AS Chemistry 9701: Ethylamine as a Base and as a Nucleophile

Ethylamine, $\text{CH}_3\text{CH}_2\text{NH}_2$ , is used in two separate experiments.

(a) In the first experiment, an excess of ethylamine is added to dilute hydrochloric acid. Give an equation for this reaction, name the salt formed, and explain, in terms of the nitrogen atom, why ethylamine reacts in this way. [3]

(b) In the second experiment, a limited amount of bromoethane, $\text{CH}_3\text{CH}_2\text{Br}$ , is added to an excess of ethylamine dissolved in ethanol. Give an equation for the main organic reaction taking place, name the type of mechanism, and give the name of the organic product formed. [3]

(c) If a large excess of bromoethane were used instead in (b), the product could react further, eventually forming a quaternary ammonium salt. Explain, in terms of the nitrogen lone pair, why this further reaction is possible. [2]

Part (a): Ethylamine as a Brønsted-Lowry base

The nitrogen atom of ethylamine carries a lone pair of electrons that is not involved in any covalent bond. This lone pair forms a new dative (co-ordinate) bond to a proton, $\text{H}^+$ , from the dilute hydrochloric acid:

$\text{CH}_3\text{CH}_2\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_3^+\text{Cl}^-$

The ionic salt formed is ethylammonium chloride, $\text{CH}_3\text{CH}_2\text{NH}_3^+\text{Cl}^-$ . Ethylamine acts as a Brønsted-Lowry base here because it is the nitrogen lone pair that accepts the proton. The molecule does not lose a proton itself.

Part (b): Ethylamine as a nucleophile

Ethylamine’s nitrogen lone pair can also attack an electron-poor carbon atom, making ethylamine act as a nucleophile. With a limited amount of bromoethane, one ethylamine molecule attacks the δ+ carbon bearing the bromine, displacing $\text{Br}^-$ ; a second molecule of ethylamine (present in excess) then reacts with the $\text{HBr}$ by-product:

$2\text{CH}_3\text{CH}_2\text{NH}_2 + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{NHCH}_2\text{CH}_3 + \text{CH}_3\text{CH}_2\text{NH}_3^+\text{Br}^-$

The mechanism is nucleophilic substitution, and the main organic product is diethylamine, $\text{CH}_3\text{CH}_2\text{NHCH}_2\text{CH}_3$ . A secondary amine, since the nitrogen atom is now bonded to two ethyl groups instead of one.

Part (c): Why further alkylation is possible with excess bromoethane

Diethylamine, like ethylamine, still has a lone pair of electrons on its nitrogen atom, because forming the new C–N bond in part (b) used one of nitrogen’s bonding positions but left the lone pair untouched. This lone pair can act as a nucleophile again and attack a further molecule of bromoethane, converting the secondary amine into a tertiary amine (triethylamine). The tertiary amine still has a lone pair too, so with a large excess of bromoethane the process can repeat once more, using the last lone pair to form a bond to a fourth ethyl group and giving a quaternary ammonium salt, $\text{N}(\text{CH}_2\text{CH}_3)_4^+\text{Br}^-$ , which has no lone pair left and so cannot react further.

Final answers

(a) $\text{CH}_3\text{CH}_2\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_3^+\text{Cl}^-$ ; ethylammonium chloride; the nitrogen lone pair accepts a proton, so ethylamine acts as a base.
(b) $2\text{CH}_3\text{CH}_2\text{NH}_2 + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{NHCH}_2\text{CH}_3 + \text{CH}_3\text{CH}_2\text{NH}_3^+\text{Br}^-$ ; nucleophilic substitution; diethylamine.
(c) The secondary (then tertiary) amine still has a nitrogen lone pair, which can act as a nucleophile towards further bromoethane, eventually giving a quaternary ammonium salt with no lone pair left.