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This master-class consolidates every organic mechanism you have met in lessons 3 to 7 of this course into a single framework. Rather than treating free-radical substitution, electrophilic addition, nucleophilic substitution, elimination, and esterification as five disconnected procedures, we will see them as five families of electron-pushing. Each family is diagnosable from a small set of clues — the substrate's degree of substitution, the reagent's character (electrophile, nucleophile, base, radical initiator), the solvent's polarity and protic/aprotic nature, and (where given) the experimentally-determined rate equation. By the end of this lesson you will be able to look at any A-Level synthesis question, identify the mechanism family in seconds, draw the curly-arrow scheme cleanly, predict the major product (with stereochemistry and regiochemistry), and justify your answer. We will also pin down the underlying pattern that all five families share: bonds break and form because electrons flow from regions of high electron density to regions of low electron density.
Spec mapping (AQA 7405): This master-class consolidates §3.3 broadly — specifically §3.3.2 (alkanes — free-radical substitution, recap from lesson 3), §3.3.3 (halogenoalkanes — nucleophilic substitution SN1 and SN2, elimination E1 and E2, recap from lesson 6), §3.3.4 (alkenes — electrophilic addition, recap from lesson 4), §3.3.5 (alcohols — dehydration as E1 elimination and esterification as addition-elimination, recap from lesson 7), and §3.3.10 (synthesis — choosing the correct mechanism for a given target). The kinetic diagnostic — using a rate equation to identify the rate-determining step and hence distinguish SN1 from SN2 or E1 from E2 — draws on §3.1.9 (rate equations and orders of reaction, A2 kinetics). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Recognising the mechanism family from arrow patterns, substrate, reagent, and solvent is AO1 (recall and apply). Drawing the mechanism cleanly — curly arrows from the correct origin to the correct target, every intermediate shown explicitly, charges and lone pairs labelled where needed — is AO2 (apply skills). Predicting products (including stereochemistry from SN2 inversion or SN1 racemisation, and regiochemistry from Markovnikov or Zaitsev rules), choosing the correct mechanism for given conditions, and justifying the choice using rate-equation evidence are AO3 (analyse, evaluate, synthesise). Mechanism questions appear on every AQA Paper 2 and frequently on Paper 3; they typically carry 6 to 14 marks per question and are the highest-yielding topic per minute of revision in organic chemistry.
A mechanism is a story told in arrows. Every curly arrow represents the movement of electrons, not the movement of atoms. Get the arrows right and the rest of the mechanism follows automatically.
Full-headed arrow (double-barbed): represents the movement of a pair of electrons (two electrons together). Used in polar mechanisms — that is, in every mechanism except free-radical substitution. An arrow with a full head must start at a region of high electron density (a lone pair, a σ bond, or a π bond) and end at an atom or at the position where a new bond will form.
Half-headed arrow (single-barbed, "fish-hook"): represents the movement of a single electron. Used only in radical mechanisms (free-radical substitution and addition polymerisation). Fish-hook arrows always come in pairs, because a covalent bond contains two electrons and homolysis splits them one each way.
Arrows start at a source of electrons. The source must be one of: a lone pair (e.g. on O, N, halide ion), a σ bond (e.g. the C–H bond in elimination, or the C–X bond in heterolysis), or a π bond (e.g. the C=C bond of an alkene attacking an electrophile). Arrows never start at a positive charge or at an empty orbital — electrons flow into those, not out of them.
Arrows end at a destination that can accept electrons. The destination is either an atom with an empty or partly-empty orbital (e.g. the δ+ carbon of a polar bond, or a fully cationic carbon) or the midpoint of a new bond being formed. An arrow that ends on an atom is forming a new bond to that atom; an arrow that ends on an existing bond is contributing to the breaking of that bond.
Electrons are conserved at every step. Count arrows in and arrows out for every atom; the formal charges must balance across the equation. If a mechanism produces a missing or surplus electron pair, an arrow is wrong.
