4.1.3(a)-(b), 4.1.3(e) - Alkene bonding, shape and reactivity

Alkenes and unsaturation

A hydrocarbon is a compound containing carbon and hydrogen only. An alkene is a hydrocarbon containing at least one carbon-carbon double bond, C=C.

The word unsaturated is used because the molecule contains a multiple bond. For this lesson, the key sign of unsaturation is the C=C bond itself. Ethene, $\ce{H2C=CH2}$, and propene, $\ce{CH3CH=CH2}$, are alkenes because each molecule contains a C=C bond and contains only carbon and hydrogen.

A useful way to read an alkene formula is to locate the C=C first. The rest of the molecule may contain C-H or C-C single bonds, but the C=C is the feature that controls this lesson's bonding, shape and reactivity.

Worked example: identifying an alkene

Butane, $\ce{CH3CH2CH2CH3}$, is a hydrocarbon, but it is not an alkene because it has no C=C bond. But-1-ene, $\ce{CH2=CHCH2CH3}$, is an alkene because it is a hydrocarbon and it contains a C=C bond.

Sigma and pi bonding in C=C

A C=C double bond is not simply "two of the same bond". It contains one sigma bond and one pi bond.

The pi bond uses a different overlap geometry.

In the C=C bond, the sigma bond lies directly along the line joining the two carbon nuclei. The pi bond is different: it comes from the sideways overlap of adjacent p orbitals, producing electron density above and below the plane of the bonding atoms.

[DIAGRAM: asset_name: Lesson 4.1.3a-b,e: Alkene bonding, shape and reactivity - diagram 01; asset_slug: 04_01_03a_alkene_bonding_shape_and_reactivity__diagram_01; recommended_method: drawn_chem; description: Combined ethene C=C model showing the sigma bond directly between the carbon atoms, adjacent p orbitals overlapping sideways above and below the bonding carbons to form a pi bond, and a separate trigonal planar 120 degree angle cue around one carbon.]

Exam questions often want the model, not just the symbol. Writing C=C tells someone that there is a double bond, but it does not by itself explain how the pi bond forms. For the pi bond model, the important phrase is sideways overlap of adjacent p orbitals.

Worked example: describing ethene's double bond

In ethene, $\ce{H2C=CH2}$, the two carbon atoms are joined by a C=C bond. That C=C contains one sigma bond from overlap directly between the carbon atoms and one pi bond from sideways overlap of adjacent p orbitals above and below the bonding carbon atoms.

Counting σ and π bonds

Propene, $\ce{CH3CH=CH2}$, has six C–H connections and two C–C connections. Each connection contributes one σ bond, so there are eight σ bonds. The one C=C contributes one extra π bond. Do not count a double bond as two σ bonds. For any larger structure, count the carbon skeleton and all hidden C–H bonds before adding one π bond for each C=C.

Restricted rotation

The pi bond explains why a C=C bond cannot rotate freely.

For the pi bond to exist, the adjacent p orbitals must stay lined up so that they overlap sideways. If one end of the double bond rotated relative to the other, the p orbitals would stop overlapping properly. In effect, rotation would require the pi bond to be broken.

That is why the C=C bond has restricted rotation. The sigma bond framework is still present, but the pi bond locks the two ends of the double bond into a fixed relationship unless enough energy is supplied to break the pi bond.

This idea will matter in the next alkene lesson, where fixed positions around C=C can lead to stereoisomerism. Here the cause matters: rotation about the C=C would disrupt the π bond.

Worked reasoning: why C-C and C=C behave differently