JEE Main Chemistry question

A bis-aryl propargylic diacid substrate reacts with 2 moles of NaNH₂ to give A. Identify A.

(A)The corresponding bis-sodium carboxylate with the alkyne deprotonated at both terminal positionsCorrect
(B)A mono-deprotonated, mono-acid salt
(C)A fully protonated diacid, unreacted
(D)A decarboxylated bis-alkynyl product

Answer: (A)

Solution

Sodium amide (NaNH₂) is a very strong, non-nucleophilic base that readily deprotonates both the acidic terminal alkyne C–H bonds (≡C–H, pKₐ ≈ 25) and the carboxylic acid O–H bonds (pKₐ ≈ 4–5) present in the substrate; with 2 moles of NaNH₂ available, the two most acidic protons (the carboxylic acid O–H groups, being

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Question ID VK-149244