Why Rotating a Structural Formula Does Not Create a New Isomer
Rotating a structural formula changes its appearance, not the molecule's connectivity. Learn how to compare atom connections, check carbon valence, and distinguish genuine structural isomers from different conformations and geometric isomers.
Turn a structural formula upside down, rotate it by 90°, or redraw it from the opposite end. Have you created a new molecule?
No. Rotating a structural formula does not create a new isomer because it does not change which atoms are bonded to which. It changes the appearance of the drawing, not the molecular structure.
But there is a catch. Two molecules can have the same molecular formula and look almost identical while having genuinely different connections. And some molecules have different spatial arrangements even though their connectivity is unchanged.
The key is to distinguish three things: the orientation of a drawing, the connectivity of atoms, and the three-dimensional arrangement of a molecule.
0101 Why a rotated drawing is still the same molecule
Consider 2-methoxypropane, with the condensed structural formula:
\[ \ce{CH3-O-CH(CH3)2} \]
You could write the same structure from the opposite direction:
\[ \ce{(CH3)2CH-O-CH3} \]
These are two representations of the same molecule.
The oxygen is bonded to the same two carbon atoms in both drawings. The central carbon is still bonded to oxygen, two methyl groups, and one hydrogen.
Nothing has been rearranged.
Now compare those representations with:
\[ \ce{CH3-O-CH2-CH2-CH3} \]
This is 1-methoxypropane.
It has the same molecular formula as 2-methoxypropane:
\[ \ce{C4H10O} \]
However, the oxygen is now connected to a terminal carbon of a three-carbon chain rather than the middle carbon.
That is a genuine structural difference. The compounds are structural isomers.
| Comparison | 2-methoxypropane | 1-methoxypropane |
|---|---|---|
| Molecular formula | \(\ce{C4H10O}\) | \(\ce{C4H10O}\) |
| Oxygen’s carbon neighbours | Methyl carbon and middle carbon | Methyl carbon and terminal carbon |
| Connectivity | Different | Different |
| Same molecule? | No | No |
The important question is not whether the drawings look different. It is whether the atoms have different neighbours.
A structural formula is a little like a railway map. Turning the map upside down does not change which stations are connected by tracks. However, changing the tracks does change the network.
The analogy has a limit: chemical bonds have different types and carbon has specific bonding requirements. We need to check those as well.
If you want to revise the broader categories of chain, position, and functional-group isomerism, see Structural Isomerism in Organic Compounds for HSC Chemistry.
0202 A model investigation: what actually changes when you rotate a molecule?
We can test the difference using a ball-and-stick molecular model.
This is a model investigation, not a chemical reaction. Its purpose is to test whether changing the orientation of a representation changes the connections it describes.
Setting up the model
Build a model of 2-methoxypropane using:
- Four carbon atoms.
- One oxygen atom.
- Ten hydrogen atoms.
- Single-bond connectors.
Place the oxygen between a methyl carbon and the central carbon of the three-carbon group.
The central carbon must also connect to two other methyl carbons and one hydrogen.
Before rotating the model, make a prediction.
If you turn the entire molecule through 180°, should the oxygen become attached to a different carbon?
No. The model turns as one connected object. The oxygen retains exactly the same two carbon neighbours.
The observation
Rotate the entire model.
You will see that:
- The molecule faces a different direction.
- The branches appear in different positions.
- The oxygen may now appear on the opposite side of the model.
- Every atom is still connected to exactly the same neighbours.
The observation supports a simple conclusion: changing the orientation of the molecule does not change its connectivity.
What could make the original test misleading?
Suppose you only compare the outlines of two models.
The same carbon skeleton can look different when viewed from different directions. Also, carbon atoms are identical, so an unlabelled model can make it difficult to track individual connections.
A better method is to mark the atoms temporarily and record which atoms each one is bonded to.
Now build 1-methoxypropane and compare the connections.
The oxygen is bonded to a methyl carbon and a terminal carbon of the propyl chain. Its carbon neighbours are no longer the same as in 2-methoxypropane.
To change one model into the other, you must disconnect and reconnect bonds. Turning the model cannot achieve this.
