JEE Advanced 2024 · Paper 1 · Chemistry Q12

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ChemistryInteger+4Question 46 of 51

In the following reaction sequence, the major product P is formed.

$\ce{H-C#C-(CH2)15-CO2Et ->[\text{i) } Hg^{2+}, H3O+][\text{ii) Zn-Hg/HCl; iii) } H3O+, \Delta] P}$

Glycerol reacts completely with excess P in the presence of an acid catalyst to form Q. Reaction of Q with excess NaOH followed by the treatment with $\ce{CaCl2}$ yields Ca-soap R, quantitatively. Starting with one mole of Q, the amount of R produced in gram is _____.

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

Reaction scheme: ethyl ester of a terminal alkyne, H-C≡C-(CH₂)₁₅-CO₂Et (the (CH₂) unit in brackets with subscript 15), reacts with i) Hg²⁺, H₃O⁺; ii) Zn-Hg/HCl; iii) H₃O⁺, Δ (reagents written above the arrow) to give P.

Your answer

Answer and worked solution

Official answer: 909

Steps:

$\ce{Hg^2+/H3O+}$ hydrates the terminal alkyne to a methyl ketone, $\ce{CH3CO(CH2)15CO2Et}$ .
Clemmensen reduction gives $\ce{CH3CH2(CH2)15CO2Et}$ .
$\ce{H3O+}$ , $\Delta$ hydrolyses the ester to stearic acid, $\ce{C17H35COOH}$ (P, 18 carbons).

Q is glyceryl tristearate. Saponification of 1 mol Q gives 3 mol $\ce{C17H35COO-}$ , and $\ce{CaCl2}$ precipitates $\ce{(C17H35COO)2Ca}$ : 1.5 mol.

Molar mass: $\ce{C17H35COO-} = 216 + 35 + 32 = 283$ , so the salt is $40 + 2\times283 = 606\,\mathrm{g\,mol^{-1}}$ .

Mass $= 1.5\times606 = 909\,\mathrm{g}$ .

Topics:Acid derivatives Alkynes Soaps and detergents

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JEE Advanced 2024 · Paper 1, Chemistry question 12