Alcohols, Phenols and Ethers · Lesson 9 of 12
Carbon–Carbon Bond Formation and Other Phenol Reactions
“Connect phenoxide activation with carboxylation and formylation, then distinguish reduction of C–O from quinone-forming oxidation.”
• Explain how phenoxide reacts with weak electrophiles. • Write Kolbe–Schmitt and Reimer–Tiemann transformations with their work-ups. • Distinguish salicylic acid from salicylaldehyde structurally. • Predict zinc-dust reduction and phenol oxidation products.
Phenoxide is an activated aromatic nucleophile
A phenol molecule contains a weakly acidic proton and an activated aromatic ring. Removing that proton with NaOH gives phenoxide, whose oxygen can donate electron density into the ring even more strongly. Under suitable conditions the electron-rich ring can make a new C–C bond with a weak carbon electrophile. The base therefore changes more than solubility: it generates a species with a different reactivity. The subsequent acid work-up returns ionic products to their neutral forms.
Two useful reactions introduce a single carbon at a ring position next to OH. Carbon dioxide provides the carboxyl carbon in the Kolbe–Schmitt reaction. Chloroform provides the formyl carbon in the Reimer–Tiemann reaction. Both predominantly give ortho products under the stated conventional conditions, but the functional groups formed are different. Distinguishing COOH from CHO is essential for predicting subsequent reactions.
Kolbe–Schmitt carboxylation gives salicylic acid
First form sodium phenoxide with NaOH. Heat dry sodium phenoxide with CO₂ under pressure, commonly described near 400 K and several atmospheres, then acidify. Carbon dioxide has an electrophilic carbon because oxygen draws electron density away from it. The activated phenoxide ring undergoes carboxylation, predominantly at an ortho position under the sodium-salt conditions. The product salt is sodium salicylate; acidification gives 2-hydroxybenzoic acid, commonly salicylic acid.
The net salt formula looks as if the original phenoxide has already regained an OH proton while the carboxyl group is ionic. Proton transfers during the sequence produce this arrangement; the balanced salt-to-salt equation conserves all atoms. In more strongly basic mixtures the oxygen protonation state can differ. The named isolated acid must be assigned only after the acidification step, not merely because CO₂ has entered the reaction.
Problem
Why does phenol become a seven-carbon acid rather than a six-carbon acid in Kolbe–Schmitt carboxylation?
- 1.The original ring contains six carbons and remains intact. CO₂ supplies one new carbon that becomes the carbonyl carbon of –COOH.
- 2.The product o-HOC₆H₄COOH therefore contains seven carbons. The original oxygen stays attached to its ring carbon as phenolic OH.
- 3.It is not benzoic acid without OH, and it is not a carbonate ester made only at oxygen. The transformation being taught is ring C–C bond formation followed by work-up.
Reimer–Tiemann formylation gives salicylaldehyde
Phenol treated with CHCl₃ and aqueous NaOH under heated alkaline conditions gives predominantly 2-hydroxybenzaldehyde after work-up. This is salicylaldehyde, o-HOC₆H₄CHO. Base both forms phenoxide and generates a reactive one-carbon species from chloroform. Deprotonation of CHCl₃ gives CCl₃⁻, which can lose Cl⁻ to produce dichlorocarbene, :CCl₂. This neutral electron-deficient carbon species reacts with the activated ring; it is not a positively charged free CHO⁺ ion.
Ring reaction and rearomatisation lead to an ortho-dichloromethyl-substituted phenoxide arrangement. Hydrolysis of its –CHCl₂ group converts that carbon into –CHO. A later acid work-up protonates the phenoxide to salicylaldehyde. The intermediate must contain a ring–CHCl₂ bond; drawing two chlorines on the original OH-bearing ring carbon would change the mechanism and the final connectivity.
The new aldehyde carbon is supplied by chloroform. All three chlorine atoms end in chloride in the overall balance. Salicylaldehyde and salicylic acid both have seven carbons and an ortho OH, but salicylaldehyde has CHO and two oxygens overall, whereas salicylic acid has COOH and three oxygens. Oxidising the aldehyde can connect the products, but they are not two names for one substance.
