0
$\begingroup$

They have slightly different $\mathrm{p}K_\mathrm{a}$ values of 9.9 (phenol) vs 10.2 (p-methoxyphenol). They both have the same number of resonance structures too.

My guess is that the resonance structure for p-methoxyphenol with the lone pair on the para carbon contributes less to the stabilisation of the negative charge since it's in an electron rich area (near the lone pairs of the methoxy group oxygen).

Is this a valid explanation? If not, what is the true reason?

$\endgroup$
2
$\begingroup$

Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be.

You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of course). This has the effect of making the ring richer in electrons. It will thus tend to less receive additional electrons, and thus stabilise less the formal negative charge. (However, keep in mind that the ring does still stabilise the negative charge, making p-methoxyphenol still much more acidic than alcohols with pKa values around 15)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.