How to Check Atom Conservation in Multi-Step Organic Synthesis

An organic synthesis can produce the correct functional group and still fail a basic atom check. Suppose you want to make ethyl propanoate, \(\ce{CH3CH2COOCH2CH3}\). The target molecule contains five carbon atoms, but your starting material, 1-bromopropane, contains only three.

Where do the other two carbon atoms come from?

Before looking at the reaction pathway, predict whether they come from the oxidising agent, the acid catalyst, or another organic reactant. They come from ethanol, which supplies the ethyl group during esterification. Neither sulfuric acid nor the oxidising agent supplies those two carbons.

This is the key to checking atom conservation across a multi-step organic synthesis: trace every carbon-containing fragment back to a starting material, check the other atoms at each reaction step, and make sure every transformation is chemically possible.

01Start with the target molecule, not the first reaction

When checking a multi-step synthesis, it is tempting to follow the reaction arrows from left to right and count atoms as you go.

A better approach is to start at the final product.

Write its structural formula, identify its functional group, and count its carbon atoms. Then work backwards to identify the compounds that could supply those atoms.

This is particularly useful for esters and amides because their formation involves joining two organic molecules.

Example: Where do the carbons in an ester come from?

Consider ethyl propanoate:

\[ \ce{CH3CH2COOCH2CH3} \]

An ester contains two recognisable sections:

• The propanoate section, \(\ce{CH3CH2COO-}\), contains three carbon atoms.
• The ethyl section, \(\ce{-CH2CH3}\), contains two carbon atoms.

The ester is formed from propanoic acid and ethanol:

\[ \ce{CH3CH2COOH + CH3CH2OH} \]

The two molecules combine, releasing water:

\[ \ce{CH3CH2COOH + CH3CH2OH} \]

\[ \ce{<=> CH3CH2COOCH2CH3 + H2O} \]

The carbon count is therefore:

\[ 3+2=5 \]

No carbon atoms have appeared during the reaction. The five atoms were already present in the two reactants.

This gives you an important rule:

When two organic molecules combine, count the carbon atoms in both reactants, not just the molecule you consider the main starting material.

The same reasoning applies when an amine and a carboxylic acid form an amide. Both organic reactants may contribute carbon atoms to the final product.

02Build a carbon ledger for every reaction step

A carbon ledger is a simple table that records how many carbon atoms are present in each organic compound and where they came from.

It is particularly useful when a pathway contains several intermediates.

Consider the following transformations:

Notice that the first two reactions retain the three-carbon skeleton.

The final reaction is different. It joins the three-carbon acid fragment to the two-carbon alcohol fragment.

What should you check for each type of reaction?

The carbon count is only one part of the audit. You also need to identify what happens to the other atoms.

This table helps you identify which atoms should remain in the organic product and which should appear in other products.

For example, when 1-bromopropane reacts with aqueous sodium hydroxide, the bromine atom does not disappear. It leaves as part of a salt.

\[ \ce{CH3CH2CH2Br + NaOH -> CH3CH2CH2OH + NaBr} \]

The oxygen atom in the alcohol comes from hydroxide. The bromine atom ends up in sodium bromide.

A correct atom audit accounts for both.

Distinguish reagents, reactants, and catalysts

This is another common source of mistakes.

A reagent can supply atoms to a product, but not every chemical written above an arrow becomes part of the product.

For example:

• Aqueous \(\ce{NaOH}\) supplies the hydroxide group during substitution.
• Acidified \(\ce{K2Cr2O7}\) provides oxidising equivalents during oxidation.
• Concentrated \(\ce{H2SO4}\) acts as an acid catalyst during esterification.
• Heat supplies energy, not atoms.

Ethanol in an esterification reaction is a reactant. Its carbon atoms become part of the ester.

Sulfuric acid is a catalyst. It is not the source of the ester’s carbon atoms.

Always identify the actual role of each substance before deciding where its atoms belong.

05A balanced equation does not automatically mean a valid pathway

Here is a misconception that can cost marks in an HSC explanation.

