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

This page is the drill hall for the parent topic. We will hit every kind of question phenol chemistry can throw at you: acidity comparisons in both directions, degenerate cases (meta substituents that "do almost nothing"), the extreme limit (picric acid), a numeric $p K_{a} \to K_{a}$ conversion, product-prediction for all three named reactions, and one nasty exam twist that mixes them.

Before any example, two reminders that unlock the whole page:

The scenario matrix

Here is the full list of case-classes this topic can test. Each worked example below is tagged with the cell(s) it covers.

#	Case class	What makes it tricky	Covered by
A	EWG at ortho AND para → more acidic	charge is helped by resonance; ortho adds sterics/H-bond	Ex 1
B	EDG at ortho AND para → less acidic	charge is fought	Ex 1
C	Degenerate position: substituent at meta	resonance can't reach → only weak inductive	Ex 2
D	Limiting case: many EWG (picric acid)	acidity pushed to mineral-acid range	Ex 3
E	Numeric: $p K_{a} \Rightarrow K_{a} \Rightarrow$ pH of a solution	plug numbers, watch units	Ex 4
F	Kolbe–Schmidt product & metal/position switch	Na→ortho vs K→para	Ex 5
G	Reimer–Tiemann product & the carbene trap	reactive species is :CCl₂, not CHCl₃	Ex 6
H	Fries temperature switch (o vs p)	hot = ortho, cold = para	Ex 7
I	Word / real-world problem (aspirin route)	translate story → reaction	Ex 8
J	Exam twist: combine reactions + rank several phenols	multi-step reasoning	Ex 9

Example 1 — Cases A & B: EWG vs EDG, at BOTH ortho and para

Worked example Rank by acidity: phenol,

p-nitrophenol, o-nitrophenol, p-cresol (p-methylphenol), o-cresol (o-methylphenol) Which gives up its H⁺ most readily — and does moving the group from para to ortho change things?

Forecast: (guess now) Which is strongest? Which weakest? And do you expect o-nitrophenol to be more or less acidic than p-nitrophenol?

Step 1 — Draw the phenoxide charge map. The figure below shows the six-carbon ring (blue), the $O^{-}$ hanging off the top carbon (yellow), and red dots marking where the negative charge lands — the two ortho carbons (next to O) and the one para carbon (opposite O). The two meta carbons are marked with white ×'s: no charge sits there. Why this step? Acidity is decided by the anion, so we must see where the charge sits before judging any substituent.

Step 2 — Nitro (EWG) at para: $- NO_{2}$ sits ON a red-dot (charged) carbon. By resonance it accepts the minus charge onto its own oxygens → anion more stable → more acidic ( $p$ -nitrophenol $p K_{a} \approx 7.1$ ). Why this step? An EWG at o/p directly drains charge off carbon (bad host) onto electronegative O (great host).

Step 3 — Nitro at ortho: same resonance drain, PLUS an H-bond that slightly opposes it. The ortho carbon is also a red-dot (charge-bearing) site, so resonance withdrawal still works and dominates. But there is a subtlety: in the neutral acid, the –OH hydrogen can form an intramolecular hydrogen bond to the neighbouring nitro oxygen. Because this H-bond stabilises the neutral acid (making it happier to keep its H⁺), it very slightly lowers acidity. Net effect: resonance withdrawal wins, so o-nitrophenol is still a strong acid ( $p K_{a} \approx 7.2$ ) — but its $p K_{a}$ is a touch higher than $p$ -nitro's (7.1), and that tiny gap is exactly the H-bond's fingerprint. Why this step? This is the ortho half of Case A that the para example alone misses: at ortho you must also weigh proximity effects (an H-bond that stabilises the neutral molecule and therefore mildly reduces acidity), not just resonance.

Step 4 — Methyl (EDG) at para and ortho. $- CH_{3}$ donates electron density into the already-negative ring → destabilises the anion → less acidic. p-cresol $p K_{a} \approx 10.3$ ; o-cresol $p K_{a} \approx 10.3$ as well (the small alkyl group's donation is similar from either position). Why this step? This is Case B at both positions: piling more negative charge onto an anion holds the H⁺ tighter, whether the donor is ortho or para.

