JEE Advanced 2021 Paper 1 · Q05 · Hydrocarbons

Correct answer:1.62

Solution

Step 1: Mg$_2$C$_3$ + H$_2$O gives propyne, P = CH$_3$-C$\equiv$CH. Molar mass of P = 40 g mol$^{-1}$, and 4.0 g = 0.1 mol of P.

Step 2: P $\xrightarrow[\text{MeI}]{\text{NaNH}_2}$ Q at 75% yield. The terminal alkyne is deprotonated by NaNH$_2$ and alkylated with MeI to give but-2-yne, Q = CH$_3$-C$\equiv$C-CH$_3$. Moles of Q = 0.75 \* 0.1 = 0.075 mol; but used calculation: 0.1 \* 3/4 = 0.075 (the FIITJEE solution treats 75% as 3/4).

Step 3: 3 Q $\xrightarrow[\text{873 K}]{\text{red-hot iron tube}}$ R at 40% yield. Three molecules of but-2-yne cyclotrimerise to hexamethylbenzene, R = 1,2,3,4,5,6-hexamethylbenzene. Moles of R formed = (0.075/3) \* 0.4 = 0.01 mol.

Step 4: Molar mass of R (C$_{12}$H$_{18}$) = 12(12) + 18(1) = 162 g mol$^{-1}$.

Therefore x = mass of R = 0.01 mol \* 162 g mol$^{-1}$ = 1.62 g.

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