How to Draw Alcohol Isomers with the Same Molecular Formula

Suppose you are given the molecular formula \(\ce{C4H10O}\). You know the molecule contains four carbon atoms, ten hydrogen atoms, and one oxygen atom. But where should the oxygen go?

There are four different alcohol structural isomers with this formula:

• Butan-1-ol
• Butan-2-ol
• 2-methylpropan-1-ol
• 2-methylpropan-2-ol

The trick is to build the possible carbon skeletons first, then attach the hydroxyl group (\(\ce{-OH}\)) to each genuinely different carbon position. Carbon’s four-bond limit and the symmetry of the carbon skeleton tell you which structures are valid and which drawings are duplicates.

Before working through the method, consider something.

If you draw butan-1-ol from left to right, then redraw it from right to left, have you created a new alcohol isomer?

No. You have only reversed the drawing. Every atom is still connected to exactly the same neighbours.

However, moving the \(\ce{-OH}\) group from the first carbon to the second carbon changes the connections between the atoms. That produces a different structural isomer.

This distinction is the foundation of drawing alcohol isomers correctly.

01Understand what must stay fixed

An alcohol contains a hydroxyl functional group, \(\ce{-OH}\). This group consists of an oxygen atom bonded to a hydrogen atom, with the oxygen also bonded to a carbon atom.

For a neutral, saturated, open-chain alcohol containing one hydroxyl group, the general molecular formula is:

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

For example, when \(n=4\):

\[
\ce{C4H10O}
\]

This formula tells you how many atoms you must use. It does not tell you how those atoms are connected.

There is another important limitation: ethers can have the same general molecular formula.

For example, ethoxyethane has the formula \(\ce{C4H10O}\), but its oxygen is bonded to two carbon atoms rather than to a carbon and a hydrogen.

\[
\ce{CH3CH2-O-CH2CH3}
\]

It is an ether, not an alcohol.

Therefore, when asked to draw alcohol isomers, make sure every structure contains an actual \(\ce{-OH}\) group.

Check carbon and oxygen valency

Before drawing anything, remember these usual valencies for neutral organic molecules:

A single bond contributes one to an atom’s bond count.

Consider butan-2-ol:

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

The second carbon is bonded to:

• The first carbon.
• The third carbon.
• The oxygen in the hydroxyl group.
• One hydrogen atom.

That gives it four bonds.

The oxygen is bonded to the carbon and the hydrogen in \(\ce{-OH}\), giving it two bonds.

Both atoms satisfy their usual valencies.

This is an important check throughout the process. A structure can contain the correct number of atoms and still be chemically invalid if an atom has too many bonds.

02Use carbon skeletons to generate different alcohols

The most reliable approach is to separate the problem into two decisions:

1. What different carbon skeletons are possible?
2. Where can the hydroxyl group attach to each skeleton?

A carbon skeleton describes how the carbon atoms are connected, without considering the oxygen or the attached hydrogens.

For a saturated, open-chain alcohol, you can think of starting with an alkane and replacing one of its carbon-bound hydrogen atoms with an \(\ce{-OH}\) group.

Let’s apply this to \(\ce{C4H10O}\).

Step 1: Draw every possible carbon skeleton

There are two different carbon skeletons containing four carbon atoms.

Straight-chain skeleton: butane

\[
\ce{C-C-C-C}
\]

Branched skeleton: 2-methylpropane

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

The first has four carbon atoms in a continuous chain. The second has a central carbon connected to three other carbon atoms.

You now have two different carbon frameworks to investigate.

Step 2: Find the different positions for the hydroxyl group

This is where symmetry becomes important.

Start with the straight-chain skeleton:

\[
\ce{C-C-C-C}
\]

You might initially think that the hydroxyl group could occupy four different positions.

However, positions 1 and 4 are equivalent. Reversing the chain makes them identical.

Likewise, positions 2 and 3 are equivalent.

Therefore, there are only two distinct positions for the hydroxyl group:

• An end carbon.
• An internal carbon.

These produce butan-1-ol and butan-2-ol.

Now consider the branched skeleton, 2-methylpropane.

The three terminal methyl carbons are equivalent. Attaching \(\ce{-OH}\) to any one of them produces the same molecule.

The central carbon is different, so attaching \(\ce{-OH}\) there produces another alcohol.

This gives two more structures.

Step 3: Check the four resulting alcohols

The complete set is:

All four have the molecular formula:

\[
\ce{C4H10O}
\]

But their atom-to-atom connections are different.

