# Directing effect in Intramolecular Friedel Crafts Alkylation

This reaction should cause ring closure at either the $$o-$$ or $$p-$$ position relative to the methyl substituent on the ring (as methyl is an $$o,p-$$directing group), giving the following products:

The question is: Which one of these is the major product? Is substitution preferred at the ortho position or para position, and why?

A few more examples of these types of questions:

This would give which of the following?

• Also, I'm not too sure of the hydride shift you did in 2. won't a six-membered ring be more favoured? Just curious. Jan 3, 2020 at 5:28
• @HahaHahaha: six-membered ring formation would favorable only when it could form. I don't think that the unstable 1° carbocation would survive as intermediatre Jan 3, 2020 at 5:33
• @RahulVerma Do you have any references? I find it hard to digest the fact that 6 membered ring is not favored here. Mar 2 at 5:48

Electron giving groups like $$-OCH_3$$ (by resonance) and $$-CH_3$$ (by hyperconjugation) are Ortho-Para directing.

But, in cases where there is no increased stability by H-bonding or other factors at ortho position, the para substituted product will give major yield.

This can simply be explained by stating that there is less steric hindernace on the para position compared to the ortho position. Which is further emphasized when you take this example:

So, the answer is para-substituted in both cases.

References

Aromatic reactivity

• Reference for this? Jan 3, 2020 at 8:48
• There you go @Waylander Jan 3, 2020 at 14:12