Question

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).

Hint:

To convert a primary amine to a quaternary ammonium salt, 3 moles of alkyl halide are required.

Answer 1

Correct Answer: 3

Step 1: Reaction: $Ph-CH_2NH_2 + 3CH_3Br \rightarrow [Ph-CH_2N(CH_3)_3]^+ Br^-$.
Step 2: Molar mass of product ($C_{10}H_{16}NBr$): $(10 \times 12) + (16 \times 1) + 14 + 80 = 120 + 16 + 14 + 80 = 230$ g/mol.
Step 3: Moles of product $= 23/230 = 0.1$ mol.
Step 4: Moles of $CH_3Br$ needed $= 3 \times 0.1 = 0.3$ mol.
Step 5: $0.3 = 3 \times 10^{-1} \implies n = 3$.