How strongly amines act as bases

A deeper blue

Take three beakers. Put a tenth of a mole of ammonia, methylamine and aniline into each litre of water. Add universal indicator. Ammonia gives a blue-green colour. Methylamine gives a deeper blue. Aniline gives almost no change.

All three are bases, and all three use a lone pair on nitrogen to take a proton from water. The difference is how far each reaction goes. We measure that with a number.

Kb, pKb and pKa

For an amine $\text{RNH}_{2}$ in water:

$\text{RNH}_{2}\ +\ \text{H}_{2}\text{O}\ \rightleftharpoons \ \text{RNH}_{3}^{+}\ +\ \text{OH}^{-}$

A larger $K_b$ means a stronger base. A smaller $\text{p}K_b$ means a stronger base. At 298 K, the $\text{p}K_a$ of the ion $\text{RNH}_{3}^{+}$ and the $\text{p}K_b$ of the amine add up to 14. So a strong base has a weak conjugate acid.

For a solution of concentration $c$, let $x$ be the concentration of $\text{OH}^-$ at equilibrium. Then $K_b = \dfrac{x^2}{c - x}$. This is a quadratic equation in $x$, and its positive root is $x = \dfrac{-K_b + \sqrt{K_b^2 + 4K_bc}}{2}$. Then $\text{pOH} = -\log_{10} x$ and $\text{pH} = 14 - \text{pOH}$.

Keep the concentration at $0.10\ \text{mol/L}$ and move the pKb from 3.38 to 9.38. The pH falls by about three units, not by six. Why? Because $[\text{OH}^-]$ depends on the square root of $K_b$, and the pH changes by half the change in pKb.

Practise with new numbers

The values in water

These are the $\text{p}K_b$ values that NCERT gives, for water at 298 K. A smaller number is a stronger base.

• Ammonia: 4.75.
• Methyl series: dimethylamine 3.27, methylamine 3.38, trimethylamine 4.22.
• Ethyl series: diethylamine 3.00, triethylamine 3.25, ethylamine 3.29.
• Phenylmethanamine (benzylamine): 4.70.
• Aniline: 9.38.

So in water the order is:

• Methyl series: $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$.
• Ethyl series: $(\text{C}_2\text{H}_5)_2\text{NH} > (\text{C}_2\text{H}_5)_3\text{N} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3$.

The two series differ. The tertiary amine is below the primary one for methyl groups, and above it for ethyl groups. Let us see why.

Three effects that act together

1. The alkyl group pushes electrons. An alkyl group gives electron density to the nitrogen through the inductive effect. The lone pair is then more ready to take up a proton. This effect alone gives the order tertiary, secondary, primary, ammonia. It is the order you find in the gas phase, where there is no water.

2. Water stabilises the ion. In water, the ion $\text{RNH}_{3}^{+}$ is held by hydrogen bonds from its N-H bonds to water molecules. This lowers its energy and pulls the equilibrium to the right. A primary ion has three N-H bonds. A secondary ion has two. A tertiary ion has one. So this effect gives the opposite order: primary, then secondary, then tertiary.

3. Crowding. Large alkyl groups get in the way, both of the proton and of the water that surrounds the ion. The effect grows from methyl to ethyl groups.

The three effects are of similar size. In the methyl series, the loss of solvation beats the gain from the third methyl group, so the tertiary amine falls below the primary one. In the ethyl series, the larger push of three ethyl groups keeps the tertiary amine above the primary one. So the order cannot be predicted from one effect alone. Learn the observed orders.

The solvent matters

The order depends on the conditions. In the gas phase, and in non-polar solvents, there is no hydrogen-bond solvation, and the order follows the inductive effect: tertiary, secondary, primary, ammonia. In water, the orders above hold. When a question asks for basic strength without naming a solvent, the answer expected is the aqueous one. Check for the words in water or in the gas phase.

Aniline and the ring

Aniline is far weaker than ammonia. Its $\text{p}K_b$ is 9.38, so $K_b$ is about $4 \times 10^4$ times smaller than that of ammonia.

Look at the lone pair. In aniline it sits next to the pi system of the ring. It spreads into the ring, as in the resonance forms of the next lesson. So it is less ready to take up a proton. The anilinium ion has no lone pair, so it has no such resonance. The free amine is stabilised, and the ion is not. That shifts the equilibrium to the left.

