4.3.5 · D5Halides and Oxygenated Derivatives
Question bank — Phenols — acidity (resonance stabilization), Kolbe-Schmidt, Reimer-Tiemann, Fries rearrangement
Look at the ring above: the blue clouds mark ortho and para — the only carbons that carry the resonance-borne negative charge. An EWG placed there can drink that charge (through-conjugation, curved arrows). An EWG at meta (grey) sits off the pi-pathway and can only tug weakly through sigma-bonds (inductive).
True or false — justify
True or false: Phenol is acidic because the O–H bond is unusually weak.
False. Acidity is about conjugate-base stability, not bond weakness — phenoxide spreads its charge into the ring by resonance, so losing H⁺ is easy. (See Acid Strength and Conjugate Base Stability (pKa).)
True or false: Phenol is a stronger acid than acetic acid.
False. Phenoxide delocalizes charge onto carbon atoms (poor at holding negative charge), while carboxylate shares it between two electronegative oxygens — so acetic acid (, lower = stronger) beats phenol ().
True or false: m-nitrophenol is roughly as acidic as p-nitrophenol.
False. The para nitro accepts negative charge by through-conjugation (resonance reaches ortho/para — see the blue clouds in the figure above); the meta nitro is off that pi-pathway and only helps by weak inductive pull, so p-nitrophenol is much more acidic.
True or false: A methyl group on the ring makes phenol more acidic.
False. is electron-donating (an EDG); it pushes electron density toward the already-negative phenoxide, destabilizing it, so p-cresol () is less acidic than phenol.
True or false: Picric acid (2,4,6-trinitrophenol) is strong enough to rival a mineral acid.
True. Three EWG (electron-withdrawing) nitro groups at o/p positions each drink up negative charge by resonance and induction, crashing the to about .
True or false: In Kolbe–Schmidt, phenol itself (not phenoxide) attacks CO₂.
False. You must first make sodium phenoxide with NaOH; the extra negative charge is what makes the ring nucleophilic enough to attack CO₂. (See the mechanism sketch below.)
True or false: The Reimer–Tiemann reaction adds CHCl₃ directly to the ring.
False. The true electrophile is dichlorocarbene generated in situ by base; CHCl₃ without strong base does nothing (see Carbenes and Dichlorocarbene).
True or false: Fries rearrangement introduces a brand-new acyl group from outside.
False. The acyl group is already present as the phenolic ester (ArO–COR); Fries just moves it from oxygen onto the ring carbon — atom count unchanged.
True or false: Kolbe–Schmidt and Reimer–Tiemann install the same functional group.
False. Kolbe–Schmidt gives a carboxylic acid (–COOH); Reimer–Tiemann gives an aldehyde (–CHO).
Left to right: Kolbe–Schmidt (electrophile = CO₂ → –COOH), Reimer–Tiemann (electrophile = :CCl₂ carbene → –CHO), Fries (electrophile = acylium from an ester → –COR ketone). The red arrow always points at the ortho carbon that attacks.
Spot the error
"Ethoxide and phenoxide both delocalize charge, phenol is just a bit better." — what's wrong?
Ethoxide has no delocalization; its charge is stuck on one oxygen. Only phenoxide spreads charge, and that difference (not a "bit better") is the whole reason phenol is ~× more acidic.
"Fries rearrangement is just Friedel–Crafts acylation done directly on phenol." — fix it.
Fries starts from the phenyl ester, not free phenol. AlCl₃ activates the ester, ejects an acylium, then that does the acylation intramolecularly/intermolecularly (see Friedel-Crafts Acylation).
"A nitro group anywhere on the ring acidifies phenol equally by resonance." — fix it.
Only ortho/para nitro groups sit on carbons that carry the resonance-borne negative charge; a meta nitro is off the conjugation path and helps only weakly by induction.
"CHCl₃ has three chlorines, so Reimer–Tiemann product has extra chlorine." — fix it.
The chlorines leave: the ortho carbon first gets a –CHCl₂ (benzal chloride) intermediate, which hydrolyzes to –CHO. No chlorine remains in salicylaldehyde.
"Since salicylic acid has both –OH and –COOH, aspirin acetylates the –COOH." — fix it.
