# What is the mechanism for this nucleophilic conjugate addition?

What is the mechanism for this reaction-

My working so far:

But my final product doesn't match with the one given. Where did I go wrong (if at all)? How should I have proceeded?

• Firstly, there is no way the -SH proton will survive in ethoxide. You have to take it off first. Apart from that, your starting is probably correct. However, after the initial addition, you probably wouldn't attack the carbonyl with the sulfur as that would form a disgusting four-membered ring with a positive charge on the sulfur that you can't get rid of. At this stage, you have to remember that you still have EtO- in your solution, and there are two ester groups in the molecule. What is likely to happen under these conditions? – orthocresol Nov 3 '15 at 11:42
• @orthocresol will it deprotonate to form a conjugated carbonyl for the S to attack? – justbehappy Nov 3 '15 at 11:52
• If you deprotonate a carbonyl group, you are forming a nucleophilic enolate. The sulfur is also nucleophilic. You need an electrophile.... somewhere..... is there one in the molecule? – orthocresol Nov 3 '15 at 11:56
• Please see my edit- does this happen? @orthocresol – justbehappy Nov 3 '15 at 12:06
• I can only see the deprotonation, did something happen with the upload? Anyway, it is a deprotonation that happens, but there's a subtle difference between the two esters. The other ester has a sulfur next to it, whereas the one that you deprotonated doesn't. Would a sulfur stabilise, or destabilise, a negative charge? – orthocresol Nov 3 '15 at 12:11

To figure out what's going on in an organic reaction, there are a few things you really should look out for:

1. what functional group(s) are present on the molecule(s);
2. what conditions the reaction is being run under;
3. which of the functional group(s) are most reactive under the given reaction conditions.

If you apply these steps to every stage of the reaction, then deducing the mechanism will be easier, and you also save yourself a lot of time by not having to memorise a bunch of mechanisms that you have been previously taught. If those mechanisms were taught by a competent lecturer, you should be able to apply the three steps above to see exactly why those reactions happen.

# The solution (courtesy of @Jan)

What functional groups are present on the molecule?

That's easy.

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What are the conditions?

That's easy too. $\ce{EtO-}$ is strongly basic, but it can also act as a decent nucleophile.

Which of the functional groups are most reactive in base?

Well, to begin with, the thiol is acidic, so it will be deprotonated easily. No question about that.

Now you have a couple of electrophiles - the two esters - as well as a couple of nucleophiles: the thiolate anion, as well as the ethoxide anion. A hard base such as ethoxide is not likely to react in a conjugate addition. Maybe it might react directly with the esters at the carbonyl carbon? But wait! Both esters are ethyl esters $\ce{RCO2Et}$. So, you can use ethoxide to attack those esters all you like, but you won't get anywhere doing it - you'll just get back the same product. In fact, that's exactly why we use ethoxide instead of a different base like methoxide, $\ce{MeO-}$: to prevent transesterification.

So, let's talk about the thiolate anion instead. It's pretty nucleophilic, and you also have an electrophilic α,β-unsaturated carbonyl, so these can react:

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Now, you have a nucleophilic enolate, as well as an electrophilic ester on the other side. They can immediately react with each other to form the product that you are given:

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This reaction has a name - it is called a Claisen condensation, but that's not important! The important part is that you realise that it is just the reaction of a nucleophile with an electrophile.

# An alternative (my original answer)

There might actually be another possible product, though. The enolate that you formed above could pick up a proton from somewhere (either ethanol, or perhaps another thiol molecule) to give the conjugate addition product, which is exactly the same as the one you drew. However, your initial addition product had a positive charge, which is never a good idea in a basic medium. Under basic conditions, you should always deprotonate first and reprotonate at the end of any reaction.

If this were to be the case, you should probably go through the three steps again:

Functional groups?

That's still pretty easy. You have a sulfide and two esters.

$\hspace{45 mm}$

Conditions?

Still basic.

Most reactive functional group?

Well, the sulfide is indeed nucleophilic and you could use it to attack either one of the electrophilic carbonyl groups. But there are a few things that don't really favour this:

1. Sulfur is a pretty soft nucleophile. It might love to react with $\ce{CH3I}$, but probably wouldn't like a carbonyl so much. In fact, this is the reason why the initial step consisted of conjugate addition instead of direct addition.
2. There's no more hydrogen on sulfur that can be taken off, so if sulfur does attack the carbonyl, it will be stuck with a positive charge. This makes the sulfur a pretty good leaving group, which means that any addition to the carbonyl group will be highly reversible.
3. If sulfur attacks the carbonyl group further away from it, it will form a four-membered ring. This is not particularly favourable. If it attacks the carbonyl group nearer to it, it will form a three-membered ring, which is even worse!

So, the sulfur is stuck where it is. You are better off thinking about the reactivity of the carbonyl group in base. You should know that they can be deprotonated to form enolates. Does it matter which carbonyl group is deprotonated? Of course it does. This molecule is not symmetric, so the two carbonyl groups are not equivalent. This obviously leads to the question of which carbonyl group should be deprotonated. Let's look at the possible anions formed from the deprotonation:

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The first of these two is more stable than the other, because the electronegative sulfur atom is capable of withdrawing electron density from the negatively charged carbon. (Actually, there are several theories as to why sulfur stabilises the carbanion, but I won't discuss that here. See this paper, for example.) So, it is that first conjugate base that will be preferentially formed.

At this stage, you have a good nucleophile (the anion) as well as a good electrophile (the carbonyl on the other side). So, what's left is to react them in the same Claisen condensation as above:

$\hspace{25 mm}$

There we go, side product.