When Can an Alkene Show Cis-Trans Isomerism?

Imagine drawing two versions of propene and labelling them cis-propene and trans-propene. The drawings might look convincing because a C=C double bond cannot rotate freely. But propene cannot actually form a cis-trans pair: one carbon of its double bond is attached to two identical hydrogen atoms.

The rule is straightforward: an alkene can show geometrical isomerism only when each carbon in the C=C bond is attached to two different substituents. Both carbons must pass this test. To use cis-trans terminology, you also need a matching group on each carbon that can be compared across the double bond. When all four substituents are different, E/Z notation provides the more precise description.

Before looking at the reasoning, predict which of these alkenes can show cis-trans isomerism:

• But-1-ene: \(\ce{CH2=CHCH2CH3}\)
• But-2-ene: \(\ce{CH3CH=CHCH3}\)

But-1-ene fails because its terminal double-bond carbon has two hydrogen atoms. But-2-ene passes because each double-bond carbon has one hydrogen and one methyl group.

01The structural test comes from carbon’s valence

To understand why this rule works, start with carbon’s usual valency of four.

A carbon atom normally forms four covalent bonds. In an alkene, each carbon in the C=C bond already has a double bond to the other carbon. That double bond accounts for two of its four bonds, leaving two more bonds available for other atoms or groups.

These two remaining attachments are called substituents.

Consider a general alkene:

\[
R^1R^2C=CR^3R^4
\]

Each carbon has two substituents:

• The left carbon is attached to \(R^1\) and \(R^2\).
• The right carbon is attached to \(R^3\) and \(R^4\).

For geometrical isomerism to occur, both of these conditions must be satisfied:

\[
R^1 \ne R^2
\]

\[
R^3 \ne R^4
\]

In other words, the two groups attached to the left carbon must be different, and the two groups attached to the right carbon must also be different.

Notice that you compare groups attached to the same carbon. You do not compare the left-hand groups with the right-hand groups.

What counts as different substituents?

A substituent is the entire group attached to the alkene carbon, not just the first atom in that group.

For example:

• A hydrogen atom and a methyl group (\(\ce{CH3}\)) are different.
• A methyl group (\(\ce{CH3}\)) and an ethyl group (\(\ce{CH2CH3}\)) are different.
• Two methyl groups are identical, even if they are drawn in different positions.

This last point is particularly important for branched alkenes. Two groups do not become different simply because they are drawn on opposite sides of a structural formula.

If either carbon has two identical substituents, the alkene cannot show geometrical isomerism.

02Why does a double bond prevent rotation?

You might be wondering why the positions of the substituents matter in the first place.

A carbon-carbon single bond can usually rotate without breaking the bond. This allows attached groups to change their relative positions.

A carbon-carbon double bond behaves differently.

It contains two types of covalent bond:

• A sigma (\(\sigma\)) bond, formed by direct overlap along the line joining the carbon atoms.
• A pi (\(\pi\)) bond, formed by sideways overlap of parallel p orbitals.

The sideways overlap creates electron density above and below the plane of the carbon atoms. Twisting one carbon relative to the other disrupts this overlap.

As a result, rotation around a C=C double bond requires substantially more energy than rotation around an ordinary C-C single bond.

One way to picture this is to imagine a hinge with a cross-brace. The hinge represents the possibility of movement around a single bond, while the cross-brace represents the extra restriction created by the pi bond.

The analogy is not exact: there is no physical brace inside the molecule. The restriction comes from the overlap of the p orbitals.

This restricted rotation means that different arrangements of substituents around a C=C bond can remain distinct molecules.

03What do cis and trans actually mean?

Once both alkene carbons have two different substituents, their relative positions can produce two geometrical arrangements.

The terms cis and trans describe the positions of matching substituents relative to the double bond.

• Cis: The matching substituents are on the same side of the C=C bond.
• Trans: The matching substituents are on opposite sides of the C=C bond.

But-2-ene is a useful example.