Step 5 — Order (most → least acidic). $p -nitro (7.1) ≳ o -nitro (7.2) > phenol (10.0) > p -cresol (10.3) \approx o -cresol (10.3)$

Verify: Nitro (EWG) phenols cluster near 7 (both positions strengthen, para just edging out ortho because of the ortho H-bond); phenol 10.0 in the middle; cresols (EDG) near 10.3 (both positions weaken). Numbers and the "EWG down, EDG up" logic agree ✓.

Rank order (strong→weak)

p-nitro (7.1) ≳ o-nitro (7.2) > phenol (10.0) > o-/p-cresol (~10.3)

Why is o-nitrophenol slightly LESS acidic than p-nitrophenol?

The ortho intramolecular H-bond stabilizes the neutral acid, making it hold its H⁺ a little tighter, so its pKa is nudged up (7.2 vs 7.1).

Why does o-CH₃ lower acidity?

A methyl donates electrons into the negative phenoxide from any o/p position, destabilizing the anion.

Example 2 — Case C (degenerate): meta vs para nitrophenol

Worked example Which is more acidic:

m-nitrophenol ( $p K_{a} = 8.4$ ) or p-nitrophenol ( $p K_{a} = 7.1$ )?

Forecast: Same nitro group, same ring — why should position matter at all?

Step 1 — Recall where the phenoxide charge lands: only ortho and para. In the left ring of the figure the nitro group is para (a green arrow shows resonance reaching it); in the right ring the nitro group is meta, sitting on a carbon that carries no phenoxide charge. A meta carbon is never a charge-bearing site in the resonance picture. Why this step? This is the degenerate case: at meta, the nitro group cannot use resonance ("through-conjugation") to grab the charge — there is no negative charge there to grab.

Step 2 — So m-nitro helps only by the inductive effect (pulling electron density through the σ-bonds, a weaker, distance-dependent tug). Why this step? Inductive-only stabilisation is much weaker than resonance + inductive together, so m-nitrophenol is stabilised less.

Step 3 — Compare. p-nitro = resonance + inductive; m-nitro = inductive only. $p -nitrophenol (7.1) > m -nitrophenol (8.4)$ Both are still more acidic than plain phenol (10.0) because even inductive withdrawal helps.

Verify: Ordering by $p K_{a}$ : $7.1 < 8.4 < 10.0$ . The para (full resonance) beats meta (inductive only) beats phenol (nothing). Chemistry and numbers agree ✓.

Example 3 — Case D (limit): picric acid

Worked example Explain why 2,4,6-trinitrophenol (picric acid,

$p K_{a} \approx 0.4$ ) is almost as strong as a mineral acid.

Forecast: How can a phenol, normally $p K_{a} \approx 10$ , reach the strength of dilute HCl?

Step 1 — Locate the three nitro groups: 2, 4, 6 = two ortho + one para. All three sit on charge-bearing carbons. Why this step? This is the limiting case of "stack EWGs on o/p": every single position that carries the phenoxide charge also has a resonance-accepting nitro group.

Step 2 — Add up the stabilisation. Three groups each draining charge onto their O atoms = the negative charge is spread across the ring and six extra nitro-oxygens. Why this step? The more (electronegative) atoms share the charge, the lower the anion energy → the more freely H⁺ leaves.

Step 3 — Compare to the series. Each added o/p nitro drops $p K_{a}$ dramatically: $phenol 10.0 \to 1 NO_{2} 7.1 \to 2 NO_{2} 4.0 \to 3 NO_{2} 0.4$

Verify: The drops are roughly $- 2.9, - 3.1, - 3.6$ per nitro — same order of magnitude, all negative, ending near $0$ (mineral-acid territory). Consistent, cumulative ✓.

Why is picric acid so strong?

Three nitro groups at 2,4,6 (all o/p) each pull the phenoxide charge onto their oxygens by resonance — maximum delocalization.

Example 4 — Case E (numeric): $p K_{a} \to K_{a} \to$ pH

Worked example Phenol has

$p K_{a} = 10.0$ . (a) Find $K_{a}$ . (b) Acetic acid has $p K_{a} = 4.76$ ; how many times stronger (larger $K_{a}$ ) is acetic acid than phenol? (c) Estimate the pH of a $0.10 M$ phenol solution.

