Meet benzene

Spec 6.1.1 (a) (b) (f)

By the end of this lesson I can…

describe the Kekulé model and the delocalised model of benzene
explain THREE pieces of evidence for the delocalised model
explain why benzene does not react with bromine like an alkene does

Watch first (MaChemGuy)

The 5 things you must know

Benzene is C₆H₆: six carbon atoms in a flat, six-sided ring.
Kekulé drew alternating double bonds. The REAL ring has a circle: the electrons are delocalised (shared round the whole ring).
All six C–C bonds are the SAME length: 0.139 nm.
Benzene is MORE stable than Kekulé expected — by 152 kJ mol⁻¹.
Benzene does NOT react with bromine water. Alkenes DO.

Memory hook: Kekulé = alternate. Circle = share. • One-three-nine: every bond is 0.139!

Benzene is 152 kJ mol⁻¹ more stable than the Kekulé structure predicts.

Do now — tap to check

Draw the displayed formula of ethene. What TWO types of bond join the carbon atoms?

A double bond = one σ (sigma) bond + one π (pi) bond. The π bond is formed by sideways overlap of p-orbitals.

What is an electrophile?

An electron-pair acceptor — a species attracted to a region of high electron density.

What do you SEE when bromine water is added to an alkene? What type of reaction is it?

Orange → colourless. Electrophilic addition.

Exit ticket — tap to check

State the C–C bond length in benzene and explain what it tells us. [2]

All 0.139 nm; between a single (0.153 nm) and a double bond (0.134 nm) → no alternating single/double bonds.

Explain why the enthalpy change of hydrogenation of benzene is evidence for delocalisation. [3]

Kekulé structure (3 C=C) predicts 3 × (−120) = −360 kJ mol⁻¹; actual is −208 kJ mol⁻¹; 152 kJ mol⁻¹ less exothermic → benzene more stable than expected → electrons delocalised.

Why does benzene not decolourise bromine water? [2]

π electrons are delocalised → lower electron density between carbons → cannot polarise / induce a dipole in Br₂ → no electrophile forms.

Check yourself

What is the formula of benzene?

C₆H₆

How long is each C–C bond?

0.139 nm (all the same)

Does benzene decolourise bromine water?

No — it has no localised C=C bond to polarise Br₂

Stretch ★ STRETCH: what makes a ring aromatic?

Cyclooctatetraene, C₈H₈, is a ring with four C=C bonds. Is it aromatic? What about naphthalene (two fused rings, 10 π electrons) and pyridine?

Answer: C₈H₈: 8 π electrons = 4n, not 4n + 2 → NOT aromatic; it is tub-shaped, not planar, and reacts like an alkene. Naphthalene: 10 = 4(2) + 2 → aromatic. Pyridine: 6 (the N lone pair is NOT in the ring) → aromatic. Pyrrole: aromatic — its N lone pair IS part of the ring, making 6.

This lesson’s files

Worksheet 1 Worksheet 1 (greyscale)Slides Uni section U1 Mark your work →

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Naming aromatic compounds

Spec 6.1.1 (c)

By the end of this lesson I can…

name compounds with ONE group on the ring
use 'phenyl' when the ring is the add-on
number the ring to name compounds with TWO or more groups

Watch first (MaChemGuy)

The 5 things you must know

ONE group on the ring → group + benzene: chlorobenzene, nitrobenzene, methylbenzene.
FOUR special names to learn: phenol (OH), phenylamine (NH₂), benzoic acid (COOH), benzaldehyde (CHO).
TWO or more groups → number the ring, 1 to 6.
Use the LOWEST numbers you can.
Write the groups in ALPHABETICAL order: bromo, chloro, methyl, nitro.

Memory hook: Lowest numbers. Alphabetical letters. • PhenOL has the OH.

Do now — tap to check

Name CH₃CHClCH₃.

2-chloropropane

Name CH₃CH(OH)CH₃.

propan-2-ol

What is the empirical formula of benzene, C₆H₆?

Exit ticket — tap to check

Name C₆H₅CH₂CH₃. [1]

ethylbenzene

Draw 2,4-dinitromethylbenzene. [1]

CH₃ on C1; NO₂ on C2 and C4.