Suppose a student proposes oxidising propan-1-ol directly into propanone.

They write:

\[ \ce{CH3CH2CH2OH + [O] -> CH3COCH3 + H2O} \]

Count the atoms.

The equation is atom-balanced.

So is the reaction valid?

No.

Propan-1-ol is a primary alcohol. Its standard oxidation pathway produces propanal, followed by propanoic acid if oxidation continues.

Propanone is formed by oxidising a secondary alcohol, such as propan-2-ol.

The tempting mistake is to look at the molecular formula and assume that any compound with the correct number of atoms is an acceptable product.

But propanal and propanone are structural isomers. Both have the molecular formula \(\ce{C3H6O}\), yet their functional groups occupy different positions.

The proposed equation passes an atom-count check but fails the chemical reaction check.

This gives you two separate tests:

1. Atom conservation: Are all atoms accounted for?
2. Chemical feasibility: Can the stated reactants actually produce the proposed products under the given conditions?

You need both.

06Experimental check: How could you trace an individual carbon atom?

So far, we have checked atom conservation using structural formulas and reaction equations.

But how could a chemist experimentally test whether a particular carbon atom ends up in the predicted position in a final product?

One approach is isotope labelling.

Carbon has different isotopes. Carbon-13, written \({}^{13}\mathrm{C}\), is a stable isotope of carbon with one more neutron than the more common carbon-12 isotope.

It behaves chemically like carbon-12 but can be distinguished using analytical instruments.

Designing a tracer experiment

Imagine an experiment using 1-bromopropane in which the carbon attached to bromine is labelled with carbon-13.

This starting compound can be written as:

\[ \mathrm{CH_3CH_2{}^{13}CH_2Br} \]

The proposed pathway is:

1. Convert the labelled 1-bromopropane into labelled propan-1-ol using aqueous sodium hydroxide.
2. Oxidise the labelled alcohol to labelled propanoic acid.
3. React the acid with ordinary ethanol to produce ethyl propanoate.
4. Analyse the purified final product.

At each stage, a sample could be collected to examine the labelled carbon.

The prediction is that the labelled carbon will remain in the same carbon position within the original three-carbon chain.

During substitution:

\[ \ce{CH3CH2CH2Br -> CH3CH2CH2OH} \]

The labelled carbon remains attached to the hydroxyl group.

During oxidation, that carbon becomes the carboxyl carbon:

\[ \ce{CH3CH2CH2OH -> CH3CH2COOH} \]

Finally, during esterification, it becomes the carbonyl carbon of ethyl propanoate.

\[ \mathrm{CH_3CH_2{}^{13}COOCH_2CH_3} \]

The labelled carbon has changed its bonding environment, but it is still the same carbon atom.

Why one measurement may not be enough

A first approach might be to use mass spectrometry.

If the final product contains the labelled carbon, its molecular mass will be one mass unit greater than the corresponding unlabelled molecule.

An enhanced \(M+1\) signal would support the conclusion that the labelled carbon is present in the final product.

However, there is a limitation.

Mass spectrometry identifies the mass of the molecule or its ions. It does not, by itself, tell you which of the five carbon positions contains the labelled atom.

A revised approach is to use carbon-13 nuclear magnetic resonance spectroscopy (\({}^{13}\mathrm{C}\) NMR).

In an NMR instrument, the sample is placed in a strong magnetic field. A radiofrequency pulse excites the carbon nuclei, and a detector records their resonance signals.

Different carbon environments produce signals at different chemical shifts.

The predicted observation is an enriched signal corresponding to the ester carbonyl carbon.

This provides evidence about the position of the labelled atom, rather than simply confirming that it exists somewhere in the molecule.

There is still a limitation: analysing only the final ester does not prove the exact sequence of every intermediate reaction. Samples of the intermediate compounds would need to be analysed as well.

This is an illustrative tracer investigation, not a claim that isotope labelling is a required HSC practical. The important distinction is between checking the total number of atoms and experimentally investigating where individual atoms end up.

07A practical method for checking any multi-step synthesis

When you encounter an unfamiliar synthesis question, use this sequence.

