Question

Draw both major organic product(s) that are obtained when 4-chloro-2-methyltoluene is treated with sodium amide followed by treatment with H3O+.

Answer 1

I attached the answer as an image. I also drew in the two most acidic hydrogens.

Step-by-step explanation:

This goes through the 'benzyne' intermediate, meaning it does an E2-looking reaction by expelling a leaving group (chloride) from the adjacent part of the ring using the amide as a strong base. The triple-bonded benzyne has absurd bond angle strain, and is vulnerable to a good nucleophile like an amide ion, and the resultant sp2 anion is then reprotonated by the acid. I didn't draw in the acid-base reaction in step one, or the spectator ion (sodium).

Answer 2

The both major organic products are 3-amino-2-methyltoluene and 4-amino-2-methyltoluene.

The reaction of 4-chloro-2-methyltoluene with sodium amide is an E2 reaction. The abstraction of hydrogen ion by a base and the loss of chloride ion occurs simultaneously leading to the formation of a benzyne intermediate.

This benzyne intermediate is attacked by the amide ion and H3O+ to yield the products; 3-amino-2-methyltoluene and 4-amino-2-methyltoluene. The detailed mechanism of the reaction is shown in the images attached below.

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