What the model investigation establishes
The model demonstrates the difference between:
- Rotating a connected structure.
- Reconnecting atoms to create a different structure.
It does not measure bond energies, reaction rates, or the actual geometry of a molecule. Those require additional chemical models and experimental evidence.
It does, however, give us a reliable way to think about structural formulae.
0303 Carbon valence: the constraint that prevents random rearrangement
There is another reason you cannot simply move branches around until a drawing looks different.
Carbon normally forms four covalent bonds in the neutral organic structures considered here.
This is called carbon’s valency of four.
For bond counting:
- A single bond counts as one.
- A double bond counts as two.
- A triple bond counts as three.
The total bond order around an ordinary neutral carbon must equal four.
If some of those bonds are already shown, the remaining bonding capacity is taken up by hydrogen atoms.
We can express this as:
\[ H_{\mathrm C}=4-\sum b \]
where:
- \(H_{\mathrm C}\) is the number of hydrogen atoms bonded to the carbon.
- \(b\) represents the bond order of each bond already attached to that carbon.
For example, consider a carbon attached to three other carbon atoms by single bonds.
\[ H_{\mathrm C}=4-(1+1+1)=1 \]
It has one hydrogen.
Now consider a carbon attached to one carbon by a double bond and two other atoms by single bonds.
\[ H_{\mathrm C}=4-(2+1+1)=0 \]
It has no hydrogen atoms.
This gives us a useful way to check whether a proposed structural formula is chemically possible.
Consider the central carbon in 2-methylpropan-2-ol:
\[ \ce{CH3C(OH)(CH3)CH3} \]
That carbon has:
- One single bond to the first methyl group.
- One single bond to the second methyl group.
- One single bond to the third methyl group.
- One single bond to oxygen.
Its total bond order is four. It has no hydrogen atoms.
You cannot add another branch to that carbon while keeping all four existing bonds. That would give it five single bonds, which is not a valid ordinary neutral carbon structure.
This is why carbon valence matters when checking whether a proposed rearrangement is possible.
One important distinction: passing the valence test does not prove that two structures are the same molecule. It only establishes that the proposed bonding arrangement is chemically valid under the assumptions being used.
0404 Three different operations that students often confuse
The word rotation can describe several different things in chemistry.
They do not all have the same consequences.
| Operation | What changes? | Does it create a new structural isomer? |
|---|---|---|
| Rotate the drawing on the page | Orientation of the representation | No |
| Rotate the whole molecule in space | Orientation of the molecule | No |
| Rotate around a carbon-carbon single bond | Three-dimensional conformation | No |
| Change which atoms are bonded together | Connectivity | Potentially, if the molecular formula stays the same |
| Change the arrangement around a carbon-carbon double bond | Spatial arrangement | Can create geometric stereoisomers |
The third row deserves some attention.
Rotating around a single bond
A carbon-carbon single bond generally allows rotation.
Imagine the bond as a connection between two parts of a molecule. One part can turn relative to the other without breaking the connection.
This changes the molecule’s conformation, meaning its temporary three-dimensional shape.
However, the atoms remain connected in the same order.
For the ordinary structural-isomer comparisons in HSC Chemistry, these different conformations are not counted as different structural isomers.
A double bond behaves differently.
Why a double bond is different
A carbon-carbon double bond contains:
- One sigma (\(\sigma\)) bond.
- One pi (\(\pi\)) bond.
The pi bond forms through sideways overlap of p orbitals. Twisting the atoms around the double bond disrupts this overlap.
Consequently, rotation around a carbon-carbon double bond is strongly restricted under ordinary conditions.
This allows certain alkenes to exist as distinct geometric isomers, such as cis-but-2-ene and trans-but-2-ene.
Notice the distinction:
- Rotating the entire structural formula does not change the geometric isomer.
- Changing the relative arrangement of groups around the double bond can produce a different geometric isomer.
For a more detailed explanation of when this type of isomerism is possible, see When Can an Alkene Have Cis-Trans Isomers? The Structural Test.
0505 Worked examples: applying the distinction
Worked example: Are these two ether formulae the same molecule?
