How to Check if Two Skeletal Formulas Show the Same Molecule

Two students draw butan-2-ol. One puts the \(\ce{-OH}\) group on the second carbon from the left; the other puts it on the second carbon from the right. Their sketches look different, but have they drawn different compounds?

No. Both drawings show the same carbon-oxygen connections. One is simply written in the opposite direction.

To check whether two skeletal formulas represent the same molecule, compare which atoms are connected to which, the types of bonds between them, and any explicitly shown spatial arrangement. The shape, orientation, and numbering of a drawing can change without changing the molecule.

The important distinction is between redrawing a molecule and rearranging its bonds.

01The first test: does the connectivity change?

Imagine building a molecule using carbon balls and small connecting rods.

Each ball represents an atom, and each rod represents a covalent bond. You can turn the finished model around, look at it from another direction, or rotate parts of it around single bonds. None of those actions changes which atoms are joined.

But if you disconnect a rod and attach it to a different carbon, you have changed the connectivity.

That is the central idea behind comparing skeletal formulas.

A molecule can be represented as a network:

• Each atom is a point in the network.
• Each bond joins two points.
• Single, double, and triple bonds must remain distinguishable.
• The identity of every atom, including oxygen or other explicitly labelled elements, must be preserved.

Two skeletal formulas have the same connectivity if you can match every atom in the first drawing to an atom of the same element in the second, while preserving all these connections and bond types.

Chemists call molecules with the same molecular formula but different atom-to-atom connections constitutional isomers, also known as structural isomers.

Why matching the outlines is unreliable

Suppose you trace two skeletal drawings onto transparent sheets and try to make them overlap.

You might predict that identical molecules should produce identical outlines. But skeletal drawings are not exact measurements of bond lengths or bond angles. The same molecular connections can be drawn with different zigzag angles, branch directions, and orientations.

The first test is therefore incomplete: it compares the artwork rather than the chemical structure.

A better test is to build the two structures separately, identify corresponding atoms, and compare their connections.

For example, consider:

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

and

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

In the first representation, the methyl branch is on carbon 2 when counting from the left. In the second, it appears on carbon 3.

Reverse the direction of the second chain, however, and the connections match exactly. Both represent 2-methylbutane.

This model test is a way to reason about structural drawings. It is not a laboratory test identifying the contents of a sample. Two drawings can be equivalent even when neither has been experimentally verified.

There is also a limitation: simple flexible models do not capture every restriction on three-dimensional arrangement. We will return to that when considering double bonds and stereochemistry.

02A reliable method for comparing skeletal formulas

Instead of trying to recognise the entire shape at once, work through four checks.

Step 1: Check the atoms and molecular formula

Count the carbon atoms, explicitly labelled elements, and implied hydrogen atoms.

Every unlabelled endpoint or corner of a skeletal formula normally represents a carbon atom. Hydrogens attached to carbon are usually omitted.

If the molecular formulas are different, the structures cannot represent the same molecule.

But matching formulas prove much less than students sometimes assume.

For example:

\[
\ce{CH3CH2CH2OH}
\]

and

\[
\ce{CH3CHOHCH3}
\]

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

The first is propan-1-ol, while the second is propan-2-ol. The hydroxyl group is attached to a different position in the carbon network.

They are different structural isomers.

A matching molecular formula is a useful first filter, not a final identity test.

Step 2: Find an atom or bond that is easy to recognise

Start with a distinctive feature of the structure:

• An oxygen, nitrogen, or halogen atom.
• A carbon-carbon double or triple bond.
• A carbonyl group, \(\ce{C=O}\).
• A ring or a carbon bonded to three or four other carbons.

Use that feature as an anchor.

For an alcohol, trace the carbon attached directly to \(\ce{-OH}\). For an alkene, identify the two carbons joined by the double bond and examine what is attached to each.

If the anchor has different neighbours in the two structures, the connectivity is different.

Step 3: Follow every path away from the anchor

This is where many difficult questions are decided.

A branch pointing upwards in one drawing and downwards in another may represent exactly the same connection. What matters is the length and composition of the path.

For a branched carbon, compare what is attached to it:

• Is one branch a methyl group, \(\ce{-CH3}\), or an ethyl group, \(\ce{-CH2CH3}\)?
• Does a path reach oxygen after one carbon or after two?
• Is the next carbon part of a double bond?
• Does a path close back on itself to form a ring?

