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Deciding more electrophilic centre between ester and thioester for SN type reaction
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Karen J Gray
Deciding more electrophilic centre between ester and thioester for SN type reaction
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.
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)?More
Since the ovelap with $\ce {O}$ is very good, its resonance hybrids 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.More
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.
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.
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