Nitrogen Compounds: Question 3

Worked solution

Part (a): Amines as Brønsted-Lowry bases

A Brønsted-Lowry base is a proton (H⁺) acceptor. Methylamine, $\text{CH}_3\text{NH}_2$, can act as a base because its nitrogen atom carries a lone pair of electrons that is not involved in any covalent bond. This lone pair can form a new dative (co-ordinate) bond to an $\text{H}^+$ ion, so methylamine accepts a proton from water:

$\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-$

The equilibrium lies mostly to the left, but the small amount of $\text{OH}^-$ produced makes an aqueous solution of methylamine weakly alkaline, in the same way that aqueous ammonia is weakly alkaline.

Part (b): Why methylamine is a stronger base than ammonia

Both ammonia and methylamine act as bases through the lone pair on nitrogen, but the strength of that base behaviour depends on how available (electron-rich) the lone pair is.

In methylamine, the methyl group is a weak electron-donating group (it has a positive inductive effect, or +I effect): it pushes electron density along the C–N bond and onto the nitrogen atom. This makes the nitrogen lone pair more electron-dense, so it forms a bond to an incoming $\text{H}^+$ ion more readily than the lone pair on ammonia’s nitrogen (which has no such electron-donating group attached).

So the order of increasing base strength is: $\text{ammonia} < \text{methylamine}$

Part (c): Why phenylamine is a much weaker base than ammonia

In phenylamine, $\text{C}_6\text{H}_5\text{NH}_2$, the nitrogen atom is bonded directly to the benzene ring. The lone pair on this nitrogen atom is close enough to the ring’s delocalised pi electron system that it partly overlaps with (delocalises into) it, spreading some of the electron density from the nitrogen lone pair around the ring.

This has two consequences that both weaken phenylamine as a base compared with ammonia:

1. The nitrogen lone pair is less concentrated on the nitrogen atom, so it is less able to bond to an incoming proton.
2. If the lone pair were used to bond to $\text{H}^+$, it would have to withdraw from the ring, breaking the delocalisation and losing the extra stability (resonance energy) it provides. This makes protonation energetically less favourable.

For both reasons, phenylamine is a much weaker base than ammonia, which has no ring to delocalise its lone pair into. Overall, for these three species, base strength increases in the order:

$\text{phenylamine} < \text{ammonia} < \text{methylamine}$

Final answers

• (a) Methylamine’s nitrogen lone pair accepts a proton: $\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-$.
• (b) Ammonia < methylamine; the electron-donating methyl group increases the electron density of the nitrogen lone pair, making it more available to accept a proton.
• (c) Phenylamine is a much weaker base than ammonia because its nitrogen lone pair delocalises into the benzene ring, reducing its availability to accept a proton and making protonation energetically unfavourable.