For a chain, try tracing from the opposite end. For a ring, try following the ring in either direction.

A single matching section is not enough. Every connection must match under one consistent correspondence.

Step 4: Use carbon valence to reject impossible connections

A neutral carbon atom in the ordinary organic structures considered here forms four bonds in total, counting a double bond twice and a triple bond three times.

The number of implied hydrogens on a carbon can be calculated using:

\[
H_{\mathrm C}=4-\sum b
\]

where \(H_{\mathrm C}\) is the number of hydrogens attached to that carbon, and \(\sum b\) is the sum of the bond orders already shown around it. These are atom and bond counts, so they have no physical units.

Consider a carbon already joined to three other carbons by single bonds.

Its bond-order sum is:

\[
H_{\mathrm C}=4-(1+1+1)=1
\]

It has one implied hydrogen.

Now consider a carbon joined to one carbon by a double bond and to two other carbons by single bonds:

\[
H_{\mathrm C}=4-(2+1+1)=0
\]

It has no implied hydrogens.

This matters because you cannot freely move branches or functional groups without checking whether the new arrangement still satisfies carbon’s valence.

If the displayed bond-order sum exceeds four for an ordinary neutral carbon, the proposed drawing is not a valid structure of that kind. It is not simply another isomer.

For a refresher on identifying hidden carbon and hydrogen atoms before comparing structures, see How to Read Skeletal Formulas and Write Condensed Formulas.

05Connectivity is not the whole story: spatial arrangement

There is one important limitation to the method so far.

Two molecules can have identical atom-to-atom connections yet differ in their three-dimensional arrangement.

Consider but-2-ene:

\[
\ce{CH3CH=CHCH3}
\]

A carbon-carbon single bond generally allows rotation. A carbon-carbon double bond restricts rotation because rotating one side relative to the other would disrupt the bonding that forms the double bond.

Predict what happens if the two methyl groups attached to the double-bonded carbons are placed on opposite sides rather than the same side.

The connectivity remains unchanged, but the spatial arrangement is different.

These are the geometric isomers cis-but-2-ene and trans-but-2-ene. They are distinct stereoisomers.

A similar extra check is needed when wedge and dashed bonds indicate three-dimensional arrangements around a tetrahedral carbon. A reflected drawing can represent a different enantiomer rather than just another viewpoint.

For the structural comparisons in this article, we assume no stereochemistry is specified unless the question explicitly says otherwise.

So the complete decision rule is:

This is why a drawing that appears to be a mirror image needs more care when stereochemical information is present.

06Questions and solutions

The first question tests straightforward chain reversal. The later questions require you to check functional groups, ring closures, carbon valence, and molecular symmetry together.

For Questions 1-4, assume ordinary neutral organic compounds and no specified stereochemistry.

Question 1

Two students draw branched alkanes.

• A shows a four-carbon chain with a methyl group attached to carbon 2, counting from the left.
• B shows a four-carbon chain with a methyl group attached to carbon 3, counting from the left.

Both contain only single bonds.

Are the two structures the same molecule? Give their systematic name and molecular formula.

Solution 1

They represent the same molecule: 2-methylbutane, with molecular formula \(\ce{C5H12}\).

Writing their condensed structures gives:

A:

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

B:

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

Reverse the direction in which B is read. Its branch now appears at carbon 2, and every carbon has the same neighbours as in A.

The structure contains five carbons. Its hydrogen count is:

\[
\begin{aligned}
N_{\mathrm H}
&=3+1+3+2+3\\
&=12
\end{aligned}
\]

That is 12 hydrogen atoms per molecule, giving \(\ce{C5H12}\).

A common mistake is to treat left-to-right numbering as a permanent property of the molecule. It is only a drawing convention.

Question 2

Three proposed skeletal formulas are represented by the following condensed structures.

A:

\[
\ce{CH3CH2C(=O)OH}
\]

B:

\[
\ce{CH3C(=O)OCH3}
\]

C:

\[
\ce{HOOCCH2CH3}
\]

All three contain three carbon atoms and two oxygen atoms.

Determine:

(a) Which structures represent the same molecule?

(b) Which structures are constitutional isomers?