Its structural formula is:

\[
\ce{CH3CH=CHCH3}
\]

Each carbon in the double bond is attached to one hydrogen and one methyl group.

This allows two arrangements:

In cis-but-2-ene, the two methyl groups are on the same side. In trans-but-2-ene, they are on opposite sides.

Both molecules have:

• The same molecular formula.
• The same atoms connected in the same order.
• The same C=C bond position.

Only their spatial arrangements differ.

This is why cis-but-2-ene and trans-but-2-ene are geometric stereoisomers rather than structural isomers.

04Connectivity versus spatial arrangement

It is important not to confuse two different types of isomerism.

Structural isomerism occurs when molecules have the same molecular formula but different arrangements of bonded atoms.

Geometrical isomerism occurs when molecules have the same connectivity but different spatial arrangements around a restricted bond.

Compare these examples:

But-1-ene and but-2-ene both have the molecular formula \(\ce{C4H8}\). However, the double bond occupies a different position in their carbon chains.

They are positional structural isomers, not cis-trans isomers.

Similarly, having the same molecular formula is not enough to establish that two compounds are geometrical isomers. You must check their connectivity first.

For a broader explanation of how connectivity determines structural isomerism, see Structural Isomerism in Organic Compounds for HSC Chemistry.

05A quick comparison of common alkenes

The following examples show how the carbon-by-carbon test works in practice.

Notice that only one carbon needs to fail for the entire alkene to fail the test.

For example, propene has two different groups on one alkene carbon, but two identical groups on the other. That is enough to rule out geometrical isomerism.

A reliable decision process

When you encounter an unfamiliar alkene in an HSC Chemistry question, follow these steps:

1. Locate the C=C double bond.
2. Identify the two substituents attached directly to the first alkene carbon.
3. Check whether those two substituents are different.
4. Repeat the check for the second alkene carbon.
5. If either carbon has two identical substituents, geometrical isomerism is impossible.
6. If both carbons have two different substituents, geometrical isomerism is possible. Check whether cis/trans or E/Z terminology is appropriate.

The key is to check both carbons independently. Never decide based on only one end of the double bond.

06Worked examples

07When should you use cis/trans rather than E/Z?

There is one useful distinction beyond the basic structural test.

The condition for geometrical isomerism is that each alkene carbon has two different substituents. However, cis/trans terminology requires a suitable matching pair of substituents to compare.

What happens if all four substituents are different?

Consider:

\[
\ce{BrCH=C(F)Cl}
\]

The left alkene carbon is attached to bromine and hydrogen. The right alkene carbon is attached to fluorine and chlorine.

Both carbons have two different substituents. Therefore, the molecule can have two geometrical arrangements.

However, there is no matching substituent on opposite carbons that can be used for a straightforward cis/trans comparison.

The more general naming system is called E/Z notation.

In this system, substituents are ranked according to priority rules. The higher-priority group on each alkene carbon determines the configuration:

• Z: The two higher-priority groups are on the same side.
• E: The two higher-priority groups are on opposite sides.

This is a naming distinction, not a change to the underlying structural test.

The important point is that the absence of a matching pair does not automatically mean geometrical isomerism is impossible.

08Questions and solutions

Question 1

Propene has the structural formula \(\ce{CH3CH=CH2}\).

A student draws two structures and labels them cis-propene and trans-propene.

Can propene actually form these two geometrical isomers? Explain your answer using the substituents attached to each carbon of the double bond.

Solution 1

No, propene cannot show cis-trans isomerism.

The carbon atoms in the double bond have the following substituents:

• First carbon: H and \(\ce{CH3}\).
• Second carbon: H and H.

Although the first carbon has two different substituents, the second carbon has two identical hydrogen atoms.

Therefore, the second carbon fails the structural test.

Even though the double bond prevents free rotation, changing the apparent positions of the two hydrogen atoms does not produce a different molecule because the hydrogen atoms are identical.

The student’s two drawings represent the same compound, not two distinct geometrical isomers.

