§-Syllabus dot point

NSWChemistrySyllabus dot point

Inquiry Question 6: How do amines and amides form, and how do their properties differ from other organic compounds?

Investigate the structural formulae, classification, properties and formation of amines and amides

A focused answer to the HSC Chemistry Module 7 dot point on amines and amides. Classifying primary, secondary, tertiary amines, the basicity of amines, formation of amides by condensation of an amine with a carboxylic acid, and worked HSC past exam questions.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

In plain English

Think of nitrogen in these molecules like a person who can either shake hands (share electrons in a normal bond) or lend an arm for someone to hang off (donate its lone pair to grab a stray proton). An amine is just ammonia with one, two or three of its hydrogens swapped for carbon chains, so it still has that free lone pair available, which is why amines act as weak bases, just like ammonia does, and why small amines smell fishy the way ammonia-based cleaning products do. An amide is different: the nitrogen has been bolted directly onto a carbonyl carbon (a carbon double-bonded to oxygen), so that lone pair gets pulled towards the oxygen instead and is no longer free to grab a proton, which is why amides are not bases the way amines are. Making an amide is exactly like making an ester: an acid and a "partner" molecule join up and kick out a water molecule as the leftover; for esters the partner is an alcohol, for amides the partner is an amine. That single swap, alcohol for amine, is the entire difference between the two reactions, and it is also literally how your body links amino acids together into every protein you are made of.

What this dot point is asking

NESA wants you to identify and classify primary, secondary and tertiary amines, write equations for the formation of amines and amides, explain why amines are weak bases analogous to ammonia, and contrast the properties of amines and amides with other nitrogen-free organic compounds.

The answer

Amines: structure and classification

An amine has a nitrogen with at least one $N-H$ or $N-C$ bond and no carbonyl on that nitrogen. Classify by counting how many carbons are bonded to the nitrogen.

Type	Formula	Example
Primary (1 degrees)	$R-NH_2$	ethanamine $CH_3CH_2NH_2$
Secondary (2 degrees)	$R-NH-R'$	$N$ -methylethanamine $CH_3CH_2NHCH_3$
Tertiary (3 degrees)	$R-NR'-R''$	$N,N$ -dimethylethanamine $CH_3CH_2N(CH_3)_2$

Note this classification is by nitrogen substitution count, which differs from the alcohol classification (which counts substitution on the carbon bearing the OH). A primary amine simply means one carbon on the nitrogen, regardless of where on the chain.

Naming. For a primary amine, name as alkanamine with a locant for the nitrogen-bearing carbon. For secondary and tertiary, name the parent amine after the longest chain and use $N-$ locants for the other substituents.

Amides: structure and classification

An amide has a nitrogen directly attached to a carbonyl carbon: $R-CO-NR'R''$ . Classify by counting how many carbons are on the nitrogen (the same as amines, ignoring the carbonyl-attached carbon for classification purposes in many texts; HSC convention varies, but the safe call is to say "primary amide has $-CONH_2$ , secondary has $-CONHR$ , tertiary has $-CONR_2$ ").

Naming: replace the $-oic\ acid$ of the parent carboxylic acid with $-amide$ . So $CH_3CONH_2$ is ethanamide, $CH_3CH_2CONH_2$ is propanamide.

Physical properties

Boiling points. Amines with $N-H$ bonds hydrogen-bond, so primary and secondary amines boil above hydrocarbons of similar molar mass. Tertiary amines have no $N-H$ and cannot donate hydrogen bonds (though they can accept), so they boil lower than primary/secondary amines.

The $N-H \cdots N$ hydrogen bond is weaker than $O-H \cdots O$ because nitrogen is less electronegative than oxygen. So amines boil below alcohols of similar molar mass.

Amides, despite the carbonyl, have unusually high boiling points because of strong $N-H \cdots O=C$ hydrogen bonds. Ethanamide is a solid at room temperature (mp 82 degrees C), while ethanamine is a gas.

