Question
(a) Sodium phenoxide is treated with CO₂ under pressure (4–7 atm) at 400 K, followed by acidification with dilute HCl. Identify the product and name the reaction.
(b) Sodium ethoxide reacts with bromoethane. Identify the product, name the reaction, and state the mechanism type.
(JEE Main 2021 — Alcohols, Phenols and Ethers)
Solution — Step by Step
Sodium phenoxide () is an electron-rich aromatic system. When CO₂ (electrophile) attacks under pressure at 400 K, electrophilic substitution happens at the ortho position — not at oxygen. The intermediate is then acidified.
Product: Salicylic acid (2-hydroxybenzoic acid).
The group is strongly ortho/para directing. At moderate temperature, the ortho product is thermodynamically favoured because the sodium counterion chelates with the adjacent group, stabilising the product. This is the Kolbe-Schmitt reaction — not a Friedel-Crafts, and CO₂ acts as the electrophile here.
Sodium ethoxide () is the nucleophile. Bromoethane () is the electrophile. The alkoxide attacks the carbon bearing bromine in a backside attack.
Product: Diethyl ether. Mechanism: SN2 (bimolecular nucleophilic substitution).
This reaction works only with primary alkyl halides. The alkoxide must attack the alkyl halide — not the other way around. If you need to make an aryl alkyl ether like anisole (), you use sodium phenoxide + methyl iodide, never methyl phenyl oxide + phenyl halide (aryl halides don’t undergo SN2).
Why This Works
The Kolbe reaction works because phenol’s group (after deprotonation to phenoxide) pushes electron density into the ring, making the ortho and para positions nucleophilic enough to attack CO₂, a weak electrophile. Normal electrophilic substitution reagents like or are far stronger — CO₂ only works here because the phenoxide is highly activated.
Williamson synthesis relies on the hard-hard interaction between the alkoxide oxygen (hard nucleophile) and the -carbon of the alkyl halide. The SN2 mechanism means the rate depends on both the nucleophile and the substrate — a second-order reaction. That’s why tertiary alkyl halides give elimination (E2) instead of substitution: too much steric crowding for backside attack.
Together, these two reactions cover the two main routes to functionalised phenol derivatives and ethers — a high-weightage combination in JEE Main organic chemistry.
Alternative Method
For ether synthesis: Dehydration of alcohols in at 413 K also gives ethers (), but this only works for symmetrical ethers. Williamson synthesis is the go-to for unsymmetrical ethers like methyl phenyl ether (anisole) or methyl tert-butyl ether (MTBE).
For Williamson synthesis of aryl alkyl ethers: always use sodium phenoxide + alkyl halide. Never attempt aryl halide + alkoxide — aryl halides are almost inert to SN2 (sp² carbon, resonance stabilised C–X bond).
Common Mistake
Students write the Kolbe reaction product as para-hydroxybenzoic acid instead of ortho (salicylic acid). The ortho product dominates because the sodium ion forms a six-membered chelate ring stabilising the ortho transition state — a thermodynamic preference. In JEE questions, the answer is always salicylic acid (ortho), not the para isomer. Marking the para product here loses the full mark.
A second mistake appears in Williamson synthesis: using a tertiary alkyl halide and expecting SN2. tert-Butyl bromide with sodium ethoxide gives 2-methylpropene (E2 elimination) — not the ether. The question sometimes tests this by giving you a tertiary substrate and asking for the “major product” — always write the alkene, not the ether.