Aromatic compounds

In a nutshell

Benzene, C₆H₆, is the parent of the aromatic compounds (arenes). Its six π-electrons are delocalised around the ring. That is why it is unusually stable, and why it reacts by electrophilic substitution rather than addition.

This subtopic covers the evidence for that delocalised model, how to name aromatic compounds, the three substitution reactions of benzene, and why phenol reacts so much more easily than benzene.

Assumed knowledge: Basic concepts of organic chemistry (curly arrows, electrophiles), Alkenes (π-bonds, electrophilic addition of bromine).

Core content

Benzene and aromatic compounds: the Kekulé model and the delocalised model

In 1865 Kekulé proposed that benzene is a ring of six carbons with alternating single and double bonds (cyclohexa-1,3,5-triene). In this model the π-electrons are localised in three separate C=C bonds.

The modern delocalised model replaced it:

• Each carbon forms three σ-bonds: to two other carbons and to one hydrogen.
• That leaves one electron in a p-orbital on each carbon, at right angles to the plane of the ring.
• The six p-orbitals overlap sideways with their neighbours on both sides.
• This forms a delocalised π-system: rings of π-electron density above and below the plane of the carbon atoms, shared over all six carbons.

Benzene is planar, a regular hexagon, with bond angles of 120°.

Still don't get it? · the delocalised π-system

Picture six people standing in a circle. In the first version they pair up into three couples, each couple holding hands only with each other. In the second version everyone holds hands with both neighbours, so the whole circle is one connected loop.

The first version is Kekulé's model. The second is the delocalised model.

Now step by step:

1. Each carbon has four outer electrons. Three go into σ-bonds (two to carbons, one to H).
2. The fourth electron sits in a p-orbital, a dumb-bell shape sticking up above and down below the ring.
3. In Kekulé's model, each p-orbital overlaps with just one neighbour, making three fixed π-bonds.
4. In reality, each p-orbital overlaps sideways with both neighbours, so the overlap runs all the way round.
5. The six electrons are then not tied to any one pair of carbons. They are spread over the whole ring, above and below it.

Exam wording: "p-orbitals overlap (sideways) to form a delocalised π-system (ring of π-electrons) above and below the plane of the ring". For Kekulé, say the π-electrons are localised in three C=C bonds.

In equations and mechanisms you may draw benzene either way: as a Kekulé hexagon with three double bonds, or as a hexagon with a circle inside (the circle is the delocalised ring).

Benzene and aromatic compounds: evidence for the delocalised model

Three pieces of evidence show that benzene is not cyclohexa-1,3,5-triene.

1. Bond lengths. X-ray diffraction shows that all six carbon-carbon bonds are the same length, and that length lies between a C-C single bond and a C=C double bond. Kekulé's model predicts three short bonds and three long ones.

2. Enthalpy change of hydrogenation. Hydrogenating one C=C in cyclohexene releases about 120 kJ mol^-1:

C₆H₁₀ + H₂ → C₆H₁₂    ΔH = -120 kJ mol^-1

If benzene had three C=C bonds, hydrogenating it should release 3 × 120 = 360 kJ mol^-1. The measured value is far smaller:

C₆H₆ + 3H₂ → C₆H₁₂    ΔH = -208 kJ mol^-1

Both reactions end at the same product, cyclohexane. So less energy released means benzene started lower in energy: it is more stable than the Kekulé structure, by about 152 kJ mol^-1.

You do not need to recall these values: questions give them. You do need to explain what they show.

3. Resistance to reaction. If benzene had three localised C=C bonds it would react like an alkene. It does not:

• It does not decolourise bromine water at room temperature.
• It only reacts with bromine when a halogen carrier is present.
• It reacts by substitution, which keeps the delocalised ring, not by addition, which would break it.

Benzene and aromatic compounds: naming aromatic compounds

Most names use benzene as the parent, with the substituent as a prefix: chlorobenzene, bromobenzene, nitrobenzene, methylbenzene.

Some groups give a parent name of their own. Learn these:

When the ring is attached to a longer chain or a functional group that takes the name, the ring becomes the phenyl group, C₆H₅-. Examples: phenylethanone (C₆H₅COCH₃), phenylethene (C₆H₅CH=CH₂).

For two or more substituents, use locant numbers:

1. The carbon carrying the parent group (the OH in phenol, the CH₃ in methylbenzene) is carbon 1.
2. Number round the ring in the direction that gives the lowest set of numbers.
3. Use di-, tri- for repeated groups, and list different prefixes in alphabetical order.
4. If benzene is the parent, the first group alphabetically gets carbon 1 when there is a tie.

