How to Distinguish Oxidation, Substitution and Elimination

Start with one molecule: butan-1-ol, \(\ce{CH3CH2CH2CH2OH}\). You want to turn it into one of three products: butanal, 1-bromobutane, or but-1-ene.

All three products contain four carbon atoms. However, each requires a different type of reaction.

Before reading on, predict which transformation is oxidation, which is substitution, and which is elimination.

The answers are:

• Butan-1-ol to butanal: oxidation.
• Butan-1-ol to 1-bromobutane: substitution.
• Butan-1-ol to but-1-ene: elimination.

The reliable way to distinguish these reactions is to compare the structures of the reactant and product. Then use the reagents and reaction conditions to check your decision.

01The three reaction types: what actually changes?

The biggest mistake students make is classifying a reaction based only on its starting material or reagent.

Instead, look at what happens to the atoms and bonds.

1. Oxidation: the carbon becomes more oxidised

In an organic reaction map, oxidation commonly involves increasing the number of carbon-oxygen bonds or decreasing the number of carbon-hydrogen bonds.

For example, a primary alcohol can be oxidised to an aldehyde and then to a carboxylic acid.

\[ \ce{CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O} \]

Propan-1-ol becomes propanal.

With further oxidation:

\[ \ce{CH3CH2CHO + [O] -> CH3CH2COOH} \]

Propanal becomes propanoic acid.

The carbon skeleton remains intact. What changes is the chemical environment around the carbon that originally held the hydroxyl group.

A useful way to confirm oxidation is to examine oxidation numbers.

For example, the carbon attached to the hydroxyl group in propan-1-ol has an oxidation number of \(-1\). In propanal, the corresponding carbon has an oxidation number of \(+1\).

Its oxidation number increases, confirming oxidation.

Remember that adding oxygen is a useful clue, not the complete definition. Removing hydrogen can also be oxidation.

For a refresher on the formal rules, see Oxidation numbers in redox reactions: HSC Chemistry Guide.

2. Substitution: one group replaces another

Substitution occurs when an atom or functional group is replaced by another atom or group.

Consider the conversion of propan-1-ol into 1-bromopropane:

\[ \ce{CH3CH2CH2OH + HBr -> CH3CH2CH2Br + H2O} \]

The hydroxyl group, \(-OH\), is replaced by bromine.

The three-carbon skeleton remains unchanged.

Another common example is the conversion of a halogenoalkane into an alcohol:

\[ \ce{CH3CH2Br + NaOH -> CH3CH2OH + NaBr} \]

Here, the bromine atom is replaced by a hydroxyl group.

The key feature is that one group takes the place of another. You do not form a new carbon-carbon double bond.

3. Elimination: atoms are removed and a double bond forms

Elimination involves removing atoms or groups from a molecule, usually from adjacent carbon atoms.

The removal allows a carbon-carbon double bond to form.

For example, heating butan-1-ol with concentrated sulfuric acid can produce but-1-ene:

\[ \ce{CH3CH2CH2CH2OH -> CH2=CHCH2CH3 + H2O} \]

The hydroxyl group and a hydrogen atom from the neighbouring carbon are removed together as water.

The product is an alkene.

A similar reaction occurs when a halogenoalkane reacts with ethanolic sodium hydroxide:

\[ \ce{CH3CHBrCH3 + NaOH -> CH3CH=CH2 + NaBr + H2O} \]

The bromine and a neighbouring hydrogen are removed, forming propene.

Notice that elimination and substitution can begin with the same halogenoalkane. The difference is what happens to the molecule after the bromine leaves.

02A quick comparison for reaction maps

When you are unsure which reaction type an arrow represents, use this table.

The carbon-count column is particularly useful.

For the standard HSC reaction pathways discussed here, the carbon skeleton is retained. If you start with four carbons, you should still have four carbons in the organic product.

However, preserving the carbon count does not tell you which reaction occurred. You must still inspect the functional group.

03Work backwards from the desired functional group

Reaction maps often give you a target molecule and ask you to identify a suitable pathway.

