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 × 10⁻¹, 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]

Solution

Chemistry ma’am · 4 steps

Count the methyl groups on nitrogen: each one comes from its own molecule of bromomethane.

1. Moles of product. Benzyl trimethyl ammonium bromide is C₆H₅CH₂N(CH₃)₃⁺ Br⁻, that is C₁₀H₁₆NBr. Its molar mass is 120 + 16 + 14 + 80 = 230 g mol⁻¹, so 23 g is 0.1 mol, matching the 0.1 mol of benzylamine.
2. Three methylations. The nitrogen of C₆H₅CH₂NH₂ starts with two H atoms and ends with three CH₃ groups. Two CH₃Br replace the two N–H bonds, releasing HBr each time, and a third CH₃Br adds to the lone pair to make the quaternary salt.

   C₆H₅CH₂NH₂ + 3CH₃Br → C₆H₅CH₂N(CH₃)₃⁺Br⁻ + 2HBr

3. Use the ratio. 1 mol of salt needs 3 mol of CH₃Br, so 0.1 mol of salt needs 3 × 0.1 = 0.3 mol of bromomethane.
4. So. 0.3 = 3 × 10⁻¹ mol of bromomethane is consumed, so n = 3.

The common slip: 1. Counting only the one bromide ion in the product gives a 1 : 1 ratio and n = 1. The other two Br atoms leave as HBr, but those methyls still came from CH₃Br.

The idea to keep. In exhaustive methylation, every methyl on the final nitrogen needs one mole of CH₃X. A primary amine needs 3 to become a quaternary salt, a secondary needs 2, a tertiary needs 1.

Answer from the NTA final answer key. Solution by PrepWiser.