JEE Main
Chemistry
Haloalkanes and Haloarenes
2026
JEE Main 2026 (24 January Shift 2)
JEE Main Chemistry Question (2026) — Solution
Question
Grignard reagent RMgBr ( P ) reacts with water and forms a gas (Q). One gram of Q occupies 1.4 dm ^ 3 at STP. (P) on reaction with dry ice in dry ether followed by H _ 3 O ^ + forms a compound (Z). 0.1 mole of (Z) will weigh \_\_\_\_ g. (Nearest integer)
Step-by-step solution
Grignard reagent RMgBr reacts with water to produce gas Q. One gram of Q occupies 1.4 L at STP. Moles of Q = 1.4/22.4 = 0.0625 mol, so molar mass = 1/0.0625 = 16 g/mol, identifying Q as CH₄. Thus P = CH₃MgBr. When P reacts with dry ice: CH _3 MgBr + CO _2 CH _3 COOMgBr . Upon treatment with H₃O⁺, this gives Z = CH₃COOH (acetic acid). Molar mass of Z = 60 g/mol. Mass of 0.1 mol = 0.1 × 60 = 6 g.
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Related: Chemistry — Haloalkanes and Haloarenes · All PYQ Banks