End of September 2026, Year 5 of an Italian liceo scientifico (science high school): after IUPAC nomenclature and functional groups, the organic chemistry syllabus reaches the part that asks you to reason, not just to name — how and why reactions happen. In class this block is covered between late October and November, with the test between late November and December — eight to ten weeks from now. It is worth arriving with a method: science was an oral subject at the 2026 maturità (Italy’s school-leaving exam), examined by the internal commissioner, and reaction mechanisms are exactly the kind of topic you end up reasoning through aloud, months after the class test.
I take for granted hybridisation, isomerism and the nomenclature of hydrocarbons (I covered them in Organic chemistry in Year 5: hybridisation, isomerism, hydrocarbons) and functional groups with their nomenclature (in Functional groups in organic chemistry: names and properties), which flagged exactly these mechanisms as “topics for another day”: here they are.
Electrophiles and nucleophiles: who attacks whom
Every organic reaction can be read as an encounter between two species with complementary charges:
- an electrophile (literally “electron-loving”) is an electron-poor species, often with a positive or partial positive charge, that looks for a pair of electrons somewhere to complete its octet or stabilise itself;
- a nucleophile (literally “nucleus-loving”) is an electron-rich species — an anion, or a neutral molecule with lone pairs — that has a pair of electrons available to form a new bond.
The two reactions in this post both come from the same electrophile–nucleophile encounter, applied to two different situations: a double bond that attracts an electrophile (electrophilic addition) and a carbon bonded to a good leaving group that attracts a nucleophile (nucleophilic substitution).
The carbocation and its stability
At the centre of almost every mechanism in this block is the carbocation: a carbon with only three bonds and a positive charge, and therefore electron-poor (an electrophile by definition). Its stability depends on how many alkyl groups are attached to the charged carbon, because each alkyl group “donates” electron density towards the positive charge and eases it (inductive effect):
This ranking alone explains almost all the results that follow: whenever a mechanism can go through a more stable carbocation, it preferentially does.
Electrophilic addition to alkenes
The double bond of an alkene has an exposed, available cloud of electrons: it is a nucleophile, and it attracts electrophiles. In the reaction with (a hydrogen halide, for example ), the mechanism has two steps:
- The double bond attacks the hydrogen of (the electrophile), forming a new bond on one of the two carbons of the double bond and releasing the halide ion . The carbon that did not receive the hydrogen is left with only three bonds: it has become a carbocation.
- The halide ion (now a nucleophile) attacks the carbocation, forming the bond and completing the product.
The same two-step logic applies to the addition of (which gives an alcohol, under acid catalysis) and of (a halogen such as , which gives a vicinal dihalide, by a slightly different mechanism — it goes through a cyclic halonium ion instead of a free carbocation, but at this level it is enough to recognise reactants and product).
Markovnikov’s rule, explained by the carbocation
When the starting alkene is asymmetric (the two carbons of the double bond do not carry the same number of substituents), the hydrogen of can bond to two different carbons, and the two possible intermediate carbocations do not have the same stability. Markovnikov’s rule says that the hydrogen bonds to the carbon of the double bond that already has more hydrogens, so that the halogen ends up on the more substituted carbon — but the right way to remember it, instead of learning it by heart, is to think about the reason: the hydrogen takes the position that produces the more stable intermediate carbocation, and carbocation stability runs tertiary > secondary > primary, as seen above. The Markovnikov product is simply the one that goes through the more stable intermediate, so it forms faster and is the major product.
Nucleophilic substitution: SN2 and SN1
In nucleophilic substitution, a nucleophile replaces a leaving group attached to a saturated carbon (typically a halogen, in an alkyl halide ). There are two alternative mechanisms, and recognising which of the two is taking place is the heart of the test.
SN2 (bimolecular nucleophilic substitution). It happens in a single concerted step: the nucleophile attacks the carbon from the side opposite the leaving group (a “back-side” attack), while the bond to the leaving group breaks at the very moment the new bond to the nucleophile forms. Direct consequences of this mechanism:
- the rate depends on the concentration of both reactants (substrate and nucleophile): hence the name bimolecular;
- it takes place with inversion of configuration at the carbon (like an umbrella turning inside out in the wind) — if the starting carbon is a stereogenic centre, the product has the spatial configuration inverted relative to the reactant;
- it works well on primary substrates (little crowding around the carbon, so the nucleophile can approach) and requires a strong nucleophile.
SN1 (unimolecular nucleophilic substitution). It happens in two separate steps:
- The leaving group departs on its own, without waiting for the nucleophile, forming an intermediate carbocation (the slow step, which determines the rate of the reaction).
- The nucleophile attacks the carbocation, which, being planar, can be attacked from either side.
Consequences:
- the rate depends only on the concentration of the substrate (the first step is the slow one): hence the name unimolecular;
- the planar carbocation can be attacked from both faces with similar probability, so if the starting carbon was a stereogenic centre the product is an almost 1:1 mixture of the two stereoisomers (partial racemisation) rather than a clean inversion;
- it works well on tertiary substrates (the tertiary carbocation is stable enough to form), is favoured by a protic solvent (which stabilises the carbocation by solvation) and does not require a strong nucleophile.
