Given below are some nitrogen containing compounds :
Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume _______ mg of HCl.
(Given molar mass in g mol⁻¹ C : 12, H : 1, O : 16, Cl : 35.5)

Solution

Chemistry ma’am · 4 steps

Find the amine whose lone pair is not tied up with the ring, then it is a 1 : 1 mole calculation with HCl.

1. Pick the base. In aniline the N lone pair is delocalised into the ring. The NO₂ group in p-nitroaniline pulls it away even more, and in acetanilide the C=O group draws it in. In benzylamine, C₆H₅CH₂NH₂, the CH₂ group keeps N away from the ring, so its lone pair is free and it is the most basic.
2. Molar mass. Benzylamine is C₇H₉N: 7 × 12 + 9 × 1 + 14 = 107 g mol⁻¹. HCl is 1 + 35.5 = 36.5 g mol⁻¹.

   C₆H₅CH₂NH₂ + HCl → C₆H₅CH₂NH₃⁺Cl⁻

3. Moles to mass. One NH₂ group takes one HCl, so moles of HCl = 1.0/107 = 9.35 × 10⁻³ mol. Mass of HCl = 36.5/107 = 0.341 g.
4. So. 1.0 g of benzylamine consumes 341 mg of HCl.

The common slip: 392. Choosing aniline (M = 93 g mol⁻¹) as the most basic gives 36.5/93 × 1000 ≈ 392 mg, not 341.

The idea to keep. An aryl amine is weak because its lone pair spreads into the ring. Put even one CH₂ between N and the ring and it behaves like an alkyl amine, far more basic.

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