Organic chemistry in Year 5: hybridisation and isomers

It is the end of September 2026, and in the last year (Year 5) of the Italian science high school (liceo scientifico) the organic chemistry block has just started: from now to December the course moves from hybridisation and isomerism to the naming of hydrocarbons, then to functional groups and polymers. The first test on this part typically falls six to nine weeks from now, so between the beginning of November and mid-December. It is worth taking these first weeks seriously, because natural sciences was an oral subject at the 2026 maturità (Italy’s school-leaving exam), with the internal examiner: what gets fixed now — hybridisation, isomerism, names — is exactly what you end up answering questions on in June, often months after you have “closed” the topic in your notebook.

Carbon hybridisation: where the geometry comes from

Carbon has four valence electrons and always forms four bonds, but the geometry of those bonds changes depending on how its atomic orbitals combine. This is the point that most explanations skip too quickly: hybridisation is not a fact on its own, it is the explanation of why a molecule has a certain shape.

sp3 hybridisation. One $s$ orbital and three $p$ orbitals mix into four identical hybrid orbitals, arranged in space as far from one another as possible: a tetrahedron, with bond angles of about $109.5^{\circ }$. It is the hybridisation of every carbon that forms four single bonds, as in all the alkanes.

sp2 hybridisation. One $s$ orbital and two $p$ orbitals mix into three hybrid orbitals lying in a plane, at $120^{\circ }$ to one another (trigonal planar geometry); the fourth $p$ orbital, not hybridised, stays perpendicular to the plane and forms the $\pi$ bond of a double bond. It is the hybridisation of every carbon involved in a carbon-carbon double bond, as in the alkenes.

sp hybridisation. One $s$ orbital and a single $p$ orbital mix into two hybrid orbitals at $180^{\circ }$ (linear geometry); the two unhybridised $p$ orbitals stay perpendicular to each other and form the two $\pi$ bonds of a triple bond. It is the hybridisation of every carbon involved in a triple bond, as in the alkynes.

The practical rule for recognising the hybridisation of a carbon, looking at its structural formula: count how many “groups” of electrons (single, double or triple bonds, each counted as one group) leave that carbon. Four single groups → sp3. Three groups, one of which is a double bond → sp2. Two groups, one of which is a triple bond (or two double bonds) → sp. You don’t need to draw the orbitals each time: it is enough to count the bonds and recognise whether there is a double or triple bond involving that specific carbon. Hybridisation is a property of the individual atom, not of the whole molecule, and in the same molecule different carbons can have different hybridisations.

Isomerism: same formula, different molecules

Two or more compounds are isomers when they have the same molecular formula but a different structure. It is a concept that seems obvious until you have to tell the types apart, and that is exactly where test mistakes are born.

Chain isomerism. The carbon chain is arranged differently: straight or branched. Butane (${\mathrm{C}}_4{\mathrm{H}}_{10}$) exists as a straight chain (n-butane) or a branched one (isobutane, 2-methylpropane): same molecular formula, different skeleton.

Position isomerism. The carbon skeleton is identical, but a substituent or a functional group sits in a different position on the chain. Pent-1-ene and pent-2-ene both have five carbons and a double bond, but at different points of the chain.

Functional-group isomerism. The same molecular formula corresponds to two completely different functional groups. The classic example is ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$, which can be either ethanol (an alcohol) or dimethyl ether (an ether): two substances with very different chemical and physical properties behind the same molecular formula.

Stereoisomerism. Here the skeleton and the bonds are identical; only the arrangement in space changes. Two cases to keep apart:

• cis-trans (or E-Z) isomerism, which concerns double bonds: a double bond does not allow free rotation of the groups around it, so if each of the two carbons of the double bond carries two different substituents, there are two distinct spatial arrangements — cis, with the two main groups on the same side, and trans, with the two main groups on opposite sides;
• chirality, which concerns a carbon bonded to four substituents that are all different from one another (called a stereogenic centre): that carbon generates two molecules that are non-superimposable mirror images of each other, like a right hand and a left hand.

The most common mistake here is not conceptual, it is one of classification: a student who sees two slightly different structures tends to call them “chain isomers” out of habit, even when they actually differ in the position of a double bond or in their spatial arrangement. Before answering, it is always worth asking yourself: does the skeleton change? It is chain isomerism. Does only the position of something on the same skeleton change? It is position isomerism. Does the functional group itself change? It is functional-group isomerism. Does none of this change, only the orientation in space? It is stereoisomerism.