The spec's own example is 2,4-dinitromethylbenzene: a methylbenzene with NO₂ groups on carbons 2 and 4.

Electrophilic substitution: the three reactions of benzene

In electrophilic substitution, an electrophile (an electron pair acceptor) replaces a hydrogen atom on the ring. The delocalised ring is kept in the product.

Halogen carriers are iron, iron halides (FeBr₃, FeCl₃) and aluminium halides (AlBr₃, AlCl₃). Iron works because it reacts with the halogen to make the iron halide in the flask: 2Fe + 3Br₂ → 2FeBr₃.

At higher nitration temperatures, more than one NO₂ group substitutes.

Overall equations:

• Nitration: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
• Bromination: C₆H₆ + Br₂ → C₆H₅Br + HBr
• Alkylation: C₆H₆ + CH₃CH₂Cl → C₆H₅CH₂CH₃ + HCl
• Acylation: C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl

Why Friedel-Crafts matters: it forms a new C-C bond to the aromatic ring. That is one of the few ways to attach a carbon chain to a benzene ring, so it is a key step in building bigger molecules in synthesis.

Electrophilic substitution: the mechanism for nitration

First, make the electrophile. Concentrated sulfuric acid reacts with nitric acid to form the nitronium ion, NO₂⁺:

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄^- + H₂O

Then the mechanism, in two curly-arrow steps:

1. A curly arrow from the delocalised ring (the circle) to the N of NO₂⁺. The ring donates an electron pair and a new C-N bond forms.
2. This gives a positive intermediate: one carbon now carries both H and NO₂, and the delocalised ring is broken. Draw it as a horseshoe over the other five carbons, open towards the carbon with H and NO₂, with a + inside.
3. A curly arrow from the C-H bond back into the ring. The H leaves as H⁺ and the delocalised ring is restored.

Finally, the catalyst is regenerated:

H⁺ + HSO₄^- → H₂SO₄

Electrophilic substitution: the mechanism for halogenation

First, make the electrophile. The halogen carrier accepts a halide ion from the halogen. You can take the electrophile to be Br⁺:

FeBr₃ + Br₂ → FeBr₄^- + Br⁺

The mechanism is the same two steps as nitration, with Br⁺ in place of NO₂⁺: ring to Br⁺, horseshoe intermediate with +, C-H bond back into the ring.

Then the halogen carrier is regenerated, and HBr forms:

H⁺ + FeBr₄^- → FeBr₃ + HBr

The same pattern works for chlorine with AlCl₃: AlCl₃ + Cl₂ → AlCl₄^- + Cl⁺.

Electrophilic substitution: why benzene resists bromination but alkenes do not

Cyclohexene decolourises bromine at room temperature. Benzene needs a halogen carrier. The explanation is all about π-electron density:

1. In an alkene, the π-electrons are localised between two carbon atoms, so the C=C has a high electron density.
2. In benzene, the π-electrons are delocalised over six carbon atoms, so the electron density at any one place is lower.
3. Benzene therefore cannot polarise a bromine molecule enough: it cannot induce a big enough dipole in Br₂ to make it attack.
4. So benzene needs a halogen carrier to produce the stronger electrophile Br⁺.

Still don't get it? · why benzene resists bromine

Think of a torch. Point the beam at one spot and it is bright enough to read by. Spread the same light round a whole room and every spot is dim.

The π-electrons are the light.

Step by step:

1. A bromine molecule is non-polar. To react, something electron-rich has to push on its electrons and make one Br slightly positive (an induced dipole).
2. In an alkene, two π-electrons sit between just two carbons: a bright, concentrated beam. That is enough to polarise Br₂.
3. In benzene, six π-electrons are spread round six carbons: the room light. No single spot is concentrated enough to polarise Br₂.
4. So benzene needs help. The halogen carrier makes a ready-made positive electrophile, Br⁺, which does not need polarising.

Exam wording: benzene has delocalised π-electrons spread over the ring, so lower electron density; an alkene has localised π-electrons between two carbons, so higher electron density. Benzene cannot polarise Br₂ / induce a dipole, so a halogen carrier is needed.

Electrophilic substitution: predicting unfamiliar reactions

Exams can give you an aromatic reaction you have never met, with extra information such as the electrophile. Every one follows the same template:

1. Write the equation that makes the electrophile, E⁺ (it is usually given).
2. Curly arrow from the delocalised ring to E⁺.
3. Horseshoe intermediate, with +, on the carbon that now carries H and E.
4. Curly arrow from the C-H bond back into the ring, losing H⁺.
5. Show how the catalyst is regenerated (H⁺ reacts with the anion made in step 1).
6. Use directing effects to place E on a ring that already has a group.

Phenols: weak acidity

In a phenol, the -OH group is bonded directly to the aromatic ring. (C₆H₅CH₂OH is not a phenol: its OH is on a CH₂, so it is an alcohol.)

Phenol is a weak acid. It partially dissociates in water:

C₆H₅OH(aq) ⇌ C₆H₅O^-(aq) + H⁺(aq)

It is strong enough to neutralise NaOH, forming sodium phenoxide and water:

C₆H₅OH + NaOH → C₆H₅O^-Na⁺ + H₂O

It is too weak to react with carbonates, so there is no fizzing (no CO₂). Carboxylic acids are stronger and do react with carbonates:

2CH₃COOH + Na₂CO₃ → 2CH₃COONa + CO₂ + H₂O

PAG7: qualitative analysis of organic functional groups (testing for a phenol)

OCR's PAG7 uses this difference in acidity to identify a phenol in an unknown compound.

1. Add aqueous sodium carbonate. A carboxylic acid fizzes (CO₂ given off). A phenol shows no effervescence.
2. Add aqueous sodium hydroxide to a fresh sample. A phenol reacts (neutralisation): phenol is only slightly soluble in water, but it dissolves in NaOH(aq) because the ionic sodium phenoxide is soluble.
3. Weak acid + NaOH but no reaction with carbonate means a phenol.

Record observations as what you see: "no effervescence", "the solid dissolves to give a colourless solution". Not "no CO₂ was made".

Phenols: electrophilic substitution of phenol

Phenol reacts with electrophiles far more easily than benzene.

With bromine (e.g. bromine water), at room temperature and with no halogen carrier:

C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr

• Product: 2,4,6-tribromophenol. Three H atoms are substituted, not one.
• Observations: bromine water is decolourised (orange to colourless) and a white precipitate forms.

With dilute nitric acid, at room temperature, a mixture of 2-nitrophenol and 4-nitrophenol forms:

C₆H₅OH + HNO₃ → C₆H₄(OH)NO₂ + H₂O

This nitration does not need concentrated HNO₃ or a concentrated H₂SO₄ catalyst. Benzene needs both.

Phenols: why phenol reacts more easily than benzene

1. A lone pair of electrons on the oxygen sits in a p-orbital.
2. This p-orbital overlaps with the ring, so the lone pair is (partially) delocalised into the π-system.
3. This increases the electron density of the ring.
4. The ring is more susceptible to attack by electrophiles: for example it can polarise Br₂ without a halogen carrier.

Explain this only in terms of the ring's electron density and its susceptibility to attack. OCR does not want an answer based on the stability of the intermediate.

Phenols: directing effects of substituent groups

A group already on the ring decides where the next electrophile goes.

Positions 2 and 6 are equivalent, and so are 3 and 5. So "2- and 4-directing" covers 2, 4 and 6.

That is why phenol gives 2- and 4-nitrophenol, and 2,4,6-tribromophenol. Nitrobenzene with Br₂ and FeBr₃ gives 1-bromo-3-nitrobenzene.

You only need to know OH, NH₂ and NO₂. For any other group, the exam gives you the directing effect.

Why this matters for synthesis: the order of steps decides which isomer you make. Put the directing group on first, so it steers the next group to the position you want (see Worked example 4).

Worked examples

1. Calculating benzene's extra stability from hydrogenation data

Data: ΔH for cyclohexene + H₂ → cyclohexane is -120 kJ mol^-1. ΔH for benzene + 3H₂ → cyclohexane is -208 kJ mol^-1. How much more stable is benzene than the Kekulé structure?

Step 1: the Kekulé structure has three C=C bonds, so its predicted value is three times cyclohexene's:

$\Delta H_{\text{predicted}} = 3 \times (-120\ \text{kJ mol}^{-1})$

$\Delta H_{\text{predicted}} = -360\ \text{kJ mol}^{-1}$

Step 2: compare with the measured value:

$\text{difference} = (-208\ \text{kJ mol}^{-1}) - (-360\ \text{kJ mol}^{-1})$

$\text{difference} = +152\ \text{kJ mol}^{-1}$

Step 3: interpret it. Benzene is 152 kJ mol^-1 more stable than the Kekulé structure (3 significant figures, matching the data). Hydrogenating benzene releases 152 kJ mol^-1 less energy than predicted because the delocalised ring is lower in energy to start with.

