JEE Main 2026 — Apr 4, Shift 1 — Chemistry solution

JEE Main2026Apr 4, Shift 1Chemistry

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

$P$ g of the major product ( $X$ ) formed is reacted with $NaHC O_{3}$ solution 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 : $35.5$ )

Your answer

Solution

Track the transformations on benzene:

(i) Friedel-Crafts methylation gives toluene.

(ii) Chlorination with $C l_{2} /FeC l_{3}$ installs Cl ortho/para to the methyl group, giving chlorotoluene.

(iii) Side-chain oxidation by acidic $K_{2} C r_{2} O_{7}$ converts $- C H_{3}$ to $- COOH$ , yielding chlorobenzoic acid $X$ ( $Cl C_{6} H_{4} COOH$ , formula $C_{7} H_{5} O_{2} Cl$ ).

Molar mass of $X$ : $7 (12) + 5 (1) + 2 (16) + 35.5 = 84 + 5 + 32 + 35.5 = 156.5$ g mol $^{- 1}$ .

Reaction with $NaHC O_{3}$ liberates $C O_{2}$ in $1 : 1$ stoichiometry. Moles of $C O_{2}$ at STP ( $22.4$ L mol $^{- 1}$ for this calculation): $\frac{11.2}{22.4} = 0.5$ mol.

Hence $0.5$ mol of $X$ reacted, mass $P = 0.5 \times 156.5 = 78.25 \approx 78$ g.

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