Key Point: When a teacher says "draw the mechanism", they mean: every intermediate shown explicitly, every curly arrow drawn from source to destination, every charge labelled, every lone pair shown where it participates. Marks are awarded for arrows, not for the right product.
| Family | When it occurs | Key reagent(s) | Substrate | Rate equation | Stereochemistry | Regiochemistry | Worked example |
|---|---|---|---|---|---|---|---|
| 1. Free-radical substitution (FRS) | Alkane + halogen + UV | Cl₂, Br₂ + hν | Alkane | Complex (chain) | None controlled | Mixture (poor) | CH₄ → CH₃Cl |
| 2. Electrophilic addition | Alkene + polar electrophile | Br₂, HBr, H₂SO₄, H₂O/H⁺ | Alkene | Rate = k[alkene][E] | syn/anti depending on E | Markovnikov | CH₃CH=CH₂ + HBr → 2-bromopropane |
| 3a. Nucleophilic substitution SN2 | Primary RX + strong nucleophile in polar aprotic solvent | OH⁻, CN⁻, NH₃ | 1° RX | Rate = k[RX][Nu] | Inversion (Walden) | n/a | CH₃Br + OH⁻ → CH₃OH |
| 3b. Nucleophilic substitution SN1 | Tertiary RX + weak nucleophile in polar protic solvent | H₂O, ethanol | 3° RX | Rate = k[RX] | Racemisation | n/a | (CH₃)₃CBr + H₂O → (CH₃)₃COH |
| 4a. Elimination E2 | 1° or 2° RX + strong base, heated in ethanol | KOH/ethanol, NaOEt | 1° or 2° RX | Rate = k[RX][base] | Antiperiplanar | Zaitsev | CH₃CH₂Br + OH⁻ → CH₂=CH₂ |
| 4b. Elimination E1 | 3° RX or alcohol + conc. acid, heat | conc. H₂SO₄/H₃PO₄ | 3° RX or alcohol | Rate = k[substrate] | None controlled | Zaitsev | CH₃CH₂OH → CH₂=CH₂ |
| 5. Addition-elimination (esterification) | Carboxylic acid + alcohol + acid catalyst | conc. H₂SO₄ catalyst | RCOOH + R'OH | Equilibrium (Le Chatelier) | n/a | n/a | CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ |
The table above is the most useful single page in this lesson. Memorise the rows. Half of every mechanism question on Paper 2 can be answered just by matching the question's conditions to a row.
Free-radical substitution is the only mechanism in this lesson that involves single-electron arrows (fish-hooks). It is the reaction of an alkane (no π bonds, no lone pairs, no polar bonds) with a halogen (Cl₂ or Br₂) in the presence of ultraviolet light. UV light supplies the energy needed for homolysis of the halogen-halogen bond, kicking off a radical chain.
Stage 1 — Initiation: UV light homolytically cleaves the halogen bond, generating two halogen radicals.
Cl₂ → 2 Cl• (each Cl gets one electron from the σ bond)
Curly arrows: two fish-hooks, one going from the σ bond to each Cl atom. This is the only photochemical step.
Stage 2 — Propagation (two steps, chain-carrying):
Step 2a: Cl• + CH₄ → HCl + •CH₃
Curly arrows: one fish-hook from a C–H σ bond to the Cl•, one fish-hook from the same σ bond to the C atom (which becomes the methyl radical).
Step 2b: •CH₃ + Cl₂ → CH₃Cl + Cl•
Curly arrows: one fish-hook from •CH₃ to one Cl, one fish-hook from the Cl–Cl σ bond to the other Cl. The Cl• regenerated here propagates the chain.
The chain length (number of propagation cycles per initiation event) can reach 10⁴ or higher, which is why a small UV dose can convert a large quantity of alkane.
Stage 3 — Termination: any two radicals collide and combine, ending a chain.
Three possible termination products: Cl• + Cl• → Cl₂; •CH₃ + •CH₃ → C₂H₆ (ethane, a trace by-product); •CH₃ + Cl• → CH₃Cl.
Because Cl• will abstract any C–H hydrogen — primary, secondary, or tertiary — with similar (though not identical) rates, mono-chlorination of any alkane larger than methane produces a mixture of isomers. Mono-chlorination of methane itself produces a mixture of CH₃Cl, CH₂Cl₂, CHCl₃, and CCl₄ unless the halogen is the limiting reagent and is mixed in large excess of methane. FRS is therefore an industrial-scale tool but a poor laboratory synthesis when a specific product is needed.
Electrophilic addition is the characteristic reaction of alkenes — molecules containing a C=C double bond. The π electrons of the double bond are exposed above and below the plane of the alkene and are easily polarised, making the alkene a good electron donor (nucleophile). It reacts with electrophiles: molecules or ions with an electron-poor site (δ+, or fully positive).
