Functional groups in organic chemistry: names and properties

End of September 2026: in Year 5 of the liceo classico (the Italian high school focused on Latin and Ancient Greek), organic chemistry has just opened the year with hybridisation and hydrocarbons, and functional groups arrive between late October and November, with the test typically falling between November and December — so seven to ten weeks from now. At the classical high school, with 2 hours a week, this block moves fast: there is no time for reaction mechanisms in detail, and the test asks above all for nomenclature and properties. I have already written about the starting point — hybridisation, isomerism, hydrocarbons — in Organic chemistry in Year 5: hybridisation, isomerism, hydrocarbons: that post is aimed at the science-oriented liceo scientifico, but the basic theory is identical, and here I pick up where it ends, without repeating it.

What a functional group is

A functional group is an atom or a small group of atoms, other than the plain skeleton of carbon and hydrogen, that determines the chemical behaviour of an organic molecule: how it reacts, at what temperature it boils, whether it dissolves in water. Two molecules with the same functional group behave in a similar way even if their carbon chains have different lengths — which is why it makes sense to study functional groups as categories, rather than molecule by molecule.

The main functional groups

$\mathrm{R}$ stands for a generic remainder of the carbon chain (an alkyl group), $\mathrm{Ar}$ for an aromatic ring, $\mathrm{X}$ for a halogen.

The order of priority of the groups

When a molecule contains more than one functional group (rare at this level, but it can happen in a more advanced exercise), the IUPAC name uses as its suffix only the group with the highest priority, and treats all the others as prefixes. The order of priority, from highest to lowest, among the groups seen above is:

For example, a molecule that has both an $-\mathrm{OH}$ group and a $-\mathrm{COOH}$ group is named as an acid (suffix -oic), with the hydroxyl indicated by the prefix hydroxy-: that is how, for instance, lactic acid is named in systematic nomenclature.

The procedure for giving the IUPAC name

Step 1 — identify the functional group with the highest priority. It is the one that determines the main suffix.

Step 2 — find the longest carbon chain that contains the carbon of the functional group. The longest chain overall is not enough: it must include the atom (or atoms) of the main functional group.

Step 3 — number the chain starting from the end closest to the functional group, so that it receives the lowest possible number. For aldehydes and carboxylic acids the carbon of the group is almost always C1, and often the number is omitted because it cannot be anywhere else.

Step 4 — apply the suffix (replacing the final -e of the alkane name with the group’s suffix, for example propan + ol = propanol) and add any substituents as prefixes, in alphabetical order and with the position indicated by a number.

The most frequent mistake here is numbering the chain from the wrong end relative to the functional group: counting from wherever you happen to start instead of from the end that gives the group the lowest number. A butan-2-ol numbered from the wrong end becomes (wrongly) “butan-3-ol”, which simply does not exist as a correct name for that molecule.

Physical properties: hydrogen bonding decides everything

The physical property the test insists on most is the boiling point, and the explanation almost always goes through hydrogen bonding.

A hydrogen bond forms when a hydrogen atom bonded to a very electronegative atom (oxygen or nitrogen) is attracted to a lone pair of electrons on an electronegative atom of another molecule. It is weaker than a covalent bond, but much stronger than ordinary intermolecular forces (van der Waals forces), and breaking it — that is, making the substance boil — takes more energy.

A comparison between molecules of similar mass:

• an alkane (for example butane, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$) has no hydrogen bonds between its molecules: it boils at a low temperature (about $-1{}^{\circ }\mathrm{C}$);
• an ether (for example methoxyethane, with the same molecular formula as butanol) has an oxygen but no hydrogen bonded directly to it: it does not form hydrogen bonds between its own molecules (the oxygen can only accept them, not donate them), and boils at a temperature only slightly higher than an alkane of similar weight;
• an alcohol (for example butan-1-ol) has an $-\mathrm{OH}$ group: it forms hydrogen bonds between its own molecules, and boils at a distinctly higher temperature (about $118{}^{\circ }\mathrm{C}$);
• a carboxylic acid (for example butanoic acid) forms particularly strong hydrogen bonds, often in pairs of molecules (dimers), and boils at an even higher temperature (about $164{}^{\circ }\mathrm{C}$) than the corresponding alcohol.

The same logic explains solubility in water: small molecules with an $-\mathrm{OH}$ or $-\mathrm{COOH}$ group also form hydrogen bonds with water molecules, so they dissolve well; as the carbon chain (the “uncharged”, hydrophobic part) gets longer, solubility falls, because the non-polar part of the molecule weighs more and more against the polar functional group.