1. Write the target’s structural formula.

Do not rely only on its name. Draw the molecule and count its carbon atoms.

2. Identify the functional group.

Determine what reaction could produce it. For an ester, identify the carboxylic acid and alcohol fragments.

3. Trace each carbon-containing fragment backwards.

Ask which starting material supplies each part of the final molecule.

4. Check the reagents.

Identify which reagents supply atoms, which act as oxidising or reducing agents, and which are catalysts or conditions.

5. Audit every reaction separately.

Count the atoms in all reactants and products. Include water, salts, and other by-products.

6. Add the equations together.

Cancel intermediates that are produced and subsequently consumed. Check that the overall equation accounts for every atom.

7. Check that the reactions are chemically possible.

A balanced equation is not enough. Confirm that the functional groups and reaction conditions support each transformation.

If the carbon count does not work, look for a missing carbon-containing reactant or an unaccounted-for carbon-containing product.

If the atom count works but the reaction is chemically impossible, the pathway still needs correcting.

For a refresher on balancing individual chemical equations, see Conservation of Atoms in Chemical Reactions: HSC Chemistry Guide.

For a broader explanation of working backwards from functional groups and constructing reaction pathways, see How to Construct Organic Synthesis Pathways for HSC Chemistry.

08Questions and solutions

Question 1

1-bromobutane, \(\ce{CH3CH2CH2CH2Br}\), is converted into butan-1-ol using aqueous sodium hydroxide. The alcohol is then oxidised using excess acidified potassium dichromate to produce butanoic acid.

Determine the number of carbon atoms in each organic compound. Identify the source of the oxygen atom introduced during substitution and the additional oxygen required during oxidation.

Solution 1

Each organic compound contains four carbon atoms. No carbon atoms are added or removed during these reactions.

The starting compound is:

\[ \ce{CH3CH2CH2CH2Br} \]

It contains four carbon atoms.

During substitution:

\[ \ce{CH3CH2CH2CH2Br + NaOH} \]

\[ \ce{-> CH3CH2CH2CH2OH + NaBr} \]

The bromine is replaced by an OH group. The oxygen in that hydroxyl group comes from aqueous sodium hydroxide.

The resulting butan-1-ol still contains four carbon atoms.

During oxidation, the primary alcohol forms butanal and then butanoic acid:

\[ \ce{CH3CH2CH2CH2OH + 2[O]} \]

\[ \ce{-> CH3CH2CH2COOH + H2O} \]

The additional oxygen required for oxidation comes from the oxidising agent, acidified potassium dichromate.

The carbon count is four throughout:

• 1-bromobutane: 4 carbons.
• Butan-1-ol: 4 carbons.
• Butanal: 4 carbons.
• Butanoic acid: 4 carbons.

The reaction changes the functional group and the hydrogen-to-oxygen ratio, not the number of carbon atoms.

Question 2

A student wants to synthesise propyl ethanoate, \(\ce{CH3COOCH2CH2CH3}\), using 1-bromopropane as the only organic starting material.

The available reagents are:

• Aqueous sodium hydroxide.
• Acidified potassium dichromate.
• Concentrated sulfuric acid.
• Water and heat.

Assume only standard HSC organic reaction pathways are available, with no carbon-carbon bond cleavage or other unlisted carbon-skeleton transformations.

Determine whether this synthesis is possible. Identify the carbon-containing fragment that is missing and suggest an additional organic starting material that would make the synthesis possible.

Solution 2

The proposed synthesis is not possible using only the supplied starting material and reagents. The target contains five carbon atoms, but 1-bromopropane supplies only three.

First, identify the two fragments needed to form propyl ethanoate.

The ester is:

\[ \ce{CH3COOCH2CH2CH3} \]

It can be formed from:

• Ethanoic acid, \(\ce{CH3COOH}\), containing two carbon atoms.
• Propan-1-ol, \(\ce{CH3CH2CH2OH}\), containing three carbon atoms.

