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I'm looking at a problem where sodium methanethiolate and sodium methoxide, both in equal amounts, are being reacted with $\ce{CH3I}$ and the solvent is ethanol.

I understand that sodium methanethiolate is a better nucleophile than sodium methoxide when the two are reacted in equal amounts with something like methyl iodide in a polar aprotic solvent. But why is sodium methanethiolate still a better nucleophile than sodium methoxide when the solvent is polar protic, like ethanol?

I thought the ethanol would reduce the nucleophilicity of both, but would do so at by different amounts, given the different electronegativity of sulfur and oxygen. If ethanol reduces the nucleophilicity of both at the same rate, why even use ethanol as a solvent in this case?

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Thiolates are generally more nucleophilic than alcoholates. So whenever the choice happens to be between a nucleophilic sulphur and a nucleophilic oxygen, the sulphur will win.

Both the methanolate and the methanethiolate will show reduced nucleophilicity due to partial protonation, though. And of course you have introduced ethanolate as an additional nucleophile, too (which should be the least reactive, though).

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Nucleophiles can be hard or soft (as can electrophiles); a hard nucleophile tends to react preferentially with a hard electrophile, while a soft nucleophile tends to react preferentially with a soft electrophile.

In your system the methanethiolate anion, $\ce{CH3S-}$, is a soft nucleophile while methoxide, $\ce{CH3O-}$, is a relatively hard nucleophile. You will also have ethoxide ($\ce{CH3CH2O-}$) in your system. As to the electrophile, methyl iodide, $\ce{CH3I}$: it is soft and hence it preferentially reacts with the softer nucleophile in the system, methanethiolate.

A more sophisticated explanation of the hard/soft idea can be found in frontier orbital analysis.

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    $\begingroup$ Hard and soft nucleophiles and electrophiles very badly describe this case and shouldn’t be pulled into this argument. Remember that the HSAB concept originally intended to explain the formation of stable adducts of Lewis acids and bases. $\endgroup$ – Jan Jun 17 '15 at 16:05
  • $\begingroup$ @Jan thanks for the response but I disagree. I will try and find some literature to support my argument and amend my answer. $\endgroup$ – user1945827 Jun 17 '15 at 16:44

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