Acid-base chemistry of aliphatic amines weak bases pKb Kb values why stronger than aomatic amines reactions with acids primary secondary tertiary balanced neutralisation equations organic nitrogen compounds organonitrogen molecules advanced A level organic chemistry revision notes doc brown

Advanced Level Organic Chemistry Nitrogen compounds: Acid-base chemistry of amines

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Part 8. The chemistry of organic nitrogen (organonitrogen) compounds

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All my advanced level organo-nitrogen compound chemistry notes

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Part 8.4 The acid-base chemistry of aliphatic amines, their comparative strength as bases and reactions with acids

Sub-index for this page

8.4.1 The ionisation of aliphatic amine bases in water
8.4.2 The strength of aliphatic bases - database of pK_b and K_b values

8.4.3 The reaction of aliphatic bases with acids - neutralisation and salt formation

8.4.4 Conjugate acids - why are salts solutions of amine bases acidic?

More on Definition of a weak base, theory and examples of K_b, pK_b, K_w weak base CALCULATIONS

8.4.1 The ionisation of aliphatic amine bases in water

Reminders:

The term weak base refers to one that only ionises (dissociates) to a small extent in aqueous solution.

A base is defined as a proton acceptor (See Lewis and Bronsted–Lowry acid–base theories)

e.g. ammonia is a weak inorganic base in water: NH_3(aq) + H₂O_(l) == ~2% ==> NH₄⁺(aq) + OH^-_(aq)

A strong base like sodium hydroxide ionises completely: NaOH(s) + aq == 100% ==> Na⁺(aq) + OH^-(aq)

The equilibrium involved when a weak base B dissolves in water:

(i) :B_(aq) + H₂O_(l) BH⁺(aq) + OH^-_(aq)

(for any soluble base, note the lone pair on N: which makes this molecule a base, and BH⁺ is the conjugate acid - proton donor)

The B:→H bond is a dative covalent bond (or co-ordinate bond) - the proton from the acid accepts the pair of electrons donated from the base e.g. the non-bonding lone pair of electrons on the nitrogen atom of ammonia or aliphatic/ary organic bases.

(ii) :NH_3(aq) + H₂O_(l) NH₄⁺(aq) + OH^-_(aq)

(for ammonia giving an alkaline solution, the NH₄⁺ ion is the conjugate acids of the base NH₃)

ionisation of ammonia in water Kb pKb values diagram balanced equation of ionization ammonium ion hydroxide ion

(iii) CH₃CH₂NH_2(aq) + H₂O_(l) CH₃CH₂NH₃⁺(aq) + OH^-_(aq)

(for ethylamine, primary aliphatic amine base, the cation formed is called the ethylammonium ion)

ionisation of prmary aliphatic amine ethylamine in water Kb pKb values diagram balanced equation of ionization ethylammonium ion hydroxide ion

From equation (iii) for a weak organic base, the ionisation constant, the equilibrium constant (K_b) for equilibrium (iii), is given by the expression (water is a subsumed constant):

K_b =

[CH₃CH₂NH₃⁺_(aq)] [OH^–_(aq)]

–––––––––––––––––––––––––– mol dm^-3

[CH₃CH₂NH_2(aq)]

Like pH, the range of K_b is so great, it is often quoted on the logarithmic scale where
pK_b = -log₁₀(K_b/mol dm^-3) and K_b = 10^-pKb

The stronger the base, the greater the K_b and the lower the pK_b value.

More on Definition of a weak base, theory and examples of K_b, pK_b, K_w weak base CALCULATIONS

TOP OF PAGE and sub-index

8.4.2 The strength of aliphatic bases - database of pK_b and K_b values

Aliphatic amines are weak bases, but generally stronger bases than aromatic bases like phenylamine

Name	Structure	pK_b	K_b/moldm^-3	Solubility in water	Comments Ammonia and a primary aromatic amine are added for comparison purposes
ammonia	NH₃	4.75	1.78 x 10^-5	very high	Inorganic base
phenylamine	C₆H₅NH₂	9.38	4.16 x 10^-10	relatively low, 3.6g/100g water	Aromatic organic amine base, much weaker than simple aliphatic amine bases listed below.
methylamine	CH₃NH₂	3.38	4.17 x 10^-4	~miscible	Primary aliphatic amine base ...
ethylamine	CH₃CH₂NH₂	3.27	5.37 x 10^-4	~miscible	Primary aliphatic amine base
propylamine	CH₃CH₂CH₂NH₂	3.36	4.37 x 10^-4	~miscible	Primary aliphatic amine base
butylamine	CH₃CH₂CH₂CH₂NH₂	3.39	4.07 x 10^-4	~miscible	Primary aliphatic amine base
Three for comparison
propylamine	CH₃CH₂CH₂NH₂	3.36	4.37 x 10^-4	~miscible	A series of three structurally isomeric molecules (based on C₃H₉N) in the sequence: primary ==> secondary ==> tertiary aliphatic amines
N-ethylmethylamine	CH₃CH₂NHCH₃	3.24	5.75 x 10^-4	very high
trimethylamine	(CH₃)₃N	4.20	6.31 x 10^-5	very high

Comments on the data table

(a) The simpler aliphatic amines are stronger bases than the simplest aromatic amines.

pK_b(aliphatic) < pK_b(aromatic) and K_b(aliphatic) > K_b(aromatic)
In the case of aromatic amines, the benzene ring has an minus inductive effect (-I electron shift) making the non-bonding electron pair on the nitrogen is less energetically favourable for protonation.