Question 2

Consider the following three alkenes:

A: But-2-ene, \(\ce{CH3CH=CHCH3}\)

B: But-1-ene, \(\ce{CH2=CHCH2CH3}\)

C: 2-methylpropene, \(\ce{(CH3)2C=CH2}\)

Identify which compounds can show geometrical isomerism.

For each compound, identify the substituents attached to both alkene carbons and explain why the structure passes or fails the test.

Solution 2

Only A, but-2-ene, can show geometrical isomerism.

The reasoning for each compound is as follows.

Compound A: But-2-ene

\[
\ce{CH3CH=CHCH3}
\]

• First alkene carbon: H and \(\ce{CH3}\).
• Second alkene carbon: H and \(\ce{CH3}\).

Both carbons have two different substituents.

Therefore, but-2-ene can form cis and trans isomers.

Compound B: But-1-ene

\[
\ce{CH2=CHCH2CH3}
\]

• First alkene carbon: H and H.
• Second alkene carbon: H and \(\ce{CH2CH3}\).

The first carbon has two identical hydrogen atoms.

Therefore, but-1-ene cannot show geometrical isomerism.

Compound C: 2-methylpropene

\[
\ce{(CH3)2C=CH2}
\]

• First alkene carbon: \(\ce{CH3}\) and \(\ce{CH3}\).
• Second alkene carbon: H and H.

Both carbons have identical substituents.

Therefore, 2-methylpropene cannot show geometrical isomerism.

The important distinction is that a carbon can fail the test even when the other carbon has two different substituents. Both ends must pass independently.

Question 3

Two compounds have the same molecular formula, \(\ce{C6H12}\).

Compound P:

\[
\ce{CH3CH2CH=CHCH2CH3}
\]

Compound Q:

\[
\ce{CH3C(CH3)=CHCH2CH3}
\]

A student claims that P and Q must be cis-trans isomers because they have the same molecular formula.

Explain why this conclusion is incorrect.

Also determine whether each compound can independently show geometrical isomerism.

Solution 3

P and Q are structural isomers, not cis-trans isomers. P can show geometrical isomerism, but Q cannot.

Start by examining their connectivity.

Compound P has a continuous six-carbon chain with the double bond between carbons 3 and 4.

Compound Q has a branched carbon structure, with an additional methyl group attached to one of the alkene carbons.

Although both compounds have the molecular formula \(\ce{C6H12}\), their atoms are connected differently.

They are therefore structural isomers.

Now consider their individual ability to show geometrical isomerism.

Compound P:

\[
\ce{CH3CH2CH=CHCH2CH3}
\]

• First alkene carbon: H and ethyl.
• Second alkene carbon: H and ethyl.

Both carbons have two different substituents.

Therefore, P can form cis and trans geometrical isomers.

Compound Q:

\[
\ce{CH3C(CH3)=CHCH2CH3}
\]

• First alkene carbon: methyl and methyl.
• Second alkene carbon: H and ethyl.

The first carbon has two identical methyl groups.

Therefore, Q cannot show geometrical isomerism.

This question exposes an important misconception: matching molecular formulas establish a possible relationship between structural isomers, but they do not establish geometrical isomerism.

You must examine the connectivity and then apply the double-bond test to each molecule.

Question 4

Consider the following two alkenes.

A:

\[
\ce{CH3CH=C(CH3)2}
\]

B:

\[
\ce{CH3CH=C(CH3)CH2CH3}
\]

A student argues that the groups attached to the right-hand carbon in both structures are simply carbon-containing groups, so both alkenes must fail the geometrical isomerism test.

Is this reasoning correct?

Determine which alkene can show cis-trans isomerism. Explain why the entire substituent must be considered rather than just the first atom.

Solution 4

Only compound B can show cis-trans isomerism. The student’s reasoning incorrectly treats different carbon-containing groups as identical.

Consider compound A first.

\[
\ce{CH3CH=C(CH3)2}
\]

The substituents are:

• Left alkene carbon: H and methyl.
• Right alkene carbon: methyl and methyl.

