Q.Phenol (a benzene ring bearing -OH) is treated with
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Start your 14-day free trial to unlock the full solution →This is the Kolbe-Schmitt reaction run with KOH: phenol forms potassium phenoxide, and with the larger K+ ion the CO2 adds mainly at the para position, so after acidification the product X is 4-hydroxybenzoic acid (para), not salicylic acid (ortho).
Step 1. Phenol + KOH gives potassium phenoxide (C6H5-O- K+). The negative charge on oxygen is delocalised into the ring, raising the electron density at the ortho and para carbons and making the ring nucleophilic toward CO2.
Step 2. Under the classic Kolbe-Schmitt conditions (about 400 K and 3-7 atm) the phenoxide reacts with CO2 (the electrophile implied by these conditions). The regiochemistry is set by the counter-ion: the small sodium ion chelates and locks CO2 next to the -O-, giving the ortho product; but with the larger potassium ion this ortho chelation is weaker, so carboxylation occurs preferentially at the para position.
Step 3. This gives potassium 4-hydroxybenzoate as the intermediate salt.
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