How to Find All Chain Isomers of Small Alkanes

Take six carbon atoms and join them using single C-C bonds. You can make a straight chain, or move a connection to create a branch. How many genuinely different open-chain alkanes can you build before you start drawing the same molecule again?

For \(\ce{C6H14}\), there are exactly five chain isomers:

• Hexane
• 2-methylpentane
• 3-methylpentane
• 2,2-dimethylbutane
• 2,3-dimethylbutane

The reliable way to find them all is to work through possible longest carbon chains, add the remaining carbon atoms as branches, check carbon valence, and eliminate structures that have identical connectivity.

The crucial distinction is that a different drawing is not necessarily a different molecule. You need to change which carbon atoms are connected, not simply change the direction or appearance of the drawing.

01What actually makes a new chain isomer?

Chain isomers are structural isomers with:

• The same molecular formula.
• The same functional group (for alkanes, no functional group other than the saturated carbon framework).
• Different arrangements of carbon-to-carbon connections.

Consider pentane and 2-methylbutane.

Their condensed structural formulas are:

\[ \ce{CH3CH2CH2CH2CH3} \]

and

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

Both have the molecular formula \(\ce{C5H12}\), but their carbon skeletons are different.

In pentane, all five carbon atoms form one continuous chain. In 2-methylbutane, one carbon is connected to three other carbon atoms, creating a branch.

Connectivity matters more than appearance

Imagine drawing 2-methylbutane from left to right:

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

Now reverse the direction:

\[ \ce{CH3CH2CH(CH3)CH3} \]

These are the same molecule.

The methyl branch appears at carbon 2 in the first representation and carbon 3 in the second. However, reversing the direction of the drawing does not change which carbon atoms are connected.

The correct IUPAC name is 2-methylbutane because the chain must be numbered to give the substituent the lowest possible number.

This is where students often accidentally inflate their isomer count. They count different orientations as different structures.

A useful analogy is a road network. Turning a map upside down does not create new roads or change which intersections connect. You have changed the appearance of the map, not the network.

There is one limitation to this analogy: carbon atoms have a strict maximum number of bonds, while road intersections do not.

What about three-dimensional arrangements?

A skeletal formula is a simplified representation of a molecule. Its zigzag pattern does not mean that the molecule is locked into that exact shape.

Rotation around a carbon-carbon single bond can change the molecule’s conformation without changing its connectivity.

These conformations are not new chain isomers.

Some branched alkanes can also have stereoisomers, including optical isomers. Those are a separate type of isomer and are not counted as additional carbon skeletons here.

For a more detailed method of comparing two existing drawings, see How to Check if Two Skeletal Formulas Show the Same Molecule.

02Carbon valence sets the limits

Before trying to construct every possible arrangement, remember one important rule:

A neutral carbon atom forms four covalent bonds in ordinary organic structures.

In an alkane, every carbon-carbon bond is a single bond. Any remaining bonding positions are occupied by hydrogen atoms.

This gives us a simple way to check whether a proposed structure is possible.

Let \(d\) represent the number of carbon atoms directly connected to a particular carbon atom.

The number of hydrogen atoms attached to that carbon is:

\[ H_{\mathrm C}=4-d \]

For example:

A carbon connected to four other carbon atoms has no hydrogen atoms attached. This is perfectly valid.

A carbon connected to five other carbon atoms is not valid. It would have five single covalent bonds, exceeding carbon’s valence.

A useful model-building test

You can investigate the possible structures using a molecular model kit.

The apparatus consists of identical carbon centres, each with four bonding positions, and connectors representing single covalent bonds.

Start by connecting the carbon centres into one continuous, loop-free arrangement. Use exactly the required number of carbon atoms. Any unused bonding positions are filled by hydrogen atoms.

Before changing the model, make a prediction: if you turn the whole model around, have you created a new molecule?

No. Every carbon still has the same carbon neighbours.

Now move a connector so that it joins a different carbon. If the resulting arrangement has different connectivity and still obeys carbon valence, you may have created a new chain isomer.

The model demonstrates an important distinction: rotating or repositioning a structure changes its appearance, while reconnecting atoms can change its identity.

A molecular model kit is a representation, not an experiment measuring actual bond lengths or proving the identity of an unknown sample. Its purpose here is to make the connections easier to inspect.

Why does every open-chain alkane have the same general formula?