A compound has a benzene ring with –OH on carbon 1 and –Cl on carbons 2 and 4. Give its systematic name. [1]

2,4-dichlorophenol

Check yourself

Name C₆H₅Cl.

chlorobenzene

Name C₆H₅OH.

phenol

Name the picture: A, B, C, D.

A 1,2-dimethylbenzene; B 2-bromophenylamine; C 4-chlorobenzoic acid; D 3-nitrophenol

Stretch ★ STRETCH: ortho, meta and para

Give the o/m/p name AND the IUPAC name for 1,4-dichlorobenzene and for 3-nitrophenol. Why can't o/m/p be used for a ring with THREE groups?

Answer: p-dichlorobenzene = 1,4-dichlorobenzene; m-nitrophenol = 3-nitrophenol. o/m/p only describes the relationship between TWO groups — with three, it becomes ambiguous, so locants are needed.

This lesson’s files

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Chemistry Student — benzene structure and naming

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Electrophilic substitution: nitration

Spec 6.1.1 (d)(i) (e)

By the end of this lesson I can…

write the equation and conditions for the nitration of benzene
write the equation for making the electrophile, NO₂⁺
draw the full mechanism with curly arrows and explain the role of the catalyst

Watch first (MaChemGuy)

The 5 things you must know

Reagents: concentrated nitric acid + concentrated sulfuric acid (the catalyst).
Temperature: 50 °C in a water bath. Hotter → a second NO₂ goes on.
The electrophile is NO₂⁺, the nitronium ion.
Make it: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
Product: nitrobenzene + water. Catalyst back: H⁺ + HSO₄⁻ → H₂SO₄

Memory hook: Ring gives, ring gets back. • Fifty for one NO₂, seventy for two.

Where the marks are — exactlya mark each · 4–5 a paper = a grade

Mechanism with the three mark points numbered — Tap to enlarge.

Curly arrow from the delocalised ring to N of NO₂⁺Tail ON the circle — that is where the electron pair is — next to the carbon being attacked. Head touching the N atom of NO₂⁺ — the atom that carries the + (never an O). An arrow that starts on the hexagon edge (a σ bond), ends on the wrong atom, or points the other way scores nothing.

The intermediate, drawn exactlyOpen the circle into a HORSESHOE that stops at the two carbons either side of the carbon holding the H and the NO₂ — it must not reach that carbon. Write + INSIDE the horseshoe (the charge is spread over the other five carbons; never on the carbon with the H, never on the H itself). Draw H AND NO₂ on the SAME carbon.

Curly arrow from the C–H bond into the ring, and H⁺ leavesTail on the MIDDLE of the C–H bond (a bond is breaking — never on the H atom). Head pointing into the ring, towards the +, to re-form the circle. Products: nitrobenzene + H⁺.

Making the electrophileHNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

Regenerating the catalyst (H₂SO₄)H⁺ + HSO₄⁻ → H₂SO₄

The 7 ways people lose these marks

Arrow 1 ends on an O of NO₂⁺ instead of the N.
Arrow 1 drawn backwards (from NO₂⁺ to the ring) — the electrons come FROM the ring.
Circle left complete in the intermediate — it must be a horseshoe that stops either side of the carbon holding the H and the new group.
+ missing, or written on the carbon with the H / on the H itself — it goes INSIDE the horseshoe.
Arrow 2 starts on the H — it must start on the C–H BOND.
No H drawn where the new group joined — H and the new group must be on the SAME carbon.
Products missing: write the product AND H⁺ (and, if asked, the equation that regenerates the catalyst).

Kekulé structures are allowed throughout. Then arrow 1 starts on a C=C double bond, and the intermediate still needs the + and the H.

Six common curly-arrow mistakes — Six ways to lose a mark — each drawing loses exactly one of the three. Tap to enlarge.

Do now — tap to check

Define 'electrophile'.

An electron-pair acceptor.

What does a curly arrow show?

The movement of a PAIR of electrons (from where the pair is, to where it ends up).

Why does benzene not undergo addition reactions easily?

Addition would destroy the stable delocalised π-system (losing about 152 kJ mol⁻¹ of stability).