Forecast: Roughly — is acetic acid tens, thousands, or hundreds-of-thousands of times stronger? And is a phenol solution acidic like lemon juice, or almost neutral?

Step 1 — Definition. $p K_{a} = - lo g_{10} K_{a}$ , so $K_{a} = 1 0^{- p K_{a}}$ . Why this step? $p K_{a}$ is a compressed (log) scale; to ratio two acids or compute a pH we must undo the log and work with real $K_{a}$ values.

Step 2 — Phenol $K_{a}$ . $K_{a} = 1 0^{- 10.0} = 1.0 \times 1 0^{- 10}$ .

Step 3 — Ratio. Strength ratio $= \frac{K _{a} ( acetic )}{K _{a} ( phenol )} = 1 0^{- 4.76} /1 0^{- 10.0} = 1 0^{(10.0 - 4.76)} = 1 0^{5.24}$ . Why this step? Dividing powers of ten = subtracting exponents; the difference in $p K_{a}$ is the log of the strength ratio.

Step 4 — Evaluate the ratio. $1 0^{5.24} \approx 1.7 \times 1 0^{5}$ . So acetic acid is about 170 000 times stronger than phenol.

Step 5 — pH of $0.10 M$ phenol. For a weak acid, $[H^{+}] \approx K_{a} \cdot C$ (valid because so little dissociates that $C$ barely changes). $[H^{+}] \approx (1.0 \times 1 0^{- 10}) (0.10) = 1.0 \times 1 0^{- 11} = 3.16 \times 1 0^{- 6} M$ $pH = - lo g_{10} (3.16 \times 1 0^{- 6}) \approx 5.5$ Why this step? The question of the matrix is "so what does $K_{a}$ mean for a real beaker?" — the pH turns the abstract constant into a measurable number.

Interpretation: pH ≈ 5.5 is only mildly acidic (pure water is 7). So even at a fairly concentrated 0.10 M, phenol barely acidifies the water — exactly what a weak acid ( $p K_{a} 10$ ) should do. Contrast: 0.10 M acetic acid would give pH ≈ 2.9.

Verify: $K_{a} = 1 0^{- 10}$ ✓; ratio $1 0^{5.24} \approx 1.74 \times 1 0^{5}$ ✓; $[H^{+}] = K_{a} C = 3.16 \times 1 0^{- 6}$ giving pH $\approx 5.5$ (checked in VERIFY). Units: $[H^{+}]$ in mol/L, pH dimensionless ✓.

K_{a}

of phenol (

p K_{a} = 10

)

1.0 \times 1 0^{- 10}

Acetic acid is how many × stronger than phenol?

about

1.7 \times 1 0^{5}

(from

Δ p K_{a} = 5.24

)

pH of 0.10 M phenol

≈ 5.5 (only mildly acidic — it's a weak acid)

Example 5 — Case F: Kolbe–Schmidt with metal switch

Worked example Sodium phenoxide + CO₂ (4–7 atm, 125 °C) then H⁺. (a) Name the product. (b) What changes if we use

potassium phenoxide at higher temperature?

Forecast: Guess the product and its substitution position before reading.

Step 1 — Identify nucleophile & electrophile. Phenoxide ring = electron-rich nucleophile; the carbon of $O=C=O$ = electrophile. Why this step? Every named phenol reaction is "electron-rich ring meets an electrophile" — name both first (Electrophilic Aromatic Substitution).

Step 2 — Position with Na⁺. The phenoxide oxygen chelates $Na^{+}$ , holding the incoming carboxylate at the neighbouring (ortho) carbon → ortho substitution. Why this step? The metal ion is the "steering wheel": small Na⁺ chelates tightly → ortho.

Step 3 — Product (Na). After acid work-up: salicylic acid (2-hydroxybenzoic acid, a –COOH ortho to –OH).

Step 4 — Potassium, hotter. Larger K⁺ chelates poorly and higher temperature favours the less-hindered site → para-hydroxybenzoic acid. Why this step? Change the steering wheel (metal + temperature), change the destination.

