Alkene structure — Alkenes & Alkynes

Module 6 was about carbons bearing leaving groups. From here the course shifts to functional groups that react because of what they have rather than what they can lose, and the alkene is the first of them. An alkene's pi bond is a pool of exposed, loosely held electron density — which makes the alkene a nucleophile, and reverses the role carbon has been playing for the last chapter.

sp² carbons and the pi bond

Each carbon of a C=C is sp² hybridized: three sp² orbitals form sigma bonds in a trigonal planar arrangement 120° apart, and one unhybridized p orbital is left over, perpendicular to that plane. The two leftover p orbitals overlap side-by-side above and below the sigma framework, forming the pi bond.

Some numbers worth having. A C=C is about 134 pm long against 154 pm for C–C, and worth roughly 147 kcal/mol against 83. The sigma component accounts for about 83 of that, so the pi bond is worth about 65 kcal/mol — strong enough to matter, and weak enough to be the part that breaks first in every reaction in this chapter.

Why the double bond is flat and rigid

A sigma bond rotates freely because its overlap is cylindrically symmetric about the bond axis. A pi bond cannot: rotating one carbon 90° relative to the other would destroy the side-by-side p-orbital overlap entirely. Every atom directly attached to a C=C is therefore locked in the same plane, and the two ends cannot swap.

Rotating about a C=C costs roughly 65 kcal/mol — the entire strength of the pi bond — against essentially zero for a C–C single bond. That is why alkene geometry is configurational, a permanent structural identity, rather than conformational, a freely changing shape. Cis and trans alkenes are different compounds you can put in separate bottles, not two shapes of one compound.

Naming alkene geometry: cis/trans and E/Z

Why alkene geometry is worth naming at all. Because the pi bond forbids rotation, the two arrangements are not conformations a molecule flits between — they are separate compounds, with different boiling points and different dipole moments. Where the two groups on a carbon are not simply "the same or different", cis/trans stops working and E/Z takes over, ranking each carbon's two groups by the same CIP priorities you used for R and S.Identical formula, identical connectivity — and two different substances you can bottle separately, because turning one into the other means breaking the π bond, and that costs about 65 kcal/mol. Rotation about a SINGLE bond is free. Rotation about a DOUBLE bond is not. That is the whole difference.

Cis and trans work when each alkene carbon carries one hydrogen and one other group: cis has the two substituents on the same side, trans on opposite sides. That covers many cases and fails as soon as a carbon carries two different non-hydrogen groups.

The general system is E/Z, and it reuses the CIP priority rules from Module 5 unchanged. Rank the two groups on each alkene carbon independently. If the two higher-priority groups are on the same side, the alkene is Z (German zusammen, together); on opposite sides, E (entgegen, opposite).

E is not always trans. The two systems agree only when the priorities happen to fall the same way. In 1-bromo-1-chloropropene the priorities on one carbon are bromine over chlorine, and the E/Z label can come out opposite to what a casual "trans" reading suggests. When both non-hydrogen groups differ, use E/Z and rank them properly.

Degrees of unsaturation

Each ring or pi bond removes exactly two hydrogens relative to the saturated formula CₙH₂ₙ₊₂, which lets you read structural information straight off a molecular formula:

DoU = (2C + 2 + N − H − X) / 2

Halogens count like hydrogens; oxygen is ignored entirely, because inserting an oxygen into a chain changes no hydrogen count.

C₄H₈ gives DoU = (8 + 2 − 8)/2 = 1: one ring or one pi bond, consistent with an alkene or a cyclobutane. C₆H₆ gives 4 — three pi bonds and a ring, which is benzene. This is the first thing to compute when a spectroscopy problem hands you a molecular formula in Module 14, because it tells you immediately whether to look for a ring, a carbonyl or an aromatic system.

Alkene stability: more substituted is more stable

Substituted alkenes are more stable, for the same reason substituted carbocations are: the alkyl groups next door donate electron density into the pi system. The measurement behind this ladder is neat — hydrogenate each isomer and they all give the same alkane, so whichever releases the least heat started out lowest. Two things in this course rest directly on the ordering: Zaitsev's rule in Module 6, and which alkene an elimination hands you here.The same hyperconjugation that stabilises a carbocation stabilises a π bond: each alkyl group next door feeds electron density into it. This is why Zaitsev elimination gives the more substituted alkene, and why hydrogenation heats of the isomers come out in exactly this order.

Heats of hydrogenation measure alkene stability directly: hydrogenate two different alkenes to the same alkane, and whichever released less heat started out lower in energy.

Alkene	Substitution	ΔH°hyd
1-butene	mono	−30.3
cis-2-butene	di (cis)	−28.6
trans-2-butene	di (trans)	−27.6
2-methyl-2-butene	tri	−26.9
2,3-dimethyl-2-butene	tetra	−26.6

Two trends. More alkyl substitution means more stable, by roughly 1–1.5 kcal/mol per substituent, because the alkyl groups donate electron density into the pi system by hyperconjugation — the same effect that stabilizes carbocations, acting on a neutral molecule. And trans beats cis by about 1 kcal/mol, because cis substituents on the same side crowd each other.

This ordering is what Zaitsev's rule is ultimately reporting. When E1 or E2 could give either of two alkenes, the more substituted one is more stable, and under thermodynamic control that is the one you get.

Worked example — ranking by stability

Order 1-pentene, cis-2-pentene, trans-2-pentene and 2-methyl-2-butene.

Count substituents first: 1-pentene is monosubstituted, both 2-pentenes are disubstituted, 2-methyl-2-butene is trisubstituted. Then break the tie between the 2-pentenes on cis versus trans.

Least to most stable: 1-pentene < cis-2-pentene < trans-2-pentene < 2-methyl-2-butene.

Bridgehead alkenes and Bredt's rule

A double bond must be planar, and in a small bridged bicyclic system a bridgehead carbon cannot become planar without impossible strain. Bredt's rule says a double bond cannot be placed at the bridgehead of a small bridged ring system. This occasionally rules out what would otherwise be the Zaitsev product of an elimination, and it is a reminder that geometric feasibility is checked before stability.

What carries forward

The pi bond's exposed electrons make the alkene a nucleophile, which is the premise of every reaction in the next two sections. Alkene stability determines elimination product ratios throughout Modules 6 and 8. E/Z notation is how the stereochemical outcome of every addition is reported. And degrees of unsaturation is the first calculation you will make in every structure-determination problem in Module 14.