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

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.

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):

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:

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:

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.

2. 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:

Or written as the overall condensation, releasing water:

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.

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.