Verify: Group introduced = –COOH (CO₂ carries acid-oxygens ✓). Na → ortho (salicylic acid, the aspirin precursor) ✓; K/hot → para ✓. Consistent with parent table.

Kolbe–Schmidt on Na phenoxide gives

salicylic acid (2-hydroxybenzoic acid), –COOH ortho to –OH

What steers ortho vs para in Kolbe–Schmidt?

the metal ion + temperature — Na⁺ chelation → ortho; K⁺/hotter → para

Example 6 — Case G: Reimer–Tiemann and the carbene trap

Worked example Phenol + CHCl₃ + aq. NaOH, heat, then H⁺. (a) Product? (b) A student writes "CHCl₃ adds directly to the ring." Correct them.

Forecast: Which group is installed, –CHO or –COOH? And where?

Step 1 — Base makes the real reagent. NaOH removes CHCl₃'s H → $: CCl_{3}^{-}$ , which loses $Cl^{-}$ → dichlorocarbene $: CCl_{2}$ . Why this step? The actual electrophile is generated in situ; without it there is no reaction (see Carbenes and Dichlorocarbene).

Step 2 — Ring attacks the carbene at ortho. Electron-rich ortho carbon of phenoxide bonds to the electron-deficient $: CCl_{2}$ → an Ar–CHCl₂ (benzal chloride) intermediate.

Step 3 — Hydrolyse. NaOH converts –CHCl₂ → –CHO; acid work-up gives salicylaldehyde (2-hydroxybenzaldehyde).

Step 4 — Correct the student. CHCl₃ alone does nothing to the ring; the reactive species is :CCl₂, and strong base is essential to form it. Why this step? This is the classic trap — the reagent you see is not the reagent that reacts.

Verify: Group introduced = –CHO (CHCl₃'s single C–H → aldehyde ✓, contrast Kolbe's –COOH). Position ortho ✓. Reactive species = dichlorocarbene ✓.

Reimer–Tiemann installs which group, from what species?

–CHO (salicylaldehyde), via dichlorocarbene :CCl₂ generated from CHCl₃ + base

Example 7 — Case H: Fries temperature switch

Worked example Phenyl acetate (

$C_{6} H_{5} -O-CO-CH_{3}$ ) + anhydrous AlCl₃. Give the major product at (a) low temperature, (b) high temperature.

Forecast: Same ester, same Lewis acid — what does temperature decide?

Step 1 — What Fries does. AlCl₃ pops the acyl group $CH_{3} CO^{+}$ off the ester oxygen; it re-attaches to the ring (Friedel–Crafts acylation, Friedel-Crafts Acylation) at o or p of the freed phenol. Why this step? Fries just moves the acyl group from O to a ring carbon — net atoms unchanged.

Step 2 — Low temperature → para. The bulky acyl group goes to the less-crowded para position. Why this step? At low temperature the reaction settles on the least-hindered site; para is far from the –OH so there is no steric clash, making it the kinetically favoured product when we do not drive the system hard.

Step 3 — High temperature → ortho. The ortho product is stabilised by an intramolecular hydrogen bond between the –OH and the ketone C=O. Why this step? At high temperature the more thermodynamically stable (H-bonded) ortho product wins. Memory hook: "Hot = ortho."

Answers: (a) low T → 4′-hydroxyacetophenone (para); (b) high T → 2′-hydroxyacetophenone (ortho).

Verify: Group introduced = –CO–CH₃ (a ketone, not acid/aldehyde ✓). Starts from an ester ✓. Hot→ortho, cold→para ✓ (matches parent).

Fries: hot vs cold selectivity?

Hot → ortho (H-bond stabilized), cold → para (less hindered)

Fries must start from a

phenolic (aryl) ester — it moves the acyl group from O onto the ring

Example 8 — Case I (word problem): the aspirin route

Worked example A factory has phenol and wants to make

aspirin. Outline the two-step route and identify each reagent's job.

Forecast: Which named reaction gets you the –COOH, and what turns –OH into the acetyl of aspirin?

Step 1 — Phenol → sodium phenoxide → salicylic acid by Kolbe–Schmidt (CO₂, pressure, then H⁺). Why this step? Aspirin is acetylsalicylic acid, so first we need salicylic acid — the –COOH ortho to –OH (see Aspirin and Salicylic Acid (applications)).

