JEE Advanced 2026 — Paper 1 — CHEM-III solution

JEE Advanced2026Paper 1CHEM-III

Treatment of buta-1,3-diyne with NaNH $_{2}$ (2 equivalents), followed by reaction with excess of trans-CH $_{3}$ –CH=CH–CH $_{2}$ –Br gives $X$ as the major product. The maximum number of carbon atoms that are collinear (in a straight line) in $X$ is ____.

Your answer

Solution

Acid–base step. Buta-1,3-diyne (HC $\equiv$ C–C $\equiv$ CH) has two terminal acetylenic protons. Two equivalents of NaNH $_{2}$ deprotonate both ends to give the bis-acetylide $^{-} C \equiv C-C \equiv C^{-}$ .

Alkylation step. The bis-acetylide performs S $_{N}$ 2 displacement on the allylic bromide twice (excess), giving a symmetrical product with the chain

CH $_{3}$ –CH=CH–CH $_{2}$ –C $\equiv$ C–C $\equiv$ C–CH $_{2}$ –CH=CH–CH $_{3}$ (with trans geometry on both alkene ends).

Collinearity analysis. The two internal triple bonds together with the four sp carbons and the two sp $^{3}$ CH $_{2}$ carbons adjacent to them lie on the same straight line (sp–sp–sp–sp axes are collinear, and the CH $_{2}$ carbons attached at each end of the diyne system are also on that axis). Counting the carbons that lie on the diyne axis: the 4 sp carbons of the two triple bonds plus the 2 sp $^{3}$ CH $_{2}$ carbons attached at each end gives 6 collinear carbons. The CH=CH–CH $_{3}$ allylic ends are bent away from the axis (sp $^{2}$ geometry, bond angle 120°), so they are not collinear.

Answer = $6.00$ .

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