JEE Advanced 2021 Paper 1 · Q06 · Hydrocarbons

Correct answer:3.20, 3.90

Solution

From the same scheme: P (0.1 mol propyne) undergoes Hg$^{2+}$/H$^+$ hydration at 333 K with 100% conversion (Markovnikov hydration of an alkyne) giving S = acetone (propan-2-one), 0.1 mol. S then undergoes aldol condensation with Ba(OH)$_2$/heat at 80% (two molecules of acetone combine), giving T = mesityl oxide / 4-methylpent-3-en-2-one, moles = 0.5 \* 0.1 \* 0.8 = 0.04 mol. T treated with NaOCl (haloform reaction on the methyl ketone) at 80% gives U.

The haloform attack on T affords either the sodium carboxylate / carboxylic acid CH$_3$-C(CH$_3$)=CH-COOH (4-methyl-pent-2-enoic acid) of molar mass 100 g mol$^{-1}$, or, depending on whether one accepts the carboxylate Na salt CH$_3$-C(CH$_3$)=CH-COONa of molar mass 122 g mol$^{-1}$.

Moles of U = 0.04 \* 0.8 = 0.032 mol. Mass of U: - Using molar mass 100: y = 0.032 \* 100 = 3.20 g. - Using molar mass 122: y = 0.032 \* 122 = 3.90 g.

The official answer therefore lies in the range 3.20-3.90 g; both 3.20 and 3.90 are credited.

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