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The basic strength is determined by the ability of an ion or molecule to accept a proton. How do I know whether RSH is more stable than ROH? (R is an alkyl group)

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    $\begingroup$ My recent answer to this question is quite relevant, with the small difference that one hydrogen gets swapped for an alkyl group. Just remember that stronger acids create weaker conjugate bases. $\endgroup$ – Nicolau Saker Neto Mar 26 '15 at 12:42
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    $\begingroup$ I think it's the other way around. Thiols are more acidic than alcohols so the conjugate base of a thiol is a weaker base than the alcohol conjugate base. $\endgroup$ – RobChem Mar 26 '15 at 15:31
  • $\begingroup$ It isn't. ${}$ $\endgroup$ – orthocresol Jan 6 at 15:23
  • $\begingroup$ It may also be helpful to point out that the strength of a base is only as you define it for a Brønsted-Lowry base. $\endgroup$ – Zhe Jan 6 at 15:35
  • $\begingroup$ You may be confusing "more basic" with "more nucleophilic". Nucleophilicity has a kinetic component, so it tends to be favored by the nucleophile being more polarizable like sulfur would be versus oxygen. $\endgroup$ – Oscar Lanzi Jan 7 at 15:20
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RobChem has already pointed out in his comment that your assumption is not quite correct.

Take a look at $\mathrm{p}K_\mathrm{a}$ values for $\ce{ROH}$ and $\ce{RSH}$ in water from the CRC Handbook of Chemistry and Physics and/or other online sources, such as this or that.

\begin{array}{lrr} \mathbf{R} & {\ce{\mathbf{OH}}} &\ce{\mathbf{SH}}\\ \hline \ce{H} & 15.7 &7.0\\ \ce{Et} & 15.9 & 10.6\\ \ce{(H3C)3C} & 18.0 & 11.7\\ \ce{C6H5} & 9.9 & 6.6\\ \end{array}

In all the cases above, the $\mathrm{p}K_\mathrm{a}$ value for $\ce{RSH}$ is smaller than that of $\ce{ROH}$.

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$\ce{RSH}$ is a better acid than $\ce{ROH}$ (as the $\mathrm{p}K_\mathrm{a}$ of $\ce{RSH}$ is lower than the $\mathrm{p}K_\mathrm{a}$ of $\ce{ROH}$, shown in a previous answer). This means that $\ce{RSH}$ dissociates into $\ce{RS-}$ and $\ce{H+}$ more, i.e. the equilibrium below lies more to the right.

$\ce{RSH + H2O <=> RS- + H3O+}$

To compare between $\ce{ROH}$ and $\ce{RSH}$, we must look at the difference, which is the atom: $\ce{S}$ or $\ce{O}$. As they are in the same group, we look at their polarisability. As $\ce{S}$ is larger and hence more polarisable, it means that the negative electron is more stable in $\ce{RS-}$ than in $\ce{RO-}$. Hence the conjugate base, $\ce{RS-}$ is more stable than $\ce{RO-}$, and so $\ce{RSH}$ is a stronger acid than $\ce{ROH}$.

Conversely, $\ce{RS-}$ is a worse base than $\ce{RO-}$.

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Other answers gave focused on the thermodynamics involved. But if we are focused on nucleophilic activity, which has a kinetic component, then indeed sulfur bases are more nucleophilic.

This source explains that sulfur is less electronegative than oxygen so its electrons are "more available", which I take to mean more polarizable. More polarizability would lead to faster reaction therefore more nucleophilic for the sulfur bases.

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