# In the reversible reactions of acyl substitution, how do backward reactions happen by being against the forward reaction drives?

Consider this mechanism for acyl substitution by nucleophilic addition-elimination:

To start with, the nucleophile having a lone pair of electrons attacks carbon (because of oxygen being more electronegative?) and forms the conformation/arrangement of atoms as shown in the image.

But (there is an equilibrium sign) doesn't this indicate that the atoms are going back to the initial state against their original tendency which was responsible for the formation of the product in the first place? For example, in second step, what would make a nucleophile attached to a positive carbon site, leave to form back the reactants of step one?

Which force would make nucleophile again leave the carbon?

Edit (25/5/17): Above reaction mechanism seems to be supported by "isotope labeling experiments". See Nucleophilic acyl substitution -- Wikipedia.

Edit (26/5/17): For more clarification: (These comments were present below the deleted answer of Berry Holmes)

Are the intermediates given in any mechanism experimentally verified by any of the techniques (though they may be having short life span?)? I will search for it, if you know it, it will be helpful.

I have found one such experiment: http://pubs.acs.org/doi/abs/10.1021/ac60221a003?journalCode=ancham ; I am still seeing the reasoning behind every action of atoms in the reaction using even your answer, I will say you after completion of that process.

Let us consider only nucleophillic addition, i.e. first two steps; nucleophile in the second step is the only product, there may be energy for it due to temperature, this kinetic energy can be used to collide with either of the reactants in the first step. If it collides with nuclephile containing compound :Nu-H, Nu+ in second step may form bond with :Nu-H from by being attached with carbon of second step, this seems to not form the first reactants back again.

If we consider another possibility of collision of second reactant with the acyl compound, the attack of Nu+ of second step goes to oxygen of acyl compound, here I am now not seeing any easy way of the initial products being formed again. Even if energy profile diagram of the above reaction has less energy of activation for the back reaction, which could be easily attainable from the initial given energy for forward reaction, I am not seeing any way of this energy being used to form back the reactants.

There seems to be one possibility for the compound in second step to produce the reactants back. The negative charge in second step of O- might produce tendency for its lone pair to form double bond with carbon at the center, this needs to make -(Nu+-H) group leave the central carbon. We need to now explore on how -(Nu+-H) group is a better leaving group than other groups, and even on the time taken by oxygen's lone pair to form double bond, to know on whether this happens before H of -(Nu+-H) attaches to L.

Image Source: Solomons, Fryhle and Snyder's "Organic Chemistry" [International Student Version (India) 11th edition (Page No. 784)].

• Does anyone know who gave acyl substitution mechanism? I am searching for it, if you know it and can say it, it will be helpful. Knowing the person might help in reading his reasoning for the above mechanism. – Immortal Player May 25 '17 at 15:01
• I am also searching on how reactions in mechanism are tested as either reversible or not. If you know it and can say it, it will be helpful. – Immortal Player May 25 '17 at 15:21
• You can award the bounty now if you wish, by clicking on the +50 button below the up/downvote buttons. I cannot assign a bounty for you. – orthocresol May 26 '17 at 15:38
• @orthocresol: Thank you for replying. It seems that I can only award it after 7 hours, as indicated when I try to award it. I will do it later itself. – Immortal Player May 26 '17 at 16:22

You seem to be confusing a number of different concepts here. Let's try address a few of them.

1) As to the original question, addition-elimination reactions are reversible (indeed all reactions are reversible in some sense, via the principle of microscopic reversibility, it's just a matter of scale), depending on the partners being exchanged. If, for example, you were to add water to an acyl chloride (that is, "L" in your figure is Cl and "Nu" in your figure is OH), the reaction is, for all intents an purposes, irreversible. The resulting Cl is such a bad nucleophile that it's much happier just being a $\ce{Cl-}$.

If, instead, water was being added to a methyl ester (that is, "L" is OMe) the reaction would be reversible (also incredibly slow without a catalyst, but that's a separate issue).

The best way to look at this is to look at the tetrahedral intermediate. It is unstable and it will collapse back down to a carbonyl. It must expel one of the groups bound to the carbonyl carbon in order to do that. If there is an obvious best choice (that is, if there is one path that lowers the total system energy much more) for leaving group (like $\ce{Cl-}$) then that group will always be the one expelled and the reaction really only goes in the direction in which the leaving group has left. If there isn't an obvious choice (like between OH and OMe, where the energy of the system is similar regardless of which one is bound and which is free), either one could be expelled and the reaction is reversible.

It's important to keep in mind that chemicals don't have an agenda. They don't always go in the direction of your reaction arrows. In reality, chemicals try to react in all possible ways but only those ways that are accessible according to the physical realities of thermodynamics are successful. If only one product is thermodynamically favorable, the reaction will appear irreversible. If many products are thermodynamically favorable (and the reaction barriers are low enough), the reaction will appear reversible.

2) Yes, the carbonyl carbon is the most electrophilic site because the more electronegative oxygen is drawing electron density away from that carbon.

3) In addition to evidence from isotope labelling studies, as you mentioned, there is considerable evidence for this mechanism in the form of molecular beam studies, in which a microscopic jet of nucleophiles is shot at different angles at the acyl group.

I'm not sure I understand what you are saying after the paragraph with the ACS paper, but hopefully the above answers your questions.

• Thank you for the answer. I might need to learn QM and particle physics to know exact and precise reasoning on each action of atoms in the reaction. I have now stopped thinking on this. – Immortal Player May 26 '17 at 15:10
• I will come back to this later, after reading physics. – Immortal Player May 26 '17 at 15:18