Why Alkenes Have Geometric Isomers but Single Bonds Rotate

Take but-2-ene, \(\ce{CH3-CH=CH-CH3}\). You can draw it with the two methyl groups on the same side of the double bond, or with one methyl group on either side. Both structures have exactly the same atoms connected in exactly the same order. Yet they represent two different molecules: cis-but-2-ene and trans-but-2-ene.

Why can’t one simply rotate around the double bond and turn into the other?

The answer comes down to the type of bonding. An ordinary carbon-carbon single bond contains one sigma (\(\sigma\)) bond, which allows relatively easy rotation. A carbon-carbon double bond contains an additional pi (\(\pi\)) bond. Rotating around the double bond disrupts the sideways orbital overlap responsible for that extra bond.

This is why alkenes can have geometric isomers, while ordinary single bonds generally produce different conformations of the same molecule.

Before going further, make a prediction. If you rotate one carbon around the C-C bond in ethane, have you created a new molecule? No. The atoms remain connected in exactly the same way. You have simply changed their spatial arrangement. The distinction becomes clearer when we examine what each type of bond actually allows.

01First, distinguish connectivity from spatial arrangement

There are three related ideas that you need to keep separate.

Connectivity describes which atoms are bonded to which. It tells you the order in which atoms are joined, regardless of how the molecule is drawn in three dimensions.

Conformation describes the different spatial arrangements a molecule can adopt through rotation around single bonds. These arrangements can usually change without breaking any bonds.

Configuration describes a particular spatial arrangement that cannot be changed simply by rotating around ordinary single bonds. Changing the configuration requires a more substantial change to the bonding arrangement.

Geometric isomers are examples of configurational stereoisomers.

Consider the difference:

The key point is that two different drawings do not automatically represent two different isomers.

If you rotate a single bond and redraw the molecule, you have usually produced another conformation. If you change the arrangement of groups around a carbon-carbon double bond, you can produce a different geometric isomer.

02How carbon’s valence determines the arrangement

To understand why double bonds behave differently, start with carbon’s bonding capacity.

Carbon normally forms four covalent bonds in the neutral organic molecules considered here.

Look at ethane:

\[ \ce{CH3-CH3} \]

Each carbon forms four single bonds:

• Three bonds to hydrogen atoms.
• One bond to the other carbon atom.

Each carbon therefore has a tetrahedral arrangement of its four bonding directions.

Now compare but-2-ene:

\[ \ce{CH3-CH=CH-CH3} \]

Each carbon involved in the double bond has:

• One double bond to the other carbon.
• One single bond to hydrogen.
• One single bond to a methyl group.

The double bond uses two of carbon’s four bonding contributions. The remaining two are supplied by its single bonds to the other groups.

Each alkene carbon therefore has three bonding directions in its sigma-bond framework. These are arranged approximately \(120^\circ\) apart, producing trigonal planar geometry.

There is an important consequence.

The groups attached to each carbon of the double bond occupy defined positions within the planar alkene arrangement. In but-2-ene, each carbon has one hydrogen and one methyl group.

Because these groups are different, two distinguishable arrangements are possible:

• Both methyl groups on the same side: cis-but-2-ene.
• Methyl groups on opposite sides: trans-but-2-ene.

If either double-bonded carbon instead had two identical groups, exchanging their positions would not produce a distinguishable geometric isomer.

This is the structural requirement for geometric isomerism. For a more detailed application of that test, see When Can an Alkene Have Cis-Trans Isomers? The Structural Test.

The important distinction for this article is that carbon’s valence determines which groups are attached and how they are arranged, while the type of bond determines whether those arrangements can freely interconvert.

03Why ordinary carbon-carbon single bonds rotate

An ordinary carbon-carbon single bond consists of one sigma (\(\sigma\)) bond.

A sigma bond forms through the overlap of atomic orbitals along the line joining the two carbon nuclei. This is called head-on overlap.

The electron density responsible for the bond is concentrated around the internuclear axis.

Here is the useful feature: the sigma bond can maintain its overlap when one carbon rotates around that axis.

Imagine a spindle connecting two rotating sections. One section can turn while the connection remains intact.

That is approximately what happens when you rotate around an ordinary C-C single bond.

What actually changes during rotation?

Consider ethane again:

\[ \ce{CH3-CH3} \]

You can rotate one \(\ce{CH3}\) group relative to the other around the central carbon-carbon bond.

This changes the relative positions of the six hydrogen atoms.

Two important arrangements are:

• Staggered conformation: the hydrogen atoms on one carbon sit between those on the other carbon when viewed along the C-C bond.
• Eclipsed conformation: the hydrogen atoms on one carbon line up with those on the other carbon.

Both arrangements have exactly the same connectivity. Neither requires breaking the C-C sigma bond.

They are simply different conformations of ethane.

Does this mean single bonds rotate with absolutely no energy cost?

No. This is where the word freely needs a little care.

Rotation around a single bond can involve an energy barrier. Different arrangements experience different interactions between their electron clouds and attached atoms.

For example, eclipsed ethane is less stable than staggered ethane because of increased torsional interactions.

However, the energy barrier is sufficiently accessible for ordinary acyclic carbon-carbon single bonds to rotate readily under normal conditions.

The molecule can move between these conformations without breaking its carbon-carbon bond.

That is why ordinary single-bond rotation does not usually produce isolable geometric isomers.

04Why a carbon-carbon double bond cannot rotate freely

Now we reach the important difference.

A carbon-carbon double bond contains two components:

1. One sigma (\(\sigma\)) bond.
2. One pi (\(\pi\)) bond.

The sigma bond forms through head-on orbital overlap, just as it does in a single bond.

The pi bond forms differently. It comes from the sideways overlap of parallel p orbitals on the two carbon atoms.

This creates electron density above and below the line joining the carbon nuclei.

What happens if you try to rotate a double bond?

Imagine holding one carbon atom fixed and attempting to rotate the other around the C=C bond axis.

The sigma bond can remain intact because its overlap is along the axis.

The pi bond cannot maintain its overlap in the same way.

As one carbon rotates, its p orbital becomes increasingly misaligned with the p orbital on the other carbon. The sideways overlap decreases.

At a twist of \(90^\circ\), the p orbitals are perpendicular, so the sideways overlap responsible for the pi bond disappears in the idealised orbital model.

You have effectively disrupted the pi component of the double bond.

This requires a substantial energy input. It is not comparable to the relatively easy rotation around an ordinary single bond.

Consequently, the two arrangements of groups around an alkene remain distinct under ordinary conditions.

This is the physical reason geometric isomers can exist.

The comparison to remember

Notice that restricted rotation does not mean rotation is impossible under every conceivable condition. Sufficient energy or suitable photochemical conditions can allow double-bond isomerisation.

The important HSC distinction is that a double bond does not permit the easy, continuous rotation associated with an ordinary single bond.

For a closer look at the underlying carbon geometries, see Carbon-Carbon Bond Shapes: Single, Double and Triple Bonds.

05The historical puzzle: Why are maleic acid and fumaric acid different?

There is a useful historical example that shows why this distinction mattered to chemists.

Maleic acid and fumaric acid have the same molecular formula:

\[ \ce{C4H4O4} \]

They also have the same connectivity:

\[ \ce{HOOC-CH=CH-COOH} \]

Yet they are different substances.

Maleic acid has its two carboxyl groups on the same side of the double bond. Fumaric acid has them on opposite sides.

The problem with the original structural formulas

In nineteenth-century chemistry, structural formulas were commonly drawn as flat arrangements of symbols and bonds.

These formulas were useful for showing connectivity. However, they did not always explain why compounds with the same connectivity could exist as different substances.

If the C=C bond in maleic acid could rotate freely, its two carboxyl groups could change from the same-side arrangement to the opposite-side arrangement.

Why, then, did chemists find two distinct acids rather than one compound rapidly changing between two shapes?

Van’t Hoff’s proposed explanation

In 1874, Dutch chemist Jacobus Henricus van’t Hoff proposed a three-dimensional model of molecular structure.

For a carbon-carbon double bond, his model represented the two tetrahedral carbon centres as sharing an edge.

This arrangement restricted the relative positions of the groups attached to the two carbons.

To construct the two possible arrangements:

1. Represent each carbon using a tetrahedral arrangement of its four bonding directions.
2. Join the two tetrahedral carbon centres along a shared edge to represent the double bond.
3. Attach a hydrogen atom and a carboxyl group to each carbon.
4. Arrange the two carboxyl groups on the same side or on opposite sides.

The result is two different spatial arrangements.

Van’t Hoff’s model provided an explanation for why maleic acid and fumaric acid could exist as distinct substances.

The Science Museum Group preserves stereochemical models illustrating this historical explanation. Its collection record describes the tetrahedral carbon centres and the different arrangements of the carboxyl groups.

Science Museum Group: Stereochemical models of fumaric and maleic acids

It is worth separating the historical model from the modern explanation. Van’t Hoff’s model described the three-dimensional arrangement of atoms. It was not a direct measurement of p-orbital overlap. Modern orbital theory explains the bonding reason for the restricted rotation.

What did the different physical properties tell chemists?

Maleic acid and fumaric acid have markedly different reported thermal properties.

These are reference values, not measurements from one shared experiment. The maleic acid value is reported by PubChem, while the fumaric acid value is given in the International Organisation of Vine and Wine monograph.

The difference between the reported values is:

\[ \begin{aligned} \Delta T &= 287-132.5\\ &=154.5^\circ\text{C} \end{aligned} \]

The difference is approximately \(154.5^\circ\text{C}\).

This is a substantial difference between the thermal properties of two compounds with the same molecular formula and connectivity.

However, be careful about what this calculation tells you.

A melting point depends on how molecules pack together in a crystal and the intermolecular attractions within that solid. It is not a direct measurement of the energy needed to rotate a C=C bond.

The different properties support the conclusion that maleic acid and fumaric acid are distinct substances. Their cis/trans arrangements explain how that distinction can exist despite identical connectivity.

The historical development therefore followed a useful chain of reasoning:

• The existing structural formulas showed the same connectivity.
• The compounds displayed different properties.
• A three-dimensional model provided two possible fixed arrangements.
• The restricted rotation of the double bond explained why those arrangements could remain distinct.

This is a good example of how a better structural model can explain observations that a simpler representation cannot.

06Worked examples

07Questions and solutions

Question 1

Ethane and ethene have the following structures:

\[ \ce{CH3-CH3} \]

\[ \ce{CH2=CH2} \]

a. Which molecule allows relatively free rotation around its carbon-carbon bond?

b. Explain why the other molecule cannot freely rotate around its carbon-carbon bond.

c. Explain why ethene does not have cis-trans geometric isomers, despite containing a double bond.

Solution 1

Ethane allows relatively free rotation around its carbon-carbon bond, whereas ethene has restricted rotation because twisting its double bond disrupts the pi overlap. Ethene also cannot exhibit geometric isomerism because each double-bonded carbon has two identical hydrogen atoms.

In ethane, the carbon-carbon bond contains only one sigma bond.

Rotation around this bond does not destroy the head-on orbital overlap, so the molecule can change conformation without breaking the bond.

In ethene, the carbon-carbon double bond contains:

• One sigma bond.
• One pi bond.

The pi bond depends on sideways overlap between parallel p orbitals. Rotation disrupts this overlap, making free rotation energetically unfavourable.

For part c, geometric isomerism requires two different groups attached to each carbon of the double bond.

In ethene, both groups attached to each carbon are hydrogen atoms.

Swapping the positions of two identical hydrogen atoms does not produce a distinguishable arrangement.

Therefore, ethene has restricted rotation but does not have cis-trans isomers.

