Why Terminal Alkenes Usually Cannot Have Cis-Trans Isomers

Put propene, \(\mathrm{CH_2=CHCH_3}\), on paper. The carbon-carbon double bond cannot freely rotate, so you might expect to be able to draw two arrangements: one with the \(\mathrm{CH_3}\) group above the double bond and another with it below.

Would those be cis-trans isomers?

No. They represent the same molecule. The carbon at the end of the double bond is attached to two identical hydrogen atoms. Swapping which hydrogen appears above or below the double bond does not create a different spatial arrangement of the molecule.

This is the reason ordinary terminal alkenes cannot have cis-trans isomers. The important issue is not simply where the double bond sits. It is whether each carbon in the double bond has two different groups attached to it.

## Why terminal alkenes usually cannot have cis-trans isomers

### Start with the structure of a terminal alkene

A terminal alkene has its carbon-carbon double bond at the end of its carbon chain. In an ordinary, unsubstituted hydrocarbon, its structure can be represented as:

\[
\mathrm{R-CH=CH_2}
\]

Here, \(\mathrm{R}\) represents the rest of the carbon chain.

Look closely at the two carbons in the double bond:

- The first carbon is attached to \(\mathrm{R}\) and H.
- The second carbon is attached to H and H.

The second carbon therefore has two identical substituents.

[IMAGE: Annotated structural diagram of propene, CH2=CH-CH3, showing the carbon-carbon double bond horizontally, the two hydrogen atoms attached to the terminal CH2 carbon, and the hydrogen and methyl group attached to the other alkene carbon. Clearly highlight the identical H/H attachments and show the molecular connectivity.]

### The carbon-valence rule explains why

Carbon normally forms four covalent bonds. A carbon-carbon double bond counts as two bonds towards each carbon's valency.

For the terminal carbon in propene:

- Two bonds come from the C=C double bond.
- One bond connects it to hydrogen.
- One bond connects it to another hydrogen.

That gives carbon its four bonds:

\[
2+1+1=4
\]

There is no room for a different fourth attachment. In an ordinary terminal alkene, the end carbon is therefore \(\mathrm{CH_2}\).

This creates the problem for cis-trans isomerism: both groups attached to that carbon are hydrogen atoms. They are chemically identical, so exchanging their positions cannot produce a different geometrical isomer.

**The key rule:** For an alkene to exhibit geometrical isomerism, each carbon of the C=C bond must have two different groups attached to it.

If either carbon has two identical groups, the requirement is not satisfied.

### Why the terminal carbon must have two hydrogen atoms

The reason becomes clearer if you count carbon's bonds rather than memorising a rule about terminal alkenes.

Consider but-1-ene:

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

The first carbon is at the end of the chain. It has only one neighbouring carbon, connected through the double bond.

Count its bonds:

| Connection | Number of bonds |
|---|---:|
| Double bond to the next carbon | 2 |
| Bond to first hydrogen | 1 |
| Bond to second hydrogen | 1 |
| **Total** | **4** |

Carbon has reached its usual valency of four.

There is no additional carbon atom attached to this end of the molecule, so the two remaining bonds are to hydrogen atoms. That is why the terminal group is \(\ce{CH2}\), rather than \(\ce{CH}\).

This conclusion assumes an ordinary, neutral, open-chain hydrocarbon without isotopic substitution or another atom replacing one of the terminal hydrogens.

[IMAGE: Annotated structural diagram of but-1-ene showing the terminal carbon, its double bond to the adjacent carbon, and its two C-H bonds. Include bond counts of 2 + 1 + 1 = 4 around the terminal carbon, with the adjacent carbon and ethyl chain also visible.]

Now consider what happens if the carbon chain branches.

Take 2-methylbut-1-ene:

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

The second carbon of the double bond is attached to a methyl group and an ethyl group. These are different.

However, the first carbon still has two hydrogen atoms. Adding a branch to the neighbouring carbon does not change that.

The molecule still cannot have geometrical isomers.