The required carbon count is:

\[ 2+3=5 \]

1-bromopropane can be converted into propan-1-ol through substitution:

\[ \ce{CH3CH2CH2Br + NaOH} \]

\[ \ce{-> CH3CH2CH2OH + NaBr} \]

This supplies the three-carbon propyl fragment.

However, there is no two-carbon organic starting material available to produce ethanoic acid.

Sodium hydroxide, potassium dichromate, sulfuric acid, and water contain no carbon atoms. Heat cannot create carbon atoms either.

An additional two-carbon starting material, such as ethanol, would solve the problem. Ethanol could be oxidised to ethanoic acid:

\[ \ce{CH3CH2OH + 2[O] -> CH3COOH + H2O} \]

The ethanoic acid could then react with propan-1-ol to form propyl ethanoate.

The important lesson is that identifying the correct functional-group reactions is not enough. You must also have a source for every carbon-containing fragment in the target.

Question 3

A chemist prepares propanoic acid by quantitatively oxidising 1-bromopropane through propan-1-ol.

The resulting propanoic acid is mixed with ethanol and undergoes esterification.

The initial quantities are:

• 0.18 mol of 1-bromopropane.
• 0.12 mol of ethanol.

After the reaction, the mixture contains 0.10 mol of ethyl propanoate.

Assume that:

• Conversion of 1-bromopropane to propanoic acid is quantitative.
• No side reactions occur.
• The only carbon-containing substances in the final mixture are ethyl propanoate, propanoic acid, and ethanol.
• The stated product amount is the amount present in the reaction mixture, before separation.

Calculate:

a) The amount of propanoic acid remaining.

b) The amount of ethanol remaining.

c) The total amount of carbon atoms, in moles, before and after esterification.

Solution 3

The mixture contains 0.08 mol of propanoic acid and 0.02 mol of ethanol after esterification. The total carbon inventory is 0.78 mol of carbon atoms both before and after the reaction.

The key is the 1:1 mole ratio in esterification.

The reaction is:

\[ \ce{CH3CH2COOH + CH3CH2OH} \]

\[ \ce{<=> CH3CH2COOCH2CH3 + H2O} \]

One mole of propanoic acid reacts with one mole of ethanol to produce one mole of ester.

Part a: Calculate the remaining propanoic acid.

Initially, there are 0.18 mol of propanoic acid because oxidation is quantitative.

The reaction produces 0.10 mol of ester, so it consumes 0.10 mol of propanoic acid.

\[ n_{\text{remaining acid}} =0.18-0.10 \]

\[ =0.08\text{ mol} \]

Therefore, 0.08 mol of propanoic acid remains.

Part b: Calculate the remaining ethanol.

The initial amount of ethanol is 0.12 mol.

The esterification reaction consumes 0.10 mol of ethanol.

\[ n_{\text{remaining ethanol}} =0.12-0.10 \]

\[ =0.02\text{ mol} \]

Therefore, 0.02 mol of ethanol remains.

Part c: Check the total carbon inventory.

Each mole of propanoic acid contains three moles of carbon atoms.

Each mole of ethanol contains two moles of carbon atoms.

Before esterification:

\[ n(\mathrm C) =(0.18\times3)+(0.12\times2) \]

\[ =0.54+0.24 \]

\[ =0.78\text{ mol of C atoms} \]

Now calculate the carbon atoms after esterification.

Ethyl propanoate contains five carbon atoms per molecule. The remaining propanoic acid contains three, and the remaining ethanol contains two.

\[ n(\mathrm C) =(0.10\times5) +(0.08\times3) +(0.02\times2) \]

\[ =0.50+0.24+0.04 \]

\[ =0.78\text{ mol of C atoms} \]

The total is unchanged.

This is an important distinction: not every carbon atom from the starting materials ends up in the desired ester. Some remain in unreacted propanoic acid and ethanol.

Atom conservation applies to the entire reaction mixture, not just the isolated product.

Question 4

A researcher investigates the synthesis of ethyl propanoate using an isotopically labelled starting material.

