JEE Main Chemistry question
The mechanism of ester formation in acidic medium is as follows:
R–C(=O)–OH --H⁺, Step (i)--> [HO⁺=C(OH)–R ↔ HO–C(OH)=R⁺] --+R'OH, −H⁺, Step (ii)--> R–C(OH)(OR')(OH) --H⁺, Step (iii)--> R–C(OH₂⁺)(OR')(OH) --−H₂O, Step (iv)--> R–C(OR')=OH⁺ --−H⁺, Step (v)--> R–C(=O)–OR'
Which is the slowest step in the above mechanism?
Answer: (B)
Solution
In the Fischer esterification mechanism, protonation and deprotonation steps (i, iii, v) are fast proton-transfer equilibria, and loss of water (step iv) from the positively charged tetrahedral intermediate is also comparatively fast once that intermediate forms. The
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