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

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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?

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    $\begingroup$ Maybe because the negative charge is more stable on $\ce{S}$ compared to $\ce{O}$. $\endgroup$ – Apoorv Potnis Dec 27 '17 at 6:07
  • $\begingroup$ 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. $\endgroup$ – Ritwik Das Dec 27 '17 at 6:09
  • $\begingroup$ Sulfur makes a better leaving group than oxygen. That makes nucleophilic attack faster at the sulfur-bound site. $\endgroup$ – Oscar Lanzi Dec 27 '17 at 14:28
  • $\begingroup$ @OscarLanzi Yeah this makes sense. $\endgroup$ – Ritwik Das Dec 27 '17 at 16:15
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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:

reaction mechanism

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.

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  • $\begingroup$ @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-. $\endgroup$ – orthocresol May 13 '18 at 17:33

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