Two students write the following structures:
\[ A:\quad \ce{CH3-O-CH(CH3)2} \]
\[ B:\quad \ce{(CH3)2CH-O-CH3} \]
Determine whether these are two structural isomers or two representations of the same compound.
Step 1
Both structures contain four carbon atoms, ten hydrogen atoms, and one oxygen atom.
Their molecular formula is:
\[ \ce{C4H10O} \]
This means they could be the same compound or structural isomers. We need another check.
Step 2
In structure A, oxygen is bonded to:
- One methyl carbon.
- One carbon bonded to two methyl groups and one hydrogen.
In structure B, oxygen has exactly the same two carbon neighbours.
Step 3
In both structures, the central carbon is bonded to:
\[ \ce{O},\quad \ce{CH3},\quad \ce{CH3},\quad \ce{H} \]
It has four single bonds.
Step 4
Every atom has the same bonding environment in both representations.
Therefore, both formulae represent 2-methoxypropane.
They are not structural isomers. One is simply written from the opposite direction.
Worked example: Does moving the hydroxyl group create a new isomer?
Consider two alcohols:
\[ A:\quad \ce{CH3C(OH)(CH3)CH3} \]
\[ B:\quad \ce{HOCH2CH(CH3)2} \]
A student claims that B is just A after rotating the structural formula.
Determine whether the claim is correct.
Step 1
Both structures contain four carbon atoms, ten hydrogen atoms, and one oxygen atom.
Therefore, both have the molecular formula:
\[ \ce{C4H10O} \]
Step 2
In structure A, the oxygen is bonded to a central carbon.
In structure B, the oxygen is bonded to a terminal \(\ce{CH2}\) carbon.
This is already a warning that the connectivity may have changed.
Step 3
In A, the central carbon has three carbon-carbon single bonds and one carbon-oxygen single bond.
\[ \sum b=1+1+1+1=4 \]
It has no hydrogen atoms.
In B, the central carbon is bonded to two methyl groups, one \(\ce{CH2OH}\) group, and one hydrogen.
Its bond-order total is also four.
The terminal carbon bonded to oxygen has two hydrogen atoms, one carbon-carbon bond, and one carbon-oxygen bond.
It also satisfies carbon’s valency of four.
Both structures are chemically valid.
Step 4
Structure A is 2-methylpropan-2-ol.
Structure B is 2-methylpropan-1-ol.
The hydroxyl group is attached to a different carbon atom. That is a genuine connectivity change.
Conclusion: These are position isomers, not two rotated drawings of the same molecule.
The important lesson is that moving a functional group to another carbon is not the same as rotating the drawing. The actual attachment point has changed.
0606 Questions and solutions
For all questions, assume ordinary neutral organic compounds unless stated otherwise. Count single bonds as one, double bonds as two, and triple bonds as three.
Question 1
Two students draw propan-1-ol.
Student A writes:
\[ \ce{CH3CH2CH2OH} \]
Student B writes:
\[ \ce{HOCH2CH2CH3} \]
Are these different structural isomers or two representations of the same compound?
Explain your reasoning using atom connectivity.
Solution 1
They are two representations of the same compound: propan-1-ol.
Both structures contain a three-carbon chain with a hydroxyl group attached to a terminal carbon.
In Student A’s representation, the formula is written from the methyl end towards the hydroxyl group.
In Student B’s representation, it is written in the opposite direction.
The oxygen is bonded to the same terminal carbon in both cases. The carbon-carbon connections are also identical.
Both have molecular formula:
\[ \ce{C3H8O} \]
No atom has gained or lost a neighbour.
The tempting mistake is to assume that reversing the written order changes the position of the hydroxyl group. It does not. The same molecule can be written from either end.
Question 2
Consider two compounds:
\[ A:\quad \ce{CH3CH2-O-CH2CH3} \]
\[ B:\quad \ce{CH3-O-CH2CH2CH3} \]
Both contain only carbon, hydrogen, and oxygen.
(a) Determine the molecular formula of each compound.
(b) Determine whether they represent the same molecule or structural isomers.
(c) Identify the exact connectivity difference that prevents them from being identical.
Solution 2
The compounds are structural isomers with molecular formula \(\ce{C4H10O}\).
