Geometrical Isomerism in Alkenes

Introduction

But-2-ene can be drawn with both CH₃ groups on the same side of C=C or on opposite sides. These drawings have the same formula and connectivity, yet they are different stereoisomers. The double bond prevents ordinary free rotation, so one arrangement cannot become the other merely by turning the page or rotating a single bond elsewhere.

Core explanation

A carbon-carbon double bond consists of a sigma bond and a pi bond. The pi bond is formed by overlap of p orbitals on the two alkene carbons. Twisting one end substantially would disrupt this overlap, so rotation around C=C requires a substantial energetic change rather than the low-barrier rotation possible around many single bonds. The two sides of the double bond can therefore retain distinguishable substituent arrangements.

Two conditions are needed for ordinary alkene geometrical isomerism. First, there must be a C=C bond with restricted rotation. Second, each double-bond carbon must carry two different substituents. In CH₃CH=CHCH₃, each alkene carbon bears H and CH₃; two arrangements exist. In CH₂=CHCH₃, the terminal carbon bears two identical H substituents. Swapping sides at that carbon changes no identifiable configuration, so propene has no cis/trans pair.

For but-2-ene, cis means the two methyl groups lie on the same side of the double bond; trans means they lie on opposite sides. Both are C₄H₈ and both have the same bond network. They are non-mirror-image stereoisomers, a diastereomeric pair. Their three-dimensional shapes and dipole arrangements differ, so some physical properties differ even though many bond types are shared.

The cis/trans labels are convenient when the same recognizable group occurs on both alkene carbons, or when an unambiguous reference pair is specified. More general alkenes may have four different substituents, making “same side of what?” unclear. The E/Z priority system resolves that case and is introduced next. Even for simple examples, draw the groups attached directly to each C=C carbon before using a label.

Do not mistake conformations for geometrical isomers. Rotation about a neighbouring C–C single bond can change a molecule's appearance without changing the side relationship across the C=C. Conversely, rotating the drawing in the plane does not turn cis into trans. A bond-breaking or high-energy photo/thermal process may interconvert alkene geometrical isomers, but ordinary free single-bond rotation does not.

The existence of a geometrical pair matters in synthesis. An elimination reaction can make E and Z alkenes in different amounts, affected by mechanism, substrate geometry and product stability. A reaction scheme that specifies a single alkene geometry must justify selectivity or purification. If the alkene has identical groups on one double-bond carbon, no E/Z outcome can be assigned, regardless of how it was prepared.

To count possibilities in a simple molecule with one qualifying double bond, first hold connectivity fixed. There are at most two geometrical configurations at that bond. If a stereogenic carbon exists elsewhere, additional stereoisomers may occur, but they are a separate source of stereochemical variation. Avoid announcing “two isomers” as the entire molecular count without checking the rest of the structure.

Step-by-step reasoning

Circle C=C and label the two substituents on its left carbon and its right carbon. If either side has duplicate substituents, stop: this double bond cannot create a geometrical pair. If both sides differ, draw the substituents above and below a fixed horizontal C=C and exchange one side to obtain the alternative. Decide whether cis/trans is unambiguous; otherwise use E/Z.

Visual explanation

Draw two horizontal C=C lines. Above both carbons of the first put CH₃ and below both put H: cis-but-2-ene. On the second put CH₃ above left but below right: trans-but-2-ene. Alongside draw CH₂=CHCH₃ and circle the two H atoms on terminal carbon to show why no distinct swap exists.

Real-world analogy

Two people holding a rigid bar with signs fixed to its ends cannot exchange the signs from one side to the other by twisting the bar without breaking its rigid connection. A freely rotating rope would allow reorientation. The double bond behaves more like the rigid bar because its pi overlap resists rotation.

Real-world example

In an alkene-separation lab, cis- and trans-but-2-ene have the same molecular mass but can be treated as distinct substances because their geometries and some properties differ. A product report that merely says “but-2-ene” may be incomplete if the reaction gives a mixture or the target specifies one arrangement.

Why?

Why is propene excluded? The terminal alkene carbon has two identical hydrogens, so swapping them does not create an observable new arrangement. Why is but-2-ene included? Each alkene carbon has one H and one CH₃, making the same-side versus opposite-side methyl relation meaningful.

Common misconception

"Every molecule with a C=C bond has cis/trans isomers." A terminal alkene CH₂= often fails because one alkene carbon has two H atoms. The substituent test must be applied separately to both alkene carbons before drawing or naming geometrical isomers.

Worked example

Question: Which can show alkene geometrical isomerism: but-1-ene, but-2-ene or 2-methylpropene?

Reasoning: But-1-ene has terminal CH₂= and two H on one alkene carbon. But-2-ene has H and CH₃ on each alkene carbon. 2-Methylpropene has both a terminal CH₂= side and two CH₃ substituents on the other alkene carbon.

Answer: Only but-2-ene has a cis/trans geometrical pair among the three.

Quick check

1. What substituent condition must hold on each carbon of an alkene for E/Z isomerism? Answer: Each C=C carbon must be attached to two different substituents.

Exam focus

Inspect substituents at each alkene carbon before naming cis/trans or E/Z. Keep the bond graph fixed and vary only spatial arrangement. Draw both candidates when asked to count. If all four substituents are not simply comparable by a common group, use priority-based E/Z notation instead of an ambiguous cis/trans label.

Advanced insight

Geometrical isomerism reflects a high rotation barrier rather than absolute impossibility of interconversion. Light can excite a pi bond and enable isomerization, an important principle in photochemistry and biological light sensing. Under ordinary dark conditions, however, alkene geometries are stable enough to count as distinct configurational stereoisomers.

Summary

Alkene geometrical isomers share connectivity but differ across a rotation-restricted C=C bond. They exist only if each alkene carbon bears two different substituents. But-2-ene has cis and trans forms; propene and 2-methylpropene do not. Draw substituent positions explicitly and reserve E/Z notation for the general priority-based case.

Practice questions

1. Why does ethene not have cis/trans isomers? Answer: Each double-bond carbon has two identical H substituents. 2. Are cis- and trans-but-2-ene constitutional isomers? Answer: No. They share connectivity and are geometrical stereoisomers. 3. What bond component resists free rotation around C=C? Answer: The pi-bond orbital overlap resists twisting. 4. Does rotating a drawing of cis-but-2-ene on paper make it trans? Answer: No. A page rotation does not change the relative side of the two methyl groups.