Consider the following sequence of reactions to give the major product (X) P g of the major product ( X ) formed is reacted with NaHCO 3 soluion to liberate a gas which occupied 11.2 dm 3 at STP. P = \ \ \ \ . g. (Given molar mass in g mol -1 H : 1, C : 12 , O : 16, Cl : 35.5)

Chemistry · Carboxylic Acids and Derivatives

JEE Main 2026 — 4 April, Morning Shift — Question 64

Consider the following sequence of reactions to give the major product (X) P g of the major product ( X ) formed is reacted with $\mathrm{NaHCO}_{3}$ soluion to liberate a gas which occupied $11.2 \mathrm{dm}^{3}$ at STP. $\mathrm{P}=$ $\_\_\_\_$ . g. (Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12$ , $\mathrm{O}: 16, \mathrm{Cl}: 35.5)$

Answer: 78

Numerical answer — enter this value.

Step-by-step solution

Moles of $\mathrm{P}=\frac{11.2}{22.4}=0.5$ mole $\mathrm{M}_{\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{Cl}}=156.5 \mathrm{gm}$ $\mathrm{W}_{\mathrm{P}\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{Cl}\right)}=\frac{156.5}{2}=78.25 \approx 78$ grams

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Chemistry
Chapter: Carboxylic Acids and Derivatives
Topic: Chemical Properties of Carboxylic Acids