(a) Both structures contain four carbon atoms and one oxygen atom.
Counting the hydrogen atoms gives ten in each structure.
Therefore:
\[ \boxed{\ce{C4H10O}} \]
(b) Both are ethers, but their connectivity is different.
Structure A is ethoxyethane.
Structure B is 1-methoxypropane.
(c) Examine the atoms directly bonded to oxygen.
In A, oxygen is bonded to two \(\ce{CH2}\) carbons. Each of these is connected to a \(\ce{CH3}\) group.
In B, oxygen is bonded to a \(\ce{CH3}\) carbon on one side and a \(\ce{CH2}\) carbon on the other. That second carbon connects to another \(\ce{CH2}\), followed by a terminal \(\ce{CH3}\).
The carbon paths around oxygen are different.
The tempting approach is to see that both structures contain four carbons and one oxygen and conclude that they must be identical. But matching molecular formulae only establishes that they could be isomers. It does not establish identical connectivity.
Question 3
Two students propose the following structures for compounds with molecular formula \(\ce{C4H8}\):
\[ A:\quad \ce{CH2=CHCH2CH3} \]
\[ B:\quad \ce{CH3CH=CHCH3} \]
Student A argues that B is simply A after rotating the structural formula and numbering the chain from the opposite end.
(a) Is this argument correct?
(b) Calculate the bond-order total and implied hydrogen count for each carbon atom in the carbon-carbon double bond of both structures.
(c) Identify the relationship between the compounds.
Solution 3
The argument is incorrect. These are position isomers: but-1-ene and but-2-ene.
(a) Rotating the drawing cannot change which pair of carbon atoms is joined by the double bond.
In A, the double bond is between carbon 1 and carbon 2.
In B, it is between carbon 2 and carbon 3.
Reversing the direction of the chain does not remove this difference.
(b) Calculate the bond-order totals.
For A, \(\ce{CH2=CHCH2CH3}\):
The terminal alkene carbon has one double bond and two carbon-hydrogen single bonds.
\[ \sum b=2+1+1=4 \]
Therefore:
\[ H_{\mathrm C}=4-4=0 \]
The two hydrogen atoms are already included in the displayed \(\ce{CH2}\) group.
The second alkene carbon has one double bond, one carbon-carbon single bond, and one carbon-hydrogen single bond.
\[ \sum b=2+1+1=4 \]
It has one hydrogen.
For B, \(\ce{CH3CH=CHCH3}\):
Each alkene carbon has:
- One carbon-carbon double bond.
- One carbon-carbon single bond.
- One carbon-hydrogen single bond.
Therefore:
\[ \sum b=2+1+1=4 \]
Each alkene carbon has one hydrogen.
All the carbon atoms satisfy their normal valency.
(c) The compounds are position isomers because the position of the double bond changes while the carbon skeleton and molecular formula remain the same.
The important trap is assuming that changing the direction of a drawing can move a double bond to another carbon-carbon connection. It cannot.
Question 4
A reference drawing shows trans-but-2-ene.
Its carbon-carbon double bond is horizontal:
- The methyl group attached to the left alkene carbon is above the double bond.
- The methyl group attached to the right alkene carbon is below the double bond.
Two alternative drawings are proposed.
- Drawing A has a vertical double bond. Its methyl group on the upper alkene carbon points right, while its methyl group on the lower alkene carbon points left.
- Drawing B also has a vertical double bond, but both methyl groups point right.
For each alternative:
(a) Determine whether it represents the same geometric arrangement as the reference or a different geometric isomer.
(b) Explain why rotating the entire drawing cannot convert cis-but-2-ene into trans-but-2-ene.
(c) Explain the role of the double bond in maintaining the different arrangements.
Solution 4
Drawing A represents the same trans arrangement as the reference, while drawing B represents cis-but-2-ene.
(a) Consider drawing A first.
The entire reference drawing can be rotated through 90° to obtain A.
The double bond changes from horizontal to vertical, but the methyl groups remain on opposite sides of it.
Therefore, A is the same geometric isomer.
Now consider B.
Both methyl groups are on the same side of the vertical double bond.
