Structure, bonding and reactivity — AQA A-Level Chemistry
Test yourself on Structure, bonding and reactivity with AQA A-Level practice questions.
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Structure, bonding and reactivity explained
Alkenes contain only carbon and hydrogen, so they are hydrocarbons, and they contain at least one carbon–carbon double bond, so they are unsaturated.
Read the full explanation
Unsaturated has a precise meaning: the molecule can add extra atoms across that double bond without losing anything, unlike a saturated alkane, which can only substitute. A single double bond gives the general formula CₙH₂ₙ, two hydrogens fewer than the alkane with the same number of carbons. The practical consequence is the test for unsaturation: shaking with bromine water at room temperature turns it from orange to colourless as bromine adds across the double bond, while an alkane leaves it orange. Say colourless, not clear, because the alkane mixture is also clear.
Bonding in alkenes involves a double covalent bond, a centre of high electron density.
The carbon–carbon double bond is made of two different bonds. One is a σ bond from head-on overlap of orbitals directly between the two carbons; the other is a π bond, formed by sideways overlap of p orbitals, giving electron density above and below the plane of the molecule. Those π electrons are exposed and held less tightly than σ electrons, so the double bond is a region of high electron density that attracts electrophiles, species that accept a pair of electrons. That is the whole basis of alkene chemistry. Two other consequences follow: the three groups around each double-bond carbon lie in a plane at about 120°, and rotation about the double bond is restricted, which is why E-Z stereoisomers exist.
Your focus
- Define unsaturated in terms of the carbon–carbon double bond and the ability to undergo addition.
- Describe what is seen when bromine water is shaken with an alkene and with an alkane.
- Deduce from a molecular formula whether a hydrocarbon could contain one carbon–carbon double bond.
Show all 6 objectives
- Describe how the π bond in an alkene arises from sideways overlap of p orbitals.
- Explain why alkenes are attacked by electrophiles rather than by nucleophiles.
- Deduce whether a given alkene shows E-Z isomerism, referring to restricted rotation about the double bond.
Structure, bonding and reactivity exam tips
Marking Points
- one mark for defining unsaturated as containing a carbon–carbon double bond
- one mark for saying an unsaturated molecule can undergo addition reactions
- one mark for the general formula CₙH₂ₙ for an alkene with one double bond
- one mark for the bromine water observation, orange to colourless, at room temperature
- one mark for describing the π bond as sideways overlap of p orbitals above and below the plane
- one mark for stating that the high electron density attracts electrophiles
- one mark for defining an electrophile as an electron pair acceptor
- one mark for linking restricted rotation about the C=C to the existence of E-Z stereoisomers
Examiner Tips
- 💡Give both colours in any observation answer, before and after.
- 💡Count the hydrogens against CₙH₂ₙ to check a structure you have drawn is really an alkene.
- 💡Remember hydrocarbon means carbon and hydrogen only, so a molecule containing oxygen is not an alkene.
- 💡Whenever you write "high electron density", follow it immediately with "attracts electrophiles" to secure the link.
- 💡If a question asks for the origin of stereoisomerism in an alkene, the phrase wanted is restricted rotation about the C=C.
- 💡Draw the π electron density above and below the plane, not around the whole molecule.
Common Mistakes
- writing that bromine water "goes clear", when the required word is colourless
- defining unsaturated as containing any double bond, when it must be carbon to carbon
- giving the general formula as CₙH₂ₙ₊₂, which is the alkane formula
- saying "decolourises bromine water" without stating the starting colour
- saying the double bond attracts nucleophiles, reversing the whole of alkene chemistry
- describing the C=C as two identical bonds rather than one σ and one π bond
- claiming free rotation about the double bond, which would remove E-Z isomerism
- giving the bond angle around the double bond as 109.5° instead of about 120°