# Regioselectivity in nitration of a 1,2-disubstituted aromatic compound

For 1,1,4,4-tetramethyl-1,2,3,4-tetrahydronaphthalene:

What would be the products following a standard nitration with $$\ce{NO2+}$$?

My reasoning is thus:

Since the substituent group is alkyl then it is ortho- or para- directing. But surely there are steric hindrance factors to be considered due to the bulk of the ring which would mean para- is favoured instead of ortho. However, when I try to draw out the structure it seems the arrows are meta- instead!

The molecule is symmetrical so the two positions on the far side of the alkyl groups are exactly the same. Similarly, the two positions closest to the alkyl groups are the same. For the far positions, each one is para with respect to one alkyl group and meta with respect to the other. This is not a contradiction because ortho, meta and para are purely relative terms.

I suspect that you are right that substitution will mostly occur at the far sites. The directing effect of the alkyl groups is essentially the same for every site because they are all ortho or para to one group and meta to another. Therefore, the steric effect of the bulky methyl groups adjacent to the near positions will disfavour substitution there.

Rewording my five-year-old comments into an answer:

The idea of "meta" or "para" is a bit different here, since you have not just one, but two alkyl groups. Being meta to one means ortho/para to the other, and vice versa. Since there are two substituents to start off with, you have to consider both their directing effects separately and add them up.

• The carbons labelled a are ortho to one alkyl group (stabilises the cationic intermediate a lot), but meta to another (stabilised a bit).

• On the other hand, the carbons labelled b are meta to one alkyl group (stabilised a bit), but para to another (stabilised a lot).

So, purely based on electronics, both positions are almost equally favoured. Now you turn to sterics to help you decide: as you said, b is favoured because it is less hindered.