This is a useful exam distinction: branching elsewhere in the molecule cannot fix identical groups attached to the terminal alkene carbon.

### Connectivity versus spatial arrangement

There is another distinction worth getting right.

Connectivity describes which atoms are bonded to which. Spatial arrangement describes how those connected atoms are positioned relative to one another.

Moving a double bond along a carbon chain changes its position in the structure. It does not simply produce another spatial arrangement of the same molecule.

Consider two compounds:

\[
\begin{aligned}
\text{Pent-1-ene: }&\ce{CH2=CH-CH2-CH2-CH3}\\
\text{Pent-2-ene: }&\ce{CH3-CH=CH-CH2-CH3}
\end{aligned}
\]

Both have the molecular formula \(\ce{C5H10}\), but their double bonds occupy different positions.

Pent-1-ene cannot have geometrical isomers because its terminal carbon has two hydrogen atoms.

Pent-2-ene can have geometrical isomers because:

- One double-bonded carbon is attached to H and \(\ce{CH3}\).
- The other is attached to H and \(\ce{CH2CH3}\).

Both carbons have two different groups.

[IMAGE: Side-by-side structural comparison of pent-1-ene and the two possible geometrical arrangements of pent-2-ene. Show pent-1-ene with its terminal CH2 group, and show the two pent-2-ene arrangements with the hydrogen atoms on the same side and opposite sides of the C=C bond. Clearly distinguish the change in double-bond position from the change in spatial arrangement.]

Pent-1-ene and pent-2-ene are positional structural isomers, not cis-trans isomers of each other. The two forms of pent-2-ene, however, are geometrical isomers because their connectivity is unchanged while their spatial arrangements differ.

For more practice with the general structural test, see [When Can an Alkene Have Cis-Trans Isomers? The Structural Test](https://tutorgum.com/learn/resources/when-can-an-alkene-have-cis-trans-isomers/).

### Does the double bond still restrict rotation?

Yes. A terminal alkene still contains a C=C double bond, which strongly restricts rotation.

A double bond consists of a sigma (\(\sigma\)) bond and a pi (\(\pi\)) bond. Twisting the carbon atoms relative to one another disrupts the sideways overlap responsible for the pi bond.

However, restricted rotation alone does not guarantee geometrical isomerism.

Think of it this way: the double bond prevents the atoms from freely rearranging, but the groups attached to those atoms must still be different enough to create distinguishable arrangements.

In a terminal alkene, the two hydrogen atoms on the terminal carbon are identical. There is no second distinguishable geometrical arrangement to name.

The important distinction is:

- Restricted rotation makes geometrical isomerism possible.
- Different substituents on both alkene carbons are needed for distinct geometrical isomers.

For a refresher on why carbon forms four bonds, see [HSC Chemistry: Alkane, Alkene and Alkyne Formulae](https://tutorgum.com/learn/resources/hsc-chemistry-hydrocarbon-homologous-series-and-formulae/).

## Worked examples

### Worked example: Can 2-methylbut-1-ene have geometrical isomers?

Determine whether 2-methylbut-1-ene, \(\ce{CH2=C(CH3)CH2CH3}\), can exhibit cis-trans isomerism. Explain your reasoning using the groups attached to each carbon of the double bond.

[IMAGE: Structural formula of 2-methylbut-1-ene, showing the terminal CH2 carbon, the adjacent alkene carbon, and its methyl and ethyl branches. Keep all bonds and attachments clearly visible.]

**Step 1**: Identify the two carbon atoms forming the double bond.

They are the first and second carbon atoms in the main chain.

**Step 2**: Identify the groups attached to the first carbon.

The terminal carbon is \(\ce{CH2}\). Its two groups are:

- Hydrogen
- Hydrogen

These groups are identical.

**Step 3**: Identify the groups attached to the second carbon.

The second carbon is attached to:

- A methyl group, \(\ce{CH3}\)
- An ethyl group, \(\ce{CH2CH3}\)

These groups are different.

**Step 4**: Apply the geometrical isomerism requirement.

The second carbon passes the test, but the first carbon does not.

