Difficulty: Intermediate | Prerequisites: Chapter 3 (stereochemistry, CIP priority rules), Chapter 4 (conformational analysis)
Big picture: Alkenes are the first functional group you study in depth after alkanes, and they open the door to nearly every reaction mechanism in the rest of the course. This chapter focuses on structure, not reactions: how the C=C double bond forms, why it restricts rotation, and how to name and assign stereochemistry to alkene isomers. If you are comfortable with hybridisation, Lewis structures, and CIP priority from Chapter 3, you are ready. If those feel shaky, revisit them first, because E/Z assignment reuses the exact same ranking rules.
Alkenes contain a carbon-carbon double bond made of one sigma bond and one pi bond. The pi bond locks the molecule flat and prevents rotation, which creates geometric (cis/trans) isomers. When more than two substituents sit on the double bond, you assign configuration using the E/Z system and CIP priority rules.
Alkane
A saturated hydrocarbon containing only C–C single bonds. General formula: C_nH_(2n+2). All carbons are sp3-hybridised. In simple terms, these are the "fully loaded" hydrocarbons with the maximum number of hydrogens.
Alkene
An unsaturated hydrocarbon containing at least one C=C double bond. General formula: C_nH_(2n). The double-bond carbons are sp2-hybridised. Think of it as an alkane that has lost two hydrogens to form a double bond.
Sigma (σ) bond
A bond formed by head-on overlap of orbitals along the internuclear axis. In alkanes, the C–C σ bond comes from overlap of two sp3 orbitals. In alkenes, the C–C σ bond comes from overlap of two sp2 orbitals. In simple terms, this is the stronger, more direct bond that allows rotation.
Pi (π) bond
A bond formed by side-on overlap of two unhybridised 2p orbitals, above and below the plane of the molecule. The pi bond is the second bond in a double bond. Think of it as the "extra" bond that locks the double bond flat and prevents rotation.
Bond dissociation energy (BDE)
The energy required to homolytically break a bond in the gas phase. A higher BDE means a stronger bond. In simple terms, it tells you how much energy you need to snap a bond apart.
Cis isomer
A geometric isomer of a 1,2-disubstituted alkene where the two identical (or similar) substituents are on the same side of the double bond.
Trans isomer
A geometric isomer of a 1,2-disubstituted alkene where the two identical (or similar) substituents are on opposite sides of the double bond.
Z (zusammen)
The E/Z designation meaning "together." The two highest-priority substituents (one from each carbon of the double bond) are on the same side. Think of Z as "zee zame zide" as a mnemonic.
E (entgegen)
The E/Z designation meaning "opposite." The two highest-priority substituents are on opposite sides of the double bond.
CIP priority rules (Cahn-Ingold-Prelog)
The ranking system used for both R/S (Chapter 3) and E/Z assignments. Atoms are ranked by atomic number: higher atomic number = higher priority. Ties are broken by moving outward to the next set of attached atoms. In simple terms, heavier atoms win. If there is a tie, keep looking further along the chain until you find a difference.
Degree of unsaturation (index of hydrogen deficiency)
A count of how many pairs of hydrogens are "missing" compared with the fully saturated formula. Each double bond or ring adds one degree of unsaturation. Formula: (2C + 2 + N – H – X) / 2 for a molecule C_xH_yN_nX_x. In simple terms, it tells you how many double bonds and/or rings a molecule has.
Alkane bonding (ethane as model)
Each carbon is sp3-hybridised (four equivalent sp3 orbitals, tetrahedral geometry, ~109.5° bond angles)
The C–C bond is a single σ bond formed by head-on overlap of two sp3 orbitals
Six C–H σ bonds form from overlap of C sp3 orbitals with H 1s orbitals
Free rotation around the C–C bond (rotation barrier is only ~3 kcal/mol)
Alkene bonding (ethene as model)
Each double-bond carbon is sp2-hybridised (three sp2 orbitals in a trigonal planar arrangement, ~120° bond angles, plus one unhybridised 2p orbital perpendicular to the plane)
The C=C double bond consists of one σ bond (sp2-sp2 head-on overlap) plus one π bond (side-on overlap of the two 2p orbitals)
Four C–H σ bonds form from overlap of C sp2 orbitals with H 1s orbitals
Rotation around the C=C bond is restricted (rotation barrier is ~63 kcal/mol) because rotating would break the π bond
Why the π bond matters
The π bond is weaker than the σ bond. You can estimate its strength from the BDE data: the total C=C BDE is 172 kcal/mol, and the C–C σ BDE is 90 kcal/mol, so the π bond contributes roughly 172 – 90 = 82 kcal/mol. That 82 kcal/mol is enough to lock the molecule flat and prevent free rotation at room temperature, which is the entire reason geometric isomers (cis/trans) exist for alkenes.
Cis/trans (for 1,2-disubstituted alkenes)
Because the C=C double bond prevents rotation, groups attached to the double-bond carbons are locked in place. When two identical or similar substituents are on the same side, the isomer is cis. When they are on opposite sides, it is trans.
Example: cis-2-butene has both methyl groups on the same side of the double bond. Trans-2-butene has the methyl groups on opposite sides. The trans isomer is generally more stable because the bulky groups are further apart (less steric strain).
E/Z system (for alkenes with two or more different substituents)
Cis/trans only works cleanly when you have two of the same substituent, one on each double-bond carbon. When you have three or four different groups, you need the E/Z system.
The procedure:
Look at each carbon of the double bond separately
Rank the two substituents on each carbon using CIP priority rules (higher atomic number = higher priority; break ties by moving outward)
If the two higher-priority groups (one from each carbon) are on the same side, the configuration is Z (zusammen, "together")
If the two higher-priority groups are on opposite sides, the configuration is E (entgegen, "opposite")
CIP priority refresher
Compare atoms directly attached to the double-bond carbon. Higher atomic number wins.
