Phenols — acidity (resonance stabilization), Kolbe-Schmidt, Reimer-Tiemann, Fries rearrangement: Exercises

Definition Two words we use everywhere on this page

Phenoxide (Ar–O⁻) = the anion you get when phenol loses its $- O H$ proton. "Ar" just means an aromatic ring (here, the benzene ring). So phenol $C_{6} H_{5} OH$ minus $H^{+}$ = phenoxide $C_{6} H_{5} O^{-}$ . This anion is the conjugate base of phenol, and its stability is what the whole acidity story is about.
$K_{a}$ and $p K_{a}$ = when an acid $HA$ sits in water it partly splits: $HA ⇌ H^{+} + A^{-}$ . The acid-dissociation constant $K_{a}$ measures how far it splits: $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}$ . A bigger $K_{a}$ means more splitting = stronger acid. Because $K_{a}$ spans huge ranges, we take a log: $p K_{a} = - lo g_{10} K_{a}$ . The minus sign flips it, so smaller $p K_{a}$ = stronger acid. See Acid Strength and Conjugate Base Stability (pKa).

One idea underneath all of it: an acid is strong exactly when its conjugate base is stable (happy without the H⁺).

Level 1 — Recognition

L1.1

State whether each is more or less acidic than phenol ( $p K_{a} = 10.0$ ), and say why in one line: (a) ethanol, (b) p-nitrophenol, (c) p-cresol (p-methylphenol), (d) acetic acid.

Recall Solution

(a) ethanol — LESS acidic ( $p K_{a} \approx 16$ ). Its conjugate base ethoxide keeps the negative charge stuck on one oxygen; nothing spreads it. Phenoxide spreads charge into the ring by resonance, so phenol wins.
(b) p-nitrophenol — MORE acidic ( $p K_{a} \approx 7.1$ ). $- NO_{2}$ is electron-withdrawing at para; it soaks up the extra negative charge of phenoxide → more stable base → stronger acid.
(c) p-cresol — LESS acidic ( $p K_{a} \approx 10.3$ ). $- CH_{3}$ pushes electrons in (donating), making the already-negative phenoxide even less happy → weaker acid.
(d) acetic acid — MORE acidic ( $p K_{a} \approx 4.76$ ). Its base (carboxylate) spreads charge over two electronegative oxygens; phenoxide only spreads onto carbons, which hate negative charge. Oxygen beats carbon.

L1.2

Match reagent → product functional group: CO₂/pressure, CHCl₃/NaOH, aryl ester + AlCl₃.

Recall Solution

CO₂ + pressure (on sodium phenoxide) → –COOH (Kolbe–Schmidt).
CHCl₃ + NaOH → –CHO (Reimer–Tiemann, via dichlorocarbene).
Aryl ester + AlCl₃ → –COR ketone (Fries rearrangement).

Mnemonic (spell out the product, not a scrambled letter):

Kolbe → Carboxylic acid (has CO₂ built in).
Reimer → aldehyde (from the CH of CHCl₃).
Fries → ketone (the acyl R–CO– lands on the ring).

Level 2 — Application

L2.1

Rank by increasing acidity (least → most): phenol, 2,4-dinitrophenol, p-cresol, p-nitrophenol, picric acid.

Recall Solution

Count electron-withdrawing groups (EWG) at o/p; more EWG = more stable phenoxide = stronger acid. $- CH_{3}$ is donating, so it goes below phenol.

$10.3 p -cresol < 10.0 phenol < 7.1 p -nitrophenol < 4.0 2, 4 -dinitrophenol < 0.4 picric acid$

Why picric acid (2,4,6-trinitrophenol) is so extreme: it carries three nitro groups, all at ortho/para positions. All three drink up the phenoxide's negative charge by resonance at once — the stabilisation stacks. That cumulative through-conjugation drops the $p K_{a}$ to $\approx 0.4$ , as strong as a mineral acid.

L2.2

Sodium phenoxide is treated with CO₂ at ~125 °C under 5 atm, then acidified. Draw/name the product and identify which ring carbon the new group ends up on and why.

Recall Solution

Product: salicylic acid (2-hydroxybenzoic acid) — a $- COOH$ installed ortho to $- OH$ .

Mechanism in brief (an EAS-type attack where the ring is the nucleophile):

Phenoxide's ring is electron-rich; the ortho carbon attacks the electrophilic carbon of $O=C=O$ .
Ortho is favoured because the phenoxide oxygen chelates the $Na^{+}$ , holding the incoming carboxylate right next door.
Rearomatise, tautomerise → sodium salicylate; acidify → salicylic acid.