Solubility. Small amines (up to about C4) are very soluble in water through hydrogen bonding. Aliphatic amines have a characteristic ammonia-like or fishy smell. Decaying flesh produces low-molar-mass amines such as putrescine ( $H_2N(CH_2)_4NH_2$ ) and cadaverine ( $H_2N(CH_2)_5NH_2$ ), responsible for the smell.

Amines as weak bases

The lone pair on nitrogen can accept a proton, making amines bases (analogous to ammonia):

R-NH_2 + H_2O \rightleftharpoons R-NH_3^+ + OH^-

Alkyl groups donate electron density to the nitrogen, making the lone pair more available; this makes aliphatic amines slightly stronger bases than ammonia ( $K_b$ for ethylamine is about $5 \times 10^{-4}$ , versus $1.8 \times 10^{-5}$ for ammonia). Aromatic amines like aniline are weaker than ammonia because the lone pair is delocalised into the ring.

Amines react with acids to form ammonium salts, just as ammonia does:

CH_3CH_2NH_2 + HCl \rightarrow CH_3CH_2NH_3^+Cl^-

This salt formation is the basis of pharmaceutical formulations: many drugs (codeine, morphine, ephedrine) are basic amines administered as their water-soluble hydrochloride salts.

Amine basicity and amide non-basicity, at a glance

Amines are weak bases because the nitrogen lone pair can accept a proton, exactly like ammonia: $R-NH_2 + H_2O \rightleftharpoons R-NH_3^+ + OH^-$ .
Alkyl groups increase basicity (electron donation makes the lone pair more available); aliphatic amines have $K_b$ values around $10^{-3}$ to $10^{-4}$ , larger than ammonia's $1.8 \times 10^{-5}$ .
Aryl groups decrease basicity (the lone pair delocalises into the ring); aniline is a weaker base than ammonia.
Amides are NOT basic in the HSC sense: the nitrogen lone pair is drawn towards the adjacent carbonyl (delocalised over the $C=O$ ), so it is not available to accept a proton.
$pOH = -\log_{10}[OH^-]$ and $pH + pOH = 14.00$ at 25 degrees C, same relationships as for any weak base calculation.

Formation of amines

The HSC scope includes two pathways:

Reaction of a haloalkane with ammonia (substitution). Heat a haloalkane with concentrated ammonia in ethanol under pressure:

CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_2 + HBr

The reaction is hard to stop at the primary amine; further substitutions give secondary, tertiary amines and a quaternary ammonium salt. Excess ammonia favours the primary amine.

Reduction of an amide. Heat an amide with $LiAlH_4$ in dry ether to give the corresponding amine. This pathway is mentioned in some HSC references but the haloalkane route is more commonly tested.

Formation of amides

An amide forms by condensation of a carboxylic acid with ammonia or an amine. The initial salt loses water on heating:

R-COOH + R'-NH_2 \rightarrow R-COO^-R'-NH_3^+ \xrightarrow{\Delta} R-CO-NH-R' + H_2O

Or written as the overall condensation, releasing water:

R-COOH + H-NHR' \rightarrow R-CO-NHR' + H_2O

This is the same kind of condensation as esterification (acid plus alcohol gives ester plus water), but the alcohol is replaced by an amine and the product is an amide. The reaction is the laboratory basis for forming the amide bond (also called the peptide bond when between amino acids), which links amino acids into proteins.

Amides in polymers

The amide linkage is the repeating unit in polyamides such as nylon 6,6 (made from 1,6-diaminohexane and hexanedioic acid) and proteins (made from amino acids). See the polymers dot point for full equations.

An owned illustrative bar chart shows how base strength (as $K_b$ ) changes with the substituents on nitrogen, which is exactly the reasoning tested in comparison questions:

Model calculations and structures

Worked example 1: pH of a weak amine base solution.

Calculate the pH of a 0.20 mol L $^{-1}$ solution of ethylamine, $K_b = 5.6 \times 10^{-4}$ .

