JEE Challenger
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ChemMedMain 2026 • 4th Apr - Shift 1

Mass of Organic Product Reacting to Liberate Gas at STP

Consider the following sequence of reactions to give the major product (X)(X)

P gP\text{ g} of the major product (X)(X) formed is reacted with NaHCO3\text{NaHCO}_3 solution to liberate a gas which occupied 11.2 dm311.2\text{ dm}^3 at STP. P=‾ gP = \underline{\quad}\text{ g}. (Given molar mass in g mol−1\text{g mol}^{-1} H:1\text{H}: 1, C:12\text{C}: 12, O:16\text{O}: 16, Cl:35.5\text{Cl}: 35.5)

Question Diagram 1
Official Numerical Answer78

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsOxidation and Reduction (Organic)Electrophilic Aromatic Substitution

Step-by-Step Solution

To determine the mass PP of the major product (X)(X), we analyze the given sequence of reactions step-by-step:

Step 1: Reaction Sequence

  1. Friedel-Crafts Alkylation: Benzene reacts with methyl chloride (CH3Cl\text{CH}_3\text{Cl}) in the presence of anhydrous AlCl3\text{AlCl}_3 to form toluene: C6H6+CH3Cl→anhydrous AlCl3C6H5CH3+HCl\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{anhydrous AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}

  2. Electrophilic Aromatic Substitution (Chlorination): The methyl group (−CH3-\text{CH}_3) is an ortho/para-directing group. Due to steric hindrance at the ortho-position, the major product is pp-chlorotoluene (4-chlorotoluene): C6H5CH3+Cl2→FeCl3Cl-C6H4-CH3(major product)\text{C}_6\text{H}_5\text{CH}_3 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{Cl-C}_6\text{H}_4\text{-CH}_3 \quad (\text{major product})

  3. Oxidation of Side Chain: pp-chlorotoluene is oxidized by K2Cr2O7/H2SO4\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4, which selectively oxidizes the benzylic methyl group into a carboxylic acid group (−COOH-\text{COOH}): Cl-C6H4-CH3→K2Cr2O7/H2SO4Cl-C6H4-COOH(X)\text{Cl-C}_6\text{H}_4\text{-CH}_3 \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4} \text{Cl-C}_6\text{H}_4\text{-COOH} \quad (X)

Thus, the major product (X)(X) is 4-chlorobenzoic acid (C7H5ClO2\text{C}_7\text{H}_5\text{ClO}_2).

Step 2: Reaction with NaHCO3\text{NaHCO}_3

4-chlorobenzoic acid reacts with sodium bicarbonate (NaHCO3\text{NaHCO}_3) to liberate carbon dioxide (CO2\text{CO}_2) gas according to the equation: Cl-C6H4-COOH+NaHCO3⟶Cl-C6H4-COONa+H2O+CO2↑\text{Cl-C}_6\text{H}_4\text{-COOH} + \text{NaHCO}_3 \longrightarrow \text{Cl-C}_6\text{H}_4\text{-COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow

From the stoichiometry of the reaction: 1 mole of X produces 1 mole of CO2 gas.\text{1 mole of } X \text{ produces } 1 \text{ mole of } \text{CO}_2 \text{ gas.}

Step 3: Molar Calculations

  • Volume of CO2\text{CO}_2 at STP: 11.2 dm3=11.2 L11.2\text{ dm}^3 = 11.2\text{ L}

  • Moles of CO2\text{CO}_2 liberated: nCO2=11.2 L22.4 L mol−1=0.5 moln_{\text{CO}_2} = \frac{11.2\text{ L}}{22.4\text{ L mol}^{-1}} = 0.5\text{ mol}

  • Therefore, moles of major product (X)(X) reacted: nX=0.5 moln_{X} = 0.5\text{ mol}

  • Molar mass of (X)(X) (C7H5ClO2\text{C}_7\text{H}_5\text{ClO}_2): Molar Mass=(7×12)+(5×1)+35.5+(2×16)=84+5+35.5+32=156.5 g mol−1\text{Molar Mass} = (7 \times 12) + (5 \times 1) + 35.5 + (2 \times 16) = 84 + 5 + 35.5 + 32 = 156.5\text{ g mol}^{-1}

  • Mass PP of major product (X)(X): P=nX×Molar Mass=0.5 mol×156.5 g mol−1=78.25 g≈78 gP = n_X \times \text{Molar Mass} = 0.5\text{ mol} \times 156.5\text{ g mol}^{-1} = 78.25\text{ g} \approx 78\text{ g}

Answer: 78

Mass of Organic Product Reacting to Liberate Gas at STP | Chemistry PYQ Solution - JEE Challenger