Step 1 — Electrophilic attack: the π electrons of the C=C bond attack the electrophile (e.g. the δ+ end of HBr, or one end of an Br–Br molecule polarised by the approaching alkene). This forms a new C–E bond at one carbon and leaves the other carbon as a carbocation.
Curly arrows: one full-head arrow from the C=C π bond to the electrophile's δ+ atom; one full-head arrow from the E–X σ bond to the leaving X⁻ (heterolysis). Two arrows in this step.
Step 2 — Nucleophilic capture: the leaving group X⁻ (now a free anion) attacks the carbocation from the opposite side, forming the second new C–X bond.
Curly arrow: one full-head arrow from a lone pair on X⁻ to the carbocation carbon.
Propene (CH₃CH=CH₂) reacts with HBr. The δ+ H attacks the C=C double bond. The π electrons can attack the H from either terminal carbon, giving two possible carbocations:
The secondary carbocation is more stable than the primary (more alkyl groups donate electron density through hyperconjugation and induction), so the reaction proceeds preferentially via the secondary cation. In step 2, Br⁻ captures the carbocation to give 2-bromopropane (CH₃CHBrCH₃), the Markovnikov product — the major isomer.
Markovnikov's rule (modern form): in the electrophilic addition of HX to an asymmetric alkene, the H atom adds to the carbon already bearing more H atoms, and the X atom adds to the carbon already bearing fewer H atoms. The modern justification is via carbocation stability: the more substituted carbocation is more stable, so it forms preferentially.
When Br₂ (rather than HBr) is the electrophile, the carbocation intermediate is replaced by a cyclic bromonium ion — a three-membered ring with a positively charged Br bridging the two former alkene carbons. The bromonium ion is opened by Br⁻ attacking from the opposite face, giving anti addition of two Br atoms (one above the plane, one below). This stereochemistry is testable: bromination of cyclohexene gives only trans-1,2-dibromocyclohexane, not the cis isomer. The bromonium ion is the reason. AQA does not require the full bromonium mechanism for all questions, but A* candidates should know it.
Nucleophilic substitution is the most-tested mechanism family at A-Level because the decision between SN1 and SN2 demonstrates almost every principle of physical organic chemistry: substrate effects, solvent effects, kinetic versus thermodynamic control, stereochemistry, and the rate equation as a mechanistic probe.
When it operates: primary (and sometimes methyl or secondary) halogenoalkanes with a strong nucleophile in a polar aprotic solvent (acetone, DMSO, DMF). Strong nucleophiles include OH⁻, CN⁻, RO⁻, NH₂⁻, and (less strongly) NH₃.
Mechanism — one step, concerted: the nucleophile attacks the C bearing the leaving group from the opposite side (back-side attack) at the same time as the C–LG bond breaks. The transition state has the C partially bonded to both Nu (forming) and LG (breaking), with the other three substituents in the plane of the C, like an inverted umbrella.
Curly arrows (in one transition state): one full-head arrow from the lone pair on Nu to C; one full-head arrow from the C–LG σ bond to LG.
Rate equation: rate = k[RX][Nu]. Second-order overall, first-order in each. Because the rate-determining (and only) step involves both molecules colliding, doubling either concentration doubles the rate.
Stereochemistry: inversion of configuration at C (the Walden inversion). If the starting halogenoalkane is a single enantiomer with R configuration at the chiral C, the SN2 product has S configuration. The umbrella flips. This is testable using optically active starting materials and a polarimeter.
When it operates: tertiary (and sometimes secondary) halogenoalkanes with a weak nucleophile in a polar protic solvent (water, ethanol, methanol). Weak nucleophiles include H₂O, ROH, and the alcohol's own conjugate base only in low concentration.
Mechanism — two steps:
Step 1 (rate-determining, slow): the C–LG bond ionises heterolytically. The protic solvent stabilises the resulting carbocation through ion-dipole interactions; the tertiary alkyl groups stabilise it through induction and hyperconjugation.
Curly arrow: one full-head arrow from the C–LG σ bond to LG. This forms a fully ionic carbocation intermediate.
Step 2 (fast): the nucleophile attacks the carbocation from either face. Because the carbocation is sp² hybridised (trigonal planar at the cationic carbon), there is no stereochemical preference for which face is attacked.
Curly arrow: one full-head arrow from the Nu lone pair to the carbocation carbon.
Rate equation: rate = k[RX]. First-order overall, zero-order in Nu. The Nu does not appear in the rate equation because it joins after the rate-determining step.
Stereochemistry: racemisation — if the starting halogenoalkane is a single enantiomer, the carbocation intermediate is planar and is attacked equally from both faces, producing a 50:50 mixture of enantiomers (a racemate). Optical activity is lost. This is the diagnostic test for SN1 versus SN2.
| Question | If yes → | If no → |
|---|---|---|
| Substrate tertiary? | SN1 (or E1) | Continue |
| Substrate primary? | SN2 (or E2) | Continue (secondary — both possible) |
| Nucleophile strong (OH⁻, CN⁻, RO⁻)? | SN2 | SN1 |
| Solvent polar protic (water, ROH)? | Favours SN1 | Favours SN2 |
| Rate equation = k[RX][Nu]? | SN2 confirmed | If k[RX] only, SN1 confirmed |
| Stereochemistry — inversion observed? | SN2 | If racemisation, SN1 |
Methyl bromide is primary (in fact methyl — the simplest possible substrate). Hydroxide is a strong, small, hard nucleophile. The reaction is run in water or aqueous ethanol. The C of CH₃Br is δ+ because Br is more electronegative. OH⁻ attacks the C from the side opposite to Br. The transition state has C–OH partially bonded (forming) and C–Br partially bonded (breaking), with the three H atoms in the plane. The transition state collapses as Br⁻ leaves and OH bond completes, with full inversion of the C tetrahedron (though for CH₃ this is invisible — all H's are identical).
Two curly arrows in the single transition state: OH⁻ lone pair → C (forming new bond); C–Br σ bond → Br (breaking bond). Product: CH₃OH + Br⁻.
2-bromo-2-methylpropane is tertiary. Water is a weak nucleophile but is a polar protic solvent. The mechanism is two steps.
Step 1: heterolysis of C–Br. Curly arrow: C–Br σ bond → Br. This gives the tert-butyl cation (CH₃)₃C⁺ and Br⁻. The carbocation is stabilised by three alkyl groups donating electron density and by ion-dipole solvation from water.
Step 2: water (acting as nucleophile through its O lone pair) attacks the carbocation. Curly arrow: water's O lone pair → carbocation C. This gives (CH₃)₃COH₂⁺ (oxonium intermediate).
Step 3 (deprotonation, sometimes drawn): a second water molecule removes a proton from the oxonium ion. Curly arrow: water lone pair → H of oxonium; O–H σ bond → O of oxonium. Product: (CH₃)₃COH + H₃O⁺.
If the starting halogenoalkane had been chiral (which (CH₃)₃CBr is not — it has no chiral carbon), the planar carbocation intermediate would be attacked from both faces equally, producing a racemic mixture.
Elimination is competition to substitution. The same conditions that favour SN1 (tertiary substrate, weak nucleophile, polar protic solvent) also favour E1; the same conditions that favour SN2 (primary substrate, strong base) also favour E2. Which one dominates depends on the nucleophile's basicity, the substrate, and the temperature.
When it operates: primary or secondary halogenoalkane with a strong base (concentrated KOH in ethanol, or NaOEt). High temperature favours elimination over substitution.
Mechanism — one concerted step: the base removes a proton from the β-carbon (the C adjacent to the one bearing the leaving group) at the same time as the C–LG bond breaks and the C=C π bond forms.
Curly arrows (in one transition state, three arrows): base lone pair → β-H; β-C–H σ bond → between Cα–Cβ (forms the new π bond); Cα–LG σ bond → LG.
Rate equation: rate = k[RX][base]. Second-order overall.
Stereochemistry: antiperiplanar. The β-H and the LG must be on opposite sides of the C–C bond and in the same plane (dihedral angle 180°). This stereo-requirement allows you to predict stereochemistry of E2 from acyclic substrates: only the antiperiplanar conformer reacts, so the E2 product has a defined geometry.
When it operates: tertiary halogenoalkane or tertiary alcohol with conc. H₂SO₄ or conc. H₃PO₄ at high temperature. The β-H removal can be by the conjugate base of the acid or by a passing solvent molecule.
Mechanism — two steps:
Step 1 (slow, rate-determining): heterolysis of C–LG (or in the alcohol case, protonation of OH followed by loss of H₂O). Forms a carbocation.