Exit ticket — tap to check

Write the equation for the formation of the nitronium ion. [1]

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

Draw the mechanism for the nitration of benzene, including the intermediate. [3]

Arrow from ring to N of NO₂⁺; intermediate with horseshoe and + charge, H and NO₂ on the same carbon; arrow from C–H bond into ring; nitrobenzene + H⁺.

State the conditions for nitration and explain the role of H₂SO₄. [2]

Conc. HNO₃ + conc. H₂SO₄, 50 °C (water bath). H₂SO₄ is a catalyst: makes NO₂⁺ (step 1) and is regenerated by H⁺ + HSO₄⁻ → H₂SO₄ (step 3).

Check yourself

What is the electrophile in nitration?

NO₂⁺ (nitronium ion)

What are the conditions?

conc. HNO₃ + conc. H₂SO₄, 50 °C

Which acid is the catalyst?

sulfuric acid, H₂SO₄

Stretch ★ STRETCH: why can the intermediate exist at all?

Which carbons carry the positive charge, relative to the attacked carbon? Why is step 2 (attack) slow but step 3 (loss of H⁺) fast?

Answer: Positions 2, 4 and 6 (ortho and para to the attacked carbon) — three resonance forms. Step 2 breaks aromaticity → high activation energy → rate-determining. Step 3 restores aromaticity → very favourable and fast. Evidence: nitration shows no kinetic isotope effect (C–D reacts as fast as C–H), so C–H cleavage is not in the slow step.

This lesson’s files

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Halogenation and halogen carriers

Spec 6.1.1 (d)(ii) (e) (f)

By the end of this lesson I can…

write equations for the bromination and chlorination of benzene
draw the mechanism, including making the electrophile and regenerating the catalyst
compare the reactions of benzene and alkenes with bromine

Watch first (MaChemGuy)

The 5 things you must know

Benzene + Br₂ needs a halogen carrier: FeBr₃, AlCl₃ or iron filings.
The carrier makes the electrophile Br⁺: Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻
Then the SAME 4-move mechanism — just swap NO₂⁺ for Br⁺.
Catalyst back: H⁺ + FeBr₄⁻ → FeBr₃ + HBr
Alkenes react with Br₂ with NO catalyst (addition). Benzene does not (needs a carrier; substitution).

Memory hook: The CARRIER makes the PLUS. • Alkene: adds. Arene: swaps.

Where the marks are — exactlya mark each · 4–5 a paper = a grade

Curly arrow from the delocalised ring to Br of Br⁺Tail ON the circle — that is where the electron pair is — next to the carbon being attacked. Head touching the Br⁺ ion (the spec lets you assume the electrophile is Br⁺). An arrow that starts on the hexagon edge (a σ bond), ends on the wrong atom, or points the other way scores nothing.

The intermediate, drawn exactlyOpen the circle into a HORSESHOE that stops at the two carbons either side of the carbon holding the H and the Br — it must not reach that carbon. Write + INSIDE the horseshoe (the charge is spread over the other five carbons; never on the carbon with the H, never on the H itself). Draw H AND Br on the SAME carbon.

Making the electrophileBr₂ + FeBr₃ → Br⁺ + FeBr₄⁻

Regenerating the catalyst (FeBr₃)H⁺ + FeBr₄⁻ → FeBr₃ + HBr

The 7 ways people lose these marks

Arrow 1 ends on an O of NO₂⁺ instead of the N.
Arrow 1 drawn backwards (from NO₂⁺ to the ring) — the electrons come FROM the ring.
Circle left complete in the intermediate — it must be a horseshoe that stops either side of the carbon holding the H and the new group.
+ missing, or written on the carbon with the H / on the H itself — it goes INSIDE the horseshoe.
Arrow 2 starts on the H — it must start on the C–H BOND.
No H drawn where the new group joined — H and the new group must be on the SAME carbon.
Products missing: write the product AND H⁺ (and, if asked, the equation that regenerates the catalyst).

Kekulé structures are allowed throughout. Then arrow 1 starts on a C=C double bond, and the intermediate still needs the + and the H.

Do now — tap to check

Complete the nitration mechanism on your copy: add the curly arrows, the + charge and the products.

See the full mechanism (Lesson 3).

Write the equation for the formation of NO₂⁺.

