A Level Chemistry: Organic Mechanisms and Arenes — Revision Notes

Condensed for the final weeks. For the full explanation, use the Organic Chemistry: Naming, Mechanisms and Aromatic Shape study guide — note that this study guide covers the naming, mechanisms and aromatic-shape outcomes (Topics 29) in depth, but the benzene reagents/conditions table below (Topic 30) goes beyond what that guide covers, so treat this page as the primary reference for that material. The substitution mechanisms and bond-strength reactivity (Topic 15) and Markovnikov/carbocation stability and free-radical substitution (Topic 14) covered below are AS material included here as recap and context for the A Level mechanisms that build on them, not new A Level content in themselves.

New functional groups and naming at A Level

Seven functional groups appear for the first time at A Level:

Functional group	Example	Name
Arene (benzene ring)	C₆H₆	benzene
Halogenoarene	C₆H₅Cl	chlorobenzene
Phenol	C₆H₅OH	phenol
Acyl chloride	CH₃CH₂COCl	propanoyl chloride
Amide	CH₃CH₂CONH₂	propanamide
Amino acid	HOOCCH(NH₂)H	2-aminoethanoic acid
Secondary/tertiary amine	(CH₃)₂NH	dimethylamine

Aromatic naming adds numbering around the ring to locate substituents, e.g. 3-nitrobenzoic acid (a benzoic acid with a nitro group at position 3) or 2,4,6-tribromophenol (a phenol with bromine at positions 2, 4 and 6). Secondary and tertiary amines can be recognised and drawn, but naming them systematically is not required at this level.

Curly-arrow rules — non-negotiable

1. An arrow shows the movement of a pair of electrons (a half-arrow shows one electron, in radical mechanisms).
2. It starts from a bond or a lone pair — never from an atom, never from a positive charge.
3. It ends where the pair goes — at an atom, or at the midpoint of a new bond.
4. Charges must balance on both sides.

Most mechanism marks are lost on rules 2 and 4.

The four mechanisms

Mechanism	Typical substrate	Attacking species
Free radical substitution	Alkanes + halogen, UV	Radicals (homolytic)
Electrophilic addition	Alkenes	Electrophile attacks the π bond
Nucleophilic substitution	Halogenoalkanes	Nucleophile attacks the δ+ carbon
Electrophilic substitution	Arenes	Electrophile, but the ring is restored

Why alkenes add and arenes substitute is the single most examined comparison in the topic. An alkene’s localised π bond is a region of high electron density that is expendable — addition destroys it and gains two σ bonds. Benzene’s delocalised ring is unusually stable, so addition would destroy that stabilisation; substitution preserves it.

Nucleophilic substitution — Sₙ1 vs Sₙ2

	Sₙ2	Sₙ1
Substrate	Primary	Tertiary
Steps	One, concerted	Two, via a carbocation
Transition state	5-coordinate carbon	—
Rate depends on	Both reagents	Halogenoalkane only
Stereochemistry	Inversion of configuration	Racemic mixture

Tertiary goes Sₙ1 because the carbocation formed is stabilised by three electron-donating alkyl groups (positive inductive effect) — and because the crowded carbon blocks backside attack. Primary goes Sₙ2 because the primary carbocation is too unstable to form.

Rate by halogen: C–I > C–Br > C–Cl > C–F. This follows bond enthalpy, not electronegativity. C–F is the most polar bond but by far the strongest, so fluoroalkanes are the least reactive — a favourite trap.

Addition-elimination

A fifth pattern, distinct from the four basic mechanisms above: a nucleophile first adds to a carbon (as in nucleophilic addition), and the resulting tetrahedral intermediate then eliminates a leaving group to restore a double bond, rather than simply picking up a proton. This two-stage add-then-eliminate pattern is why the overall transformation looks like substitution (one group replaces another) even though the mechanism passes through an addition intermediate first — contrast it with simple addition (nothing leaves, e.g. HCN adding to a carbonyl) and simple substitution (no addition intermediate, e.g. a nucleophile directly displacing a leaving group from a halogenoalkane).

Electrophilic addition to alkenes

The π bond attacks the electrophile. With an unsymmetrical alkene, Markovnikov’s rule applies: the major product forms via the more stable carbocation, i.e. tertiary > secondary > primary.

Don’t state Markovnikov as a rule about hydrogen — explain it through carbocation stability, which is what the mark scheme rewards.

Benzene

Structure: planar, regular hexagon, all C–C bonds 139 pm — between a single bond (154) and a double bond (134). Each carbon is sp² hybridised with one electron in a p-orbital; these overlap sideways to form a delocalised π system above and below the ring.

Three pieces of evidence for delocalisation:

1. Bond lengths are all equal — Kekulé’s alternating structure predicts two different lengths.
2. Enthalpy of hydrogenation is about 152 kJ mol⁻¹ less exothermic than 3 × cyclohexene predicts — benzene is more stable than the hypothetical alternating structure by that much.
3. Benzene resists addition and does not decolourise bromine water without a catalyst, unlike alkenes.

Key electrophilic substitutions

Nitration:      conc. HNO3 / conc. H2SO4, 55 C     electrophile NO2+
Halogenation:   Cl2 / AlCl3 (halogen carrier)      electrophile Cl+
Friedel-Crafts alkylation:  RCl / AlCl3            electrophile R+
Friedel-Crafts acylation:   RCOCl / AlCl3          electrophile RCO+

Every one proceeds by the same three-step pattern: generate the electrophile, attack by the delocalised ring to form an unstable intermediate, then lose H⁺ to restore aromaticity. That final step is the mark students most often omit.

Exam traps

• Arrows starting at an atom or a + charge instead of a bond or lone pair.
• Explaining C–X reactivity by electronegativity rather than bond enthalpy.
• Forgetting the H⁺ loss that regenerates the aromatic ring.
• Drawing benzene with alternating double bonds in a mechanism answer.
• Stating Markovnikov without justifying by carbocation stability.
• Omitting the AlCl₃ halogen carrier, or forgetting it is regenerated.

Self-test

1. Give the three rules for curly arrows.
2. Why do alkenes undergo addition but arenes substitution?
3. Which halogenoalkane hydrolyses fastest and why?
4. Give three pieces of evidence for delocalisation in benzene.
5. What is the final step of every electrophilic substitution on benzene?
6. Name the seven functional groups introduced for the first time at A Level.

Answers: 1. Arrows represent a pair of electrons; they start from a bond or lone pair, never an atom or charge; charges must balance. 2. An alkene’s localised π bond is expendable and addition gains two σ bonds; benzene’s delocalised system confers extra stability that addition would destroy, so substitution is preferred. 3. The iodoalkane — C–I has the lowest bond enthalpy, so it breaks most readily; polarity is irrelevant here. 4. All C–C bond lengths equal at 139 pm; enthalpy of hydrogenation ~152 kJ mol⁻¹ less exothermic than predicted; resistance to addition and failure to decolourise bromine water. 5. Loss of H⁺ from the intermediate, restoring the delocalised aromatic ring. 6. Arene, halogenoarene, phenol, acyl chloride, amide, amino acid, and secondary/tertiary amines.