IUPAC naming of hydrocarbons: the procedure step by step

IUPAC naming of hydrocarbons always follows the same scheme, however complicated the molecule looks:

1. Find the main chain: the longest continuous chain, the one with the most carbons. If the molecule has a double or a triple bond, the main chain must contain it, even if it is not the longest chain overall.
2. Number the chain starting from the end that gives the lowest possible number to the first point of unsaturation (double or triple bond) or, for the alkanes, to the substituents.
3. Identify the substituents (the alkyl groups that branch off the main chain) and the number of the carbon they are attached to.
4. Build the name: the substituent prefixes in alphabetical order, each with its own number (locant), followed by the name of the main chain with the suffix that indicates the type of bond — -ane for the alkanes (single bonds only), -ene for the alkenes (one double bond), -yne for the alkynes (one triple bond), with the number of the first carbon of the multiple bond in front of the suffix when the chain has more than three carbons.

For cyclic hydrocarbons the scheme is identical, with the prefix cyclo- in front of the chain name and the numbering starting from the carbon that gives the lowest locants to the substituents on the ring. For the aromatics, the parent compound is benzene (${\mathrm{C}}_6{\mathrm{H}}_6$), a six-carbon ring with a delocalised electron system that gives it a particular stability compared with an ordinary ring with three alternating double bonds: at this level it is enough to recognise it and to name its monosubstituted derivatives (for example methylbenzene, better known as toluene).

The typical mistake to avoid

The most frequent problem in tests is not ignorance of the rules, it is applying them in the wrong order: the chain is numbered before the main chain has been chosen correctly, or the first sequence of carbons you see drawn is taken as the main chain, instead of explicitly looking for the longest one that contains any double or triple bond. A second mistake, separate from the first, is confusing chain isomerism with conformations: two conformations of the same molecule (for example the staggered and eclipsed forms of ethane) are exactly the same molecule seen at different moments of free rotation around a single bond — they do not require breaking any bond to turn into each other. Two chain isomers, on the other hand, are really different compounds: to go from one to the other you would have to break and re-form covalent bonds. If the question asks “how many isomers does ${\mathrm{C}}_5{\mathrm{H}}_{12}$ have” and you answer by counting the conformations as well, the number is wrongly inflated.

Three worked examples

1. IUPAC name of a branched alkane

Given the condensed structure $\mathrm{C}{\mathrm{H}}_3\text{-}\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\text{-}\mathrm{C}{\mathrm{H}}_2\text{-}\mathrm{CH}({\mathrm{C}}_2{\mathrm{H}}_5)\text{-}\mathrm{C}{\mathrm{H}}_2\text{-}\mathrm{C}{\mathrm{H}}_3$, find the IUPAC name.

Step 1 — main chain. The longest chain, counting from one end to the other through all the carbons drawn in a row, has six carbon atoms: it is a hexane. The two groups sticking out sideways — a methyl and an ethyl — are substituents, not part of the main chain.

Step 2 — numbering. Numbering from the left: C1 (the first $\mathrm{C}{\mathrm{H}}_3$), C2 (carries the methyl), C3, C4 (carries the ethyl), C5, C6. The substituents are on carbons 2 and 4: set of locants $\{2,4\}$.

Numbering from the right, the same substituents would end up on carbons 3 and 5: set $\{3,5\}$. Comparing the two sets point by point, $\{2,4\}$ is lower than $\{3,5\}$ at the first comparison (2 against 3), so you number from the left: it is exactly the step that, if skipped or inverted, produces the wrong name.

Step 3 — name. In alphabetical order (ethyl before methyl): 4-ethyl-2-methylhexane.

