Primary and secondary amines — practice questions

Basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton.

1. In ethylamine (CH₃CH₂NH₂), the ethyl group is an electron-donating group. It exerts a positive inductive effect, pushing electron density onto the nitrogen atom. [1 mark]
2. This increases the electron density on the nitrogen, making the lone pair more available to bond with a proton. [1 mark]
3. In phenylamine (C₆H₅NH₂), the lone pair on the nitrogen atom is delocalised into the delocalised π-electron system of the benzene ring. [1 mark]
4. This delocalisation makes the lone pair less available to accept a proton. Therefore, ethylamine is a stronger base than phenylamine. [1 mark]

(a) The amine acts as a nucleophile. The initial reaction forms the salt, which is then deprotonated by another amine molecule. Overall equation for the formation of the secondary amine: CH₃CH₂CH₂NH₂ + CH₃CH₂Br → CH₃CH₂CH₂NH(CH₂CH₃) + HBr [1 mark] Note: Often shown with a second amine molecule reacting with HBr: CH₃CH₂CH₂NH₂ + CH₃CH₂Br → [CH₃CH₂CH₂NH₂(CH₂CH₃)]⁺Br⁻, which then reacts with another CH₃CH₂CH₂NH₂ to give the final products.

(b) The secondary amine has a propyl group and an ethyl group attached to the nitrogen. The name is N-ethylpropan-1-amine. [1 mark]

(c) The reaction continues until four alkyl groups are attached to the nitrogen. Since bromoethane is in excess, ethyl groups will be added. The final product is a quaternary ammonium salt where the nitrogen from the original propylamine is bonded to one propyl group and three ethyl groups. Structure: A central nitrogen atom bonded to one propyl group and three ethyl groups, with a positive charge on the nitrogen and a bromide counter-ion, Br⁻. [ (CH₃CH₂CH₂)N(CH₂CH₃)₃ ]⁺ Br⁻ [1 mark for structure, 1 mark for charges]