The starting material is \([1-{}^{13}\mathrm C]\)1-bromopropane. The carbon bonded to bromine is labelled with carbon-13.

The compound is converted into propan-1-ol, then propanoic acid, and finally ethyl propanoate.

Two analytical observations are obtained:

• Mass spectrometry shows an enhanced \(M+1\) isotopologue peak in the final product compared with an unlabelled control.
• Carbon-13 NMR shows an enriched signal in the ester carbonyl carbon environment.

Explain:

a) Where the labelled carbon should be located in each intermediate.

b) What the mass spectrometry result establishes.

c) Why carbon-13 NMR provides additional information.

Solution 4

The labelled carbon becomes the carbonyl carbon of ethyl propanoate. Mass spectrometry confirms that the labelled isotope remains in the final molecule, while carbon-13 NMR identifies its chemical environment.

Part a: Trace the labelled carbon.

Initially, the labelled carbon is the carbon attached to bromine:

\[ \mathrm{CH_3CH_2{}^{13}CH_2Br} \]

During substitution, bromine is replaced by OH:

\[ \mathrm{CH_3CH_2{}^{13}CH_2OH} \]

The labelled carbon remains at the end of the carbon chain.

During oxidation, the carbon attached to OH becomes the carboxyl carbon:

\[ \mathrm{CH_3CH_2{}^{13}COOH} \]

During esterification, that carbon remains the carbonyl carbon:

\[ \mathrm{CH_3CH_2{}^{13}COOCH_2CH_3} \]

The isotope has not changed its identity. Its bonding environment has changed.

Part b: Interpret the mass spectrometry result.

Carbon-13 is one mass unit heavier than carbon-12.

The enhanced \(M+1\) signal indicates that the final product contains the labelled carbon-13 isotope.

This supports the conclusion that the labelled carbon has been retained somewhere in the final molecule.

However, mass spectrometry alone does not establish which of the five carbon positions contains the isotope.

Part c: Interpret the carbon-13 NMR result.

Carbon-13 NMR distinguishes different carbon environments.

A carbonyl carbon in an ester has a different chemical environment from the carbon atoms in the ethyl group.

The enriched signal in the ester carbonyl environment therefore supports the predicted atom mapping.

The labelled carbon from the original 1-bromopropane has become the carbonyl carbon of ethyl propanoate.

The two analytical techniques provide different information:

• Mass spectrometry supports the presence of the labelled isotope in the product.
• Carbon-13 NMR provides information about the isotope’s position within the molecular structure.

Neither final-product measurement, by itself, proves every intermediate reaction occurred exactly as proposed. Analysing the intermediates would provide further evidence.

Question 5

Two students independently propose pathways for synthesising ethyl propanoate from \([1-{}^{13}\mathrm C]\)1-bromopropane and ethanol.

Both students claim that their equations conserve all atoms.

Proposed pathway A:

\[ \ce{1-bromopropane -> propan-1-ol} \]

\[ \ce{propan-1-ol -> propanal} \]

\[ \ce{propanal -> propanoic\ acid} \]

\[ \ce{propanoic\ acid + ethanol -> ethyl\ propanoate} \]

Proposed pathway B:

\[ \ce{1-bromopropane -> propan-1-ol} \]

\[ \ce{propan-1-ol -> propanone} \]

\[ \ce{propanone -> propanoic\ acid} \]

\[ \ce{propanoic\ acid + ethanol -> ethyl\ propanoate} \]

For pathway B, the students point out that the proposed oxidation equations can be written in atom-balanced form:

\[ \ce{CH3CH2CH2OH + [O] -> CH3COCH3 + H2O} \]

\[ \ce{CH3COCH3 + [O] -> CH3CH2COOH} \]

Both groups also obtain a final product with an enhanced \(M+1\) peak in mass spectrometry.

A further carbon-13 NMR analysis shows that the labelled carbon is in the carbonyl environment of the final ester.

Assume that only standard HSC organic reaction pathways are permitted, with no unlisted carbon-carbon bond cleavage or rearrangement.

Determine which pathway is chemically valid.