(c) The molecular formula shared by all three structures.

(d) The specific difference in oxygen connectivity that prevents all three from being identical.

Solution 2

A and C are identical representations of propanoic acid. B is methyl ethanoate, a different constitutional isomer. All three have molecular formula \(\ce{C3H6O2}\).

(a) Identical structures

A and C have the same connections.

In both, a carbonyl carbon is attached to:

• An ethyl group, \(\ce{-CH2CH3}\).
• An oxygen by a double bond.
• A hydroxyl oxygen by a single bond.

Reversing the written order does not change these bonds.

(b) Constitutional isomers

B is different from A and C.

B is an ester with a \(\ce{-C(=O)-O-C-}\) connection. Its singly bonded oxygen connects the carbonyl carbon to a methyl carbon.

A and C are carboxylic acids with a \(\ce{-C(=O)-OH}\) group.

Changing A into B requires changing actual atom-to-atom connections. It cannot be achieved by rotating a skeletal formula.

(c) Molecular formula

For propanoic acid:

\[
N_{\mathrm H}=3+2+1=6
\]

For methyl ethanoate:

\[
N_{\mathrm H}=3+3=6
\]

Both contain three carbon atoms, six hydrogen atoms, and two oxygen atoms per molecule:

\[
\ce{C3H6O2}
\]

(d) The decisive connection

In A and C, the singly bonded oxygen is directly attached to hydrogen.

In B, that oxygen is directly attached to a methyl carbon instead.

The trap is treating the two oxygen atoms as interchangeable decorations around the carbonyl group. Their connections define different functional groups and different compounds.

Question 3

Two students propose that their skeletal diagrams represent the same hydrocarbon.

• Structure A contains a five-membered carbon ring with one methyl group attached to a ring carbon.
• Structure B contains a four-membered carbon ring with an ethyl group attached to a ring carbon.

All bonds are single bonds.

One student argues that the molecules must be identical because each structure has six carbon atoms, one ring, one branching carbon, and the same numbers of carbons connected to one, two, or three other carbons.

Evaluate this claim.

In your response:

(a) Determine the molecular formula of each structure.

(b) Explain why the carbon-neighbour counts agree.

(c) Identify a structural feature that proves the two molecules cannot be made identical by rotating, reflecting, or redrawing them.

Solution 3

The molecules are different constitutional isomers, even though their molecular formulas and carbon-neighbour counts agree. Both have formula \(\ce{C6H12}\), but their ring sizes are different.

(a) Molecular formulas

For A, methylcyclopentane, the branching ring carbon is bonded to three carbons and has one hydrogen.

Four other ring carbons are \(\ce{CH2}\), and the methyl branch is \(\ce{CH3}\).

Its hydrogen count is:

\[
\begin{aligned}
N_{\mathrm H}
&=1+4(2)+3\\
&=12
\end{aligned}
\]

So its formula is \(\ce{C6H12}\).

For B, ethylcyclobutane, the ring carbon carrying the branch has one hydrogen. Three remaining ring carbons are \(\ce{CH2}\).

The ethyl branch adds one \(\ce{CH2}\) and one \(\ce{CH3}\).

Its hydrogen count is also:

\[
N_{\mathrm H}=1+4(2)+3=12
\]

Therefore B also has formula \(\ce{C6H12}\).

(b) Carbon-neighbour counts

Both structures have:

• One carbon connected to three other carbons.
• Four carbons connected to two other carbons.
• One carbon connected to one other carbon.

That explains why their hydrogen counts agree.

However, this information does not specify exactly which carbons are connected to one another.

(c) The decisive structural feature

A contains a ring of five carbon atoms.

B contains a ring of four carbon atoms.

Ring size is preserved when a molecular drawing is rotated, reflected, or rearranged without changing bonds.

To turn A into B, a carbon would have to be removed from the ring and incorporated into the side chain. That requires changing the carbon-carbon connections.

The tempting argument checks local properties around individual carbons. The missing constraint is the complete path that closes the ring.

Matching each carbon’s number of neighbours is necessary for identity, but it is not sufficient.

Question 4

Three students propose structures for a six-carbon alkene.

Every drawing contains six carbon positions and a terminal double-bonded carbon, \(\ce{=CH2}\). There are no rings or heteroatoms.