See Aspirin and Salicylic Acid (applications) — salicylic acid + acetic anhydride → aspirin.

L2.3

Predict the product of phenol + CHCl₃ + aqueous NaOH (heat), then $H^{+}$ . Name the true electrophile, and explain why the aldehyde ends up ortho rather than para.

Recall Solution

Product: salicylaldehyde (2-hydroxybenzaldehyde), a $- CHO$ ortho to $- OH$ (Reimer–Tiemann). True electrophile: dichlorocarbene $: CCl_{2}$ , made in situ: $CHCl_{3} + OH^{-} \to: CCl_{3}^{-} + H_{2} O, : CCl_{3}^{-} \to: CCl_{2} + Cl^{-}$ The electron-rich ortho carbon attacks $: CCl_{2}$ ; the resulting $- CHCl_{2}$ is hydrolysed by NaOH to $- CHO$ . (See Carbenes and Dichlorocarbene.)

Why ortho over para: in the reaction the substrate is the phenoxide ( $- O^{-}$ ). Its alkoxide oxygen sits right next to the ortho carbons and can hydrogen-bond to / hold the approaching electrophile close, steering the $: CCl_{2}$ to the neighbouring ortho position. The finished salicylaldehyde is then further stabilised by an intramolecular H-bond between its $- OH$ and the $- CHO$ oxygen. Both effects favour ortho; a little para product also forms.

Level 3 — Analysis

L3.1

The $p K_{a}$ of p-nitrophenol is 7.1 but m-nitrophenol is 8.4. Both have one nitro group. Explain the 1.3-unit gap using resonance structures. (Study the figure — it draws the two cases side by side.)

Figure — Phenols — acidity (resonance stabilization), Kolbe-Schmidt, Reimer-Tiemann, Fries rearrangement

Figure (para case, left): a curved arrow carries the phenoxide's lone-pair charge around the ring onto the para carbon, where the $- NO_{2}$ sits — its N=O bond opens to soak the charge onto an oxygen (deep-teal path). Right (meta case): the charge (orange dots) lands only on ortho/para carbons, but the $- NO_{2}$ is on a meta carbon that never carries charge, so no arrow can reach it — the plum "×" marks the broken conjugation.

Recall Solution

In the phenoxide, the negative charge can be pushed by resonance onto the ortho and para ring carbons (top row of the figure). Written as structures: $Ar-O^{-} \leftrightarrow O=ring, charge at o \leftrightarrow charge at o^{'} \leftrightarrow charge at p$ If a $- NO_{2}$ sits at para, that very charge-bearing carbon is bonded to it, so an extra resonance structure exists where the charge slides right onto a nitro oxygen: $\dots \leftrightarrow C -_{p} - NO_{2} \leftrightarrow C_{p} = N^{+} (- O^{-}) -O^{-}$ The charge finishes on electronegative oxygen — a big extra stabilisation → $p K_{a} = 7.1$ .

A meta $- NO_{2}$ sits on a carbon the resonance charge never lands on (no arrow reaches it — the "×" in the figure). It can only tug electron density weakly through the sigma framework (inductive). Weak help → less stabilisation → higher $p K_{a} = 8.4$ .

Smaller $p K_{a}$ means more acidic: $Δ p K_{a} = 8.4 - 7.1 = 1.3$ , i.e. p-nitrophenol is $1 0^{1.3} \approx 20$ times more acidic. (See Inductive vs Mesomeric Effects of Substituents.)

L3.2

Phenyl acetate ( $C_{6} H_{5} -O-CO-CH_{3}$ ) is heated with anhydrous AlCl₃ at low temperature, then at high temperature in a separate run. Predict the major product each time and justify.

Recall Solution

This is the Fries rearrangement. AlCl₃ frees an acylium ion, which does Friedel–Crafts acylation on the (now-free) phenol ring.

A note on that ion: we write it $CH_{3} C \equiv O^{+}$ , but that is shorthand for the resonance pair $CH_{3} - C + = O \leftrightarrow CH_{3} - C \equiv O +$ — the real species has the positive charge shared between C and O, not a literal, permanent triple bond. The carbon is the electrophilic (electron-poor) end that the ring attacks.

Low temperature → para-hydroxyacetophenone (less steric strain; the "cold" product).
High temperature → ortho-hydroxyacetophenone (stabilised by intramolecular H-bond between $- OH$ and the ketone $C=O$ ; the "hot" product, and it's steam-volatile so it can be separated).

Hook: Hot = Ortho.

Level 4 — Synthesis

L4.1

Design a synthesis of aspirin (acetylsalicylic acid) starting from benzene. Give reagents for every step and the reason each step works.

