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Chemistry · Carboxylic Acids and Derivatives

JEE Advanced 2024 — Paper 1 — Question 42

In the following reaction sequence, major product P\mathbf{P} is formed. Glycerol reacts completely with excess P\mathbf{P} in the presence of an acid catalyst to form Q\mathbf{Q}. Reaction of Q\mathbf{Q} with excess NaOH followed by the treatment with CaCl2\mathrm{CaCl}_{2} yields Ca - soap R\mathbf{R}, quantitatively. Starting with one mole of Q\mathbf{Q}, the amount of R\mathbf{R} produced in gram is _____\_\_\_\_\_ [Given, atomic weight: H=1,C=12, N=14,O=16,Na=23,Cl=35,Ca=40\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{Cl}=35, \mathrm{Ca}=40 ]

Question figure

Answer: 909

Numerical answer — enter this value.

Step-by-step solution

H−C≡C−(CH2)15−COOEt\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\left(\mathrm{CH}_{2}\right)_{15}-\mathrm{COOEt} (Q)

Q→CaCl2 excess NaOH[CH3−(CH2)16−COO]2CaQ \xrightarrow[\mathrm{CaCl}_{2}]{\text { excess } \mathrm{NaOH}}\left[\mathrm{CH}_{3}-\left(\mathrm{CH}_{2}\right)_{16}-\mathrm{COO}\right]_{2} \mathrm{Ca}

1 moleQ⟶3 moleCH3−(CH2)16−COO−1 \mathrm{~mole} \mathrm{Q} \longrightarrow 3 \mathrm{~mole} \mathrm{CH}_{3}-\left(\mathrm{CH}_{2}\right)_{16}-\mathrm{COO}^{-}

2CH3−(CH2)16−COO−+Ca+2⟶(CH3−(CH2)16COO)2Ca\begin{gathered} 2 \mathrm{CH}_{3}-\left(\mathrm{CH}_{2}\right)_{16}-\mathrm{COO}^{-}+\mathrm{Ca}^{+2} \longrightarrow\left(\mathrm{CH}_{3}-\left(\mathrm{CH}_{2}\right)_{16} \mathrm{COO}\right)_{2} \mathrm{Ca} \end{gathered}

Mole of R=1.5R=1.5 WtW t. of R=1.5×606R=1.5 \times 606 Wt. of R=909\mathrm{R}=909

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Chemistry
Chapter
Carboxylic Acids and Derivatives
Topic
Chemical Properties of Carboxylic Acids