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

What does 'delocalised' mean?

Electrons spread over several atoms (all six carbons), not fixed in one bond.

Exit ticket — tap to check

Write an equation to show how the electrophile is formed when Br₂ reacts with FeBr₃. [1]

Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻

Outline the mechanism for the reaction of benzene with chlorine using an AlCl₃ catalyst, including how the electrophile is generated. [4]

Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻; arrow from ring to Cl⁺; intermediate with + and horseshoe; arrow from C–H bond to ring; chlorobenzene + H⁺; H⁺ + AlCl₄⁻ → AlCl₃ + HCl.

Explain why benzene is less reactive with bromine than cyclohexene. [3]

Benzene's π electrons are delocalised → lower electron density → cannot polarise Br₂ / induce a dipole → no electrophile; alkene has localised π bond with high electron density which polarises Br₂.

Check yourself

Name three halogen carriers.

FeBr₃, AlCl₃, iron (also AlBr₃, FeCl₃)

Write the equation that makes Br⁺.

Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻

What do you SEE when bromine water is added to benzene?

No change — stays orange

Stretch ★ STRETCH: halogen carriers are Lewis acids

(a) Why does iron metal work as a 'halogen carrier'? (b) Iodination of benzene with I₂/FeI₃ barely works — suggest why. (c) Why is fluorination not done this way?

Answer: (a) 2Fe + 3Br₂ → 2FeBr₃ — the real catalyst is made in situ. (b) I⁺ is a weak electrophile; the reaction is endothermic/reversible (the HI formed reduces the product back), so an oxidising agent such as HNO₃ is added. (c) F₂ is far too reactive — the reaction is violent and uncontrolled. Fluoroarenes are made another way (via diazonium salts).

This lesson’s files

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Friedel–Crafts: building C–C bonds

Spec 6.1.1 (d)(iii) (g)

By the end of this lesson I can…

write equations for the alkylation and acylation of benzene
identify the electrophile and draw the mechanism
use the general pattern to predict products and mechanisms of unfamiliar reactions

Watch first (MaChemGuy)

The 5 things you must know

Friedel–Crafts reactions ADD CARBON to the ring: a new C–C bond.
Alkylation: haloalkane + AlCl₃ → alkylbenzene. Example: benzene + C₂H₅Cl → ethylbenzene + HCl.
Acylation: acyl chloride + AlCl₃ → a ketone. Example: benzene + CH₃COCl → phenylethanone + HCl.
Electrophiles: C₂H₅⁺ (alkylation) and CH₃CO⁺ (acylation). AlCl₃ makes them: C₂H₅Cl + AlCl₃ → C₂H₅⁺ + AlCl₄⁻
Same 4-move mechanism. Catalyst back: H⁺ + AlCl₄⁻ → AlCl₃ + HCl

Memory hook: Alkyl = a chain. Acyl = C=O. • Cross out the Cl, join the rest.

Where the marks are — exactlya mark each · 4–5 a paper = a grade

Curly arrow from the delocalised ring to C of CH₃CO⁺Tail ON the circle — that is where the electron pair is — next to the carbon being attacked. Head touching the carbonyl C of CH₃CO⁺ — the carbon that carries the + (not the O, not the CH₃). An arrow that starts on the hexagon edge (a σ bond), ends on the wrong atom, or points the other way scores nothing.

The intermediate, drawn exactlyOpen the circle into a HORSESHOE that stops at the two carbons either side of the carbon holding the H and the COCH₃ — it must not reach that carbon. Write + INSIDE the horseshoe (the charge is spread over the other five carbons; never on the carbon with the H, never on the H itself). Draw H AND COCH₃ on the SAME carbon.

Making the electrophileCH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻

Regenerating the catalyst (AlCl₃)H⁺ + AlCl₄⁻ → AlCl₃ + HCl

The 7 ways people lose these marks

Arrow 1 ends on an O of NO₂⁺ instead of the N.
Arrow 1 drawn backwards (from NO₂⁺ to the ring) — the electrons come FROM the ring.
Circle left complete in the intermediate — it must be a horseshoe that stops either side of the carbon holding the H and the new group.
+ missing, or written on the carbon with the H / on the H itself — it goes INSIDE the horseshoe.
Arrow 2 starts on the H — it must start on the C–H BOND.
No H drawn where the new group joined — H and the new group must be on the SAME carbon.
Products missing: write the product AND H⁺ (and, if asked, the equation that regenerates the catalyst).

Kekulé structures are allowed throughout. Then arrow 1 starts on a C=C double bond, and the intermediate still needs the + and the H.

Do now — tap to check

What does a halogen carrier do?

Generates the electrophile (e.g. Br⁺) by accepting a halide ion with its electron pair from Br₂.

Draw the intermediate in the bromination of benzene.

Ring with H and Br on the same carbon, horseshoe, + charge.

Name C₆H₅COCH₃.

phenylethanone

Exit ticket — tap to check

Write an equation for the reaction of benzene with 2-chloropropane in the presence of AlCl₃. [2]

C₆H₆ + CH₃CHClCH₃ → C₆H₅CH(CH₃)₂ + HCl

Identify the electrophile in the acylation of benzene with ethanoyl chloride and write an equation for its formation. [2]

CH₃CO⁺; CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻

Why are Friedel–Crafts reactions important in organic synthesis? [1]

They form a C–C bond to the ring, so carbon atoms can be added / the carbon skeleton extended.

Check yourself

What type of bond does Friedel–Crafts make?

a C–C bond to the ring

Name the product of benzene + CH₃COCl.

phenylethanone

What is the electrophile in acylation with CH₃COCl?

CH₃CO⁺

Stretch ★ STRETCH: Friedel–Crafts — the fine print

How would you make propylbenzene (a straight three-carbon chain) from benzene in TWO steps?

Answer: Acylate with CH₃CH₂COCl/AlCl₃ — acylium ions do not rearrange, and the ketone product is deactivated (no over-reaction). Then reduce C=O to CH₂: Clemmensen (Zn/Hg, HCl) or Wolff–Kishner (N₂H₄, KOH).

This lesson’s files

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Phenol

Spec 6.1.1 (h) (i) (j)

By the end of this lesson I can…

describe phenol as a weak acid: reacts with NaOH but NOT with carbonates
write equations for phenol with bromine and with dilute nitric acid
explain why phenol is more reactive than benzene

Watch first (MaChemGuy)

The 5 things you must know

Phenol = an OH group DIRECTLY on the ring. (OH on a side chain = an alcohol.)
Phenol is a WEAK acid: it reacts with NaOH (yes) but NOT with Na₂CO₃ (no fizz).
Phenol + bromine water → white precipitate of 2,4,6-tribromophenol. NO catalyst needed.
Phenol + DILUTE nitric acid → 2-nitrophenol and 4-nitrophenol.
Why more reactive than benzene? A lone pair on the O goes into the ring → more electron density.

Memory hook: Phenol: NaOH yes, carbonate no. • Phenol + Br₂: no catalyst, three Br, white solid.

Phenol + bromine water: decolourised, white precipitate, no catalyst.

Do now — tap to check

What is a weak acid?

An acid that only PARTIALLY dissociates in water.

What do you observe when a carboxylic acid is added to sodium carbonate solution?

Fizzing / effervescence — CO₂ gas is given off.

Name C₆H₅OH.

phenol

Exit ticket — tap to check

Write an equation for the reaction of phenol with sodium hydroxide. [1]

C₆H₅OH + NaOH → C₆H₅O⁻Na⁺ + H₂O

State what you would SEE when bromine water is added to phenol, and write the equation. [3]

Bromine decolourised (orange → colourless); white precipitate. C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr (2,4,6-tribromophenol).

Explain why phenol reacts more readily than benzene with electrophiles. [3]

O lone pair in a p-orbital is delocalised into the ring → increases electron density → ring more susceptible to attack / can polarise Br₂.

Check yourself

Does phenol react with Na₂CO₃?

Name the product of phenol + excess bromine water.

2,4,6-tribromophenol (white precipitate)

Which acid nitrates phenol?

DILUTE nitric acid (no H₂SO₄ needed)

Stretch ★ STRETCH: why is phenol acidic at all?

