Question

Ortho,para-bromoanisole + `NaNH_2` + Liquid `NH_3` =? How do you predict the product?

Answer 1

Here's how I would do it.

Explanation:

These are the reaction conditions for generating benzyne intermediates.

The reaction of ortho-bromoanisole with potassium amide in liquid ammonia (b. p. -33 °C) is extremely rapid.

Step 1. The amide ion attacks the `"H"` atom that is ortho to `"C3"`, generating a carbanion.

Step 1

Step 2. Loss of `"Br"^"-"` to form a benzyne intermediate.

Step 2

The elimination is by an E2cb pathway.

Step 3. Addition of `"NH"_2^"-"`

The strain caused by a triple bond in a benzene ring can be relieved by a nucleophilic addition (`"Ad"_"N"`) of `"NH"_2^"-"`.

The methoxy group is electron-withdrawing by induction, so the nucleophile will attack `"C3"` to place the carbanion as close as possible to the methoxy group.

Step 3

Step 4. Protonation of the carbanion.

Step 4

The product is meta-methoxyaniline.

para-Bromoanisole

The reaction with para-bromoanisole also follows a benzyne mechanism.

Step 5

The nucleophile can attack either end of the triple bond, and the methoxy group is far enough away that its inductive effects are minimal.

Step 6

The product is a mixture of para- and meta-methoxyaniline.