This is why they are structural isomers, also called constitutional isomers.

Step 4: Identify primary, secondary and tertiary alcohols

You can also classify an alcohol by counting how many other carbon atoms are directly attached to the carbon bearing the hydroxyl group.

For our four examples:

• Butan-1-ol: primary.
• Butan-2-ol: secondary.
• 2-methylpropan-1-ol: primary.
• 2-methylpropan-2-ol: tertiary.

Notice that the classification depends on the carbon attached to \(\ce{-OH}\), not simply on whether the overall carbon skeleton is branched.

03Worked examples

03Connectivity versus spatial arrangement

There is one more distinction worth understanding.

Two structures can have the same molecular formula and the same atom-to-atom connections but differ in their three-dimensional arrangement.

These are called stereoisomers.

Consider butan-2-ol:

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

The second carbon is attached to four different groups:

• Hydrogen.
• Hydroxyl.
• Methyl (\(\ce{-CH3}\)).
• Ethyl (\(\ce{-CH2CH3}\)).

This carbon is a chiral centre. It can have two different three-dimensional arrangements that are mirror images of one another.

These mirror-image forms are called enantiomers.

They have:

• The same molecular formula.
• The same atom-to-atom connectivity.
• Different three-dimensional arrangements.

This is different from moving the hydroxyl group from carbon 1 to carbon 2.

For example, butan-1-ol and butan-2-ol are structural isomers because the oxygen is connected to a different carbon.

The two enantiomers of butan-2-ol, on the other hand, have identical connectivity.

For the alcohol counts in this article, we have counted constitutional isomers, not individual stereoisomers. Only count stereoisomers separately when the question specifically asks you to consider three-dimensional arrangements.

Also remember that rotating a molecule around an ordinary carbon-carbon single bond changes its conformation, not its constitutional structure.

04Common mistakes when drawing alcohol isomers

A particularly useful habit is to check the carbon directly attached to oxygen before accepting a structure.

Ask yourself:

1. How many carbon atoms are attached to this carbon?
2. How many bonds does it already have?
3. Does it have a hydrogen that can be replaced by \(\ce{-OH}\)?
4. Is this attachment position genuinely different from the others?

These checks prevent most errors before you finish drawing.

06Questions and solutions

The following questions move from straightforward alcohol identification to more demanding problems involving symmetry, valency, and three-dimensional structure.

Question 1

A saturated, open-chain alcohol has the molecular formula \(\ce{C3H8O}\).

Draw all distinct alcohol constitutional isomers and write their condensed structural formulas.

For each structure, identify whether it is a primary or secondary alcohol.

Solution 1

There are two distinct alcohol constitutional isomers: propan-1-ol and propan-2-ol.

The condensed structural formulas are:

Propan-1-ol

\[
\ce{CH3CH2CH2OH}
\]

Propan-2-ol

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

Both have the molecular formula \(\ce{C3H8O}\).

The carbon skeleton is a straight chain of three carbon atoms. The two end positions are equivalent because reversing the chain does not change its connectivity.

The middle carbon is a separate position.

In propan-1-ol, the carbon attached to \(\ce{-OH}\) is bonded to one other carbon. It is therefore a primary alcohol.

In propan-2-ol, the carbon attached to \(\ce{-OH}\) is bonded to two other carbons. It is therefore a secondary alcohol.

The tempting mistake is to count the two ends of the chain separately. They produce the same structure, so there are only two distinct alcohols.

Question 2

A student proposes the following structures as possible alcohol isomers with the molecular formula \(\ce{C4H10O}\).

• A: \(\ce{CH3CH2CH2CH2OH}\)
• B: \(\ce{HOCH2CH2CH2CH3}\)
• C: \(\ce{CH3CH(OH)CH2CH3}\)
• D: \(\ce{(CH3)2CHCH2OH}\)
• E: \(\ce{(CH3)3COH}\)
• F: \(\ce{CH3CH2OCH2CH3}\)
• G: \(\ce{CH3C(OH)(CH3)CH2CH3}\)

Determine:

(a) Which structures represent distinct alcohol constitutional isomers with the formula \(\ce{C4H10O}\)?

(b) Which structures are duplicate representations of the same molecule?

(c) Which structure is an ether rather than an alcohol?

(d) Which structure has the wrong number of carbon atoms for the specified molecular formula?

Solution 2

The four distinct alcohol constitutional isomers are A/B together, C, D and E.

B is a duplicate of A, F is an ether, and G contains five carbon atoms rather than four.

Let’s examine each group.