Recall Solution

Benzene → benzenesulfonic acid: benzene + fuming $H_{2} SO_{4}$ (or conc. $SO_{3} / H_{2} SO_{4}$ ), warm. This installs a $- SO_{3} H$ handle we can later swap for $- OH$ .
Alkali fusion → sodium phenoxide → phenol: fuse the sulfonic acid with molten $NaOH$ at ~300 °C, then acidify ( $H^{+}$ ) → phenol. (Modern industry instead uses the cumene process: benzene + propene/ $H^{+}$ → cumene, then air $O_{2}$ oxidation and acid cleavage → phenol + acetone. Either route gives phenol; we need the $- OH$ to make the ring electron-rich.)
Phenol → sodium phenoxide: phenol + $NaOH$ . This maximises ring nucleophilicity for the next step.
Kolbe–Schmidt: sodium phenoxide + $CO_{2}$ , ~5 atm, ~125 °C, then $H^{+}$ → salicylic acid (2-hydroxybenzoic acid). The $- COOH$ goes ortho (Na⁺ chelation steers it).
Acetylation: salicylic acid + acetic anhydride (trace $H_{2} SO_{4}$ ) → aspirin — the phenolic $- OH$ is esterified to $- OCOCH_{3}$ , the $- COOH$ is untouched.

Net: benzene → (sulfonate → fuse) → phenol → phenoxide → salicylic acid → aspirin. (Applications: Aspirin and Salicylic Acid (applications).)

L4.2

Starting from phenol, make salicylaldehyde, then propose a rationale for why Reimer–Tiemann gives ortho-CHO and not the meta isomer.

Recall Solution

Making salicylaldehyde: phenol + CHCl₃ + aq. NaOH, heat, then $H^{+}$ (Reimer–Tiemann).

Why ortho (or para), never meta: the reactive site is chosen by where the phenoxide's electron density is highest. Resonance places the negative/partial-negative density on the ortho and para ring carbons (never meta). The electrophile $: CCl_{2}$ therefore bonds only at ortho or para. Ortho dominates in practice partly due to H-bonding stabilisation and directing by the alkoxide. Meta is electronically forbidden — that carbon never carries the reactive lone-pair density.

Level 5 — Mastery

L5.1

You are given four bottles labelled only: phenol, benzoic acid, cyclohexanol, picric acid. Using $p K_{a}$ reasoning and simple tests, distinguish all four. Order them by acidity with numbers.

Recall Solution

Acidity order (increasing $p K_{a}$ = decreasing acidity): $0.4 picric acid > 4.2 benzoic acid > 10.0 phenol > \approx 16 cyclohexanol$ Tests, from a simple solubility ladder:

NaHCO₃ (weak base): only acids stronger than carbonic acid ( $p K_{a} \approx 6.3$ ) dissolve with fizz → picric acid and benzoic acid react; phenol and cyclohexanol don't. This splits the four into two pairs.
Distinguish picric vs benzoic: picric acid is intensely yellow (and stains skin); benzoic acid is white. Also picric is a far stronger acid.
NaOH (strong base): phenol dissolves (forms phenoxide); cyclohexanol essentially does not. This separates the second pair.
Confirm phenol with neutral FeCl₃ → violet colour (phenols give coloured complexes; alcohols don't).

L5.2

Predict and rank the acidity of these three isomers of nitrophenol and give a numeric acidity ratio between the most and least acidic, using their $p K_{a}$ values ( $o = 7.2$ , $m = 8.4$ , $p = 7.1$ ).

Recall Solution

Order (most → least acidic): para ( $7.1$ ) ≳ ortho ( $7.2$ ) ≫ meta ( $8.4$ ).

para and ortho both allow direct resonance of the phenoxide charge into $- NO_{2}$ ; they are close, ortho slightly weaker here due to steric/H-bond effects on the neutral acid.
meta lacks through-conjugation → only inductive help → weakest.

Acidity ratio, most vs least acidic: $\frac{K _{a} ( para )}{K _{a} ( meta )} = 1 0^{p K_{a} (meta) - p K_{a} (para)} = 1 0^{8.4 - 7.1} = 1 0^{1.3} \approx 20.$ The para isomer is about 20 times more acidic than the meta.

Recall One-line self-check

Why smaller $p K_{a}$ = stronger acid ::: $p K_{a} = - lo g_{10} K_{a}$ , where $K_{a}$ is the acid-dissociation constant $\frac{[ H ^{+} ] [ A ^{-} ]}{[ HA ]}$ ; a bigger $K_{a}$ (more dissociation) gives a smaller $p K_{a}$ , so smaller $p K_{a}$ means stronger acid.