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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.

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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 Nβˆ’HN-H or Nβˆ’CN-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) IMATH_7 ethanamine IMATH_8
Secondary (2 degrees) IMATH_9 IMATH_10 -methylethanamine IMATH_11
Tertiary (3 degrees) IMATH_12 IMATH_13 -dimethylethanamine IMATH_14

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 Nβˆ’N- locants for the other substituents.

Amides: structure and classification

An amide has a nitrogen directly attached to a carbonyl carbon: Rβˆ’COβˆ’NRβ€²Rβ€²β€²R-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 βˆ’CONH2-CONH_2, secondary has βˆ’CONHR-CONHR, tertiary has βˆ’CONR2-CONR_2").

Naming: replace the βˆ’oicΒ acid-oic\ acid of the parent carboxylic acid with βˆ’amide-amide. So CH3CONH2CH_3CONH_2 is ethanamide, CH3CH2CONH2CH_3CH_2CONH_2 is propanamide.

Physical properties

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

The Nβˆ’Hβ‹―NN-H \cdots N hydrogen bond is weaker than Oβˆ’Hβ‹―OO-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 Nβˆ’Hβ‹―O=CN-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 (H2N(CH2)4NH2H_2N(CH_2)_4NH_2) and cadaverine (H2N(CH2)5NH2H_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):

Rβˆ’NH2+H2Oβ‡ŒRβˆ’NH3++OHβˆ’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 (KbK_b for ethylamine is about 5Γ—10βˆ’45 \times 10^{-4}, versus 1.8Γ—10βˆ’51.8 \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:

CH3CH2NH2+HClβ†’CH3CH2NH3+Clβˆ’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.

Formation of amines

The HSC scope includes two pathways:

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

CH3CH2Br+NH3β†’CH3CH2NH2+HBrCH_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.

  1. Reduction of an amide. Heat an amide with LiAlH4LiAlH_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:

Rβˆ’COOH+Rβ€²βˆ’NH2β†’Rβˆ’COOβˆ’Rβ€²βˆ’NH3+β†’Ξ”Rβˆ’COβˆ’NHβˆ’Rβ€²+H2OR-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:

Rβˆ’COOH+Hβˆ’NHRβ€²β†’Rβˆ’COβˆ’NHRβ€²+H2OR-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.

Common traps

Confusing amine and amide classifications. Amine: βˆ’NH2-NH_2, no adjacent carbonyl. Amide: βˆ’CONH2-CONH_2 or βˆ’CONR2-CONR_2, has a carbonyl. The two are not interchangeable.

Comparing amine basicity with alcohol. Amines are weak bases; alcohols are essentially neutral. Mention the lone pair on nitrogen.

Forgetting the loss of water in amide formation. The condensation releases water, just like esterification. Without that, your equation is unbalanced.

Claiming tertiary amines hydrogen bond as donors. They have no N-H, so they cannot donate. They can still accept hydrogen bonds via the nitrogen lone pair.

Drawing an N-H on a tertiary amine. A tertiary amine has three carbons on N and zero hydrogens.

In one sentence

Amines Rβˆ’NH2R-NH_2 (primary), R2NHR_2NH (secondary), R3NR_3N (tertiary) are weak bases that hydrogen bond (except tertiary) and form from haloalkanes plus ammonia, while amides Rβˆ’CONH2R-CONH_2 form by condensation of a carboxylic acid with an amine, releasing water, and contain the amide linkage that builds nylon and proteins.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

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

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 Nβˆ’HN-H bond, so molecules can hydrogen bond. However, nitrogen is less electronegative than oxygen, so Nβˆ’Hβ‹―NN-H \cdots N hydrogen bonds are weaker than Oβˆ’Hβ‹―OO-H \cdots O hydrogen bonds. Intermediate boiling point.

Ethanol has a polar Oβˆ’HO-H bond. Oβˆ’Hβ‹―OO-H \cdots O hydrogen bonds are stronger than Nβˆ’Hβ‹―NN-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 β†’

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.

CH3COOH+NH3β†’CH3COONH4β†’Ξ”CH3CONH2+H2OCH_3COOH + NH_3 \rightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O

Or written directly as the overall condensation:

CH3COOH+NH3β†’CH3CONH2+H2OCH_3COOH + NH_3 \rightarrow CH_3CONH_2 + H_2O

The product ethanamide CH3CONH2CH_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.

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