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The following reaction has been taken from my study material. This shows that the benzene ring is undergoing a nucleophilic aromatic substitution via benzyne mechanism (presence of a strong base like $\ce{NaNH2}$). My question is:

Why is substitution preferred at the ortho position?

The following link explains the mechanism quite well and gives an example of ortho-chlorotoluene undergoing a similar reaction. Since both proceed via the same benzyne intermediate, why is there a difference in the products formed by meta-chlorotoluene? Or is my study material wrong in saying that o-aminotoluene is a major product? Any help regarding this is appreciated!

Attaching the related reaction's image: enter image description here

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The Me group has no directing effect in the deprotonation so there are 2 protons (o & p) that can be removed with equal probability by the NaNH2 to give the benzyne. These give 2 possible benzynes (2 and 3 relative to Me). The nucleophile will add with equal facility to either end of either benzyne giving a mixture of 3 isomers in 1:2:1 ratio. The major isomer is the m- aminotoluene because it can arise from both of the possible benzynes.

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  • $\begingroup$ Ok that makes a lot of sense. So that means that m-aminotoluene and not o-aminotoluene is the major product, which means that the first reaction is wrong. $\endgroup$ – Aniruddha Deb Nov 19 '19 at 10:03

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