Therefore, 2-methylbut-1-ene cannot exhibit cis-trans isomerism.

Notice that the methyl and ethyl groups are different. That does not rescue the molecule because both alkene carbons must have two different groups attached.

### Worked example: Why are pent-1-ene and pent-2-ene different?

Two compounds have the molecular formula \(\ce{C5H10}\):

- Pent-1-ene: \(\ce{CH2=CHCH2CH2CH3}\)
- Pent-2-ene: \(\ce{CH3CH=CHCH2CH3}\)

Determine which compound can exhibit geometrical isomerism. Explain why the two compounds should not be described as cis-trans isomers of each other.

**Step 1**: Examine pent-1-ene.

The first carbon of its double bond is attached to two hydrogen atoms.

It therefore fails the structural requirement for geometrical isomerism.

**Step 2**: Examine pent-2-ene.

The first carbon of its double bond is attached to H and \(\ce{CH3}\).

The second carbon is attached to H and \(\ce{CH2CH3}\).

Both pairs contain different groups.

**Step 3**: Determine which compound can have geometrical isomers.

Pent-2-ene can have two geometrical arrangements because both alkene carbons have different substituents.

Pent-1-ene cannot.

**Step 4**: Distinguish the relationship between the two compounds.

Both compounds have the molecular formula \(\ce{C5H10}\), but the double bond is in a different position.

They are positional isomers, not cis-trans isomers.

The distinction is important in HSC questions: sharing a molecular formula does not automatically make two compounds geometrical isomers.

## Questions and solutions

### Question 1

Consider 3-methylbut-1-ene:

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

Can this compound exhibit cis-trans isomerism? Identify the two groups attached to each carbon of the double bond to justify your answer.

[IMAGE: Clear structural formula of 3-methylbut-1-ene, showing the terminal double bond and the methyl branch on the third carbon. Do not show an answer or geometrical isomer labels.]

### Solution 1

No, 3-methylbut-1-ene cannot exhibit cis-trans isomerism.

The first carbon of the double bond is attached to two hydrogen atoms. These are identical groups.

The second carbon is attached to a hydrogen atom and the remaining carbon chain, which are different groups.

Although the second carbon passes the test, the first carbon fails it.

Both alkene carbons must have two different substituents for geometrical isomerism to occur. The branch on the third carbon does not change the two hydrogen atoms attached to the terminal carbon.

### Question 2

Two compounds have the molecular formula \(\ce{C6H12}\):

- Hex-1-ene: \(\ce{CH2=CHCH2CH2CH2CH3}\)
- Hex-2-ene: \(\ce{CH3CH=CHCH2CH2CH3}\)

A student claims that these compounds are cis-trans isomers because they have the same molecular formula.

Explain why the student's claim is incorrect. Identify which compound can exhibit geometrical isomerism and justify your answer.

[IMAGE: Side-by-side structural representations of hex-1-ene and hex-2-ene. Show all six carbon atoms and the location of each double bond. Do not show the cis or trans forms of hex-2-ene.]

### Solution 2

The student's claim is incorrect. Hex-1-ene and hex-2-ene are positional isomers, and only hex-2-ene can exhibit geometrical isomerism.

Both compounds contain six carbon atoms and twelve hydrogen atoms, so they have the same molecular formula.

However, their double bonds occupy different positions.

In hex-1-ene, the terminal carbon has two hydrogen atoms. It fails the geometrical isomerism requirement.

In hex-2-ene, the two double-bonded carbons have these groups:

| Alkene carbon | First group | Second group |
|---|---|---|
| Carbon 2 | H | \(\ce{CH3}\) |
| Carbon 3 | H | \(\ce{CH2CH2CH3}\) |

Both carbons have two different groups attached, so hex-2-ene can have geometrical isomers.

The key mistake is confusing positional isomerism with geometrical isomerism. Changing the location of a double bond changes the connectivity of the molecule, whereas geometrical isomerism involves different spatial arrangements without changing connectivity.