This question highlights an important distinction: restricted rotation makes geometric isomerism possible, but it does not guarantee that a molecule can exhibit it.

Question 2

Maleic acid and fumaric acid have the same molecular formula, \(\ce{C4H4O4}\), and the same connectivity:

\[ \ce{HOOC-CH=CH-COOH} \]

The following reference thermal values are provided:

The fumaric acid value is reported with decomposition.

a. Calculate the difference between the reported values.

b. Identify which compound has the cis arrangement and which has the trans arrangement.

c. Explain why these compounds can exist separately despite having the same connectivity.

d. Explain why their different thermal properties do not directly measure the rotational energy barrier of the C=C bond.

Solution 2

The difference between the reported values is \(154.5^\circ\text{C}\). Maleic acid has the cis arrangement, while fumaric acid has the trans arrangement. Their restricted double-bond rotation allows these different configurations to remain distinct.

Part a

Subtract the lower reference value from the higher value:

\[ \begin{aligned} \Delta T &= 287-132.5\\ &=154.5^\circ\text{C} \end{aligned} \]

Part b

Maleic acid has its two carboxyl groups on the same side of the C=C bond. It is the cis isomer.

Fumaric acid has its two carboxyl groups on opposite sides. It is the trans isomer.

Part c

Both compounds have the same molecular formula and connectivity, but their carboxyl groups occupy different positions relative to the double bond.

The C=C bond contains a pi bond, which restricts rotation because twisting disrupts sideways orbital overlap.

As a result, the two spatial arrangements do not freely interconvert under ordinary conditions.

They exist as distinct geometric isomers.

Part d

The thermal properties of a solid depend on its crystal structure, molecular packing, and intermolecular attractions.

The cis and trans arrangements influence how the molecules pack together, contributing to their different thermal behaviour.

However, the temperature difference does not directly tell us how much energy is needed to twist a C=C bond.

That rotational restriction is explained by the bonding within an individual molecule, rather than by the melting behaviour of the bulk solid.

The distinction between molecular bonding and intermolecular forces is important when interpreting experimental evidence.

Question 3

But-1-ene and cis-but-2-ene both have the molecular formula \(\ce{C4H8}\).

Their condensed structural formulas are:

\[ \ce{CH2=CH-CH2-CH3} \]

and

\[ \ce{CH3-CH=CH-CH3} \]

A student claims that they must be geometric isomers because they have the same molecular formula and contain a carbon-carbon double bond.

Evaluate this claim. Identify the type of isomerism involved and explain which pair of compounds would actually be geometric isomers.

Solution 3

The student’s claim is incorrect. But-1-ene and cis-but-2-ene are structural isomers because their double bonds occur at different positions in the carbon chain. Cis-but-2-ene and trans-but-2-ene, however, are geometric isomers.

The molecular formulas are identical:

\[ \ce{C4H8} \]

But the connectivity differs.

In but-1-ene, the double bond connects the first and second carbon atoms.

In but-2-ene, the double bond connects the second and third carbon atoms.

The position of the double bond changes which atoms are connected through a double bond.

Therefore, these compounds are structural isomers, not geometric isomers.

For geometric isomerism, the connectivity must remain unchanged.

Cis-but-2-ene and trans-but-2-ene satisfy this requirement. Both have the same carbon chain and the same double-bond position. Only the relative positions of the two methyl groups differ.

The tempting mistake is to assume that the same molecular formula automatically means the same connectivity.

It does not. Always compare the actual bonding arrangement before deciding which type of isomerism is present.

Question 4

Consider 1,2-dimethylcyclopropane, which contains a three-carbon ring with a methyl group attached to each of two neighbouring ring carbons.

It can exist in two arrangements:

• Both methyl groups point towards the same face of the ring.
• The methyl groups point towards opposite faces of the ring.

All carbon-carbon bonds in the ring are single bonds.

Explain why these two arrangements can remain distinct even though the molecule contains no carbon-carbon double bond.

