JEE Advanced 2024 Paper 1 · Q12 · Alkyne Hydration / Clemmensen / Hydrolysis / Soap Formation

Question

In the following reaction sequence, the major product $\mathbf{P}$ is formed:

Ethyl ester of 17-octadecynoate, $H{-}C{\equiv}C{-}(CH_2)_{15}{-}CO_2Et$ (a terminal alkyne bearing an ethyl ester group at the far end of a $15$-carbon chain), is treated sequentially with: (i) $Hg^{2+},\,H_3O^+$; then (ii) $Zn{-}Hg/HCl$; then (iii) $H_3O^+,\,\Delta$. The major product is $\mathbf{P}$.

Glycerol reacts completely with excess $\mathbf{P}$ in the presence of an acid catalyst to form $\mathbf{Q}$. Reaction of $\mathbf{Q}$ with excess $	ext{NaOH}$ followed by treatment with $	ext{CaCl}_2$ yields Ca-soap $\mathbf{R}$ quantitatively. Starting with one mole of $\mathbf{Q}$, the amount of $\mathbf{R}$ produced in grams is ______.

[Given, atomic weight: $H = 1$, $C = 12$, $N = 14$, $O = 16$, $Na = 23$, $Cl = 35$, $Ca = 40$]

SolveKar AI · Accepted Answer

Step (i) Markovnikov hydration of terminal alkyne: $HC{\equiv}C{-}(CH_2)_{15}{-}COOEt 	o CH_3{-}CO{-}(CH_2)_{15}{-}COOEt$ — wait, with $Hg^{2+}/H_3O^+$ the terminal alkyne gives the methyl ketone $CH_3{-}C(=O){-}(CH_2)_{15}{-}COOEt$. (ii) $Zn(Hg)/HCl$ (Clemmensen) reduces the ketone to $CH_3{-}CH_2{-}(CH_2)_{15}{-}COOEt = CH_3(CH_2)_{16}COOEt$. (iii) $H_3O^+/\Delta$ hydrolyses the ester → $CH_3(CH_2)_{16}COOH$ (stearic acid) = $\mathbf{P}$.
Glycerol (3 -OH) + 3 $\mathbf{P}$ → tristearin $\mathbf{Q}$ = $C_3H_5(OOC{-}C_{17}H_{35})_3$.
Saponification of 1 mol $\mathbf{Q}$ with NaOH gives 3 mol sodium stearate; with $CaCl_2$ → $\mathbf{R}$ = $Ca(C_{17}H_{35}COO)_2$, calcium stearate. From 3 mol Na-soap → $3/2$ mol of Ca-soap.
Molar mass of Ca-stearate: $Ca + 2(C_{17}H_{35}COO) = 40 + 2(17 	imes 12 + 35 	imes 1 + 12 + 32) = 40 + 2(204 + 35 + 12 + 32) = 40 + 2 	imes 283 = 606$ g/mol.
Mass of $\mathbf{R}$ = $1.5 	imes 606 = \mathbf{909}$ g.

JEE Advanced 2024 Paper 1 · Q12 · Alkyne Hydration / Clemmensen / Hydrolysis / Soap Formation

Solution