Your response must explain:

a) Why atom-balanced equations do not necessarily establish a valid synthesis.

b) How the primary alcohol oxidation pathway distinguishes the two proposals.

c) What the \(M+1\) mass spectrometry result can establish.

d) What the NMR result establishes, and whether it proves the exact sequence of intermediate reactions.

Solution 5

Pathway A is the chemically valid pathway under the stated HSC reaction rules. Pathway B can pass a formal atom-count check but contains transformations that are not valid standard oxidation reactions.

The crucial distinction is the position of the carbonyl group and the structure of the starting alcohol.

Part a: Explain why atom balance is insufficient.

Both proposed equations for the first oxidation step can be balanced.

For example:

\[ \ce{CH3CH2CH2OH + [O] -> CH3COCH3 + H2O} \]

contains three carbon atoms, eight hydrogen atoms, and two oxygen atoms on each side.

The equation is atom-balanced.

However, this only shows that the numbers of each type of atom match. It does not establish that the proposed reaction can actually occur under the stated conditions.

The molecular formula does not describe the complete molecular structure.

Part b: Compare the oxidation pathways.

Propan-1-ol is a primary alcohol.

Its standard oxidation pathway is:

\[ \ce{CH3CH2CH2OH ->[oxidation] CH3CH2CHO} \]

Further oxidation gives:

\[ \ce{CH3CH2CHO ->[oxidation] CH3CH2COOH} \]

The correct intermediate is propanal.

Propanone is produced by oxidising a secondary alcohol, not propan-1-ol.

This is where pathway B fails.

There is a second problem. Its proposed conversion of propanone into propanoic acid would require a change in the carbonyl position and molecular connectivity. That is not a standard oxidation pathway for ketones in the HSC reaction set.

The tempting argument is that propanal and propanone have the same molecular formula:

\[ \ce{C3H6O} \]

But they are structural isomers.

Propanal contains a terminal aldehyde group, while propanone contains a ketone group within the carbon chain.

A correct pathway must account for this structural difference, not just the molecular formula.

Part c: Interpret the mass spectrometry result.

The enhanced \(M+1\) peak indicates that the carbon-13 isotope is present in the final ester.

It does not distinguish between a labelled carbonyl carbon and a labelled carbon in the ethyl or methyl portions of a molecule.

Therefore, the mass spectrometry result cannot independently establish which proposed intermediate pathway was followed.

Part d: Interpret the NMR result.

The carbon-13 NMR result identifies the labelled carbon in the ester carbonyl environment.

This is consistent with pathway A.

In pathway A, the carbon initially attached to bromine becomes the carbon attached to OH in propan-1-ol. It then becomes the carbonyl carbon in propanal and remains the carbonyl carbon in propanoic acid and the final ester.

In pathway B, the proposed conversion of propan-1-ol into propanone would place the original terminal labelled carbon in a methyl group, rather than at the ketone carbonyl.

The proposed subsequent conversion to propanoic acid would require that carbon to change its position within the carbonyl-containing structure. No standard HSC oxidation reaction supports this transformation.

However, the final NMR result does not independently prove that pathway A was the exact sequence used. It identifies the labelled carbon’s final environment, not the entire reaction history.

To establish the intermediate pathway experimentally, samples of the intermediate compounds would also need to be collected and analysed.

The central conclusion: Atom conservation is a necessary condition for a valid synthesis, but it is not sufficient. You must also check the functional group, molecular structure, reaction type, and reagent conditions at every stage.

Once your atom ledger passes, the next question is whether the pathway is practical. Reagent availability, reaction conditions, yield, purity, and waste all matter when designing a chemical synthesis.

That is the next step beyond checking conservation: deciding whether a chemically valid pathway is also a sensible synthesis.

Syllabus scope note (2026): NSW HSC Chemistry students are still following the Chemistry Stage 6 (2017) syllabus. The Chemistry 11-12 Syllabus (2025) begins implementation in 2028. The isotope-tracing discussion is an explanatory extension, not a claim that isotope labelling is a prescribed practical. See NESA’s syllabus implementation timeline.