For an acyclic alkane containing \(n\) carbon atoms, the molecular formula is:

\[ \ce{C_nH_{2n+2}} \]

We can understand this directly from carbon valence.

A connected, loop-free carbon skeleton containing \(n\) carbon atoms must have \(n-1\) carbon-carbon bonds.

Each carbon-carbon bond uses one bonding position on each of its two carbon atoms.

Therefore, the total number of bonding positions used by carbon-carbon bonds is:

\[ 2(n-1) \]

There are \(4n\) bonding positions available across all carbon atoms.

The remaining positions must be occupied by hydrogen atoms:

\[ \begin{aligned} H_{\text{total}}&=4n-2(n-1)\\ &=4n-2n+2\\ &=2n+2 \end{aligned} \]

This gives the general formula for an open-chain alkane.

For example, if \(n=6\):

\[ H=2(6)+2=14 \]

So every valid open-chain alkane containing six carbon atoms has the formula \(\ce{C6H14}\), regardless of how branched its carbon skeleton is.

This is a useful check when constructing structures. If your proposed six-carbon alkane has a different number of hydrogen atoms, something is wrong.

03A repeat-proof method for finding chain isomers

Use the following method whenever you are asked to find all chain isomers of a small alkane.

Step 1: Identify the number of carbon atoms

Start with the molecular formula.

For example:

\[ \ce{C6H14} \]

You need exactly six carbon atoms.

Do not start drawing random branches before establishing the number of carbon atoms available.

Step 2: Draw the longest possible continuous chain

Begin with every carbon atom in one straight chain.

For \(\ce{C6H14}\), this gives hexane.

You have found your first structure.

Step 3: Shorten the longest chain and introduce branches

Now consider structures with a longest continuous chain containing one fewer carbon atom.

Attach the remaining carbon atoms as branches in every possible position.

Continue reducing the longest chain and distributing the remaining carbon atoms.

The key is to use the longest continuous carbon path, not simply whichever chain happens to look horizontal in your drawing.

Step 4: Remove equivalent structures

Check whether a structure has already been counted.

Try:

• Reversing the direction of the main chain.
• Rotating or flipping the drawing.
• Comparing the carbon atoms connected to each branching carbon.
• Comparing the lengths and connections of the branches.

If every carbon-to-carbon connection can be matched between two drawings, they represent the same molecule.

Step 5: Check carbon valence

For every proposed structure, check that:

• Each carbon has four bonds in total.
• No carbon has more than four carbon neighbours.
• All remaining bonding positions are occupied by hydrogen atoms.
• The final molecular formula matches the original formula.

Step 6: Check that your list is complete

Make sure you have considered each possible longest-chain length.

This is what prevents you from missing a structure simply because it has a less obvious arrangement.

The important improvement over random drawing is that you now have a systematic way to organise the possibilities.

06Questions and solutions

The following questions progress from standard HSC applications to more demanding problems involving carbon connectivity, valence, and equivalent structures.

Question 1

An alkane has the molecular formula \(\ce{C4H10}\).

Determine the number of possible chain isomers and give the IUPAC name and condensed structural formula of each.

Solution 1

There are exactly two chain isomers of \(\ce{C4H10}\): butane and 2-methylpropane.

Start with the four-carbon continuous chain:

\[ \ce{CH3CH2CH2CH3} \]

This is butane.

Now rearrange the carbon skeleton to produce a branched structure:

\[ \ce{CH3CH(CH3)CH3} \]

This is 2-methylpropane.

The central carbon is attached to three other carbon atoms and one hydrogen atom, so it obeys carbon’s valence of four.

Both structures contain four carbon atoms and ten hydrogen atoms.

No other distinct carbon skeleton is possible. A different orientation of either structure does not create another chain isomer.

Question 2

A student attempts to construct a highly branched isomer of \(\ce{C6H14}\).

They draw six carbon atoms, with one central carbon connected by single bonds to the other five carbon atoms. All five outer carbon atoms are terminal groups.

Is this a valid alkane structure?

If not, describe the maximum number of carbon neighbours a single carbon can have and give an example of a valid \(\ce{C6H14}\) structure containing such a carbon.

Solution 2

The proposed structure is invalid because the central carbon has five single covalent bonds, exceeding its maximum valence of four.

A carbon atom can have a maximum of four carbon neighbours in an ordinary neutral alkane.

