What is the most likely site of iodination in an iodination of salicylamide experiment and why?

Answer 1

I guess you could end up with either a 1,2,3-trisubstituted or 1,2,4-trisubstituted ring, but the main product in my experiment was 2-hydroxy-5-iodobenzamide (a 1,2,4-trisubstituted ring) determined by the ortho/para-directing hydroxyl group at the C1 carbon, coupled with the meta-directing amide at the C2 (both directing substitution to the C4 position), and then narrowed down to a specific position by the 816.28 cm^-1 strong peak that was on the fingerprint region of my IR spectra that is characteristic of a 1,2,4-trisubstituted ring.

Also, if you draw out the carbocation intermediates you can see that the positive charge would not end up on the C2, which is the most likely scenario when you consider the C2 is already partially positive due to the electron withdrawing amide that is attached to it.

Answer 2

You have to determine your electron donating and withdrawing substituent and once that is done identify where the BEST position is for the iodine based on the para, meta and ortho position.

Note:

Donators (Activators) prefer a para OR ortho addition

Withdrawers (Deactivators) prefer the meta position

Once that is done you'll find Iodine goes into the meta of the deactivator and para position of the activator POSITION 5!

Answer 3

The most likely site of iodination in an iodination of salicylamide experiment is the ortho position relative to the hydroxyl group on the benzene ring. This is because the ortho position is more reactive due to electronic effects that stabilize the intermediate formed during the iodination reaction.