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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.