Step 2 — Salicylic acid + acetic anhydride → aspirin (acetylsalicylic acid). Why this step? Acetic anhydride acetylates the phenolic –OH (esterification), converting the irritant –OH into an acetate ester; the –COOH stays as the acid group.

Answer: phenol $Kolbe-Schmidt$ salicylic acid $(CH_{3} CO)_{2} O$ aspirin.

Verify: Aspirin keeps salicylic acid's –COOH and swaps –OH → –O–CO–CH₃. Atoms accounted for; route uses exactly one named phenol reaction (Kolbe–Schmidt) + one esterification ✓.

Two-step aspirin route from phenol?

Kolbe–Schmidt (CO₂, pressure, H⁺) → salicylic acid; then acetic anhydride acetylates the –OH → aspirin.

Example 9 — Case J (exam twist): rank a mixed set

Worked example Rank these by increasing

$p K_{a}$ (i.e. list the strongest acid first, weakest last): ethanol, phenol, p-nitrophenol, picric acid, acetic acid.

Forecast: Put them in order in your head before checking — where does acetic acid fall relative to p-nitrophenol?

Step 1 — Bucket each by its anion type (where does the charge live?).

Ethanol → ethoxide: charge stuck on one O, no delocalisation.
Phenol → phenoxide: charge into the ring, sitting mostly on carbons (poor hosts for negative charge).
p-nitrophenol → its phenoxide: ring delocalisation + the para nitro drains charge onto its oxygens.
Acetic acid → acetate: charge shared equally over two equivalent O's (electronegative, love the charge).
Picric acid → its phenoxide: three o/p nitros drain charge onto six nitro-oxygens — maximum spread. Why this step? Acidity tracks anion stability, and stability tracks "how well is the charge spread, and onto O (great) or onto C (poor)?" Sorting into buckets makes the ranking mechanical (Acid Strength and Conjugate Base Stability (pKa)).

Step 2 — Order the buckets from best-stabilised anion (strongest acid) to worst (weakest acid). Best spread → worst: picric (6 O's) > acetate (2 equal O's) > p-nitrophenoxide (ring C's + nitro O's) > phenoxide (ring C's only) > ethoxide (one O, localized). Why this step? More electronegative atoms sharing the charge = lower anion energy = the H⁺ leaves more freely = stronger acid = lower $p K_{a}$ .

Step 3 — Attach the numbers and write increasing $p K_{a}$ . $0.4 picric < 4.76 acetic < 7.1 p -nitrophenol < 10.0 phenol < 16.0 ethanol$

Step 4 — Spotlight the twist. Notice acetic acid beats p-nitrophenol even though p-nitrophenol also has resonance. The reason: carboxylate shares its charge over two equivalent oxygens, whereas p-nitrophenoxide still dumps a big share on ring carbons — oxygen is the better host, so the carboxylate anion is more stable. Why this step? This is the trap the examiner sets: "has resonance" is not enough; what atoms hold the charge decides who wins.

Verify: Strictly increasing $p K_{a}$ : $0.4 < 4.76 < 7.1 < 10 < 16$ ✓. Bucketing (O-sharing > C-sharing > localized) matches the numeric order, and the acetic-vs-p-nitrophenol twist is confirmed by $4.76 < 7.1$ ✓.

Increasing-

p K_{a}

order of the mixed set

picric (0.4) < acetic (4.76) < p-nitrophenol (7.1) < phenol (10) < ethanol (16)

Which is the stronger acid, acetic acid or p-nitrophenol, and why?

Acetic acid — acetate spreads charge over two equivalent electronegative oxygens, better than phenoxide dumping charge on ring carbons.

Recall One-line recap of the whole matrix

Acidity lever = "can the phenoxide's minus charge spread onto oxygen?" EWG at o/p = yes (stronger), EDG or meta = no/weak (weaker), stack three nitros = mineral-acid strength. Reactions: Kolbe→–COOH(ortho, Na), Reimer–Tiemann→–CHO(via :CCl₂), Fries→ketone (hot=ortho, from an ester).