more congested, each adjacent to a bulky tert-amyl group and a methoxy group. Two large substituents are accommodated much more readily than a third, despite excess alcohol. The formula and NMR identify this sterically favored disubstitution; they do not imply that a Friedel-Crafts reaction always stops after two alkylations.