How is a Barrierless Reaction Confirmed in Quantum Chemistry?

Question

Background:

Some quantum chemistry papers explore potential energy surfaces by characterizing critical points with an ab initio method. Reactants, products, intermediates, and transition states are found through optimization calculations, and then confirmed using frequency calculations. Intermediates connected by a transition state can be confirmed by intrinsic reaction coordinate (IRC) or some other minimal energy path (MEP) calculation. Here's an example potential energy surface from a paper online:

Question:

For any given intermediate and reactant/product pair, how does one confirm (1) a barrierless connection or (2) a lack of connection? A connection with a barrier is simple: a transition state must exist and connect back to the two species. That's like TS3 connecting H + SO$_2$ and HSO$_2$ above. But it's not clear how something like SO--HO and OH + SO can be confirmed, or the lack of connection between HOSO and OH + SO can be confirmed.

My thoughts:

(1) Intuitively, an optimization that does very small steps and always goes downhill from the reactant/product to the intermediate of interest may be sufficient to confirm the barrierless connection. At the same time, this feels like a bit of a slippery slope.. If there is a downhill route going from OH+SO to TS', then this could be used to confirm that OH+SO can barrierlessly form HOSO (see crude cyan-drawn path below).

(2) Intuitively, confirming a lack of connection would require some sort of constrained optimization starting from the intermediate which shows that all possible degrees of freedom lead to other intermediates or products.

((Here, I am just assuming we stay on a single, simple, adiabatic potential energy surface. Here is a similar question with smaller scope: What is a barrier-less reaction in Quantum Chemistry?))

Answer 1

The absence of TS is sufficient to confirm a barrierless path. On the scheme the transition between SO...OH and SO + OH is barrierless because SO+OH is not a minimum on PES. This configuration lies on the slope near the SO...OH minimum or corresponds to a big plate area on the PES*, depending on particular orientation and distance between two molecules, so there is no saddle point between them.

*because there are a lot of possible configurations for SO+OH system with relatively equal energies. Slope corresponds to interacting SO+OH (Van der Waals), plate area corresponds to nearly not interacting molecules

In my opinion it is not correct to consider any existing path as locked on the one side, a system just may have lower probability to follow it starting from a "bad" side. This situation may correspond to a high barrier or very narrow path which is not good because it requires concentrating energy only on (for example) one vibrational degree of freedom. If I understood correctly, here transition SO+OH -> HOSO considered as locked, which in my opinion is not correct, I see two possibilities for this transformation:

1. stage by stage go backward from SO+OH (slope or plate area) -> SO..HO (minimum) -> TS' (saddle point) -> HOSO (minimum), with barrier because we have TS on this path
2. directly from SO+OH to HOSO, which in my opinion should correspond to narrow path case because requires quick, relatively hard collision O atom of OH to S atom of SO, because molecules prefer another orientation (partial charges are bigger on H (OH) and O(SO) and corresponding molecular orbitals’ energies should be closer), so this situation may be possible only if two molecules directly collide and rotation of molecules is freezed (which doesn't seems to be highly probable).

P.S. I may not be right, I will be glad to discuss my point of view