Groups on the ring

Put a group on the ring of aniline, and the ring changes how far the lone pair is held.

• Groups that give electrons make the amine stronger: $\text{-}\text{CH}_{3}$, $\text{-}\text{OCH}_{3}$, $\text{-}\text{NH}_{2}$.
• Groups that pull electrons make it weaker: $\text{-}\text{NO}_{2}$, $\text{-}\text{Cl}$, $\text{-}\text{CN}$, $\text{-}\text{COOH}$.
• Position matters. A nitro group in the para position, by resonance, pulls harder than in the meta position, where it acts only through the inductive effect. So 4-nitroaniline is weaker than 3-nitroaniline.
• The ortho effect. Almost any group in the ortho position, even one that gives electrons, makes aniline weaker. The group gets in the way of the proton and of the water around the ion. So 2-methylaniline is slightly weaker than aniline, although 4-methylaniline is stronger.

The $\text{p}K_a$ values of the ions, in water, are about 5.3 for 4-methoxyaniline, 5.1 for 4-methylaniline, 4.6 for aniline, 4.0 for 4-chloroaniline, 2.5 for 3-nitroaniline and 1.0 for 4-nitroaniline. The larger the $\text{p}K_a$ of the ion, the stronger the amine.

An amide, such as acetanilide, is hardly basic at all. The lone pair on its nitrogen is spread onto the C=O group, so it is not free to take a proton.

Salts of amines

An amine neutralises acids and gives a salt, an alkylammonium salt:

$\text{CH}_{3}\text{NH}_{2}\ +\ \text{HCl}\ \rightarrow \ \text{CH}_{3}\text{NH}_{3}^{+}\text{Cl}^{-}$

The salt is ionic and dissolves in water. Add a strong base, and the free amine comes back:

$\text{CH}_{3}\text{NH}_{3}^{+}\text{Cl}^{-}\ +\ \text{NaOH}\ \rightarrow \ \text{CH}_{3}\text{NH}_{2}\ +\ \text{NaCl}\ +\ \text{H}_{2}\text{O}$

This lets you separate an amine from a neutral compound. Shake the mixture with dilute acid, and the amine goes into the water as its salt. Separate the layers, add alkali to the water, and the amine returns.

An aqueous amine also precipitates the hydroxides of many metals, as ammonia does:

$3\text{RNH}_{2}\ +\ 3\text{H}_{2}\text{O}\ +\ \text{FeCl}_{3}\ \rightarrow \ \text{Fe}(\text{OH})_{3}\ +\ 3\text{RNH}_{3}^{+}\text{Cl}^{-}$

The brown solid is iron(III) hydroxide.

Try it in Python

The program turns each $\text{p}K_b$ into a pH for a $0.100\ \text{mol/L}$ solution, and ranks the bases.

Common mistakes

• Reading pKb the wrong way round. A smaller pKb is a stronger base.
• Using the gas phase order in water. In water, tertiary methylamine is below the primary one.
• Calling every alkylamine stronger than every other. Order depends on the solvent and on crowding.
• Forgetting the ortho effect. A methyl group in the ortho position weakens aniline slightly.
• Giving the pOH as the pH. For a base, pH is 14 minus pOH.

Remember these

• $\text{p}K_a + \text{p}K_b = 14$ at 298 K. Smaller pKb, stronger base.
• In water: $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$ and $(\text{C}_2\text{H}_5)_2\text{NH} > (\text{C}_2\text{H}_5)_3\text{N} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3$.
• In the gas phase: tertiary, secondary, primary, ammonia.
• Aniline is far weaker than ammonia, since its lone pair spreads into the ring.
• Electron-giving groups on the ring raise basic strength, and electron-pulling groups lower it.

References

• NCERT Chemistry Class 12, the chapter Amines: basic character of amines, and the table of pKb values.
• OpenStax Organic Chemistry, Chapter 24 Amines and Heterocycles, sections on the basicity of amines and of substituted arylamines.
• The pKa values of the substituted anilinium ions are standard textbook values (for example in the CRC Handbook of Chemistry and Physics).

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which is the stronger base in water: methylamine (pKb 3.38) or trimethylamine (pKb 4.22)? By how many times?

2. Why is aniline a weaker base than cyclohexylamine?

3. Arrange 4-nitroaniline, aniline and 4-methylaniline in increasing order of basic strength.

4. An amine and a neutral compound are dissolved in ether. How do you separate them?