Problem
A synthesis needs o-HOC₆H₄COOH in one case and o-HOC₆H₄CHO in another. Select reagents.
- 1.For the carboxylic acid, form sodium phenoxide, use heated pressurised CO₂, then acidify: Kolbe–Schmitt carboxylation.
- 2.For the aldehyde, use CHCl₃ and aqueous NaOH under heated conditions, then acidify: Reimer–Tiemann formylation.
- 3.In both cases the ring gains one carbon, but the reagent carbon enters at a different oxidation level. Replacing CO₂ with CHCl₃ without changing the predicted functional group would miss the central distinction.
Zinc dust removes oxygen from the ring
Zinc takes up oxygen as zinc oxide while the aromatic organic product becomes benzene. The phenolic oxygen is removed rather than merely deprotonated. The net hydrogen count is conserved: phenol and benzene both contain six hydrogens, but their hydrogen attachments differ. This special reduction must not be used as evidence that phenols readily undergo ordinary ROH + HX substitution at aromatic carbon.
Oxidation gives a quinone rather than an ordinary ketone alcohol product
Phenol can be oxidised with chromic acid, for example generated from dichromate in sulphuric acid, to a conjugated diketone, p-benzoquinone. Its systematic ring description is cyclohexa-2,5-diene-1,4-dione. The product has two C=O bonds at opposite ring positions and two C=C bonds. It is not a benzene ring that retains an intact aromatic circle plus two extra double-bonded oxygens; that drawing would misrepresent bond orders and valence.
Ordinary alcohol oxidation forms C=O at a saturated OH-bearing carbon by removing its hydrogen. Phenol has a different conjugated aromatic system and follows a different oxidation pattern. Slow air oxidation of phenols can give dark mixtures containing quinone-related products; colour change in an aged sample does not prove that only one pure quinone is present. Reaction observation, product identity and mixture composition must be distinguished.
Quick Quiz
Quiz
What is the carbon source for COOH in Kolbe–Schmitt carboxylation?
Explanation: Carbon dioxide adds one new carbon to the activated aromatic ring.
What is the principal conventional product after Reimer–Tiemann work-up?
Explanation: Chloroform-derived carbon becomes an ortho CHO group.
Which reactive species helps explain the chloroform/base reaction?
Explanation: Base-induced chloroform deprotonation and chloride loss generate an electron-deficient neutral carbon species.
What accompanies benzene in phenol reduction with zinc?
Explanation: The removed oxygen is transferred to zinc.
Which describes p-benzoquinone correctly?
Explanation: Quinone formation changes the ring bond pattern; valence must remain valid.
Practice Problems
- Write both operations in preparing salicylic acid from sodium phenoxide.
- Account for carbon and chlorine in Reimer–Tiemann formylation.
- Distinguish salicylic acid, salicylaldehyde and phenyl methanoate by connectivity.
- Why is zinc-dust reduction different from adding NaOH to phenol?
- Give the quinone product of phenol oxidation and explain why a benzene aromatic circle is unsuitable in its structural drawing.
- Why does acidification appear after both major named phenol transformations?
Heat C₆H₅ONa with pressurised CO₂ to give predominantly o-HOC₆H₄COONa. Then add acid: o-HOC₆H₄COONa + HCl → o-HOC₆H₄COOH + NaCl. CO₂ introduces a carbon; acidification changes protonation and permits isolation of the neutral acid.
Key Takeaways
• Phenoxide activates ring positions toward suitable carbon electrophiles. • Kolbe–Schmitt reaction uses CO₂ and gives salicylic acid after acidification. • Reimer–Tiemann reaction uses CHCl₃/base and gives salicylaldehyde after work-up. • Dichlorocarbene and a dichloromethyl stage explain the chloroform-derived carbon. • COOH and CHO are different functional groups and oxidation levels. • Zinc-dust heating transfers phenolic oxygen to ZnO and gives benzene. • Phenol oxidation can give p-benzoquinone with a changed conjugated ring structure.