E/Z isomerism and the CIP rules

There is no rotation about a C=C double bond, so the groups are locked in place. E/Z isomerism exists when each of the two doubly-bonded carbons carries two different groups.

Cahn-Ingold-Prelog priority rules:

• On each carbon, compare the atoms directly attached. The one with the higher atomic number has priority. (Br > Cl > O > N > C > H)
• If those atoms are the same, work outwards to the next atoms until you find a difference.

Then: the two higher-priority groups on the same side → Z (from zusammen, together). On opposite sides → E (from entgegen, opposite).

cis/trans is not the same as E/Z. cis/trans only works when there is an identical group on each carbon. E/Z always works — which is why it replaced it.

How is priority decided?

Which one shows E/Z?

Optical isomerism and chirality

A carbon with four different groups attached is a chiral centre (a stereocentre, often marked with an asterisk). The molecule and its mirror image are then non-superimposable — like your left and right hands.

The two forms are enantiomers (optical isomers). They are identical in every ordinary physical and chemical property — same melting point, same density, same reactions with achiral reagents. They differ in only two ways:

• They rotate the plane of plane-polarised light in opposite directions but by the same amount.
• They interact differently with other chiral molecules — which is why one enantiomer of a drug may be therapeutic and the other useless or harmful (the thalidomide tragedy).

A racemic mixture (racemate) is a 50:50 mixture of the two enantiomers. The rotations cancel exactly, so it is optically inactive.

Spot the chiral centre

Chiral, achiral or E/Z?

Tap a molecule, then the box that describes its stereochemistry.

Stereochemistry tells you the mechanism

Because enantiomers are made or destroyed by the geometry of an attack, the optical activity of the product is powerful evidence for a mechanism.

• S_N2 (primary halogenoalkanes): the nucleophile attacks from the opposite side to the leaving group, in one step. The stereocentre is inverted — like an umbrella turning inside out. A single enantiomer in gives a single (inverted) enantiomer out.
• S_N1 (tertiary halogenoalkanes): the C–X bond breaks first, giving a planar carbocation. The nucleophile can then attack from either face with equal probability, so the product is a racemic mixture.
• Nucleophilic addition to an unsymmetrical carbonyl: the C=O carbon is planar (trigonal, 120°), so CN⁻ attacks from either side with equal probability — the product is again a racemate, and is optically inactive.

Why is the product optically inactive?

The structure of benzene

Kekulé proposed alternating single and double bonds. Three pieces of evidence disprove that and support a delocalised ring:

• Bond lengths. All six C–C bonds are identical, at 0.139 nm — intermediate between a C–C single bond (0.154 nm) and a C=C double bond (0.134 nm). Kekulé requires two different lengths.
• Enthalpy of hydrogenation. Cyclohexene → cyclohexane releases −120 kJ mol⁻¹. Three isolated double bonds should therefore release −360. Benzene actually releases only −208 kJ mol⁻¹ — it is 152 kJ mol⁻¹ more stable than Kekulé predicts. That gap is the delocalisation (resonance) energy.
• Reactivity. Benzene does not decolourise bromine water and resists addition — it undergoes substitution instead, preserving the stable delocalised ring.

The real structure: each carbon is sp² with a p orbital perpendicular to the ring; the six p orbitals overlap sideways into a ring of delocalised π electron density above and below the planar hexagon. Bond angles are 120°.

Your turn — delocalisation energy

What the bond lengths prove

Electrophilic substitution — nitration

The delocalised π ring is electron-rich, so it attracts electrophiles. But benzene substitutes rather than adds, because substitution restores the stable delocalised system.

Nitration — concentrated HNO₃ with concentrated H₂SO₄ as catalyst, at 50 °C (above 50 °C you get dinitration):

The mechanism, in two steps: the π electrons attack the electrophile, forming an unstable intermediate in which delocalisation is broken over one carbon (drawn as a horseshoe with a + charge). Then H⁺ is lost from that carbon, and the delocalised ring is restored.

Why it matters: nitration is the gateway to dyes, explosives and — via reduction of the nitro group — to aromatic amines.

Your turn — relative molecular mass

Your turn — percentage yield

Halogenation and Friedel-Crafts

All these reactions share the same mechanism — the only variable is how the electrophile is made.

• Halogenation: Br₂ alone is not electrophilic enough. A halogen carrier (AlBr₃, FeBr₃ or iron filings) polarises it: Br₂ + AlBr₃ → Br^δ+···AlBr₄^δ−, generating Br⁺.
• Friedel-Crafts alkylation: CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻, giving methylbenzene.
• Friedel-Crafts acylation: CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻, giving phenylethanone, C₆H₅COCH₃. Acylation is the more useful synthesis, because the alkylated product is more reactive than benzene and gets attacked again.

The AlCl₃ is a catalyst, not a reagent: it is regenerated in the final step when the AlCl₄⁻ takes the H⁺ back off the ring.

The catalyst's job

Benzene vs cyclohexene

Reagents for the ring

Tap the reagent on the left, then the aromatic product it gives with benzene.

The big ideas to know

That is the whole of OA1. Press Finish to see your score.