Where marks go:

• Forgetting the factor of 3 (comparing -208 with -120).
• Saying benzene is "less stable because it releases less energy". Less released means more stable at the start.
• Quoting a stability difference with no units.

2. Naming substituted aromatic compounds

Compound A (Br and NO₂ on alternate carbons). Benzene is the parent. Alphabetically, bromo comes before nitro, so Br takes carbon 1 and the NO₂ is on carbon 3.

Name: 1-bromo-3-nitrobenzene.

Compound B (OH, with one Cl next to it and one Cl directly opposite it). Phenol is the parent, so the OH carbon is carbon 1. Numbering towards the nearer Cl gives 2 and 4. The other way gives 4 and 6.

Name: 2,4-dichlorophenol (2,4 is lower than 4,6; "di" because there are two Cl).

Compound C (three CH₃ groups: two next to each other, and a third one carbon further round). Try each possible start and direction. The lowest set is 1,2,4 (not 1,3,4 or 1,2,5).

Name: 1,2,4-trimethylbenzene.

3. Predicting the mechanism of an unfamiliar reaction: Friedel-Crafts acylation

Information given: benzene reacts with ethanoyl chloride, CH₃COCl, in the presence of AlCl₃. The electrophile is CH₃CO⁺.

Step 1: electrophile formation (the halogen carrier takes Cl^-):

CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄^-

Step 2: use the nitration template with CH₃CO⁺ as E⁺. The curly arrow goes from the ring to the positive carbon of CH₃CO⁺ (not to the O).

Step 3: regenerate the halogen carrier:

H⁺ + AlCl₄^- → AlCl₃ + HCl

Step 4: overall equation and product:

C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl   (product: phenylethanone)

4. Using directing effects to predict products and plan a synthesis

(a) Nitrobenzene + Br₂ with FeBr₃. NO₂ is electron-withdrawing, so it is 3-directing. Product: 1-bromo-3-nitrobenzene (+ HBr).

(b) Phenylamine + excess bromine water. NH₂ donates electron density, like OH in phenol, so the ring is activated and NH₂ is 2- and 4-directing. By analogy with phenol, all three positions (2, 4 and 6) are substituted:

C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr   (product: 2,4,6-tribromophenylamine)

(c) Plan: make 1-bromo-3-nitrobenzene from benzene.

1. Nitrate first: conc HNO₃ / conc H₂SO₄, about 50 °C, gives nitrobenzene.
2. Then brominate: Br₂ with FeBr₃. The NO₂ group sends Br to carbon 3.

If you brominated first, NO₂ would not be there to steer. (A Br already on the ring actually directs to 2 and 4, but that is a group an exam would tell you about.) The order of the steps is the mark.

Common exam mistakes

• Describing delocalisation without the orbitals. "The electrons are delocalised" is not enough. Say p-orbitals overlap sideways to form a π-system above and below the plane of the ring.
• Putting the delocalised electrons "in the ring" or "between the carbons" in the plane. The π-electron density is above and below the plane.
• Saying benzene resists bromine because it is "more stable". For the comparison with alkenes the marks are about electron density: delocalised (lower) in benzene, localised between two carbons (higher) in the alkene, so benzene cannot polarise Br₂.
• Reading the hydrogenation data backwards. Benzene releases less energy than predicted, so it is more stable, not less. Remember the factor of 3 for three C=C bonds.
• Starting the first curly arrow from a single carbon or from a hydrogen. It starts from the delocalised ring (the circle) and ends at the electrophile's positive atom (the N of NO₂⁺).
• Drawing the intermediate wrongly. The horseshoe covers the other five carbons, opens towards the carbon carrying H and the new group, and has a + charge inside. A full circle, a missing +, or a horseshoe reaching the tetrahedral carbon all lose the mark.
• Starting the second arrow from the H atom. It starts from the C-H bond and goes back into the ring.
• Missing the electrophile equation for nitration. You must show NO₂⁺ forming from HNO₃ and H₂SO₄. Writing NO₂ or NO₂^- for the electrophile loses the mark.
• Calling H₂SO₄ or the halogen carrier a reagent that gets used up. Both are regenerated at the end: show H⁺ + HSO₄^- → H₂SO₄ or H⁺ + FeBr₄^- → FeBr₃ + HBr.
• Forgetting the halogen carrier for benzene, or adding one for phenol. Benzene + Br₂ needs FeBr₃ (or Fe, AlBr₃). Phenol + Br₂ does not.
• Writing monobromophenol for phenol + bromine. The product is 2,4,6-tribromophenol, with 3HBr in the balanced equation.
• Using concentrated acids to nitrate phenol. Phenol needs only dilute HNO₃, and gives a mixture of 2- and 4-nitrophenol.
• Explaining phenol's reactivity as "OH is electron-donating" with no mechanism of donation. The mark needs the lone pair on O in a p-orbital, delocalised into the ring, increasing electron density.
• Explaining phenol's reactivity using the stability of the intermediate. OCR asks only for the ring's increased electron density and susceptibility to electrophilic attack.
• Saying phenol fizzes with sodium carbonate. It does not: that is the test that tells a phenol from a carboxylic acid.
• Calling C₆H₅CH₂OH a phenol. The OH must be bonded directly to the ring.
• Naming with locants that are not the lowest set, e.g. 4,6-dichlorophenol instead of 2,4-dichlorophenol, or listing prefixes out of alphabetical order.
• Mixing up the directing effects. OH and NH₂ direct to 2 and 4; NO₂ directs to 3.