(a) 4-Nitrophenol has pKₐ 7.2. Is it a stronger or weaker acid than phenol? Explain. (b) Phenol's –OH activates the ring — what would you predict about the phenoxide ion, C₆H₅O⁻?

Answer: (a) Stronger (lower pKₐ): –NO₂ withdraws electrons and delocalises the negative charge onto its oxygens → phenoxide more stable → more dissociation. (b) O⁻ donates a full negative charge → far more electron density in the ring → even more reactive (phenol brominates faster in alkali).

This lesson’s files

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Directing groups and synthesis

Spec 6.1.1 (k) (l)

By the end of this lesson I can…

state which groups direct to positions 2,4 and which direct to position 3
predict the products of substitution reactions
plan a two-step synthesis in the RIGHT order

Watch first (MaChemGuy)

The 5 things you must know

A group already on the ring decides WHERE the next group goes.
OH and NH₂ GIVE electrons → the new group goes to position 2 or 4.
NO₂ TAKES electrons → the new group goes to position 3.
OH and NH₂ make the ring MORE reactive. NO₂ makes it LESS reactive.
In a two-step synthesis, put on FIRST the group that sends the second one to the right place.

Memory hook: Givers go 2,4. Takers go 3.

Where the second group goes depends on the first.

Do now — tap to check

Which is more reactive with bromine: phenol or benzene? Why?

Phenol — O lone pair donated into the ring increases electron density.

Name the product of phenol + excess bromine water.

2,4,6-tribromophenol

Name C₆H₅NH₂.

phenylamine

Exit ticket — tap to check

Classify –OH, –NO₂ and –NH₂ as 2,4-directing or 3-directing. [1]

–OH and –NH₂: 2,4-directing. –NO₂: 3-directing.

Draw the organic product(s) of: (a) phenol + dilute HNO₃ (b) nitrobenzene + Br₂/FeBr₃. [2]

(a) 2-nitrophenol and 4-nitrophenol. (b) 3-bromonitrobenzene.

Suggest a two-step synthesis of 1-bromo-4-nitrobenzene from benzene. [2]

Step 1: Br₂/FeBr₃ → bromobenzene. Step 2: conc. HNO₃/H₂SO₄, 50 °C → 1-bromo-4-nitrobenzene (+ 2-isomer). Br is 2,4-directing (data would be given).

Check yourself

Where does OH send the next group?

2 and 4

Where does NO₂ send the next group?

To make 1-chloro-4-nitrobenzene, which group goes on first?

Cl (it is 2,4-directing)

Stretch ★ STRETCH: the real reason for directing effects

(a) Halogens DEACTIVATE the ring yet direct 2,4-. Explain. (b) Predict where –SO₃H directs, and why.

Answer: (a) Inductive effect: halogens are electronegative → withdraw electron density → deactivate. But their lone pairs still donate by resonance into the 2,4-intermediates → 2,4-directing. (b) 3-directing: S is δ+ and withdraws electrons; the 2- and 4-intermediates would put the + charge next to δ+ S — destabilised.

This lesson’s files

Worksheet 7 Worksheet 7 (greyscale)Slides Uni section U7 Mark your work →

Chemguide — electrophilic substitution in methylbenzene and nitrobenzene

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Summary and exam clinic

Spec 6.1.1 (a)–(l)

By the end of this lesson I can…

draw the reaction map for benzene and for phenol
tackle exam questions that combine several ideas
self-assess against every spec statement

Watch first (MaChemGuy)

This lesson’s files

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Everything for the course, in one place.

6.1.1 Aromatic compounds

Prep

Notices

How to use this site

Course map

Aromatic compounds

Meet benzene

Naming aromatic compounds

Electrophilic substitution: nitration

Halogenation and halogen carriers

Friedel–Crafts: building C–C bonds

Phenol

Directing groups and synthesis

Summary and exam clinic

Downloads

Lesson 1b — from alkenes to benzene

MaChemGuy videos

Meet benzene

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The 5 things you must know

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Naming aromatic compounds

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The 5 things you must know

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Electrophilic substitution: nitration

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The 5 things you must know

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Halogenation and halogen carriers

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The 5 things you must know

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Friedel–Crafts: building C–C bonds

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The 5 things you must know

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Phenol

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The 5 things you must know

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Directing groups and synthesis

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