Is RS- more basic than RO-?

Question

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)

Answer 1

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 $\ce{ROH}$.

Answer 2

$\ce{RSH}$ 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 $\ce{RO-}$.