A valid example containing a carbon with four carbon neighbours is 2,2-dimethylbutane:

\[ \ce{CH3C(CH3)2CH2CH3} \]

Its central carbon is connected to:

• One methyl group on the left.
• Two methyl branches.
• One ethyl group.

That gives the central carbon exactly four carbon-carbon bonds.

It has no attached hydrogen atoms.

The complete structure contains four \(\ce{CH3}\) groups, one \(\ce{CH2}\) group, and one central carbon:

\[ H=4(3)+2=14 \]

Therefore, its molecular formula is \(\ce{C6H14}\).

The important point is that you cannot repair a carbon with five carbon-carbon bonds by simply changing the number of hydrogen atoms. You must change the carbon connectivity.

Question 3

Determine all possible chain isomers of \(\ce{C7H16}\).

Organise your answer according to the length of the longest continuous carbon chain. Give the IUPAC name of every isomer and explain how you avoid counting equivalent structures.

Solution 3

There are nine distinct chain isomers of \(\ce{C7H16}\), counting carbon skeletons rather than individual stereoisomers.

The most reliable approach is to consider longest-chain lengths systematically.

Step 1: Consider a longest chain of seven carbon atoms.

There is only one arrangement: heptane.

This gives one isomer.

Step 2: Consider a longest chain of six carbon atoms.

One carbon remains as a methyl branch.

The branch can occupy carbon 2 or carbon 3. Positions 4 and 5 are equivalent to positions 3 and 2 when the chain is reversed.

This produces:

• 2-methylhexane.
• 3-methylhexane.

Two additional isomers.

Step 3: Consider a longest chain of five carbon atoms.

Two carbon atoms remain.

They can form either one ethyl branch or two separate methyl branches.

An ethyl branch can be attached to carbon 3 without creating a longer chain. This gives:

• 3-ethylpentane.

Alternatively, two methyl branches can be attached in four distinct arrangements:

• 2,2-dimethylpentane.
• 2,3-dimethylpentane.
• 2,4-dimethylpentane.
• 3,3-dimethylpentane.

The other possible numberings are duplicates when the chain is reversed.

This gives five additional isomers.

Step 4: Consider a longest chain of four carbon atoms.

Three carbon atoms remain.

To keep the longest continuous chain at four carbons, the additional carbons must be attached as methyl branches to the two internal carbons.

The valid arrangement is:

• 2,2,3-trimethylbutane.

Reversing the chain gives the same structure, not a new isomer.

A fourth methyl group on carbon 2 would create five carbon-carbon bonds at that carbon, which is impossible.

This gives one additional isomer.

Step 5: Check the total.

Every structure contains seven carbon atoms and sixteen hydrogen atoms.

The classification by longest-chain length also provides a completeness check: all possible valid arrangements have been considered without repeating a skeleton.

Question 4

An unknown open-chain alkane contains seven carbon atoms.

Its carbon-neighbour counts are:

• One carbon atom has four carbon neighbours.
• Two carbon atoms each have two carbon neighbours.
• Four carbon atoms each have one carbon neighbour.

(a) Calculate the total number of hydrogen atoms in the molecule.

(b) Determine its molecular formula.

(c) Explain why these carbon-neighbour counts do not identify a unique chain isomer. Give two different valid structures that satisfy the information.

Solution 4

The molecular formula is \(\ce{C7H16}\), but the information does not identify a unique chain isomer.

Step 1: Calculate the hydrogen contribution from each carbon type.

A carbon with four carbon neighbours has zero attached hydrogen atoms.

Each carbon with two carbon neighbours has two hydrogen atoms.

Each carbon with one carbon neighbour has three hydrogen atoms.

Therefore:

\[ \begin{aligned} H_{\text{total}} &=1(0)+2(2)+4(3)\\ &=0+4+12\\ &=16 \end{aligned} \]

Step 2: Write the molecular formula.

There are seven carbon atoms and sixteen hydrogen atoms.

\[ \boxed{\ce{C7H16}} \]

Step 3: Construct two different structures with these carbon-neighbour counts.

One valid structure is 2,2-dimethylpentane:

\[ \ce{CH3C(CH3)2CH2CH2CH3} \]

Another valid structure is 3,3-dimethylpentane:

\[ \ce{CH3CH2C(CH3)2CH2CH3} \]

Both have:

• One quaternary carbon.
• Two secondary carbons.
• Four primary carbons.

