> 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all.

I'm able to understand why reaction mechanism 1 is preferred over 2 if I agree with your statement. However, I'm having a small trouble in understanding it.

Usually I visualise a multi-step reaction as given below:

In the above image the RDS is the first step. The same amount of sand reaches the bottom irrespective of which step is the RDS.

We can consider the formation of intermediates (a)s and (b)s as constrictions. If so, even if (a)s allow the sand to flow freely, the (b)s can alter the formation of the final product.

This is where, I'm unable to understand your answer. The rest is clear to me.

Could you please explain this? Thanks!

I feel your hourglass visualization needs a little tweaking,so I will try drawing out my version of this thing and put it here in a little while,then maybe you can see what I am saying

Just give me a little time now

Ok. No problem.

@GuruVishnu Please look at this,and tell me when you're here

@YusufHasan It looks really great! One small correction: I think the labels 2(a) and 1(b) need to be swapped.

Anyway, coming to the main point: I understand the difference in size constrictions for the first step. But from the process which converts (a)s to (b)s, why is there no difference in the constrictions? Is there any specific reason for doing so?

@GuruVishnu Ah yes,you're right.. 2(a) and 1(b) has to be swapped

Yes there is a reason

But first of all,did you notice the stopcocks in green?

And what they represent?

Le Chatleir's principle. So does the size of the constriction variable and depend on the amount of the reactant (well here it's the intermediate 1(a))? So far, I thought as per the sand clock model, the size of the constriction is fixed.

Yes,their are two things at play here: the actual size of the constriction represents how favorable that step is as part of the ongoing mechanism... While the stopcock is there to slowly disallow reversibility as the reaction keeps occuring for a long time in the lab frame

At the beginning itself,you will have noticed sthg

Out of the 2 possible RDS paths, (a) side has a bigger tube,while the (b) tube is constricted

This is an indication of thermodynamic stability i.e. how favorable it would be for the reaction to go there

But when we take time of running reaction into account, then you will notice that the stopcock will stand closed much more on the mouth of the (b) tube for the RDS as compared to (a)

This will again mean that whatever sand may have initially trickled down the (b) RDS pathway,even that flow will be almost terminated by the slowly closing valve

After this step, the 1 and 2 pathways are essentially separated,as you can follow from the hourglass diagram.. They have no other coonection whatsover after the reactants vial

Now,once we come down to converting the (a)s to (b)s

The connecting tubes of course have to be much broader than the RDS tubes

Beyond that,their isn't much sense to compare the breadth of the 1(ab) tube with the 2(ab) tube,as each flow is now essentially separate.. What this means is that from stability concerns, 2(ab) should have been broader than 1(ab) slightly

But again,this comparsion in breadth wouldn't matter, as it would still be simply carrying the sand coming from the initial RDS source

Fine. I think I was wrong in thinking that sand will get accumulated in the process 2 and hence give a constant flow rate as of the first one. I understood this and now, it's clear to me why is the major product is dictated by the reaction mechanism 1. Also regarding the selectivity or specificity, I feel this reaction must be selective as I expect at least one molecule to be formed by the second pathway.

The "rate determining" step is called so precisely for this reason; for both the individual pathways 1 and 2, all we can say about the rest of the steps is that they would each be individually faster than their respective RDS

Yes,a somewhat negligible yield will probably be there by the 2nd pathway

But again, that's so with all of chemistry; no reaction has 100% yield

The major correction in your visualization was this parallel system of possible pathways and the slowly closing stopcocks as each step

In essence,you could draw n possible parallel hourglass paths from a single reactant vial

But the initial sudden inflow of sand at the start will be governed by the breadth of the 1st tube

And over time,the reversibility of the reaction, and finally directing all the sand towards one pathway and cutting off the supply for the other pathways

That would be facilitated by the stopcock

Ok. Now it's clear to me. Thanks a lot for your great explanation!

And also,in this model,it stands pretty clearly that the steps after the bifurcation cannot affect the other competing pathways unless another linking tube/crossway occurs

@GuruVishnu You're welcome,I hope it helped :)

:)