In the heterocycle synthesis below (taken from Org. Process Res. Dev. 2006, 10 (3), 398–402), a 5,6-fused heterocycle (an imidazo[1,2-a]pyrimidine) is formed by treating 2-amino-4-(trifluoromethyl)pyrimidine with the diethyl acetal of bromoacetaldehyde, with the cited paper claiming to get a 25:1 selectivity for 5 (minimal amounts of 5a also produced were able to be easily removed).
There are (as ever with heterocycle syntheses) a few different mechanisms one could draw, however, no matter which way round one draws the mechanism, the major product is always the one that arises from what appears to be the less favoured intermediates.
By less favoured intermediates, what I mean to say is that to arrive at the major product, necessarily a cation ends up closer to the electron withdrawing trifluoromethyl group than it would do if the minor product was formed.
Given that all the steps are essentially reversible, is the product distribution arising from a purely ground state thermodynamic preference for 5 over 5a, or is there a mechanism that I'm completely neglecting (sorry for the lack of pictures, I'm not able to ChemDraw right now).