Set up: $[OH^-] = \sqrt{K_b \times c} = \sqrt{(5.6 \times 10^{-4})(0.20)} = \sqrt{1.12 \times 10^{-4}} = 1.058 \times 10^{-2}\ \text{mol L}^{-1}$ .
$pOH = -\log_{10}(1.058 \times 10^{-2}) = 1.975$ .
$pH = 14.00 - 1.975 = 12.03$ (to 2 decimal places).

Marker's note: full marks need the equilibrium equation shown, the approximation stated explicitly, and the final pH rounded consistently with the data given (2 significant figures in $K_b$ and $c$ justify 2 decimal places in pH).

Worked example 2: naming and drawing a secondary amide.

Draw the structure and give the IUPAC name of the amide formed from propanoic acid and methylamine.

Propanoic acid: $CH_3CH_2COOH$ . Methylamine: $CH_3NH_2$ .
Condensation releases water and forms a C-N bond: $CH_3CH_2CONHCH_3$ .
The parent chain is propanamide (from propanoic acid); the methyl group on nitrogen is indicated with the italic locant $N$ : N-methylpropanamide.

Marker's note: markers check that the $N-$ locant (not a numeral) is used for substituents on the amide nitrogen, distinguishing it from a substituent on the carbon chain.

Worked example 3: distinguishing an amine salt from a free amine.

Explain why aqueous methylammonium chloride ( $CH_3NH_3^+Cl^-$ ) does not turn red litmus blue, while aqueous methylamine does.

Methylamine has a free nitrogen lone pair that accepts a proton from water, releasing $OH^-$ and turning red litmus blue.
In methylammonium chloride, the nitrogen has already accepted a proton (it is $CH_3NH_3^+$ ), so there is no lone pair left to react with water in the same way; the solution is close to neutral to weakly acidic (the ion is a weak acid itself), so red litmus does not turn blue.

Marker's note: the key discriminator markers look for is explicitly stating that protonation has already occurred and removed the available lone pair, not just asserting "it is a salt so it is neutral."

Examples in context

Example 1. Nylon 6,6 manufacturing at Invista Lithgow. The Lithgow Invista plant produces nylon 6,6 by condensation of hexane-1,6-diamine with hexanedioic acid (adipic acid). Each amide linkage forms with elimination of water: $-COOH + H_2N- \rightarrow -CONH- + H_2O$ . The repeating monomer is shown as $[-NH(CH_2)_6NH-CO(CH_2)_4CO-]_n$ . The amide hydrogen bonds link parallel chains, giving the fibre its tensile strength used in everything from tyre cord to carpet backing. Students predicting the amide linkage from a given diamine and diacid in HSC questions are doing exactly the chemistry the plant performs at 270 degrees C under nitrogen.

Example 2. Paracetamol production and the amide linkage. Paracetamol (acetaminophen) is the most prescribed analgesic in Australian pharmacies. Its structure contains an amide group formed from acetic acid and 4-aminophenol: $CH_3COOH + H_2N{-}C_6H_4{-}OH \rightarrow CH_3CONH{-}C_6H_4{-}OH + H_2O$ . The amide linkage is hydrolysed slowly in the body by liver enzymes, producing 4-aminophenol metabolites and explaining the dose-dependent hepatotoxicity that NSW Poisons advises clinicians about. The HSC formation-of-amide rule from a carboxylic acid plus an amine plus loss of water explains the molecule's structure cleanly.

Try this

Q1. Classify each as a primary, secondary or tertiary amine: methylamine, dimethylamine, trimethylamine. State whether each can hydrogen bond. [3 marks]

Cue. Methylamine primary (yes), dimethylamine secondary (yes), trimethylamine tertiary (no, no N-H bond).

Q2. Calculate the pH of a 0.10 mol L $^{-1}$ aqueous solution of methylamine, $K_b = 4.4 \times 10^{-4}$ . [3 marks]

Cue. Weak base ICE: $[OH^-] = \sqrt{K_b \times c} = \sqrt{4.4 \times 10^{-5}} = 6.6 \times 10^{-3}$ mol L $^{-1}$ ; $pOH = 2.18$ ; $pH = 11.82$ .