Solution 4

The two arrangements are distinct because the ring constrains the relative positions of the methyl groups. The carbon-carbon single bonds cannot rotate independently without distorting the ring structure.

In an ordinary acyclic molecule, a carbon-carbon single bond can rotate while maintaining its sigma overlap.

A three-membered ring is different.

Its carbon atoms are connected in a closed structure. Each carbon-carbon bond contributes to maintaining that ring.

If one of these bonds rotated independently, the ring would have to distort substantially to preserve its connections.

The ring therefore restricts the movement of the attached groups.

In one arrangement, both methyl groups point towards the same face of the ring. This is the cis arrangement.

In the other, they point towards opposite faces. This is the trans arrangement.

This does not contradict the behaviour of ordinary single bonds.

The important qualification is that single bonds in an unconstrained acyclic structure can rotate relatively easily. A ring introduces an additional structural constraint.

It is a useful reminder that the geometry of an entire molecule can restrict rotation even when individual bonds are single bonds.

Question 5

Consider cis-hex-3-ene:

\[ \ce{CH3CH2CH=CHCH2CH3} \]

The two ethyl groups attached to the double-bonded carbons initially lie on the same side of the C=C bond.

Now imagine rotating the single bond between the right-hand alkene carbon and the first carbon of its ethyl substituent through \(180^\circ\).

The terminal methyl group swings around this single-bond axis. In a perspective drawing, it now appears on the opposite side of the original alkene plane.

A student looks at the new drawing and concludes:

“The terminal methyl group has crossed over, so cis-hex-3-ene must have become trans-hex-3-ene.”

a. Is the student’s conclusion correct?

b. Explain which part of the ethyl substituent determines its position relative to the C=C bond for geometric isomerism.

c. Can any sequence of ordinary single-bond rotations convert cis-hex-3-ene into trans-hex-3-ene without disrupting the pi bond? Justify your answer.

Solution 5

The student’s conclusion is incorrect. Rotating the single bond changes the conformation of the ethyl substituent, but it does not change the geometric configuration around the C=C bond.

This is a more subtle distinction because the ethyl substituent contains several atoms, and those atoms do not all have to remain in the same plane.

Part a

The molecule is still cis-hex-3-ene.

The rotation occurs around an ordinary C-C single bond, so the sigma overlap remains intact.

The central double bond has not rotated, and its pi bonding arrangement remains unchanged.

Therefore, the geometric configuration has not changed.

Part b

The important feature is the bond connecting the alkene carbon directly to the first carbon of the ethyl substituent.

That directly attached carbon is the first atom in the substituent.

When the single bond connecting it to the alkene carbon is rotated, the attached ethyl group can change its three-dimensional conformation.

However, the position of the directly attached carbon relative to the C=C bond remains fixed because it lies on the rotation axis.

The terminal methyl group can move out of the original alkene plane without changing the relative arrangement of the groups directly attached to the double-bonded carbons.

This is why looking only at the position of the furthest carbon atom can lead to an incorrect conclusion.

Geometric isomerism is determined by the arrangement of the groups attached directly to the two alkene carbons, not by the position of an arbitrary atom further along a flexible carbon chain.

Part c

No. Ordinary single-bond rotations cannot convert cis-hex-3-ene into trans-hex-3-ene while preserving the pi bond.

These rotations can change the conformations of the ethyl groups, but they cannot change the relative positions of the directly attached groups around the C=C bond.

To change the geometric configuration, the double-bond arrangement itself must be disrupted sufficiently to allow the groups to adopt the opposite configuration.

That requires overcoming the rotational restriction associated with the pi bond.

The tempting mistake is to track the movement of the entire substituent as though it were a rigid object. A carbon chain is flexible, so its more distant atoms can move considerably without changing the configuration at the double bond.

This is the deeper distinction between conformation and configuration: a molecule can change its three-dimensional shape through single-bond rotation while retaining exactly the same geometric isomer.

The next useful step is applying the structural test for geometric isomerism, then using cis/trans or E/Z nomenclature to describe the arrangements unambiguously.