A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are \(\mathrm{n} \times 10^{-1}\), when \(\mathrm{n}=\) ______. (Round off to the nearest integer). [Given: Atomic masses: \(\mathrm{C}=12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}, \mathrm{N}: 14.0 \mathrm{u}, \mathrm{Br}: 80.0 \mathrm{u}\) ]

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Compounds Containing Nitrogen

A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are

n \times 1 0^{- 1}

, when

n =

______. (Round off to the nearest integer).
[Given: Atomic masses:

C = 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u

]

Answer: 3

Solution

Exhaustive methylation turns the primary amine into a quaternary ammonium salt, so each molecule of benzylamine takes up three methyl groups:

C_{6} H_{5} C H_{2} N H_{2} + 3 C H_{3} Br \to [C_{6} H_{5} C H_{2} N (C H_{3})_{3}]^{+} B r^{-} + 2 HBr

The salt has molar mass

10 (12) + 16 (1) + 14 + 80 = 230 g mo l^{- 1}

, so

23

g of it is

0.1

mol, which accounts for all the benzylamine taken. Three moles of bromomethane are used per mole of salt, so the amount consumed is

0.3

mol, that is

3 \times 1 0^{- 1}

mol, giving

n = 3

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Compounds Containing Nitrogen

Solution

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