Question

How does a Schiff reagent react with an aldehyde?

Answer 1

I assume you don't mean Schiff base, because that's the imine product of an aldehyde reacting with an amine.

A Schiff reagent reacts with an aldehyde basically to form a bright red product. I've looked into it, and though it's in Polish, the first one is the simpler one, and in that one, we can see that, according to the NMR studies of Robins, Abrams, and Pincock:

• The reaction starts with the addition of sulfurous acid
• Then you add two equivalents of the aldehyde.

Here's what I can figure out based on the differences in the reagent and the product:

1. The sulfurous acid very likely adds onto the central, most reactive double bond as `"SO"_3"H"`. This leads to counterclockwise `pi` electron conjugation all the way through to the bottom-right iminium group.
2. The reagent undergoes further reaction with sulfurous acid where the upper-right and lower-right anilines each lose one hydrogen and replace it with a `"SOOH"`.

Here's what I think the mechanism would be on steps 3-4, which is the more interesting part (and step 5 is simple enough to imagine):

3. The bisulfite that added in step 1 prompts the `"NH"_2` activating group on the left aniline ring (because it didn't react yet, and intramolecular processes are generally faster than intermolecular reactions) to donate electrons into the ring to form an iminium intermediate that bumps the bisulfite off.
4. The addition of the two aldehyde equivalents is like an "enolate" nucleophilic addition reaction (where the `"SOOH"` act like "enolates"), forming tetrahedral complexes.
5. An intramolecular proton transfer occurs, probably via `"HSO"_3^(-)`, to give the tetrahedral complexes protons.