Acidity and basicity: why carboxylic acids are acids and amines are bases

A carboxylic acid gives up the proton of the $-\mathrm{OH}$ easily because the ion formed after the loss (the carboxylate ion, $\mathrm{R}\text{-}\mathrm{CO}{\mathrm{O}}^{-}$) is stabilised by resonance: the negative charge does not stay fixed on a single oxygen, but is delocalised over both oxygens of the group. This extra stability is the reason why a carboxylic acid is distinctly more acidic than a simple alcohol, even though both have an $-\mathrm{OH}$.

An amine behaves as a base because the nitrogen has a lone pair of electrons (not engaged in bonds), available to accept a proton from an acid — exactly as ammonia, $\mathrm{N}{\mathrm{H}}_3$, does, of which amines are the organic derivatives.

The typical test reactions

Oxidation of alcohols. It is the reaction most often asked about in this block, and it depends on the type of alcohol:

• a primary alcohol (the carbon bearing the $-\mathrm{OH}$ is bonded to only one other carbon) is oxidised first to an aldehyde, then — if oxidation continues — to a carboxylic acid;
• a secondary alcohol (the carbon bearing the $-\mathrm{OH}$ is bonded to two other carbons) is oxidised to a ketone, and stops there: the ketone is not oxidised further with common reagents;
• a tertiary alcohol (the carbon bearing the $-\mathrm{OH}$ is bonded to three other carbons) is not oxidised, because there is no hydrogen on the carbon of the $-\mathrm{OH}$ to remove in the oxidation step.

Esterification. A carboxylic acid and an alcohol react to form an ester and water:

The $-\mathrm{OH}$ group of the acid and the hydrogen of the alcohol combine to give the water molecule, while the oxygen of the alcohol stays on as the bridge between the two organic parts in the ester. You are not asked to know the mechanism in detail at this level (this post does not cover nucleophilic substitution or electrophilic addition: they are separate topics), but being able to recognise the reactants and products of an esterification is almost always required.

The typical mistake, in short

Besides numbering the chain from the wrong end, the most common mistake is confusing aldehyde and ketone: both have a carbonyl group ($\mathrm{C}=\mathrm{O}$), but in the aldehyde the carbonyl is always at one end of the chain (bonded to at least one hydrogen), whereas in the ketone it is always inside the chain (bonded to two carbon groups, never to a hydrogen). Looking at whether the carbonyl is at the start of the chain or in the middle is almost always enough to avoid going wrong.

Exercise 1 — give the IUPAC name

Give the IUPAC name of: (a) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$, (b) $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, (c) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}$.

(a) Three carbons, $-\mathrm{OH}$ group on the first carbon of the chain: propane + suffix -ol, with the number 1 (the only possible position on a three-carbon chain with the $\mathrm{OH}$ at one end) → propan-1-ol.

(b) Four carbons, carbonyl group on the second carbon (inside the chain, so a ketone): butane + suffix -one, numbering from the end that gives the carbonyl the lowest number (2, not 3) → butan-2-one.

(c) Three carbons, $-\mathrm{COOH}$ group that by definition occupies C1: propane + suffix -oic, with “acid” after it → propanoic acid.

Exercise 2 — rank by boiling point

Rank from lowest to highest boiling point: pentane (${\mathrm{C}}_5{\mathrm{H}}_{12}$), pentan-1-ol, pentanoic acid. All three molecules have five carbon atoms, so similar molecular mass: the difference lies only in the functional group.

Pentane has no hydrogen bonds: it is the lowest. Pentan-1-ol forms hydrogen bonds through the $-\mathrm{OH}$: intermediate. Pentanoic acid forms stronger hydrogen bonds, often in dimers: the highest.

Exercise 3 — predict the oxidation products

Predict the product of the oxidation of (a) butan-1-ol (primary alcohol) and (b) butan-2-ol (secondary alcohol).

(a) Butan-1-ol is a primary alcohol: it is oxidised first to butanal (an aldehyde, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$), and with further oxidation to butanoic acid ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}$).

(b) Butan-2-ol is a secondary alcohol (the carbon bearing the $\mathrm{OH}$ is bonded to two carbons): it is oxidised to butan-2-one ($\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, a ketone), and stops there — there is no further oxidation with the reagents seen in class.

How to practise after this post

There is no dedicated chemistry practice test yet: you will find timed tests in other subjects at timed high-school tests. If you want to review hybridisation, isomerism and the naming of hydrocarbons first, the reference post is Organic chemistry in Year 5: hybridisation, isomerism, hydrocarbons; for other subjects see the sciences page. After functional groups come the mechanisms, in Organic reactions: electrophilic addition, SN1 and SN2; for the pH of carboxylic acids there is Acids, bases and pH in Year 4: how to calculate pH.

If you still have doubts about functional groups or nomenclature, you can book a free 30-minute introductory call to work out together where to restart.