If there is a tie, move to the next set of atoms along the chain and compare again.
Double bonds are treated as two single bonds to the same atom (a phantom duplicate). For example, a C=O is treated as C bonded to O and O bonded to C.
These are the same rules used for assigning R and S at stereocentres (Chapter 3).
Important note: E does not always equal trans, and Z does not always equal cis. They often coincide, but E/Z is defined by priority, while cis/trans is defined by whether similar groups are on the same side. When in doubt, use E/Z.
Property | Alkane (C–C) | Alkene (C=C) |
|---|---|---|
Hybridisation | sp3 | sp2 |
Bond length | 1.54 Å | 1.34 Å |
Bond dissociation energy | 90 kcal/mol | 172 kcal/mol |
Rotation barrier | ~3 kcal/mol (free) | ~63 kcal/mol (restricted) |
Geometry | Tetrahedral (~109.5°) | Trigonal planar (~120°) |
General formula | C_nH_(2n+2) | C_nH_(2n) |
Estimated π bond strength: 172 – 90 = 82 kcal/mol
Degree of unsaturation = (2C + 2 + N – H – X) / 2
Students often assume E always means trans and Z always means cis. This is true in many simple cases, but it breaks down when the substituents are all different. E/Z is based on CIP priority, not on whether groups "look" the same.
Students sometimes think the C=C double bond is exactly twice as strong as a C–C single bond. It is not. The σ component is about 90 kcal/mol and the π component is about 82 kcal/mol, so the double bond (172 kcal/mol) is less than double the single bond.
Students forget that restricted rotation is a consequence of the π bond, not the σ bond. You could rotate freely if only the σ bond were present. The π bond, formed by 2p orbital overlap above and below the plane, is what locks the geometry.
When assigning CIP priorities, students sometimes rank substituents by size or molecular weight rather than by atomic number at the first point of difference. Always compare atom by atom along the chain, using atomic number at each step.
⚠️ Expect to be given a structure and asked to assign E or Z. You will need to rank all four substituents correctly using CIP rules.
⚠️ Bond length, bond strength, and rotation barrier comparisons between C–C and C=C are classic short-answer or multiple-choice questions. Know the numbers.
⚠️ Understanding restricted rotation is foundational for every reaction mechanism involving alkenes later in the course. If you do not see why rotation is restricted, the logic of addition reactions in the next chapter will not land.
⚠️ Drawing both cis and trans isomers for a given molecular formula is a common "draw all isomers" question. Do not forget that cis and trans count as separate structures.
True or False: The C=C double bond in ethene is exactly twice as strong as the C–C single bond in ethane.
Fill in the blank: The hybridisation of a double-bond carbon in an alkene is ______.
True or False: Z configuration means the two highest-priority groups are on opposite sides of the double bond.
Fill in the blank: The rotation barrier around a C=C double bond is approximately ______ kcal/mol.
True or False: Cis-2-butene is more stable than trans-2-butene.
Answers: 1. False (172 vs. 2 x 90 = 180). 2. sp2. 3. False (Z = same side). 4. 63. 5. False (trans is more stable due to less steric strain).
Q: What type of orbital overlap forms the pi bond in ethene?
A: Side-on overlap of two unhybridised 2p orbitals, one from each sp2-hybridised carbon.
Q: Assign E or Z to the following: a double bond where carbon 1 carries –Br (higher priority) and –H, and carbon 2 carries –CH3 (higher priority) and –H, with Br and CH3 on the same side.
A: Z (zusammen). The two highest-priority groups (Br and CH3) are on the same side.
Q: Why is trans-2-butene more stable than cis-2-butene?
A: In the cis isomer, the two methyl groups are on the same side of the double bond, creating steric strain. In the trans isomer, they are on opposite sides, reducing that strain.
Q: A student says the C=C bond is twice as strong as a C–C bond because it has two bonds. What is wrong with this reasoning?
A: The σ and π components are not equal in strength. The σ bond contributes about 90 kcal/mol and the π bond about 82 kcal/mol, giving a total of 172 kcal/mol, which is less than 2 x 90 = 180 kcal/mol.
Q: Can you assign cis/trans to 2-methyl-2-butene? Why or why not?
A: No. Cis/trans requires each double-bond carbon to carry two different substituents that can be compared, and it works cleanly only for 1,2-disubstituted alkenes with matching groups. For trisubstituted or tetrasubstituted alkenes, use E/Z.
The CIP priority rules here are the same ones used in Chapter 3 for assigning R/S configuration at stereocentres. Practising E/Z assignment reinforces those skills and vice versa.
Restricted rotation around C=C sets up the logic for addition reactions in Chapter 6, where reagents add across the double bond. Understanding that the π bond is the reactive site (weaker, more exposed electron density) is essential for predicting regioselectivity and stereochemistry of those reactions.
The fatty acid and vision examples in this chapter connect alkene geometry to biochemistry. Cis double bonds create kinks in fatty acid chains that lower melting points, and the cis-to-trans isomerisation of retinal is the molecular basis of vision.
alkene, olefin, unsaturated hydrocarbon, C=C double bond, sp2 hybridisation, pi bond, sigma bond, cis-trans isomerism, geometric isomers, E/Z nomenclature, zusammen, entgegen, CIP priority rules, Cahn-Ingold-Prelog, bond dissociation energy, BDE, restricted rotation, rotation barrier, degree of unsaturation, index of hydrogen deficiency, trigonal planar geometry, 2p orbital overlap, steric strain, disubstituted alkene, CHEM 25500, Purdue organic chemistry, Dr Uyeda, Chapter 5