2. All the chain isomers of C5H12

The molecular formula ${\mathrm{C}}_5{\mathrm{H}}_{12}$ (pentane) admits exactly three chain isomers, distinguished only by the shape of the skeleton:

• n-pentane: $\mathrm{C}{\mathrm{H}}_3\text{-}\mathrm{C}{\mathrm{H}}_2\text{-}\mathrm{C}{\mathrm{H}}_2\text{-}\mathrm{C}{\mathrm{H}}_2\text{-}\mathrm{C}{\mathrm{H}}_3$, a straight chain of five carbons, no branching;
• 2-methylbutane (isopentane): $\mathrm{C}{\mathrm{H}}_3\text{-}\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\text{-}\mathrm{C}{\mathrm{H}}_2\text{-}\mathrm{C}{\mathrm{H}}_3$, a main chain of four carbons (butane) with a methyl on the second carbon;
• 2,2-dimethylpropane (neopentane): $\mathrm{C}(\mathrm{C}{\mathrm{H}}_3{)}_4$, a single central carbon bonded to four methyl groups — the most branched structure possible with five carbons in all.

From here on, the science page gathers the rest of the Year 5 programme. To practise against the clock you will find ready-made tests in the high-school practice test category: I will say straight away that that page covers maths and physics, not natural sciences — the correct answer is free, while the guided step-by-step solution unlocks with the first lesson.

3. Hybridisation in a molecule with a double and a triple bond

Given the molecule $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}\text{-}\mathrm{C}{\mathrm{H}}_2\text{-}\mathrm{C}\equiv \mathrm{CH}$ (pent-1-en-4-yne), determine the hybridisation of each carbon and the geometry around it.

Numbering the carbons from left to right:

• C1 ($\mathrm{C}{\mathrm{H}}_2=$): part of the double bond, so three groups of electrons (the double bond with C2 and two single bonds with the hydrogens) → sp2 hybridisation, trigonal planar geometry, angles of about $120^{\circ }$.
• C2 ($=\mathrm{CH}\text{-}$): this carbon is also in the double bond, with three groups (double bond with C1, single bond with the hydrogen, single bond with C3) → sp2 hybridisation, angles of about $120^{\circ }$.
• C3 ($\mathrm{C}{\mathrm{H}}_2$): four single bonds (two hydrogens, C2, C4) → sp3 hybridisation, tetrahedral geometry, angles of about $109.5^{\circ }$.
• C4 ($\mathrm{C}\equiv$): part of the triple bond, so two groups of electrons (the triple bond with C5 and the single bond with C3) → sp hybridisation, linear geometry, angle of $180^{\circ }$.
• C5 ($\equiv \mathrm{CH}$): also in the triple bond, two groups (triple bond with C4, single bond with the hydrogen) → sp hybridisation, angle of $180^{\circ }$.

Note how the hybridisation changes from carbon to carbon within the same molecule: it is not a single property of the structure, but of each individual atom, depending on exactly what kind of bonds leave it.

4. Cis or trans? (optional)

But-2-ene ($\mathrm{C}{\mathrm{H}}_3\text{-}\mathrm{CH}=\mathrm{CH}\text{-}\mathrm{C}{\mathrm{H}}_3$) can exist as a cis or a trans isomer: each carbon of the double bond carries a methyl group and a hydrogen, two substituents that differ from each other, a necessary condition for cis-trans stereoisomerism to exist. In the cis form the two methyl groups are on the same side of the double bond; in the trans form they are on opposite sides. If instead one of the two carbons of the double bond carried two identical substituents (for example two hydrogens, as in propene $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}\text{-}\mathrm{C}{\mathrm{H}}_3$), the cis-trans distinction would no longer exist: there is no way of telling “same side” from “opposite side” when the two groups on one side are identical.

How I work on this block

With students arriving at Year 5 organic chemistry, the first job is almost always to dismantle the idea that it is “a list of names to learn by heart”: hybridisation, isomerism and nomenclature are three connected tools, and once you have understood how they are used together, the naming part becomes the mechanical application of a fixed scheme, not a memory exercise. For inorganic nomenclature, which follows a different logic (oxidation numbers, not carbon chains), the reference remains this post on traditional and IUPAC nomenclature; for calculations with moles and reactions, which also come in handy here when we talk about the combustion of hydrocarbons, there is this post on the mole and stoichiometry. The actual mechanisms are the next step: Reazioni organiche: addizione elettrofila, SN1 e SN2.

If you want to arrive prepared for the November-December test, or you need a hand putting things in order before the maturità oral, you will find the full picture of the subject on the science page, the ready-made maths and physics tests in the high-school practice test category, or you can write to me directly via contacts.