# Why 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)

• 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. – Nicolau Saker Neto Mar 26 '15 at 12:42
• 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. – RobChem Mar 26 '15 at 15:31
• It isn't. ${}$ – orthocresol Jan 6 at 15:23
• 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. – Zhe Jan 6 at 15:35
• 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. – Oscar Lanzi Jan 7 at 15:20

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 $\ce{ROH}$.

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

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.