This is cis-but-2-ene, a different geometric isomer from the reference trans form.
(b) A rigid rotation changes the orientation of the entire drawing, not the relative positions of the groups around the double bond.
If the methyl groups are on opposite sides before rotation, they remain on opposite sides afterwards.
The same applies to groups on the same side.
The distinction depends on their positions relative to the C=C bond, not whether they appear above, below, left, or right on the page.
(c) The carbon-carbon double bond contains a pi bond formed by sideways orbital overlap.
Rotating around the double bond disrupts this overlap, so free rotation is strongly restricted.
Converting cis-but-2-ene into trans-but-2-ene requires changing the spatial arrangement around the double bond, rather than simply turning the whole molecule.
Both geometric isomers have the same connectivity. They differ in spatial arrangement.
This is the important exception to remember: different isomers do not always require different atom-to-atom connections.
Question 5
Consider cis-hex-3-ene:
\[ \ce{CH3CH2CH=CHCH2CH3} \]
Assume the two ethyl groups attached to the alkene carbons are on the same side of the carbon-carbon double bond.
A student makes the following argument:
- They rotate the entire structural drawing through 90°.
- They then rotate the carbon-carbon single bond between the two carbons of the right-hand ethyl group, so that its terminal \(\ce{CH3}\) group appears on the opposite side of the double-bond line in the drawing.
- They conclude that the molecule has become trans-hex-3-ene.
The carbon directly attached to the alkene on the right-hand side remains in its original position throughout the second operation.
Evaluate the student’s argument.
In your response:
(a) Explain what happens to the molecular structure during the 90° rotation.
(b) Explain whether rotating the internal single bond of the ethyl group changes the cis/trans arrangement.
(c) Identify which part of the ethyl group determines its attachment to the alkene carbon.
(d) Explain what must change to produce trans-hex-3-ene, and why an ordinary single-bond rotation cannot achieve it.
Solution 5
The student is incorrect. The molecule remains cis-hex-3-ene throughout both operations.
This question is trickier because the position of an individual atom in a drawing is not necessarily enough to determine the geometric arrangement of an entire substituent.
(a) Rotating the whole drawing through 90° changes only its orientation.
The two ethyl groups remain on the same side of the C=C bond.
The molecule is still cis-hex-3-ene.
(b) Rotating the internal single bond of the right-hand ethyl group changes its conformation.
The terminal methyl group can move to a different position relative to the rest of the molecule.
However, the carbon directly attached to the alkene remains in its original position.
The ethyl substituent is still attached to the same alkene carbon in the same geometric arrangement.
Therefore, the molecule remains cis.
(c) The relevant connection is the bond between the alkene carbon and the first carbon of the ethyl group.
That first carbon is the \(\ce{CH2}\) directly attached to the double-bonded carbon.
Rotating the bond further along the ethyl group changes the position of the terminal methyl group, not the position of the ethyl group’s attachment to the alkene.
This is the subtle mistake in the student’s reasoning. They have tracked the position of the terminal atom rather than the arrangement of the substituent attached to the double bond.
(d) To obtain trans-hex-3-ene, the two ethyl substituents must occupy opposite sides of the C=C bond.
That requires changing the spatial arrangement around the double bond.
The carbon-carbon double bond contains a pi bond, so the two sides cannot simply rotate freely around it.
The pi bonding must be disrupted before the arrangement can change and the double bond can be re-established in the other geometric form.
There is also a useful valence check.
Each alkene carbon has:
\[ \sum b=2+1+1=4 \]
The double bond, the single bond to the ethyl group, and the single bond to hydrogen already account for carbon’s four bonds.
Neither rotating the page nor rotating the terminal single bond changes that bonding arrangement.
The deeper lesson is that the position of an atom in a two-dimensional drawing, the conformation of a substituent, and the geometry around a double bond are three different things. You must track the correct structural feature rather than relying on the apparent shape of the drawing.
When comparing unfamiliar organic structures, work in this order: check the molecular formula, identify the atom-to-atom connections, check carbon valence, and then examine any specified three-dimensional arrangement. That final step is what separates ordinary structural comparisons from questions involving stereoisomerism.