Q3. Ethanoic acid is reacted with methylamine to form an amide. (a) Write the structural equation. (b) Name the amide. (c) Identify one industrially important polymer containing the amide linkage and state its monomers. [2+1+2 marks]

Cue. (a) $CH_3COOH + CH_3NH_2 \rightarrow CH_3CONHCH_3 + H_2O$ . (b) N-methylethanamide. (c) Nylon 6,6 from hexane-1,6-diamine and hexanedioic acid.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC3 marksCompare the boiling points of ethylamine (17°C), ethanol (78°C) and propane (-42°C), all of similar molar mass, in terms of intermolecular forces.

Show worked answer →

Molar masses are similar (45, 46, 44 g/mol). The differences come from the strength of hydrogen bonding.

Propane has only C-H and C-C bonds. The only IMF is dispersion. Lowest boiling point.

Ethylamine has a polar $N-H$ bond, so molecules can hydrogen bond. However, nitrogen is less electronegative than oxygen, so $N-H \cdots N$ hydrogen bonds are weaker than $O-H \cdots O$ hydrogen bonds. Intermediate boiling point.

Ethanol has a polar $O-H$ bond. $O-H \cdots O$ hydrogen bonds are stronger than $N-H \cdots N$ . Highest boiling point of the three.

Order: propane < ethylamine < ethanol. The trend follows hydrogen-bond strength, which scales with the electronegativity of the heteroatom (N 3.04, O 3.44).

Markers reward (1) identifying the IMF in each, (2) comparing N-H and O-H hydrogen bonds, (3) explaining the trend with electronegativity.

2018 HSC2 marksWrite a balanced equation for the formation of ethanamide from ethanoic acid and ammonia. State the conditions and classify the product by amide type.

Show worked answer →

Conditions: heat the carboxylic acid with ammonia (or an amine). Initially the ammonium salt forms; on continued heating, water is lost to give the amide.

CH_3COOH + NH_3 \rightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O

Or written directly as the overall condensation:

CH_3COOH + NH_3 \rightarrow CH_3CONH_2 + H_2O

The product ethanamide $CH_3CONH_2$ is a primary amide (the nitrogen carries two hydrogens and one carbonyl-attached carbon).

Markers reward (1) the balanced equation, (2) the condition (heat), (3) the primary amide classification.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFor each compound, state whether it is a primary, secondary or tertiary amine or amide, and give its IUPAC name: (a) CH3CH2NH2; (b) (CH3)3N; (c) CH3CH2CONH2.

Show worked solution →

(a) CH3CH2NH2: One carbon on nitrogen: primary amine. Name: ethanamine.
(b) (CH3)3N: Three carbons on nitrogen: tertiary amine. Name: N,N-dimethylmethanamine (trimethylamine).
(c) CH3CH2CONH2: Nitrogen attached to a carbonyl carbon with two hydrogens on nitrogen: primary amide. Name: propanamide.

Marking criteria: 1 mark per compound for the correct classification AND the correct IUPAC/common name together; a classification without the matching name earns no mark for that part.

foundation4 marksWrite a balanced equation, with conditions, for the formation of N-ethylethanamide from ethanoic acid and ethylamine. Classify the product amide and state one physical property that distinguishes it from ethylamine.

Show worked solution →

Equation with condition (heat drives off water):

CH_3COOH + CH_3CH_2NH_2 \xrightarrow{\Delta} CH_3CONHCH_2CH_3 + H_2O

Classification. The nitrogen carries one carbon from the ethyl group and one hydrogen (the other bond is to the carbonyl carbon), so this is a secondary amide.

Distinguishing property. N-ethylethanamide has a much higher boiling point than ethylamine of similar molar mass, because the amide N-H...O=C hydrogen bonding network is stronger than the N-H...N network in the amine; ethylamine is a gas near room temperature while comparable amides are typically liquids or solids.