### Question 3

Four compounds have the molecular formula \(\ce{C6H12}\):

| Compound | Condensed structural formula |
|---|---|
| A: Hex-1-ene | \(\ce{CH2=CHCH2CH2CH2CH3}\) |
| B: 2-Methylpent-1-ene | \(\ce{CH2=C(CH3)CH2CH2CH3}\) |
| C: 2-Methylpent-2-ene | \(\ce{CH3C(CH3)=CHCH2CH3}\) |
| D: 3-Methylpent-2-ene | \(\ce{CH3CH=C(CH3)CH2CH3}\) |

Identify which compound or compounds can exhibit geometrical isomerism.

For each compound that cannot, identify the specific carbon of the double bond that prevents geometrical isomerism and explain why.

[IMAGE: Four clearly drawn skeletal structures labelled A, B, C, and D, corresponding to the four compounds in the question. Show the double-bond positions, all carbon branches, and enough hydrogen attachments to make the alkene substituents unambiguous. Do not indicate which structures pass the geometrical isomerism test.]

### Solution 3

Only compound D, 3-methylpent-2-ene, can exhibit geometrical isomerism.

Each compound needs to be checked independently.

**Compound A: Hex-1-ene**

The first carbon of the double bond is attached to two hydrogen atoms.

It fails the requirement.

**Compound B: 2-Methylpent-1-ene**

The first carbon of the double bond is also attached to two hydrogen atoms.

The second carbon has a methyl group and a propyl group, which are different.

However, the first carbon fails, so geometrical isomerism is not possible.

**Compound C: 2-Methylpent-2-ene**

The second carbon of the double bond is attached to two methyl groups.

These groups are identical, so this carbon fails the requirement.

The other alkene carbon has a hydrogen atom and an ethyl group, but that is not enough.

**Compound D: 3-Methylpent-2-ene**

The first carbon of the double bond is attached to H and \(\ce{CH3}\).

The second carbon is attached to \(\ce{CH3}\) and \(\ce{CH2CH3}\).

Both pairs contain different groups, so geometrical isomerism is possible.

This question exposes an important trap: branching does not automatically create or prevent geometrical isomerism. What matters is the identity of the two groups attached to each individual alkene carbon.

### Question 4

A student draws two possible arrangements of but-1-ene, \(\ce{CH2=CHCH2CH3}\).

In the first drawing, the ethyl group attached to the double bond is shown above the C=C bond. In the second, it is shown below the C=C bond.

The student claims these are cis and trans isomers because the C=C bond cannot freely rotate.

Evaluate the student's claim. In your response:

- Explain why restricted rotation alone is insufficient.
- Identify the relevant groups attached to both alkene carbons.
- Explain why the two drawings do not represent distinct geometrical isomers.

[IMAGE: Two proposed planar arrangements of but-1-ene, with the ethyl group shown on opposite sides of the horizontal C=C bond in the two drawings. Show both hydrogen atoms on the terminal CH2 carbon in each drawing. Do not label either drawing cis or trans.]

### Solution 4

The student is incorrect. The two drawings do not represent distinct geometrical isomers of but-1-ene.

Restricted rotation around the double bond is important because it allows geometrical arrangements to remain distinct rather than freely interconverting.

However, the groups attached to both alkene carbons must also be different.

For but-1-ene:

- The terminal carbon has two hydrogen atoms.
- The adjacent carbon has a hydrogen atom and an ethyl group.

The terminal carbon therefore has identical substituents.

Swapping which hydrogen appears on which side of the double bond does not create a distinguishable molecule. The hydrogen atoms are chemically identical.

The student's argument confuses a physical property of the double bond with the structural requirement for geometrical isomerism.

A double bond restricts rotation, but it cannot make two identical groups become different groups.

### Question 5

An ordinary terminal alkene has the structure:

\[
\ce{CH2=CHCH3}
\]

Now consider an isotopically labelled version:

\[
\ce{CHD=CHCH3}
\]

Here, D represents deuterium, an isotope of hydrogen with one proton and one neutron. Assume that the isotope is explicitly distinguished from ordinary hydrogen throughout the question.

