Acidic Nature of Phenol

Phenol behaves as a distinctly stronger acid than any ordinary alcohol: it reacts with aqueous

sodium hydroxide to form the water-soluble salt sodium phenoxide (an alcohol does not react with

NaOH at all, since an alcohol is too weak an acid for hydroxide to deprotonate it to any

significant extent), and its aqueous solution is measurably, if weakly, acidic. The approximate

$\text{p}K_a$ values make the comparison quantitative: phenol, $\text{p}K_a \approx 10$, against

ethanol, $\text{p}K_a \approx 16$ -- phenol is roughly a million times more acidic than a

comparable simple alcohol.

Why: resonance stabilisation of the phenoxide ion. Acid strength is set by how favourable it is

to lose a proton, which in turn depends heavily on how stable the resulting conjugate base is. When

phenol loses its $-\text{OH}$ proton, the resulting phenoxide ion, $\text{C}_6\text{H}_5\text{O}^-$,

is not simply a localised negative charge sitting on oxygen the way an alkoxide ion,

$\text{RO}^-$, is. Because the oxygen bearing the negative charge is directly attached to the

aromatic ring, one of its lone pairs can delocalise into the ring's $\pi$ system by resonance:

additional resonance structures can be drawn with the negative charge shifted onto the ring carbons

ortho and para to the oxygen (never meta, since only the ortho/para positions are

conjugated through the ring to the oxygen in an alternating-bond resonance pathway). Spreading the

negative charge over four atoms (the oxygen plus three ring carbons) rather than concentrating it

entirely on one oxygen substantially lowers the phenoxide ion's energy relative to a

non-delocalised alkoxide ion of comparable size -- and a more stable conjugate base corresponds

directly to a stronger acid, by straightforward thermodynamics (a more stable products side pulls

the deprotonation equilibrium further to completion).

Why an alkoxide cannot do the same. An ordinary alkoxide ion, $\text{RO}^-$, formed by

deprotonating an alcohol, has no adjacent $\pi$ system for its lone pair to delocalise into -- the

carbon attached to oxygen in an alcohol is $sp^3$ and has no available $\pi$ orbital to accept

electron density by resonance. The negative charge in an alkoxide therefore stays entirely

localised on the one oxygen atom, which is a comparatively high-energy, poorly stabilised state --

hence an alcohol's much higher $\text{p}K_a$ and much weaker acidity than phenol's.

Ring substituents shift the acidity further. Because the phenoxide's stability depends on how

well the ring can accommodate extra negative charge at the ortho/para positions, any substituent

already at an ortho or para position changes phenol's acidity in a predictable direction: an …