enter image description here 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:

Product1enter image description here

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: enter image description here

This would give which of the following?

product1enter image description here

  • $\begingroup$ 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. $\endgroup$ Commented Jan 3, 2020 at 5:28
  • $\begingroup$ @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 $\endgroup$ Commented Jan 3, 2020 at 5:33
  • $\begingroup$ @RahulVerma Do you have any references? I find it hard to digest the fact that 6 membered ring is not favored here. $\endgroup$
    – DatBoi
    Commented Mar 2, 2022 at 5:48

1 Answer 1


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

enter image description here

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.

enter image description here

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:

enter image description here

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


Aromatic reactivity

  • $\begingroup$ Reference for this? $\endgroup$
    – Waylander
    Commented Jan 3, 2020 at 8:48
  • $\begingroup$ There you go @Waylander $\endgroup$ Commented Jan 3, 2020 at 14:12

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