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Friday, 9 October

JEE Main 2026 · ChemistryNumerical answerNumerical valueMediumMulti-step

JEE Main 4 April 2026, Shift 1, Chemistry Q72

Question 72 of 75 in this shift, Chemistry question 22 of 25, Section B.

Consider the following sequence of reactions to give the major product (X) benzene →(i) CH3Cl/anhydrous AlCl3; (ii) Cl2/FeCl3; (iii) K2Cr2O7/H2SO4\xrightarrow{\text{(i) } \mathrm{CH_3Cl}/\text{anhydrous } \mathrm{AlCl_3};\ \text{(ii) } \mathrm{Cl_2/FeCl_3};\ \text{(iii) } \mathrm{K_2Cr_2O_7/H_2SO_4}} (X) Major Product PP g of the major product (X) formed is reacted with NaHCO3\mathrm{NaHCO_3} solution to liberate a gas which occupied 11.2 dm311.2\ \mathrm{dm^3} at STP. P=P= ______ g. (Given molar mass in g mol−1\mathrm{g\ mol^{-1}} H:1, C:12, O:16, Cl:35.5)

The figure, in words

Benzene ring on the left of a reaction arrow with reagents (i) CH3Cl\mathrm{CH_3Cl}/anhydrous AlCl3\mathrm{AlCl_3}, (ii) Cl2/FeCl3\mathrm{Cl_2/FeCl_3}, (iii) K2Cr2O7/H2SO4\mathrm{K_2Cr_2O_7/H_2SO_4} giving (X) Major Product.

Official answer

78

NTA final key.