A and B: Same molecule

Both represent butan-1-ol:

\[
\ce{CH3CH2CH2CH2OH}
\]

and

\[
\ce{HOCH2CH2CH2CH3}
\]

The only difference is the direction in which the formula is written.

The carbon-oxygen connections are identical, so these are not two different isomers.

C: Butan-2-ol

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

The hydroxyl group is attached to the second carbon of the straight chain.

This is a different connectivity from butan-1-ol.

D: 2-methylpropan-1-ol

\[
\ce{(CH3)2CHCH2OH}
\]

This structure has a branched carbon skeleton and a terminal hydroxyl group.

It is different from both straight-chain alcohols.

E: 2-methylpropan-2-ol

\[
\ce{(CH3)3COH}
\]

The hydroxyl group is attached to the central carbon, which is bonded to three other carbon atoms.

This is a valid tertiary alcohol and the fourth structural isomer.

F: Ether

\[
\ce{CH3CH2OCH2CH3}
\]

The oxygen is bonded to two carbon atoms.

There is no hydroxyl group, so this structure is an ether rather than an alcohol.

It does have the molecular formula \(\ce{C4H10O}\), but it does not belong in the requested alcohol list.

G: Incorrect molecular formula

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

Count the carbon atoms:

• One carbon in the first \(\ce{CH3}\).
• One central carbon.
• One carbon in the methyl branch.
• One carbon in the \(\ce{CH2}\).
• One carbon in the final \(\ce{CH3}\).

There are five carbon atoms.

Its molecular formula is \(\ce{C5H12O}\), not \(\ce{C4H10O}\).

The important lesson is that a valid alcohol structure must satisfy both the functional-group requirement and the specified molecular formula.

Question 3

Consider the carbon skeleton of 2,2-dimethylpropane:

\[
\ce{C(CH3)4}
\]

A student suggests that two different alcohols can be produced from this skeleton:

• Structure A: Attach \(\ce{-OH}\) directly to the central carbon.
• Structure B: Replace one hydrogen on a terminal methyl group with \(\ce{-OH}\).

Determine which structure is chemically valid.

For the valid structure, write the condensed structural formula, give its systematic name, and classify it as primary, secondary or tertiary.

Solution 3

Only Structure B is chemically valid.

The central carbon already has four carbon-carbon bonds. Adding an oxygen bond would give it five bonds, violating carbon’s usual valency of four.

The four terminal methyl groups are equivalent, so attaching \(\ce{-OH}\) to any one of them produces the same alcohol.

Replacing one terminal carbon’s hydrogen with the hydroxyl group gives:

\[
\ce{HOCH2C(CH3)3}
\]

Its systematic name is 2,2-dimethylpropan-1-ol.

It is a primary alcohol because the carbon directly bonded to oxygen is attached to only one other carbon atom.

The tempting prediction is that the central carbon should give another isomer because it is a distinct position in the skeleton. However, being a different position does not automatically mean it can accept an additional bond.

Valency determines whether that proposed structure is possible.

Question 4

Three structures are being compared.

• A and B are wedge-and-dash drawings of butan-2-ol. Both have a central carbon bonded to \(\ce{-OH}\), H, \(\ce{-CH3}\), and \(\ce{-CH2CH3}\). They are mirror images, with the OH and H positions reversed in the wedge-and-dash representation.
• C is butan-1-ol, with condensed structural formula \(\ce{CH3CH2CH2CH2OH}\).

Determine:

(a) Whether A and B are structural isomers or stereoisomers.

(b) Whether A and C are structural isomers.

(c) Whether B and C are structural isomers.

Explain your reasoning using connectivity and spatial arrangement.

Solution 4

A and B are stereoisomers, while A and C, and B and C, are structural isomers.

The distinction is whether the atoms are connected differently or simply arranged differently in three-dimensional space.

(a) A and B: Stereoisomers

Both structures have the same molecular formula and identical atom-to-atom connections.

The carbon attached to \(\ce{-OH}\) has four different groups attached to it. It is a chiral centre.

The two structures are mirror-image arrangements that cannot be superimposed by ordinary rotation. They are enantiomers.

Their connectivity is identical, so they are not structural isomers.

(b) A and C: Structural isomers

A is a form of butan-2-ol, whereas C is butan-1-ol.

Both have the molecular formula \(\ce{C4H10O}\).

However, the hydroxyl group is attached to a different carbon in each molecule.

This changes the atom-to-atom connectivity, making them structural isomers.

(c) B and C: Structural isomers

The same reasoning applies.

