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

1 Answer
Jan 15, 2017

Here's how I would do it.


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.


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.