Geometric Isomerism and Restricted Rotation

Introduction

Butane can twist freely about its central C–C bond, and every twisted form interconverts millions of times a second. But-2-ene, with a C=C double bond in the same position, behaves very differently: it exists as two distinct compounds with different boiling points, which can be stored in separate cylinders for years without interconverting. The cause is restricted rotation . Understanding why double bonds and rings hold groups in fixed positions explains a whole family of stereoisomers, called geometric isomers, found in fuels, fats, drugs and even the chemistry of vision.

Core explanation

The σ and π bonds. A carbon-carbon double bond consists of one σ bond, formed by end-on overlap of sp² hybrid orbitals, and one π bond, formed by sideways overlap of unhybridised p orbitals lying parallel to each other above and below the plane of the molecule. The six atoms of the C=C unit and its four substituents lie in one plane.

Why rotation is restricted. Rotation about a σ bond does not change the overlap of the orbitals, so it costs very little energy; the barrier in ethane is only about 12 kJ/mol. Twisting a double bond, however, turns the two p orbitals away from each other. At 90° they are perpendicular and overlap is zero: the π bond is broken. The energy required is roughly that of the π bond, around 260–270 kJ/mol, far more than the thermal energy available at room temperature. Therefore the two ends of a C=C bond cannot rotate relative to one another under normal conditions.

The condition for geometric isomerism. Two isomers exist only if each carbon of the double bond carries two different groups. In but-2-ene, CH₃CH=CHCH₃, each carbon carries CH₃ and H, so there are two arrangements: both methyl groups on the same side, or on opposite sides. In but-1-ene, CH₂=CHCH₂CH₃, the first carbon carries two identical H atoms; swapping them produces the same molecule, so there are no geometric isomers. Likewise 2-methylpropene has no geometric isomers.

Rings. A ring prevents full rotation about its C–C single bonds, because turning one ring carbon through 180° would require breaking the ring. Substituents on different ring carbons are therefore fixed on the same face or on opposite faces. 1,2-dimethylcyclopropane exists as two geometric isomers: both methyls on the same face, or on opposite faces. Small and medium rings behave in this way; the ring atoms can flex, but they cannot move a substituent from one face to the other.

Other double bonds. The same reasoning applies to C=N double bonds in imines and oximes, and to N=N double bonds in azo compounds; in these the lone pair acts as one of the two "different groups".

Interconversion requires energy. Geometric isomers can be interconverted only by temporarily breaking the π bond, for example by absorbing ultraviolet or visible light, by heating strongly, or with a catalyst that adds and removes a group. Without such help, they remain separate compounds.

Step-by-step reasoning

To decide whether an alkene has geometric isomers:

1. Locate the C=C double bond. 2. List the two groups on the first alkene carbon. 3. If they are identical, stop: no geometric isomers. 4. Repeat for the second carbon. 5. If both carbons carry two different groups, two geometric isomers exist.

Visual explanation

Picture two p orbitals as a pair of upright paddles, one on each carbon, lying side by side so that their lobes overlap. Now twist one carbon: its paddle tilts away until, at 90°, it lies flat while the other stays upright, and the overlap disappears. The simulation shows the π-bond energy rising steeply as the twist angle increases, compared with the gentle ripple for ethane.

Real-world analogy

Imagine two people holding hands with both hands, side by side. Either can turn on the spot if they hold just one hand, like a single bond. Holding with both hands, the second connection, they cannot turn without letting go of one hand. The π bond is that second hand.

Real-world example

In the eye, the pigment rhodopsin contains retinal with a double bond in the cis arrangement. Absorbing a photon supplies enough energy to break the π bond briefly, and retinal snaps into the trans arrangement. That change of shape triggers the nerve signal that lets us see, before enzymes reset the molecule.

Why?

Why does the ring stop rotation even though its bonds are single bonds? A full rotation about one ring bond would force the atoms on either side to move independently, which would stretch and break other bonds in the ring. The ring connectivity itself acts as the constraint.

Common misconception

"Every alkene has cis and trans forms." Geometric isomers exist only when each carbon of the double bond has two different groups. Ethene, propene and but-1-ene have none.

Worked example

Question: Which of these show geometric isomerism: pent-2-ene, 2-methylbut-2-ene, 1,2-dichloroethene?

Reasoning: Pent-2-ene: C2 has CH₃ and H; C3 has C₂H₅ and H, so yes. 2-methylbut-2-ene: C2 carries two CH₃ groups, so no. 1,2-dichloroethene: each carbon has Cl and H, so yes.

Answer: Pent-2-ene and 1,2-dichloroethene show geometric isomerism; 2-methylbut-2-ene does not.

Quick check

1. Does propene, CH₂=CHCH₃, show geometric isomerism? Give a reason. Answer: No, because one carbon of the double bond carries two identical hydrogen atoms.

Exam focus

Explain restricted rotation in terms of the π bond formed by sideways p-orbital overlap, which would break on twisting. State the condition that both double-bond carbons must carry two different groups. Remember that ring compounds can also show geometric isomerism.

Advanced insight

Some double bonds have much lower rotation barriers than simple alkenes. When electron-donating groups on one end and electron-withdrawing groups on the other allow charge separation, the π bond gains single-bond character and the barrier drops. Amide C–N bonds show the opposite case: a formally single bond with partial double-bond character from resonance, giving a barrier of roughly 60–90 kJ/mol, high enough that separate signals for the two orientations can be seen by NMR.

Summary

A C=C double bond contains a π bond formed by sideways overlap of p orbitals. Rotation would break this overlap and requires about 260 kJ/mol, so substituents are locked in place at room temperature. Geometric isomers arise when each double-bond carbon carries two different groups. Rings also restrict rotation and allow substituents to be fixed on the same or opposite faces.

Practice questions

1. Explain why the π bond prevents rotation about a C=C double bond. Answer: The π bond needs parallel p orbitals; rotating one carbon twists the orbitals apart and breaks the π bond, which requires much more energy than is available at room temperature. 2. Does 1,1-dichloroethene show geometric isomerism? Explain. Answer: No, because one carbon carries two identical chlorine atoms and the other two identical hydrogen atoms. 3. Explain how 1,2-dimethylcyclobutane can exist as two geometric isomers, although it has no double bond. Answer: The ring prevents rotation about its bonds, so the two methyl groups are fixed either on the same face or on opposite faces of the ring. 4. Suggest one way to convert one geometric isomer of an alkene into the other. Answer: Supply energy, for example by absorbing ultraviolet light, which temporarily breaks the π bond and allows rotation.