Marking criteria: 1 mark for the correct structural formula of the product, 1 mark for balancing with water as the co-product, 1 mark for stating heat as the condition, 1 mark for the correct secondary amide classification with a correctly reasoned physical property.

core5 marksA 250 mL aqueous solution contains 0.0246 mol of methylamine,

K_b = 4.4 \times 10^{-4}

. Calculate the pH of the solution to 2 decimal places, assuming the approximation

[OH^-] \approx \sqrt{K_b c}

is valid.

Show worked solution →

Step 1: concentration of methylamine.

c = \frac{n}{V} = \frac{0.0246\ \text{mol}}{0.250\ \text{L}} = 0.0984\ \text{mol L}^{-1}

Step 2: set up the equilibrium.

CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-

K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} \approx \frac{[OH^-]^2}{c}

Step 3: solve for [OH-].

[OH^-] = \sqrt{K_b \times c} = \sqrt{(4.4 \times 10^{-4})(0.0984)} = \sqrt{4.330 \times 10^{-5}} = 6.580 \times 10^{-3}\ \text{mol L}^{-1}

Step 4: pOH.

pOH = -\log_{10}(6.580 \times 10^{-3}) = 2.182

Step 5: pH at 25 degrees C ( $pK_w = 14.00$ ).

pH = 14.00 - 2.182 = 11.82

Marking criteria: 1 mark for correct concentration, 1 mark for the correct $K_b$ expression with the approximation, 1 mark for correctly solving for [OH-], 1 mark for pOH, 1 mark for the final pH to 2 decimal places with the approximation validity noted (valid since $c / K_b > 100$ ).

core5 marksThe IR spectrum below is an owned illustrative spectrum of an unknown nitrogen-containing organic compound Y with molecular formula

C_2H_5NO

. Identify the functional group class of Y using two labelled peaks, give its structure and IUPAC name, and explain why Y is NOT an amine.

Show worked solution →

Reading the spectrum. Two features are labelled: a pair of medium peaks near 3350 and 3180 $cm^{-1}$ , and a strong, sharp peak near 1680 $cm^{-1}$ .

Interpretation.

The pair of N-H stretches near 3350 and 3180 $cm^{-1}$ indicates two N-H bonds, consistent with a primary nitrogen group (either a primary amine or a primary amide).
The strong sharp peak near 1680 $cm^{-1}$ is a $C=O$ stretch. A primary amine alone would show NO carbonyl peak, so the presence of both the N-H pair AND the carbonyl together is diagnostic of a primary amide, not a primary amine.

Structure. With formula $C_2H_5NO$ and a primary amide group $-CONH_2$ (accounting for $CONH_2$ ), the remaining $CH_3$ fits: $CH_3CONH_2$ , ethanamide.

Why not an amine. An amine has no carbonyl on the nitrogen; Y shows a clear carbonyl absorption near 1680 $cm^{-1}$ in addition to the N-H peaks, so the nitrogen here must be bonded to a carbonyl carbon, which defines an amide, not an amine.

Marking criteria: 1 mark for identifying the N-H pair and its region, 1 mark for identifying the carbonyl peak and its region, 1 mark for correctly reasoning to "primary amide" using BOTH features together, 1 mark for a structure consistent with $C_2H_5NO$ and the correct name ethanamide, 1 mark for explicitly explaining why the carbonyl peak rules out a simple amine.

exam6 marksNylon 6,6 is manufactured by condensing hexane-1,6-diamine with hexanedioic acid. (a) Write a structural equation for the formation of one amide linkage in this reaction, showing the small molecule released. (b) Explain, in terms of intermolecular forces, why nylon 6,6 fibres have high tensile strength. (c) Compare the amide linkage in nylon 6,6 with the peptide bond in a protein.

Show worked solution →

(a) Equation for one amide linkage.

