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A carbon–carbon double bond contains a sigma bond and a pi bond. The pi bond results from sideways overlap of p orbitals and prevents free rotation. E/Z isomerism requires two different groups on each double-bond carbon. Compare priorities using atomic number at the first point of difference. Z places the higher-priority groups on the same side; E puts them on opposite sides. Do not classify an alkene as E or Z if either carbon has two identical substituents.
Video by Eliot Rintoul. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.
Viewing focus: Assign E/Z isomers using priorities
After watching, attempt these questions before opening the answers. You can also use the written lesson below.
After watching 1: What prevents free rotation around C=C?
The pi bond. Rotation would disrupt sideways orbital overlap.
After watching 2: What does Z identify?
Higher-priority groups on the same side. Use priority rules rather than group size alone.
What you'll be able to do
A carbon–carbon double bond contains a sigma bond and a pi bond. The pi bond results from sideways overlap of p orbitals and prevents free rotation. E/Z isomerism requires two different groups on each double-bond carbon. Compare priorities using atomic number at the first point of difference. Z places the higher-priority groups on the same side; E puts them on opposite sides. Do not classify an alkene as E or Z if either carbon has two identical substituents.
The electron-rich pi bond attacks an electrophile. With HBr, the pi electron pair forms a C–H bond while the H–Br electrons move to bromine, leaving a carbocation. Bromide then attacks the positive carbon. With an unsymmetrical alkene, different carbocations are possible; the route through the more stable carbocation often gives the major product. Alkyl groups stabilise positive charge through electron donation. Write both arrows in the first step and show the charged intermediate.
Bromine water is decolourised by many alkenes as addition removes the C=C bond; a positive observation is useful evidence but must be interpreted alongside other functional groups and the test conditions. Hydrogenation uses hydrogen and a metal catalyst to form an alkane. Steam with an acid catalyst can produce an alcohol. Addition polymerisation opens many double bonds into a chain. Each application changes the pi bond while preserving a carbon skeleton that can be traced through the equation.
Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.
Common mistakes to avoid
Key takeaways
Question 1: Predict the major product from propene and HBr under ordinary electrophilic-addition conditions.
2-bromopropane is the major product.
Check the assessed scope and practical or mathematical requirements.
An external explanation or interactive activity to use alongside the lesson. It may include extension material.
Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.
Test yourself
Answer a few questions, then review anything you missed.
Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.