JEE Main Chemistry question

Which of the following intermediate species is not formed in the reaction of acrylic acid with HBr to give β-bromopropionic acid? (CH₂=CH–COOH --HBr--> BrCH₂CH₂COOH?)

(A)CH₂=CH–C(OH)=OH⁺ (protonated carbonyl, the acid's own protonation, unrelated to the addition step)
(B)CH₂=CH–C(⁺OH)=OH⁻ (a miswritten/invalid charge-separated species)
(C)CH₂=CH–C(OH)⁻=OH (an invalid carbanion-type resonance form)
(D)⁻CH₂–CH=C(OH)(OH) (the enolate-type species expected from conjugate addition at the correct position)Correct

Answer: (D)

Solution

Addition of HBr to acrylic acid proceeds via protonation/activation directing the overall outcome to the "anti-Markovnikov"-like β-bromo product because the electron-withdrawing –COOH group deactivates the alkene toward a simple cationic (Markovnikov) mechanism and instead favours the conjugate-addition pathway via

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Question ID VK-149453