WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Hydrocarbons
E–Z isomerism in alkenes
- 3 things to know
- AS Unit 2
Overview
Fifteen recall questions on E–Z isomerism in alkenes.
What you need to know
Answer each one before you reveal it.
But-2-ene exists as two isomers but but-1-ene does not. Explain.
Does 3-methylpent-2-ene show E–Z isomerism? Explain which group takes priority on each carbon.
Questions from this subtopic
Each one with the model answer underneath.
- 1
What two conditions must be met for an alkene to show E–Z isomerism?
Model answer
There must be restricted rotation, which the carbon–carbon double bond provides, and each of the two doubly bonded carbon atoms must carry two different groups; if either carbon has two identical groups the two possible arrangements are the same molecule.
- 2
Why does the double bond prevent rotation and so allow E–Z isomers to exist?
Model answer
Rotating about the double bond would require the sideways p orbital overlap of the π-bond to be broken, which needs far more energy than is available at ordinary temperatures, so the groups are locked on their own side of the double bond and the two arrangements cannot interconvert.
- 3
What do the labels E and Z actually mean?
Model answer
The higher priority group on each of the two doubly bonded carbons is identified, and the isomer is labelled Z if those two groups lie on the same side of the double bond and E if they lie on opposite sides.
- 4
How is the priority of the groups on a doubly bonded carbon decided?
Model answer
Priority is decided by the atomic number of the atom directly attached to the doubly bonded carbon, the higher atomic number taking priority; if the two attached atoms are the same, you move outwards to the next atoms along each branch until a difference is found.
- 5
But-2-ene exists as two isomers but but-1-ene does not. Explain.
Model answer
In but-2-ene each doubly bonded carbon carries a hydrogen atom and a methyl group, which are different, so the methyl groups can be on the same side or on opposite sides; in but-1-ene the end carbon carries two hydrogen atoms, so swapping them produces the identical molecule.
- 6
Identify the E and Z isomers of but-2-ene.
Model answer
On each doubly bonded carbon the methyl group outranks the hydrogen atom, so the isomer with the two methyl groups on the same side of the double bond is Z-but-2-ene and the one with them on opposite sides is E-but-2-ene.
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