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Organic Chemistry 1 · Structure and Bonding

Alkene Structure and Nomenclature

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

An is a hydrocarbon containing a carbon–carbon double bond: one sigma (σ) bond plus one pi (π) bond. The π bond is formed by side-by-side overlap of unhybridized p orbitals, which makes the double bond planar, rigid, and unsaturated. Naming requires selecting the longest chain that contains the double bond, so the double bond gets the lowest locator, and specifying (or ) stereochemistry where needed.

Why this matters

is the structural basis of "good" versus "bad" fats: mono- and polyunsaturated fatty acids contain C=C bonds whose geometry (naturally cis in most dietary fats) prevents tight packing and lowers melting point. A single cis/trans switch can change a molecule's shape and biological activity.

The college version

1. The Double Bond and Unsaturation

A C=C double bond is one σ bond plus one π bond. The σ bond comes from head-on overlap of sp² hybrid orbitals; the π bond comes from side-by-side overlap of the leftover p orbitals. The π bond locks rotation, so the six atoms attached to a C=C lie in one plane.

2. Degree of Unsaturation (Index of Hydrogen Deficiency)

The degree of unsaturation (DU, also called the index of hydrogen deficiency, IHD) counts rings and π bonds from a molecular formula:

IHD = 2C + 2 + N - H - X2

where C = number of carbons, N = nitrogens, H = hydrogens, and X = halogens (halogens are counted as if they were hydrogens). Oxygen is ignored. Each degree equals one ring OR one π bond; a double bond counts as one, a triple bond as two. A simple acyclic alkene has IHD = 1.

3. IUPAC Naming and Stereochemistry

The parent is the longest continuous chain that contains the double bond; drop the "-ane" ending and add "-ene" (ethene, propene, but-2-ene…). Number the chain so the double bond gets the lowest locator, and place that locator (the first double-bonded carbon) before the parent name. Use "-diene", "-triene" for multiple double bonds, and name/number substituents normally. For stereochemistry, compare the two groups on each double-bonded carbon by priority: if the two higher-priority groups are on the same side it is Z (German zusammen, "together"); if on opposite sides it is E (entgegen, "opposite"). cis/trans is the simpler label used when each double-bonded carbon carries one hydrogen and one non-hydrogen group.

How it works

  1. Find the longest chain that includes the double bond — that is the parent.
  2. Number the chain to give the double bond the lowest locator.
  3. Name substituents and combine them with the "-ene" parent name.
  4. If needed, assign E/Z (or cis/trans) by comparing priorities on each double-bonded carbon.
  5. To draw from a name, lay out the parent chain, place the double bond at its locator, then add substituents and stereochemistry.

Common confusions

Do not confuseWithDifference
cis/transE/Zcis/trans needs identical groups; E/Z compares priorities and always works
Degree of unsaturationNumber of double bondsIHD counts rings too; a ring and a double bond each add one
Sigma bondPi bondσ is head-on and rotatable; π is side-by-side and locks rotation
Terminal alkeneInternal alkeneTerminal has C=C at chain end; internal does not
GeminalVicinalGeminal substituents share one carbon; vicinal are on adjacent carbons

Memory aids

"Z = ze same side, E = Eenemies apart." Z (zusammen) together, E (entgegen) opposite — judged by the higher-priority groups.

Quick review

Topic Recap

Alkenes contain a C=C double bond (one σ + one π), which makes them unsaturated and planar. The degree of unsaturation counts rings and π bonds. IUPAC naming requires the longest chain that includes the double bond and the lowest possible double-bond locator, plus E/Z or cis/trans stereochemistry where relevant. Terminal versus internal alkenes and geminal versus vicinal substitution complete the classification.

Knowledge Check

  1. What is the degree of unsaturation of C4H8?
  2. Name CH3CH=CHCH3.
  3. Is (Z)-but-2-ene cis or trans?
  4. What two kinds of bonds make up a C=C double bond?
  5. Why can a double bond not rotate freely?

