Is RS- more basic than RO-?

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)




**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](http://www.diederich.chem.ethz.ch/teach/pKa_Table.pdf) or [that](http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf).


\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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