The left carbon passes the test, but the right carbon fails because both substituents are identical methyl groups.

Therefore, A cannot show geometrical isomerism.

Now consider compound B.

\[
\ce{CH3CH=C(CH3)CH2CH3}
\]

The substituents are:

• Left alkene carbon: H and methyl.
• Right alkene carbon: methyl and ethyl.

Both carbons have two different substituents.

Therefore, B can show geometrical isomerism.

There is also a matching pair of methyl groups, one attached to each alkene carbon. These groups can be used to identify the cis and trans arrangements.

The key misconception is that two substituents are not identical simply because they both begin with a carbon atom.

A methyl group contains one carbon atom, while an ethyl group contains two. Their structures are different, so they count as different substituents.

Question 5

Consider the alkene:

\[
\ce{BrCH=C(F)Cl}
\]

A student makes the following argument:

“There are no identical substituents attached to opposite ends of the double bond. Therefore, this molecule cannot have geometrical isomers.”

(a) Is the student’s conclusion correct? Explain using the substituents attached to each alkene carbon.

(b) The molecule can be described using E/Z notation. Identify the higher-priority substituent on each alkene carbon and explain which arrangement is E and which is Z.

(c) A student draws one arrangement, rotates the entire drawing through 180° on the page, and claims to have produced the other geometrical isomer. Is this correct? Explain the difference between rotating a whole molecular drawing and changing the arrangement around the double bond.

Solution 5

The student’s conclusion is incorrect. The alkene can have two geometrical isomers, even though cis/trans terminology is not appropriate. Rotating the entire drawing by 180° also does not create a new isomer.

(a) Determine whether geometrical isomerism is possible.

Start by examining the left alkene carbon.

It is attached to:

• Bromine, Br.
• Hydrogen, H.

These are different substituents.

Now examine the right alkene carbon.

It is attached to:

• Fluorine, F.
• Chlorine, Cl.

These are also different substituents.

Both alkene carbons pass the structural test.

Therefore, the molecule can have two distinct geometrical arrangements.

The student’s mistake is confusing the requirement for cis/trans naming with the requirement for geometrical isomerism itself.

The molecule has no matching substituent pair across the double bond, so a straightforward cis/trans comparison is unavailable. However, its two different groups at each carbon still allow two geometrically distinct arrangements.

(b) Assign the E and Z configurations.

E/Z notation uses the higher-priority substituent on each alkene carbon.

For this molecule, priority can be determined using atomic number.

On the left carbon:

• Bromine has atomic number 35.
• Hydrogen has atomic number 1.

Therefore, Br has higher priority.

On the right carbon:

• Chlorine has atomic number 17.
• Fluorine has atomic number 9.

Therefore, Cl has higher priority.

The two possible arrangements are:

• Z: The higher-priority groups, Br and Cl, are on the same side of the C=C bond.
• E: The higher-priority groups, Br and Cl, are on opposite sides of the C=C bond.

The important result is that matching substituents are not necessary for geometrical isomerism. They are simply useful for straightforward cis/trans naming.

(c) Consider rotating the entire drawing.

No, rotating the whole drawing by 180° does not produce the other geometrical isomer.

A whole-molecule rotation changes the viewing orientation, not the relative positions of the substituents around the double bond.

For example, if Br and Cl are initially on the same side, rotating the entire drawing moves both groups together. They remain on the same side relative to each other.

The E/Z configuration is unchanged.

To produce the other geometrical isomer, the relative arrangement around the C=C bond must change. This requires disrupting the restricted double-bond geometry rather than simply changing the viewing angle.

This is the final distinction to remember: a different drawing does not necessarily represent a different molecule. You must establish whether the connectivity or the actual spatial arrangement has changed.

The most useful next step is to apply this reasoning when comparing skeletal formulas and identifying structural versus stereochemical isomers. Once you can reliably identify the two substituents on each alkene carbon, the cis-trans test becomes a quick structural decision rather than a guessing exercise.