Answers and Rationales

  1. IHD = (2×4 + 2 − 8)/2 = 1 — one double bond (or one ring), consistent with an alkene.
  2. But-2-ene (the C=C sits between C2 and C3).
  3. cis — Z puts the two CH3 groups on the same side, which is the cis arrangement.
  4. One σ bond (head-on sp² overlap) and one π bond (side-by-side p-orbital overlap).
  5. Rotation would require breaking the side-by-side p-orbital overlap of the π bond, which locks the geometry.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Picture a chain of carbon atoms holding hands. In an alkane every carbon uses both hands to grab neighbors, so the chain is "full" of hydrogens — saturated. In an alkene, two neighboring carbons grab each other with two hands at once (a double bond). Now those carbons each have one less hand free for a hydrogen, so the molecule is "missing" hydrogens and is unsaturated, with a flat, rigid "double grip" that cannot twist.

A comparison: saturated fats (butter) have no double bonds and pack tightly into solids; unsaturated fats (olive oil) contain double bonds that kink the chain and stay liquid.

Where it stops being exact: the "two hands" picture suggests a double bond is just two identical single bonds. It is not — one bond is a σ bond (head-on overlap, free to rotate) and the other is a π bond (side-by-side overlap, locked in place). It is the π bond that makes the alkene flat and rotation-locked, which is exactly why cis/trans and E/Z stereoisomers exist.

Simple Example

Ethene, CH2=CH2, the simplest alkene. Each carbon is sp²-hybridized and trigonal planar with ~120° bond angles. The two carbons share one σ bond (sp²–sp² head-on overlap) and one π bond (side-by-side p-orbital overlap above and below the plane). There are no stereoisomers here because each carbon carries two identical hydrogens.

Worked example

Name and assign stereochemistry for (E)-3-methylpent-2-ene.

  1. Longest chain containing the double bond = five carbons → parent "pentene".
  2. Number to give the double bond the lowest locator. Left-to-right puts the C=C between C2 and C3; right-to-left would also give 2–3, so choose the direction that gives the methyl substituent the lower number (methyl on C3).
  3. Base name: pent-2-ene; add the substituent: 3-methylpent-2-ene.
  4. Stereochemistry: on C2 the two groups are CH3 and H — CH3 outranks H. On C3 the groups are CH2CH3 (ethyl) and CH3 — ethyl outranks methyl. The two higher-priority groups (CH3 on C2, ethyl on C3) are on opposite sides → E.

Degree of unsaturation check. For C6H12: IHD = (2×6 + 2 − 12)/2 = 1, consistent with one double bond (hex-1-ene) or one ring (cyclohexane). A molecule with both a ring and a double bond (cyclohexene, C6H10) has IHD = 2.

Key takeaways

  • High yield: A double bond = one σ bond + one π bond; the π bond blocks rotation.
  • High yield: Each C=C contributes one degree of unsaturation.
  • High yield: Number the chain so the double bond gets the lowest locator, not the substituents.
  • High yield: E/Z uses CIP priority; cis/trans is a special case of E/Z.
  • Terminal alkenes have the C=C at a chain end; internal alkenes are more substituted.
  • Geminal = same carbon; vicinal = adjacent carbons.
  • Degree of unsaturation is a prediction tool — verify against the drawn structure.

Keep learning

Ready to build on this? Continue to the next lesson.

Practice Organic Chemistry 1

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Define an alkene and describe its bonding (σ + π), geometry, and hybridization.
  • Calculate the degree of unsaturation (index of hydrogen deficiency) from a molecular formula and interpret it for alkenes.
  • Apply IUPAC rules to name alkenes, including parent selection, numbering, and E/Z (cis/trans) designators.
  • Classify alkenes as terminal or internal and geminal or vicinal, and convert between names and structures.

Key vocabulary

Alkene
Hydrocarbon containing a C=C double bond
Unsaturation
Fewer than the maximum number of hydrogens; a ring or π bond
Degree of unsaturation (IHD)
Count of rings + π bonds from the formula
Parent selection
Longest chain containing the double bond
Numbering
Assigning locators so the double bond is lowest
E/Z
Same/opposite side of higher-priority groups
cis/trans
Same/opposite side of identical groups on C=C
Terminal alkene
C=C at the end of a chain
Internal alkene
C=C in the middle of a chain
Geminal substitution
Two substituents on the same C=C carbon

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