Acetic anhydride acetylates the phenolic –OH (giving acetylsalicylic acid); the –COOH stays free (see Aspirin and Salicylic Acid (applications)).
"Dichlorocarbene is a nucleophile because it has a lone pair." — fix it.
has a lone pair and an empty p-orbital; the empty orbital makes it electron-deficient and electrophilic, which is why the electron-rich phenoxide ring attacks it.
Why questions
Why does Na⁺-phenoxide give ortho Kolbe–Schmidt product but K⁺ at high temp gives para?
Na⁺ chelates the phenoxide oxygen and the incoming carboxylate together, holding CO₂ next door at the ortho position; the larger K⁺ chelates poorly, and higher temperature favours the less-hindered para product.
Why does the Fries rearrangement give the ortho ketone preferentially when run hot?
The ortho hydroxy-ketone is stabilized by intramolecular hydrogen bonding (–OH to C=O), making it the thermodynamic favourite that dominates at high temperature ("hot = ortho").
Why is phenol so reactive in Electrophilic Aromatic Substitution compared to benzene?
The oxygen lone pair donates into the ring by resonance, raising electron density especially at ortho/para carbons, so electrophiles are attracted and the intermediate is stabilized — the ring is "electron-rich."
Why do EWG (electron-withdrawing) groups make phenol more acidic but deactivate the ring toward EAS?
Both stem from the same fact — an EWG pulls electron density away. That stabilizes the electron-rich phenoxide (more acidic) but starves the neutral ring of the electrons EAS needs (less reactive). See Inductive vs Mesomeric Effects of Substituents.
Why can't ethanol's charge be spread out the way phenol's can?
Ethoxide's oxygen is attached to an sp³ carbon with no adjacent π-system; there is no conjugated network to receive the charge, so it stays localized.
Why must strong base (NaOH) be present in Reimer–Tiemann, not just heat?
Base does two jobs: it deprotonates CHCl₃ to make the carbene, and it generates the phenoxide nucleophile. Without base neither reactive species forms.
Edge cases
What happens to Kolbe–Schmidt if you forget the acidic work-up?
You stop at sodium salicylate (the carboxylate salt); the free salicylic acid only appears after adding H⁺ to protonate –COO⁻.
If a phenol has both ortho positions blocked, what limits Reimer–Tiemann and Kolbe–Schmidt?
With both ortho spots occupied, substitution must go para (or fail if para is also blocked), since the electron-rich sites are exactly ortho/para.
Edge case in acidity: why is o-nitrophenol a weaker acid than p-nitrophenol despite both being ortho/para EWG?
In o-nitrophenol the –OH and the neighbouring –NO₂ form an intramolecular hydrogen bond; this ties up the O–H and stabilizes the neutral acid, so it lets go of H⁺ less readily — pushing its (≈7.2) slightly above p-nitrophenol (≈7.1). The intramolecular H-bond also explains why o-nitrophenol is steam-volatile while the para isomer (which H-bonds between molecules) is not.
Same edge case, contrast: how does the meta nitrophenol sit in this acidity ranking?
m-nitrophenol () is less acidic than both o- and p- isomers because its nitro group is off the resonance path and only pulls charge by weak induction — no through-conjugation, no intramolecular H-bond helping either way.
What is the acidity trend for the fully substituted case: does adding a fourth nitro keep helping?
Effects taper — after 2,4,6-positions are filled, remaining sites are meta (weak, inductive-only help) and severe steric/electronic saturation means each added EWG contributes diminishing acidification.
In Fries, what if no ortho/para position is available on the ester's ring?
The acylium has nowhere to land by o/p acylation, so rearrangement fails or gives only trace meta/side products — Fries relies on the –OH acting as an o/p director.
Degenerate case: is "phenol" with the –OH replaced by –O⁻ (phenoxide) still a discrete acid?
Phenoxide is the conjugate base — it has already lost H⁺, so it is basic/nucleophilic, not the acid; the acid is phenol itself.
Recall One-line self-test before you leave
Acidity = conjugate-base stability ::: Every phenol trap on this page reduces to "how well is the negative charge on phenoxide spread out (resonance) and pulled on (EWG at o/p)?"