Part 1: The double bond and E/Z stereoisomerism

Explain restricted rotation, decide whether stereoisomers are possible and apply priority rules to the actual attached atoms.

One sigma bond and one pi bond

A C=C double bond contains one σ bond from head-on overlap and one π bond from sideways overlap of parallel p orbitals. The π electron density lies above and below the plane of the bonded atoms. Around each double-bond carbon there are three bonding regions, arranged trigonal planar at about 120°.

Rotating one end would disrupt the sideways overlap, so free rotation is restricted. The exposed π electron density attracts electrophiles; the π bond is easier to disrupt in addition than the σ framework. Acyclic hydrocarbons with one C=C have general formula CnH2n.

Two different groups at each end

Stereoisomers have the same structural formula but a different spatial arrangement. E/Z isomerism needs restricted rotation and two different substituents on each carbon of the double bond. But-2-ene qualifies; but-1-ene does not because its terminal carbon has two H atoms.

For each end separately, rank the two attached groups by the atomic number of the atom directly bonded to the alkene carbon. Higher atomic number wins. If tied, compare the next atoms in descending atomic-number order until the first difference. Do not compare the total mass of the groups. Handling multiple bonds within substituent priority calculations is beyond this AS requirement.

Higher-priority groups on the same side give Z; opposite sides give E. In but-2-ene, CH₃ outranks H at each end. Cis/trans is a special case involving a matching pair of substituents across the double bond; it is not a naming system for every E/Z alkene.

Worked priority choices

At an alkene carbon attached to F and an ethyl group, F wins because atomic number 9 exceeds carbon’s 6, even though the whole ethyl group has greater mass. Between ethyl and methyl, both begin with C; the next comparison is [C,H,H] versus [H,H,H], so ethyl wins.

E/Z differences can change molecular polarity and physical properties. For 1,2-dichloroethene, the symmetric E isomer’s C–Cl dipoles cancel more effectively, whereas the Z isomer is polar. Explain properties from the whole structure rather than assuming one letter always means a higher boiling point.

Worked bond count: ethene contains five sigma bonds

Ethene, H₂C=CH₂, has four C–H single bonds and a C=C double bond. Each C–H contributes one σ bond, and C=C contributes one σ plus one π. Total: five σ bonds and one π bond. Counting the whole double bond as two σ bonds loses the orbital distinction.

The electron-pair regions around each alkene carbon are its two C–H bonds and the C=C region. Three regions give trigonal planar geometry, approximately 120°. The double bond counts as one region in shape reasoning but contains two bonding pairs. These are different counting tasks.

Test eligibility before assigning E or Z

Write down the two substituents at each end of C=C, excluding the other double-bond carbon. In CH₃CH=C(CH₃)₂, the left end has CH₃ and H, but the right end has two identical CH₃ groups. It therefore has no E/Z pair, even though rotation is restricted.

In CH₃CH=CHCH₂CH₃, each end has H and a carbon group, so pent-2-ene can have E and Z forms. The two carbon groups need not be identical to one another; the requirement is that the pair attached to each individual carbon differs.

A condensed formula such as CH₃CH=CHCH₃ establishes connectivity but usually does not specify which groups lie on the same side. Do not assign E or Z without a spatial arrangement, a drawing or the stereochemical name.

Resolve a tied first atom one layer at a time

Compare –CH₂CH₂CH₃ with –CH(CH₃)₂ at one end of an alkene. Both groups attach through carbon. For the next layer, excluding the alkene carbon, the first carbon has [C,H,H] in propyl and [C,C,H] in propan-2-yl. Compare in descending atomic-number order: the first entries tie, then C outranks H, so propan-2-yl has higher priority.

Now rank the two groups at the other end independently. Only after both local comparisons should you ask whether the two winning groups lie together (Z) or opposite (E). A large group does not “win across the whole molecule”; there is one winner at each end.

Cis/trans uses a matching pair as the reference. E/Z uses priority rankings. They coincide for but-2-ene, but it is unsafe to replace every cis label by Z without checking priorities in other structures.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources