Reactions of Phenols

Why phenol is acidic at all. Phenol ionises weakly in water, $\text{C}_6\text{H}_5\text{OH} \rightleftharpoons \text{C}_6\text{H}_5\text{O}^- + \text{H}^+$, and this small equilibrium extent is captured by an acid dissociation constant:

where: $K_a$ is the acid dissociation constant, square brackets denote equilibrium molar concentrations, and $pK_a$ is the standard logarithmic scale on which a smaller $pK_a$ means a stronger acid. Phenol's $pK_a \approx 10$ is dramatically lower than ethanol's $pK_a \approx 16$, meaning phenol is acidic enough to react with NaOH itself — alcohols never do this.

Why phenol beats every alcohol. The gap has nothing to do with the O–H bond strength and everything to do with what happens after the proton leaves. In the phenoxide ion the negative charge on oxygen is not stuck in place — it delocalises into the ring through resonance, spreading partial negative charge onto the ortho and para carbons. An alkoxide ion, $\text{RO}^-$, has no ring to delocalise into; worse, the electron-donating alkyl group actually intensifies the charge on oxygen, destabilising it further. A more stable anion means a more favourable equilibrium toward ionisation, hence a stronger acid.

Substituents tune the acidity. Because the effect is about stabilising the phenoxide anion, anything that pulls electron density away from the ring helps. Electron-withdrawing groups (–NO₂ above all, especially at ortho/*para*) increase acidity by spreading the negative charge even further through resonance and induction. Electron-donating groups (–CH₃, –OCH₃) push electron density back onto an already negative oxygen and decrease acidity.

The same lone pair drives ring substitution. The –OH oxygen lone pair that stabilises the anion also feeds into the neutral ring, making phenol a strong activator and an ortho/*para* director for electrophilic aromatic substitution — electron density piles up specifically at those positions, which is exactly where each named reaction below attacks.

Rank ethanol ($pK_a \approx 16$), phenol ($pK_a \approx 10$) and 4-nitrophenol ($pK_a \approx 7.2$) in order of increasing acidity, and find how many times more acidic 4-nitrophenol is than phenol.

Step 1. Order by $pK_a$: since a lower $pK_a$ means a stronger acid, the order of increasing acidity follows directly from decreasing $pK_a$: ethanol (16) then phenol (10) then 4-nitrophenol (7.2), giving ethanol < phenol < 4-nitrophenol — the same order the resonance argument predicts, because the para –NO₂ group pulls even more electron density off the phenoxide oxygen by resonance and induction than the ring alone does.

Step 2. Convert $pK_a$ to $K_a$ for the two acids being compared, using $K_a = 10^{-pK_a}$: $K_a(\text{phenol}) = 10^{-10}$ and $K_a(\text{4-nitrophenol}) = 10^{-7.2}$.

Step 3. Take the ratio: $\dfrac{K_a(\text{4-nitrophenol})}{K_a(\text{phenol})} = \dfrac{10^{-7.2}}{10^{-10}} = 10^{10-7.2} = 10^{2.8} \approx 630$.

Step 4. Combine with the earlier phenol-versus-ethanol ratio of $10^6$ (from the derivation above) to see the full span: 4-nitrophenol is roughly $10^{2.8}\times 10^{6} = 10^{8.8} \approx 6\times10^{8}$ times more acidic than ethanol. Answer: ethanol < phenol < 4-nitrophenol, with 4-nitrophenol about 630 times more acidic than phenol and phenol itself about a million times more acidic than ethanol.

Competitive stretch

Starting from phenol, outline the Reimer–Tiemann route to salicylaldehyde and name the reactive intermediate.

Step 1. Treat phenol with aqueous NaOH to form sodium phenoxide, which is far more electron-rich than neutral phenol.

Step 2. Add chloroform and excess concentrated NaOH; the base deprotonates $\text{CHCl}_3$ and drives α-elimination of $\text{Cl}^-$, generating dichlorocarbene, $:\text{CCl}_2$, the reactive electrophile.

Step 3. The electron-rich phenoxide ring attacks dichlorocarbene preferentially at the ortho position, and the intermediate hydrolyses under the basic aqueous conditions to an aldehyde group.

Step 4. Acidification protonates the phenoxide back to a phenol ring, giving the final product. Answer: 2-hydroxybenzaldehyde (salicylaldehyde), via the electrophile dichlorocarbene.

• Assuming phenol only reacts with metals like an alcohol — phenol is acidic enough to be neutralised by NaOH itself (sodium phenoxide plus water), a reaction alcohols never undergo.
• Reversing the substituent-acidity rule — electron-withdrawing groups (like –NO₂) increase phenol's acidity while electron-donating groups decrease it, the opposite of how such groups affect amine basicity, and the two get mixed up often.
• Writing bromination of phenol as needing a Lewis-acid catalyst the way benzene does — phenol is so activated that bromine water reacts immediately, no catalyst required, giving a white precipitate of 2,4,6-tribromophenol.
• Defaulting to the para position for Kolbe's reaction or the Reimer–Tiemann reaction — both are taught and tested primarily for the ortho product (salicylic acid, salicylaldehyde).
• Forgetting that a smaller $pK_a$ means a stronger acid — students sometimes read the numbers backwards when ranking acidity.

Ask the tutor: want help drawing the full set of resonance structures behind the phenoxide ion's stability, or working through why the Reimer-Tiemann mechanism favours ortho over para attack?