A student makes the following statement:

"Any alkene with its double bond at the end of a carbon chain can never exhibit geometrical isomerism."

Assess this statement using both structures.

For the isotopically labelled compound, determine:

- Whether each alkene carbon has two different groups attached.
- Whether geometrical isomerism is possible.
- Which groups have higher priority when assigning E/Z configurations.
- How the two possible arrangements differ.

Explain why your conclusion does not contradict the usual HSC rule for ordinary terminal hydrocarbon alkenes.

### Solution 5

The student's statement is too absolute. Ordinary propene cannot exhibit geometrical isomerism, but the isotopically labelled compound \(\ce{CHD=CHCH3}\) can.

The important difference is that the terminal carbon no longer has two identical hydrogen atoms.

**Part A: Ordinary propene**

In \(\ce{CH2=CHCH3}\), the terminal carbon has two hydrogen atoms.

They are identical, so the terminal carbon fails the structural requirement.

Propene cannot exhibit geometrical isomerism.

**Part B: Isotopically labelled propene**

In \(\ce{CHD=CHCH3}\), the first alkene carbon has:

- One ordinary hydrogen atom, H.
- One deuterium atom, D.

These are different isotopes and are distinguishable.

The second alkene carbon has:

- One hydrogen atom, H.
- One methyl group, \(\ce{CH3}\).

These groups are also different.

Both alkene carbons therefore pass the structural test.

Geometrical isomerism is possible.

**Part C: Assigning E/Z configurations**

For more complex structures, E/Z nomenclature uses priority rules to distinguish the two arrangements.

Deuterium has higher priority than ordinary hydrogen because it has a greater mass number.

On the other alkene carbon, the methyl group has higher priority than hydrogen because carbon has a higher atomic number than hydrogen.

The two possible arrangements are therefore:

- Z: The higher-priority groups, D and \(\ce{CH3}\), are on the same side of the double bond.
- E: The higher-priority groups, D and \(\ce{CH3}\), are on opposite sides of the double bond.

These are distinct geometrical isomers.

**Part D: Why this does not contradict the usual rule**

In ordinary HSC hydrocarbon questions, the terminal carbon of a terminal alkene is \(\ce{CH2}\). Its two hydrogen atoms are identical, so geometrical isomerism is not possible.

The isotope-labelled example changes that condition. Replacing one hydrogen with deuterium makes the two substituents different.

This is an extension of the usual HSC structural test, not an exception students need to assume for ordinary, unlabelled hydrocarbon structures.

The deeper principle is that geometrical isomerism depends on the actual identities of the groups attached to the double-bonded carbons, not simply on whether the double bond appears at the end of a carbon chain.

## What to remember for your HSC Chemistry exam

When you see a terminal alkene, do not immediately try to draw cis and trans forms.

Start by examining the terminal carbon and counting its bonds.

For an ordinary terminal hydrocarbon alkene:

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

The terminal carbon has two identical hydrogen atoms. It therefore cannot produce distinct cis-trans arrangements.

Use this quick comparison:

| Structure | Terminal carbon | Geometrical isomerism? |
|---|---|---|
| But-1-ene | H and H | No |
| Pent-1-ene | H and H | No |
| 2-Methylbut-1-ene | H and H | No |
| But-2-ene | Not terminal | Yes |
| Pent-2-ene | Not terminal | Yes |

Remember three separate ideas:

1. Carbon normally forms four covalent bonds. This explains why the terminal carbon in an ordinary hydrocarbon alkene is \(\ce{CH2}\).
2. Geometrical isomerism requires two different groups on each carbon of the double bond.
3. A change in double-bond position produces a positional structural isomer, not necessarily a geometrical isomer.

Once you can distinguish connectivity from spatial arrangement, you can move on to assigning E/Z configurations for more complicated alkenes. That is the next step after deciding whether geometrical isomerism is possible at all.