# Among thioester and ester which has more electrophilic carbonyl carbon?

In few reactions I’ve seen that nucleophiles preferably attack thioester’s carbonyl carbon instead of ester’s carbonyl carbon

According to me, the initiation of reaction is decided by the electrophilicity of the carbonyl compound, in esters O being more electronegative makes c=o more electrophilic, thus attack should have been at ester, but many reactions suggest that attack first occurs at thioester.

What makes thioester’s carbonyl carbon more electrophilic?

• Maybe because the negative charge is more stable on $\ce{S}$ compared to $\ce{O}$. – Apoorv Potnis Dec 27 '17 at 6:07
• Yes maybe that's the correct reason, but the initiation of reaction is decided by the electrophilicity of the carbonyl compound, in esters O being more electronegative makes c=o more electrophilic, thus attack should have been at ester, but many reactions suggest that attack first occurs at thioester. – Ritwik Das Dec 27 '17 at 6:09
• Sulfur makes a better leaving group than oxygen. That makes nucleophilic attack faster at the sulfur-bound site. – Oscar Lanzi Dec 27 '17 at 14:28
• @OscarLanzi Yeah this makes sense. – Ritwik Das Dec 27 '17 at 16:15

Reactivity of thioesters with respect to nucleophilic attack

Why is the thioester bond weaker than a regular ester bond?

Let's consider a reaction mechanism like this:

A steady-state analysis will show that the rate of formation of product is

$$r = \frac{k_1k_2[\text{ester}][\text{Nu}]}{k_{-1} + k_2}$$

• $k_1$ for the thioester is larger, because the sulfur 3p lone pair overlaps poorly with the C=O π* orbital (which is formed from 2p orbitals). Recall that esters are less electrophilic than ketones because there is an ester oxygen which has a lone pair capable of overlapping with the C=O π*. In this case, since this overlap is diminished, it turns out that thioesters are typically roughly as reactive towards nucleophilic attack as ketones.

• $k_2$ for the thioester is also larger because thiolates are better leaving groups than alkoxides.

Depending on the relative rates of $k_2$ and $k_{-1}$, though, the second point may or may not be relevant. For example if $k_2 \gg k_{-1}$, then the rate equation simplifies to $r \approx k_1[\text{ester}][\text{Nu}]$, in which case the comparison of $k_2$ is no longer useful.

• @ΑντώνιοςΚελεσίδης leaving group ability is quite thoroughly discussed on the internet and you will be able to find more info than I can give you in a comment. Broadly speaking the rule of thumb is the stronger the acid HX the better the leaving group X-. – orthocresol May 13 '18 at 17:33