# Deciding more electrophilic centre between ester and thioester for SN type reaction

My question: (though it won't alter the product)

Which carbon is attacked first, the one adjacent to sulphur or the one adjacent to oxygen?

The solution says that the carbon adjacent to sulphur is attacked first. I don't understand why.

According to me, sulphur is less electron withdrawing than oxygen (electronegativity decreases down the group), so the carbon adjacent to oxygen must be more electrophilic.

Could someone please explain as to why attack should first occur at the carbon adjacent to the sulphur atom? (or is the solution incorrect?)

P.S. Here, due to syn stereochemistry (NGP is not favourable) choice of electrophilic centre undergoing attack first doesn't influence the product at all. However, in the case where the sulphur and oxygen atoms are anti with respect to the ring, and NGP is a possibility, and then the products would greatly differ (depending on whether attack first happens on carbon adjacent to oxygen or sulphur)

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q35 in Chapter Alcohols, Ether, and Epoxides

• There are several questions to ask here. The first is why does it matter? The more technical answer involves first asking what the reaction mechanism is. Cleavage of both the thioester and the ester involve addition of hydroxide to the carbonyl and collapse of the tetrahedral intermediate. What step is the rate determining step? Your answer here impacts what factors could be relevant to the this question. – Zhe Feb 22 '18 at 15:30

## 1 Answer

There is also another effect of delocalisation of the electron pair on $\ce{O}$ and $\ce{S}$ atom adjacent to the cabonyl carbon, which is dominant over the electron-withdrawing effect ($\ce{-I }$ effect).

In both the parts i.e. $\ce{OCOCH_3}$ and $\ce{SCOCH_3}$, there are two lone pairs on each of $\ce{O}$ and $\ce{S}$, which can delocalise with the carbonyl carbon. Thus, the bond between $\ce{O}$ and $\ce{C}$, and $\ce{S}$ and $\ce{C}$, will get a partial double bond character. Due to this delocalisation, the electrophilicity of the carbonyl carbon is decreased.

When the delocalisation happens in case of $\ce{O}$, the overlap between $\ce{O}$ and $\ce{C}$ is $\ce{2p$\pi$- 2p$\pi$}$ overlap, which is a strong overlap and thus the double bond character between $\ce{O}$ and $\ce{C}$ is more. Thus, electrophilicity is reduced to a great extent. But in case of $\ce{S}$, there is a relatively weaker $\ce{3p$\pi$-2p$\pi$}$ overlap which introduces very feeble double bond character between them so, the electrophilicity is reduced to very less extent and thus the first electrophilic attack will happen at the $\ce{C}$ adjacent to $\ce{S}$. This is the reason to your question.

• While I agree with your answer, how would you account for the fact that, since the $\ce{S}$ atom is less electronegative than the $\ce{O}$ atom, it should should donate its lone pair to a greater extent than oxygen (contrary to your answer)? – Gaurang Tandon Feb 22 '18 at 15:22
• Since the ovelap with $\ce {O}$ is very good, its resonance hybrid's stabilisation energy will be very high and thus a very large fraction of the molecules of compound will undergo that stabilisation because by that the overall $\Delta$H will be negative even if we consider the low positive $\Delta$H of the lone pair of oxygen undergoing delocalisation. – Soumik Das Feb 22 '18 at 15:36