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

Jan 15, 2017

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 $\text{H}$ atom that is ortho to $\text{C3}$, generating a carbanion.

Step 2. Loss of $\text{Br"^"-}$ to form a benzyne intermediate.

The elimination is by an E2cb pathway.

Step 3. Addition of $\text{NH"_2^"-}$

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

The methoxy group is electron-withdrawing by induction, so the nucleophile will attack $\text{C3}$ to place the carbanion as close as possible to the methoxy group.

Step 4. Protonation of the carbanion.

The product is meta-methoxyaniline.

para-Bromoanisole

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

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

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