Rather than guessing the reagent, start with the functional group you need.

This gives you a practical decision process.

Step 1: Identify the functional group in the target

Look at the end product first.

• Does it contain a carbonyl group, \(\ce{C=O}\), or a carboxyl group, \(\ce{-COOH}\)? Consider oxidation.
• Does it contain a different atom or group in place of the original functional group? Consider substitution.
• Does it contain a carbon-carbon double bond that was absent in the starting material? Consider elimination.

Step 2: Compare the starting and final structures

Count the carbon atoms.

Then compare the positions of the functional groups.

Ask yourself:

• Has the carbon skeleton stayed the same?
• Has one group been replaced?
• Have atoms been removed from neighbouring carbons?
• Has a carbon become more oxidised?

These questions are more reliable than memorising isolated reaction equations.

Step 3: Check the reagents and conditions

This is particularly important when distinguishing substitution from elimination.

Consider a halogenoalkane reacting with sodium hydroxide.

• Aqueous NaOH favours substitution, producing an alcohol.
• Ethanolic NaOH with heat favours elimination, producing an alkene.

The sodium hydroxide is present in both cases. The solvent and reaction conditions influence which pathway is favoured.

If an exam question only says NaOH, check whether additional information is provided before committing to a product.

04An experiment that explains why oxidation conditions matter

There is another important detail in organic reaction maps: the same starting alcohol can produce different oxidation products depending on how the reaction is carried out.

Consider propan-1-ol.

It can be oxidised to propanal, but propanal can undergo further oxidation to propanoic acid.

How could you control which product forms?

The experimental problem

Imagine setting up a standard laboratory investigation using propan-1-ol and acidified potassium dichromate(VI).

The flask contains the alcohol and oxidising agent. Heating allows the reaction to proceed.

The dichromate solution is initially orange. As it is reduced to chromium(III) ions, it becomes green.

This colour change provides evidence that an oxidation reaction is occurring.

However, it does not tell you whether the organic product is an aldehyde or a carboxylic acid.

To control the product, you need to consider how long the intermediate remains in contact with the oxidising agent.

Design 1: Distillation to produce an aldehyde

In a distillation arrangement, vapour travels from the heated reaction flask through a distillation head and into a water-cooled condenser.

The condenser slopes down towards a receiving flask. Cooling water enters at the lower inlet and exits through the upper outlet.

Propanal is relatively volatile, so it can be distilled away as it forms.

Removing the aldehyde from the reaction mixture limits its exposure to the oxidising agent.

The expected organic product is propanal.

Design 2: Reflux to produce a carboxylic acid

Now change the apparatus.

Instead of a downward-sloping condenser, attach a vertical reflux condenser above the reaction flask.

Vapour rises from the heated mixture, condenses on the cool inner surface, and falls back into the flask.

This means the organic compounds remain in contact with the oxidising agent for longer.

With excess oxidising agent, the primary alcohol can be oxidised beyond the aldehyde stage to a carboxylic acid.

The expected product is propanoic acid.

What does the comparison tell us?

The important distinction is the control of the reaction conditions.

A simple heated flask without appropriate product control may allow volatile aldehyde to escape or may allow further oxidation to occur.

Distillation and reflux address different objectives.

Also, the orange-to-green colour change is evidence of the oxidising agent being reduced. It does not independently identify the organic product. Product identification requires additional evidence.

This is an illustrative experimental design, not a practical procedure. Acidified chromium(VI) reagents are hazardous and require appropriate laboratory controls.

A small calculation: estimating the aldehyde produced

Suppose 0.0500 mol of propan-1-ol is completely converted into propanal.

The reaction is:

\[ \ce{CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O} \]

The mole ratio between propan-1-ol and propanal is 1:1.

Therefore:

\[ n(\text{propanal})=0.0500\text{ mol} \]

The molar mass of propanal, \(\ce{C3H6O}\), is:

\[ M=3(12.01)+6(1.008)+16.00 \]

\[ M=58.08\text{ g mol}^{-1} \]

Using \(m=nM\):

\[ \begin{aligned} m&=0.0500\times58.08\\ &=2.904\text{ g} \end{aligned} \]

The theoretical propanal mass is therefore 2.90 g.