-COOH + H_2N- \rightarrow -CO-NH- + H_2O

Applied to the monomers (one repeat of the condensation):

HOOC(CH_2)_4COOH + H_2N(CH_2)_6NH_2 \xrightarrow{\Delta} [-CO(CH_2)_4CO-NH(CH_2)_6NH-]_n + H_2O

(b) Intermolecular forces and tensile strength. Each amide linkage carries a polar $C=O$ and $N-H$ that can hydrogen bond between adjacent polymer chains. Because the chains are long and regularly spaced, thousands of these $N-H \cdots O=C$ hydrogen bonds form between neighbouring chains, cross-linking them side by side. This dense hydrogen-bonding network resists chains sliding past each other under tension, giving the fibre high tensile strength compared with a polymer held together only by dispersion forces (such as polyethylene).

(c) Comparison with the peptide bond. The peptide bond in a protein is chemically the same amide linkage ( $-CO-NH-$ ), formed by condensation of a carboxylic acid group of one amino acid with the amine group of the next, releasing water. The difference is in the monomers: nylon 6,6 alternates a diamine and a diacid monomer, while a protein uses a single class of monomer (amino acids, each already containing both a $-COOH$ and an $-NH_2$ group), and the side chains (R groups) on each amino acid vary, giving proteins far greater structural and functional diversity than the uniform repeat unit of nylon.

Marking criteria: (a) 1 mark for the correct amide linkage equation, 1 mark for water as the co-product. (b) 1 mark for identifying N-H...O=C hydrogen bonding as the relevant IMF, 1 mark for linking this to resistance to chain slippage/tensile strength. (c) 1 mark for correctly identifying the peptide bond as chemically the same amide linkage, 1 mark for a valid point of difference (monomer type or side-chain diversity).

exam7 marksThe graph below shows [OH-] released over time as excess methylamine solution is progressively titrated into a fixed volume of dilute HCl, monitored by a pH probe, with the equivalence point at

t \approx 25

mL of titrant added. Assess whether the equivalence point of this titration would be expected at pH 7.00, and evaluate why an indicator choice matters for this specific acid-base pair.

Show worked solution →

(a) Assessing the equivalence pH. At equivalence, moles of $HCl$ equal moles of methylamine added, forming the salt methylammonium chloride ( $CH_3NH_3^+Cl^-$ ) in solution. $Cl^-$ is the conjugate base of a strong acid and does not hydrolyse, but $CH_3NH_3^+$ is the conjugate acid of the weak base methylamine and DOES hydrolyse water:

CH_3NH_3^+ + H_2O \rightleftharpoons CH_3NH_2 + H_3O^+

This hydrolysis releases excess $H_3O^+$ , so the equivalence point solution is acidic, meaning the equivalence pH is expected to be BELOW 7.00, not exactly 7.00 as it would be for a strong acid/strong base titration.

(b) Evaluating indicator choice. Because the equivalence point sits below pH 7 (typically in the region of pH 5 to 6 for a weak base/strong acid titration), an indicator with a colour-change range straddling neutral pH 7, such as phenolphthalein (range approximately 8.2 to 10.0), would change colour well AFTER the true equivalence point, over-estimating the volume of acid needed and introducing a systematic error. A more appropriate indicator is methyl orange (range approximately 3.1 to 4.4) or methyl red (range approximately 4.4 to 6.2), which change colour close to the actual acidic equivalence point of this weak base/strong acid combination.

Marking criteria: 1 mark for identifying that the salt formed contains the conjugate acid of a weak base, 1 mark for the correct hydrolysis equation, 1 mark for concluding the equivalence pH is below 7, 1 mark for naming phenolphthalein and explaining why it is unsuitable, 1 mark for naming a suitable alternative indicator (methyl orange or methyl red) with a justified range, 1 mark for an overall evaluative statement linking indicator range to measurement error, 1 mark for correctly reading/describing the shape of the given titration curve (steep rise near equivalence, gradual approach either side).