JEE Main 2026 — Jan 23, Shift 1 — Chemistry solution

JEE Main2026Jan 23, Shift 1Chemistry

Consider the following sequence of reactions.

Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce ______ mg of B.

(Given molar mass in g mol $^{- 1}$ : H : 1, C : 12, N : 14, O : 16, Br : 80)

A
$301$
B
$146$
C
$228$
D
$208$

Solution

Molar masses. 4-nitrotoluene C $_{7}$ H $_{7}$ NO $_{2}$ = 137 g/mol; product B (C $_{9}$ H $_{10}$ BrNO) = 108 + 10 + 80 + 14 + 16 = 228 g/mol.

Yield from 137 mg: moles of starting material $= \frac{137 \times 1 0 ^{- 3}}{137} = 1 0^{- 3}$ mol. Since the stoichiometry is 1:1, mass of B $= 1 0^{- 3} \times 228 = 0.228$ g $= 228$ mg.

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