Key definitions

• Aromatic compound: a compound containing a benzene ring.
• Delocalised π-system (in benzene): sideways overlap of the p-orbitals on all six carbon atoms, forming a ring of π-electron density above and below the plane of the ring, shared by all six carbons.
• Kekulé model: benzene as a ring of six carbon atoms with alternating single and double bonds (localised π-electrons).
• Electrophile: an electron pair acceptor.
• Electrophilic substitution: a reaction in which an electrophile replaces an atom (usually hydrogen) on an aromatic ring, keeping the delocalised ring.
• Halogen carrier: a substance such as iron, an iron halide or an aluminium halide that reacts with a halogen (or haloalkane or acyl chloride) to produce the electrophile, and is regenerated at the end.
• Phenol: a compound in which an -OH group is bonded directly to an aromatic ring.
• Electron-donating group (OH, NH₂): a group that increases the electron density of the ring and directs substitution to the 2- and 4-positions.
• Electron-withdrawing group (NO₂): a group that decreases the electron density of the ring and directs substitution to the 3-position.

Specification

• I can compare the Kekulé model of benzene with the delocalised model, in terms of p-orbital overlap forming a delocalised π-system. (6.1.1(a))
• I can explain how bond lengths, the enthalpy change of hydrogenation and resistance to reaction support the delocalised model. (6.1.1(b))
• I can name substituted aromatic compounds systematically, using locant numbers. (6.1.1(c))
• I can give reagents, conditions and equations for the nitration of benzene with concentrated HNO₃ and concentrated H₂SO₄. (6.1.1(d)(i))
• I can give reagents, conditions and equations for the halogenation of benzene with a halogen and a halogen carrier. (6.1.1(d)(ii))
• I can describe Friedel-Crafts alkylation and acylation and explain why forming a C-C bond to the ring matters in synthesis. (6.1.1(d)(iii))
• I can draw the electrophilic substitution mechanism for nitration (including the equations that make NO₂⁺) and for halogenation. (6.1.1(e))
• I can explain why benzene resists bromination compared with alkenes, using delocalised vs localised π-electron density. (6.1.1(f))
• I can interpret unfamiliar electrophilic substitution reactions and predict their mechanisms. (6.1.1(g))
• I can show the weak acidity of phenol: it neutralises NaOH but does not react with carbonates. (6.1.1(h), PAG7)
• I can describe phenol's reactions with bromine (2,4,6-tribromophenol) and with dilute nitric acid (2- and 4-nitrophenol). (6.1.1(i))
• I can explain why phenol reacts more easily than benzene, in terms of the oxygen p-orbital lone pair donated into the π-system. (6.1.1(j))
• I can state the 2- and 4-directing effect of OH and NH₂ and the 3-directing effect of NO₂. (6.1.1(k))
• I can predict substitution products using directing effects and use them to plan the order of steps in a synthesis. (6.1.1(l))

Related notes

• Basic concepts of organic chemistry
• Alkenes
• Enthalpy changes
• Acids, bases and buffers
• Carboxylic acids and esters
• Amines
• Carbon-carbon bond formation
• Organic synthesis
• Chromatography and qualitative analysis