Step 2 (fast): loss of a β-proton from the carbocation, with formation of the C=C double bond. Any of several β-positions may be deprotonated, leading to Zaitsev (most substituted alkene major) versus Hofmann (least substituted) regiochemistry.
Rate equation: rate = k[substrate]. The base (or proton acceptor) does not appear because it joins after the RDS.
Stereochemistry: none controlled. The carbocation is planar; both faces deprotonate equally; product mixture obeys Zaitsev's rule on regiochemistry but has no enantiomeric preference.
When more than one β-H is available, the most substituted alkene (most alkyl groups on the double-bond carbons) is the major product. For example, dehydration of 2-methylbutan-2-ol can give 2-methylbut-2-ene (more substituted, major, Zaitsev) or 2-methylbut-1-ene (less substituted, minor, Hofmann). The more substituted alkene is more thermodynamically stable due to hyperconjugation.
Step 1: protonation of OH by H₂SO₄. Curly arrow: OH lone pair → H of H₂SO₄; O–H bond of H₂SO₄ → O of H₂SO₄ (breaking). Forms CH₃CH₂OH₂⁺ (protonated ethanol).
Step 2: loss of water from the protonated alcohol. Curly arrow: C–O σ bond → O of water (heterolysis). Forms CH₃CH₂⁺ (primary carbocation — note that ethanol dehydration is borderline E1/E2 because the primary carbocation is unstable; in practice the loss of water and deprotonation may be concerted, giving an E2-like pathway).
Step 3: deprotonation by HSO₄⁻ or another ethanol. Curly arrow: C–H bond of CH₃ → between the two carbons (forming π bond); HSO₄⁻ lone pair → the H being removed. Forms CH₂=CH₂ (ethene) + H₂SO₄ regenerated.
The H₂SO₄ is a catalyst — regenerated at the end.
Esterification is the reaction of a carboxylic acid with an alcohol in the presence of an acid catalyst (usually conc. H₂SO₄), heated under reflux. The mechanism is addition-elimination at the carbonyl carbon. It is reversible — at equilibrium roughly 60:40 ester:reactants for many simple cases — so industrial ester synthesis displaces the equilibrium by removing the ester or the water (Le Chatelier).
Step 1: protonation of the carbonyl O of the carboxylic acid by H⁺ (from H₂SO₄). This makes the C of the carbonyl even more δ+. Curly arrow: O lone pair → H⁺.
Step 2: nucleophilic attack of the alcohol's O lone pair on the activated carbonyl C. Curly arrows: alcohol O lone pair → C; C=O π bond → O (breaking the π bond, leaving an OH on what was the carbonyl C). The C is now sp³ and bears four oxygen-containing substituents — the tetrahedral intermediate.
Step 3: proton transfer. The alcohol-derived O still bears a positive charge (it just donated its lone pair); a proton shifts from this O to one of the OH groups on the tetrahedral C. (Drawn as an intramolecular proton shuffle or via solvent.)
Step 4: loss of water. The OH group that received the proton in step 3 leaves as H₂O. Curly arrow: C–O bond → O of leaving water.
Step 5: deprotonation of the remaining O–H (now bearing a +) and reformation of the carbonyl π bond. Curly arrow: O–H σ bond → external base; O lone pair → C (reforming C=O π bond). This generates the ester product.
Step 6: catalyst recovery. The H⁺ used in step 1 is regenerated by a passing water molecule that accepted a proton in step 5.
The net process is: RCOOH + R'OH ⇌ RCOOR' + H₂O.
The mechanism is addition (nucleophile adds to the carbonyl forming a tetrahedral intermediate) followed by elimination (loss of water from the tetrahedral intermediate regenerates the C=O of the ester product). All carboxylic-acid-derivative chemistry (acyl chlorides, anhydrides, amides, esters, in either direction of hydrolysis or formation) proceeds by addition-elimination at the carbonyl carbon.
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Heat ethanoic acid and ethanol with a few drops of concentrated sulfuric acid catalyst under reflux. After 30 minutes the equilibrium contains ~65% ester, ~35% reactants. To isolate the ester (b.p. 77 °C), distil from the reaction mixture; the ester evaporates preferentially. The smell — sweet, fruity, like nail-polish remover — is the diagnostic A-Level test for ester formation.
The rate equation is a fingerprint of the rate-determining step. The molecularity of the RDS — the number of molecules that have to come together — equals the sum of the orders in the rate equation.
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