Attack on cyclic intermediate during NGP

Question

I have studied that when an epoxy ring opening takes place in acidic medium, the ring opens in a way that the most stable carbocation forms. This is due to the positive charge on the oxygen, which gives partial positive character to the carbon atoms.

A very similar intermediate is formed in NGP (neighboring group participation) reactions with sulphur as the neighboring atom (source). How is the attack carried out then? Is it decided sterically as a regular SN2, or does the logic with oxygen extend to sulphur?

Answer 1

Inductive reasoning in organic chemistry can be somewhat hard at times, given all of the exceptions that can pop up, for example the SN2-like mechanism of opening an epoxide in basic conditions versus the acidic conditions.

For the epoxide opening in base; as you know sterics wins, and in acid the stability of the reactive transition state is our reasoning for the product observed. I believe for the NGP reactivity that stabilization of transition states is an acceptable platform to reason the important question as to exactly how this reaction happens in the first place.

lets look at the NGP example you've pulled from Wikipedia
(Awesome btw! reading organic chem on wiki for fun?!)

I have a hard time fully following your question but here's the best answer I could come up with:

I see a symmetrical transition state where the reactive-intermediate charged cyclopropyl-S-heterocycle can probably just be attacked from either side in a SN2 like fashion. If you prefer charge polarization to be our main reasoning in the S-hetero cycle, given that it is a symmetrical substrate I believe the same product will form either way due to symmetry.

Did you read the part of that wiki article below that depicts NGP reactivity by an Alkene? The driving factor there appears to be pi-stabilization of the carbocation in the transition state.

This is a physical organic chemistry question that is not trivial. I applaud your curiosity.