Let X be the other carbon of the double bond.

The structures have the following connections:

• A: X is double-bonded to \(\ce{CH2}\) and single-bonded to two separate methyl groups and one ethyl group.
• B: X is double-bonded to \(\ce{CH2}\) and single-bonded to two separate ethyl groups.
• C: X is double-bonded to \(\ce{CH2}\) and single-bonded to one methyl group and one propyl group, \(\ce{-CH2CH2CH3}\).

A student claims that all three are structural isomers of \(\ce{C6H12}\) because each has six carbons and one double bond.

(a) Determine which drawings represent valid ordinary neutral carbon structures.

(b) Calculate the molecular formulas of the valid structures.

(c) Decide whether B and C represent the same molecule.

Solution 4

A is not a valid ordinary neutral-carbon structure. B and C are valid, but they represent different constitutional isomers, both with molecular formula \(\ce{C6H12}\).

(a) Carbon-valence check

In A, carbon X has:

• One double bond, contributing 2 to its bond-order total.
• Three single bonds, each contributing 1.

Therefore:

\[
\sum b=2+1+1+1=5
\]

That exceeds the permitted bond-order total of four for an ordinary neutral carbon.

The drawing is invalid as specified. Counting six carbon positions does not repair a carbon-valence violation.

In B and C, X has one double bond and two single bonds:

\[
\sum b=2+1+1=4
\]

Both structures satisfy carbon valence.

(b) Molecular formulas

B can be written as:

\[
\ce{CH2=C(CH2CH3)CH2CH3}
\]

The terminal alkene carbon has two hydrogens. X has none. Each ethyl group contributes five hydrogens.

Therefore:

\[
\begin{aligned}
N_{\mathrm H}
&=2+0+5+5\\
&=12
\end{aligned}
\]

B has formula \(\ce{C6H12}\).

C can be written as:

\[
\ce{CH2=C(CH3)CH2CH2CH3}
\]

Its hydrogen count is:

\[
\begin{aligned}
N_{\mathrm H}
&=2+0+3+2+2+3\\
&=12
\end{aligned}
\]

C also has formula \(\ce{C6H12}\).

(c) Comparing B and C

B and C are different.

Anchor the comparison at the central alkene carbon X.

In B, X is attached to two ethyl branches. Each branch contains two carbon atoms beyond X.

In C, X is attached to one methyl branch and one propyl branch. Their lengths are one and three carbon atoms.

Rotating or reversing the drawings cannot turn two branches of length two into branches of lengths one and three.

The student made two separate errors: assuming a valid structure from a carbon count alone, and assuming molecular formula equality proves identical connectivity.

The valence check rules out A. The branch-path comparison separates B and C.

Question 5

Three skeletal formulas, A, B, and C, each contain a six-membered carbon ring.

In every drawing:

• The ring carbons are temporarily labelled 1 to 6 in order around the ring, with bonds 1-2, 2-3, 3-4, 4-5, 5-6, and 6-1.
• Carbon 1 is a carbonyl carbon, double-bonded to an oxygen atom outside the ring.
• Carbon 4 is attached to a methyl group.
• One ring carbon is attached to a hydroxyl group, \(\ce{-OH}\).
• Exactly one carbon-carbon bond in the ring is a double bond.
• All other ring bonds are single bonds.

The three drawings differ as follows:

The position numbers are drawing labels, not permanent identities of the atoms. No stereochemistry is specified.

(a) Determine which drawings represent the same constitutional structure.

(b) Show that all three structures have molecular formula \(\ce{C7H10O2}\), despite their differences.

(c) A student argues that A and C must represent the same molecule because the double bond has simply moved to the opposite side of the ring and the hydroxyl group is still on carbon 5. Explain the error.

(d) Now replace the hydroxyl group in A and C with a hydrogen atom at the same ring position. Do these two new structures represent the same molecule? Justify your conclusion.

Solution 5

A and B represent the same constitutional structure. C is a different structural isomer. All three have formula \(\ce{C7H10O2}\), but removing the hydroxyl substituent and replacing it with hydrogen makes the resulting structures from A and C identical.

The key is that a reflection must preserve every connection simultaneously, not just one selected feature.

(a) Identify the identical structures

Begin with carbon 1.