B is another spatial arrangement of butan-2-ol, while C is butan-1-ol.

Their carbon-oxygen connections are different, so they are structural isomers.

The key point is that changing the position of the hydroxyl group changes connectivity. Changing the three-dimensional arrangement around a chiral carbon does not.

Question 5

Extension challenge

A saturated, open-chain alcohol has the molecular formula \(\ce{C6H14O}\).

The possible six-carbon parent skeletons are listed below. Assume this list is complete.

A student argues that there must be 30 different alcohols because there are five carbon skeletons, each containing six carbon atoms.

Explain why this reasoning is incorrect.

Determine the actual number of distinct alcohol constitutional isomers that can be produced.

Assume that:

• Exactly one hydrogen attached to carbon is replaced by \(\ce{-OH}\).
• Only chemically valid neutral structures are counted.
• Equivalent carbon positions related by molecular symmetry produce the same alcohol.
• Mirror-image stereoisomers are not counted separately.

Solution 5

There are 17 distinct alcohol constitutional isomers, not 30.

The student’s calculation assumes that all six carbon atoms in every skeleton are different possible attachment positions.

That is incorrect for two reasons:

1. Some carbon positions are equivalent because of molecular symmetry.
2. A carbon with four existing carbon-carbon bonds has no hydrogen available to replace with \(\ce{-OH}\).

We need to examine each carbon skeleton separately.

Step 1: Examine hexane.

The straight-chain skeleton is:

\[
\ce{CH3CH2CH2CH2CH2CH3}
\]

There are three distinct attachment positions:

• Carbon 1 and carbon 6 are equivalent.
• Carbon 2 and carbon 5 are equivalent.
• Carbon 3 and carbon 4 are equivalent.

Therefore, hexane produces three different alcohols.

Step 2: Examine 2-methylpentane.

Its structure is:

\[
\ce{CH3CH(CH3)CH2CH2CH3}
\]

There are five distinct attachment positions:

• The two methyl groups attached to carbon 2 are equivalent.
• The branching carbon is unique.
• The first \(\ce{CH2}\) position is unique.
• The second \(\ce{CH2}\) position is unique.
• The terminal methyl group at the opposite end is unique.

Therefore, this skeleton produces five different alcohols.

Step 3: Examine 3-methylpentane.

Its structure is:

\[
\ce{CH3CH2CH(CH3)CH2CH3}
\]

The central carbon is attached to a methyl branch and two identical ethyl groups.

This gives four distinct attachment positions:

• The central branching carbon.
• The methyl branch.
• Either of the two equivalent \(\ce{CH2}\) positions.
• Either of the two equivalent terminal methyl positions.

Therefore, this skeleton produces four different alcohols.

Step 4: Examine 2,2-dimethylbutane.

Its structure is:

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

The central carbon is attached to four carbon atoms.

It has no hydrogen atoms, so it cannot accept an additional hydroxyl bond.

The three methyl groups attached to this central carbon are equivalent. The \(\ce{CH2}\) position and the terminal methyl group are each unique.

Therefore, there are three valid hydroxyl attachment positions.

This skeleton produces three different alcohols.

Step 5: Examine 2,3-dimethylbutane.

Its structure is:

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

This molecule has a plane of symmetry through the middle of its carbon skeleton.

The two central \(\ce{CH}\) carbons are equivalent.

All four methyl groups are also equivalent.

There are therefore only two different hydroxyl attachment positions.

This skeleton produces two different alcohols.

Step 6: Add the results.

There is one final check.

Could two alcohols produced from different parent skeletons accidentally be the same molecule?

No. If you remove the hydroxyl group and replace it with hydrogen, you recover the original carbon skeleton. Different parent carbon skeletons therefore cannot produce identical alcohol connectivity.

The correct total is 17 constitutional alcohol isomers.

The important insight is that the number of carbon atoms does not tell you how many distinct hydroxyl positions exist. You must account for both molecular symmetry and the available bonding positions on each carbon.

07What to learn next

Once you can draw different alcohol isomers systematically, the next useful skill is comparing their structures and predicting how branching and hydroxyl position affect their chemical and physical properties.

For more practice with the underlying structural notation, read How to Read Skeletal Formulas and Write Condensed Formulas.

If two skeletal drawings look different but might represent the same molecule, use How to Check if Two Skeletal Formulas Show the Same Molecule.

The distinction to carry forward is simple: build the carbon skeleton, identify genuinely different attachment positions, and check every atom’s valency before accepting a structure.