Further substitution of amines (2° → 3° → quaternary)

Year 2

AQA 3.3.11

Nucleophilic substitution & elimination · amine + bromoethane · excess haloalkane (use excess ammonia to favour the 1° amine)

Reagents: amine + bromoethane · Conditions: excess haloalkane (use excess ammonia to favour the 1° amine)

Step 1 of 2

1.The lone pair on the nitrogen of ethylamine attacks the δ+ carbon bonded to bromine.
2.The C–Br bond breaks heterolytically; both electrons go to bromine, which leaves as Br⁻.

The nitrogen bonds to the ethyl group and gains a + charge; Br⁻ leaves — an alkylammonium ion.

Step 2 of 2

1.A lone pair on ammonia (or another amine) removes the proton, forming NH₄⁺.
2.The N–H bonding electrons return to the nitrogen, giving the neutral amine and its lone pair back.

The proton is removed: diethylamine (plus NH₄Br).

Product: diethylamine

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