It is the only ring carbon attached to a carbonyl oxygen, so any matching arrangement must map carbon 1 in one structure to carbon 1 in the other.

There are two ways to trace the ring from this fixed carbon:

• Continue around the ring in the original direction.
• Reverse direction, interchanging positions 2 and 6, positions 3 and 5, while keeping positions 1 and 4 fixed.

Apply the second mapping to A:

Now examine the important features.

A’s double bond 2=3 becomes 6=5 in B.

A’s methyl group on carbon 4 remains on carbon 4.

A’s hydroxyl group on carbon 5 becomes a hydroxyl group on carbon 3.

Every atom, attachment, and bond order now matches.

Therefore A and B are identical constitutional structures.

(b) Verify the molecular formulas

Each structure contains:

• Six ring carbon atoms.
• One methyl carbon.
• One carbonyl oxygen.
• One hydroxyl oxygen.

This gives seven carbon atoms and two oxygen atoms.

The hydrogen counts require more care.

For A, count hydrogens around ring carbons 1 through 6, followed by the methyl group and the hydroxyl group:

\[
\begin{aligned}
N_{\mathrm H}
&=0+1+1+1+1+2\\
&\quad+3+1\\
&=10
\end{aligned}
\]

In A, the hydroxyl-bearing carbon 5 has one carbon-bound hydrogen.

For C, the corresponding count is:

\[
\begin{aligned}
N_{\mathrm H}
&=0+2+2+1+0+1\\
&\quad+3+1\\
&=10
\end{aligned}
\]

In C, carbon 5 has no carbon-bound hydrogen because its double bond, ring single bond, and C-O single bond already use all four bond orders.

B has the same hydrogen count as A because its connections are identical under the demonstrated mapping.

All three therefore have ten hydrogen atoms per molecule:

\[
\ce{C7H10O2}
\]

This confirms that C is a possible structural isomer, not an invalid drawing or a structure with a different molecular formula.

(c) Explain why A and C are different

The student’s argument considers the double bond separately from the hydroxyl group.

That is the error.

In A, the hydroxyl group is attached to a saturated ring carbon, carbon 5, which has two single bonds to neighbouring ring carbons.

In C, the hydroxyl group is attached to carbon 5, which is itself part of the ring’s carbon-carbon double bond.

These carbon environments are different.

A reflection taking A’s double bond 2=3 to C’s double bond 5=6 would also move the hydroxyl group from position 5 to position 3.

But C’s hydroxyl group remains at position 5.

There is no single reversal of the ring that preserves the carbonyl group, the methyl group, the double bond, and the hydroxyl position simultaneously.

Consequently, A and C are different constitutional isomers.

(d) Replace the hydroxyl groups with hydrogen

The two new structures are identical.

After replacing each \(\ce{-OH}\) substituent with \(\ce{-H}\), the structures retain the same carbonyl group at carbon 1 and methyl group at carbon 4.

The only apparent difference is the ring double bond:

• Modified A has a double bond between positions 2 and 3.
• Modified C has a double bond between positions 5 and 6.

Reversing the direction around the ring exchanges positions 2 and 6, and positions 3 and 5.

This maps one double bond to the other while preserving the carbonyl and methyl attachments.

Nothing else remains to prevent the correspondence.

This is the deeper symmetry point.

A substituent can destroy a symmetry that the rest of the molecule possesses. Removing that substituent can restore the symmetry, making two previously distinct connectivity patterns equivalent after the modification.

The fair test is never whether individual features appear in plausible matching positions. It is whether one complete atom-to-atom mapping preserves every bond and attachment at once.

07What to do after the drawings match

When checking two skeletal formulas, use the molecular formula to reject obvious mismatches, carbon valence to reject impossible structures, and connectivity to establish whether the drawings describe the same constitutional structure. Then check stereochemistry if the question specifies it.

The next useful question is different: if two valid structures are constitutional isomers, how could you determine which one is actually present in a sample?

That is where experimental evidence becomes important. Infrared spectroscopy can provide information about functional groups, while NMR spectroscopy can reveal differences between atomic environments. These methods help distinguish structures that a molecular formula alone cannot separate.

A skeletal drawing tells you what structure is being proposed. Checking its connections tells you whether it is a new proposal or merely a different way of drawing the same one.