How to recognise which mechanism is taking place
| SN2 | SN1 | |
|---|---|---|
| Steps | One, concerted | Two, with a carbocation intermediate |
| Kinetics | Depends on substrate and nucleophile | Depends only on the substrate |
| Favoured substrate | Primary | Tertiary |
| Nucleophile | Strong, in good concentration | Even weak |
| Stereochemistry | Clean inversion | (Partial) racemisation |
| Solvent | Polar aprotic | Protic |
The secondary substrate is the ambiguous case: it can react by either mechanism depending on the nucleophile and the solvent — and this is often the point on which a test asks you to reason, instead of applying a table from memory.
A note on E1 and E2
Alongside substitution, the same substrate can undergo an elimination reaction (E1 or E2), in which the nucleophile behaves as a base and pulls off a hydrogen instead of attacking the carbon, forming a double bond in place of the leaving group. E1 and E2 compete with SN1 and SN2 respectively on the same substrates, and in general a strong, bulky nucleophile/base or a high temperature favours elimination over substitution. At this level it is enough to know that this competition exists, without going into the details.
Typical mistakes
| Mistake | How you notice it |
|---|---|
| Applying Markovnikov without thinking about the carbocation | Always ask which of the two intermediate carbocations is more stable: that one gives the major product |
| Confusing SN1 and SN2 by looking only at the nucleophile | Look at the substrate first (primary → SN2, tertiary → SN1): it is the most decisive factor, more than the nucleophile |
| Thinking SN1 always gives complete inversion like SN2 | SN1 goes through a planar carbocation: the product is a mixture of stereoisomers, not a clean inversion |
| Forgetting that a strong nucleophile pushes towards SN2 even on poorly reactive substrates | A strong nucleophile in high concentration pushes towards SN2 when the substrate allows it (primary or secondary) |
| Confusing electrophilic addition and nucleophilic substitution | Addition starts from a double bond (alkene) and adds atoms; substitution starts from a saturated carbon with a leaving group and replaces it |
Three worked exercises
1. Electrophilic addition with Markovnikov. Predict the main product of the reaction between propene () and .
The double bond is between C2 and C3 (numbering from the methyl): C2 carries only one hydrogen besides the methyl group, C3 carries two. If the hydrogen of bonds to C3 (the one with more hydrogens, as Markovnikov says), the intermediate carbocation forms on C2, which is bonded to two carbons (the methyl and C3): it is a secondary carbocation. If instead the hydrogen bonded to C2, the carbocation would form on C3, bonded to only one carbon: it would be primary, much less stable.
The mechanism goes through the more stable secondary carbocation: the bromine therefore ends up on C2. The main product is 2-bromopropane ().
There is not yet a dedicated chemistry practice test on this topic on the site: you can still find all the ones available for other subjects in /simulazioni/liceo/, with the correct answer free and the guided solution unlocked with your first lesson.
2. Recognising SN2. Bromomethane (, a primary substrate — in fact the limiting case, without even one substituent carbon) reacts with hydroxide ions (a strong nucleophile) in a polar aprotic solvent. Which mechanism takes place, and what is the product?
A primary substrate (the carbon bearing the bromine is not bonded to other carbons) and a strong nucleophile in good concentration: both conditions point to SN2. The mechanism is concerted, in a single step: the attacks the carbon from the side opposite the bromine, which leaves as at the same instant. The product is methanol (). Since this particular substrate has no stereogenic centre (the carbon already has three identical hydrogens), the inversion of configuration cannot be observed here — but the mechanism is still SN2, because of the substrate and the nucleophile.
3. Recognising SN1. 2-Bromo-2-methylpropane (a tertiary substrate) is dissolved in water (a protic solvent, weak nucleophile). Which mechanism takes place, and what stereochemistry do you expect?
A tertiary substrate and a protic solvent with a weak nucleophile: both conditions point to SN1. The bromine leaves on its own, forming a tertiary carbocation (stable, bonded to three methyl groups), which is then attacked by water to give, after loss of a proton, 2-methylpropan-2-ol. In this specific case the carbocation is not a stereogenic centre (it is bonded to three identical groups, the three methyls), so no real racemisation is observed — but if the starting substrate had four different substituents around the carbon, the planar carbocation would give a mixture of both stereoisomers, not a clean inversion as in an SN2.
Before the test
There is not yet a dedicated chemistry practice test on this topic on the site: you will find timed tests for other subjects in /simulazioni/liceo/. To review nomenclature and functional groups before tackling these mechanisms, the reference posts are Organic chemistry in Year 5: hybridisation, isomerism, hydrocarbons and Functional groups in organic chemistry: names and properties; for the rest of the science syllabus you will find everything in /materie/scienze/.
If these mechanisms are creating more doubts than certainties before the test, or you need a hand with the maturità oral, the first introductory call is free and lasts half an hour.