This weakens their basic character, increasing so if there are very electronegative substituents in the benzene ring.

However, in aliphatic amines, the alkyl groups have a plus inductive effect (+I electron shift) so the non-bonding electron pair on the nitrogen is more energetically favourable for protonation, making them stronger bases.

Remember we are talking about the relative strengths of weak bases, but there is a wide range in terms of K_b and pKb values (that is why the latter is on a logarithmic base 10 scale).

TOP OF PAGE and sub-index

8.4.3 The reaction of aliphatic bases with acids - neutralisation and salt formation

When ammonia or an organic amine base are neutralised, the base is protonated by the acid to give

protonation of inorganic base ammonia in water diagram balanced equation of formation of conjugate acid ammonium ion cation

ammonia == protonation ==> ammonium ion

protonation of base primary aliphatic amine ethylamine in water diagram balanced equation of formation of conjugate acid ethylammonium ion cation

ethylamine == protonation ==> ethylammonium ion

Therefore, you can write neutralisation equations with strong mineral acids, just as you do with ammonia e.g.
NH₃(aq) + HCl(aq) ===> NH₄⁺(aq) + Cl^-(aq) == evap. ==> NH₄⁺Cl^-(s)

(preparation of ammonium chloride)

2NH₃(aq) + H₂SO₄(aq) ===> 2NH₄⁺(aq) + SO₄^2-(aq) == evap. ==> (NH₄⁺)₂SO₄^2-(s)

(preparation of ammonium sulfate)

CH₃CH₂NH₂(aq) + HCl(aq) ===> CH₃CH₂NH₃⁺(aq) + Cl^-(aq) == evap. ==> CH₃CH₂NH₃⁺Cl^-(s)

(preparation of ethylammonium chloride, note the names of the alkylammonium ions in these examples)

2CH₃CH₂NH₂(aq) + H₂SO₄(aq) ==> 2CH₃CH₂NH₃⁺(aq) + SO₄^2-(aq) = evap. => (CH₃CH₂NH₃⁺)₂SO₄^2-(s)

(preparation of ethylammonium sulfate)

Don't be put off by coming across a cycloalkyl amine e.g. cyclohexylamine

Cyclohexylamine is neutralised by nitric acid to yield the salt, cyclohexylammonium nitrate preparation structural formula equation molecular structure

Cyclohexylamine is neutralised by nitric acid to yield the salt, cyclohexylammonium nitrate

Cyclohexylamine is neutralised by hydrochloric acid to yield the salt, cyclohexylammonium chloride preparation structural formula equation molecular structure

Cyclohexylamine is neutralised by hydrochloric acid to yield the salt, cyclohexylammonium chloride

Cyclohexylamine is neutralised by sulfuric acid to yield the salt, cyclohexylammonium sulfate preparation structural formula equation molecular structure

Cyclohexylamine is neutralised by sulfuric acid to yield the salt, cyclohexylammonium sulfate

Watch the balancing of the equation 2 : 1 mole ratio of base : acid

Note:

(i) Using evaporation, the salts can crystallised out from aqueous solution, as you can with ammonium salts.
(ii) If in an amine preparation, the initial product is a salt, you add strong alkali like aqueous sodium hydroxide to free the base.

CH₃CH₂NH₃⁺(aq) + OH^-(aq) ===> CH₃CH₂NH₂(aq) + H₂O(l)
Unfortunately, aliphatic amines are very soluble in water, so the separation is difficult, even using fractional distillation.
In the case of aromatic amines, they can be separated by steam distillation - see preparation of phenylamine.

TOP OF PAGE and sub-index

8.4.4 Conjugate acids - why are salts solutions of amine bases acidic?

The conjugate acids of weal bases are quite strong, so solutions of organic amine bases are acidic, with pH often <4.

You find the same with ammonium salts e.g. if put a piece of magnesium ribbon in ammonium chloride solution, it fizzes as hydrogen is evolved. Simple proof that the hydrated proton ion (H₃O⁺) was formed.

NH₄⁺(aq) + H₂O(l) ===> NH₃(aq) + H₃O⁺(aq)

Therefore you get exactly the same effect with the ethylammonium ion

CH₃CH₂NH₃⁺(aq) + H₂O(l) ===> CH₃CH₂NH₂(aq) + H₃O⁺(aq)

If the add sodium hydrogencarbonate to an ethylammonium salt solution, carbon dioxide is evolved.

You can write the equation in terms of the ethylammonium ion acting as the acid e.g.

CH₃CH₂NH₃⁺(aq) + HCO₃^-(aq) ===> CH₃CH₂NH₂(aq) + H₂O(l) + CO₂(g)

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