However, their carbon skeletons are different.

In 2,2-dimethylpentane, the quaternary carbon has three methyl groups attached directly to it, with a two-carbon chain extending in the remaining direction.

In 3,3-dimethylpentane, the quaternary carbon has two methyl groups and two separate ethyl groups attached.

The carbon-neighbour counts are identical, but the connections between the carbon atoms are not.

This demonstrates why counting primary, secondary, tertiary, and quaternary carbons is a useful check, but not a complete method for identifying an isomer.

Question 5

Three molecular model drawings, labelled A, B, and C, are proposed for the alkane \(\ce{C7H16}\).

All three satisfy the following conditions:

• They contain seven carbon atoms.
• Every carbon has four bonds in total.
• Four carbon atoms have one carbon neighbour each.
• One carbon has two carbon neighbours.
• Two carbon atoms have three carbon neighbours each.
• Their longest continuous carbon chain contains five carbon atoms.

The structures differ as follows:

• In model A, the two carbons with three carbon neighbours are separated by the carbon with two carbon neighbours.
• In model B, the two carbons with three carbon neighbours are directly bonded to each other.
• Model C is model A redrawn from the opposite end, with the two identical methyl branches at one branching carbon exchanged.

No wedge-and-dash stereochemistry is specified.

Determine:

(a) Which models represent the same molecule?

(b) Which models represent different chain isomers?

(c) Give the IUPAC names and condensed structural formulas of the distinct structures.

(d) Explain why molecular formula, carbon-neighbour counts, and longest-chain length are insufficient on their own to establish that two structures are identical.

Solution 5

Models A and C represent the same carbon skeleton, while model B is a different chain isomer.

The distinction comes from the connectivity between the two branching carbon atoms, not from the molecular formula or the overall appearance of the drawings.

Step 1: Identify the structure represented by model A.

The two carbons with three carbon neighbours are separated by one secondary carbon.

The condensed structural formula is:

\[ \ce{CH3CH(CH3)CH2CH(CH3)CH3} \]

This is 2,4-dimethylpentane.

Step 2: Identify the structure represented by model B.

The two carbons with three carbon neighbours are directly connected.

The condensed structural formula is:

\[ \ce{CH3CH(CH3)CH(CH3)CH2CH3} \]

This is 2,3-dimethylpentane.

Although both structures contain the same types and numbers of carbon atoms, the branching carbons have different connections.

Step 3: Compare models A and C.

Model C is simply model A drawn in the opposite direction, with two identical methyl groups exchanged.

The methyl groups are chemically identical. Swapping their positions in the drawing does not change which atoms are connected.

Therefore, A and C are the same molecule.

Step 4: Explain why model B is different.

Consider the shortest path between the two carbons that each have three carbon neighbours.

In model A, those two carbons are separated by one carbon atom. They are two carbon-carbon bonds apart.

In model B, they are directly bonded. They are one carbon-carbon bond apart.

This separation is a property of the connectivity. You cannot change it by rotating the molecule, reversing the drawing, or choosing a different direction for the parent chain.

Therefore, A and B cannot represent the same carbon skeleton.

Step 5: Check the hydrogen count.

Each structure contains:

• Four primary carbons, contributing \(4 \times 3=12\) hydrogen atoms.
• One secondary carbon, contributing \(1 \times 2=2\) hydrogen atoms.
• Two tertiary carbons, contributing \(2 \times 1=2\) hydrogen atoms.

The total is:

\[ H=12+2+2=16 \]

Both therefore have the molecular formula \(\ce{C7H16}\).

The key conclusion is that matching molecular formulas, carbon-neighbour counts, and longest-chain lengths does not guarantee identical connectivity.

You must also compare how the branching carbons and their surrounding carbon groups are connected.

07What to do after finding the isomers

Once you can generate every valid carbon skeleton without repeating one, you can move on to comparing their physical properties.

Different carbon skeletons produce different molecular shapes. That can influence how closely neighbouring molecules interact and, consequently, their boiling points.

For the next step, read Why Branched and Straight-Chain Isomers Boil Differently.

The important thing is to get the structures right first. If you count the same molecule twice or include an impossible carbon arrangement, any comparison of the isomers that follows will start from the wrong list.