This assumes complete conversion, a 1:1 reaction ratio, and no product loss during collection.

The calculation also reinforces an important point: the organic product still contains three carbon atoms. Oxidation changes the functional group, not the carbon-chain length in this reaction.

05Worked examples

06Questions and solutions

Question 1

1-bromopropane reacts with aqueous sodium hydroxide.

Write the balanced equation, identify the major organic product, and classify the reaction.

Solution 1

The product is propan-1-ol, formed through a substitution reaction.

Aqueous sodium hydroxide supplies hydroxide ions, which replace the bromine atom.

The balanced equation is:

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

There are three carbon atoms on both sides of the equation.

The carbon skeleton remains unchanged, and the bromine atom is replaced by an OH group.

This is substitution, not elimination, because no carbon-carbon double bond forms.

It is also not oxidation because the transformation does not involve an overall increase in the oxidation state of the carbon skeleton.

Question 2

Propan-2-ol is converted into two different products using two separate reaction pathways.

Pathway A: Concentrated phosphoric acid and heat produce propene.

Pathway B: Acidified potassium dichromate produces propanone.

(a) Write an equation for each reaction.

(b) Classify both reactions.

(c) A student argues that Pathway A must be oxidation because hydrogen is removed from the alcohol. Evaluate this argument using oxidation numbers.

Solution 2

Pathway A is elimination, while Pathway B is oxidation.

The fact that hydrogen is removed does not automatically make a reaction an oxidation reaction.

Pathway A: Elimination

Propan-2-ol undergoes dehydration, losing water to form propene.

\[ \ce{CH3CH(OH)CH3 -> CH3CH=CH2 + H2O} \]

The hydroxyl group and a hydrogen atom from an adjacent carbon are removed.

A carbon-carbon double bond forms, identifying the reaction as elimination.

Pathway B: Oxidation

Propan-2-ol is oxidised to propanone.

\[ \ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O} \]

The carbon attached to OH becomes a carbonyl carbon.

Its oxidation number increases from \(0\) to \(+2\).

This confirms oxidation.

Evaluating the student’s argument

The student has noticed that hydrogen is lost, which is a reasonable starting observation.

However, the loss of hydrogen alone does not establish that a reaction is oxidation.

We can compare the total oxidation numbers of carbon in both pathways.

For propan-2-ol, \(\ce{C3H8O}\):

• Hydrogen has an oxidation number of \(+1\).
• Oxygen has an oxidation number of \(-2\).
• The molecule is neutral.

Therefore, the sum of the carbon oxidation numbers is:

\[ x+8(+1)-2=0 \]

\[ x=-6 \]

For propene, \(\ce{C3H6}\):

\[ x+6(+1)=0 \]

\[ x=-6 \]

The total carbon oxidation number has not changed.

In contrast, propanone, \(\ce{C3H6O}\), has a total carbon oxidation number of:

\[ x+6(+1)-2=0 \]

\[ x=-4 \]

The total increases from \(-6\) to \(-4\), confirming oxidation.

The important distinction is that dehydration removes water and forms a double bond. Oxidation changes the oxidation state of the carbon framework overall.

Question 3

You are given 1-bromobutane, \(\ce{CH3CH2CH2CH2Br}\).

Your target product is butanoic acid, \(\ce{CH3CH2CH2COOH}\).

The available reagents are:

• Aqueous sodium hydroxide.
• Ethanolic sodium hydroxide.
• Acidified potassium dichromate(VI).
• Heat.

Design a two-stage reaction pathway to produce butanoic acid.

Write the equations, classify both reactions, and explain why using ethanolic sodium hydroxide in the first stage would be a poor choice.

Solution 3

The correct pathway is substitution followed by oxidation.

The first stage must produce butan-1-ol. The second stage oxidises this primary alcohol to butanoic acid.

Stage 1: Substitution

React 1-bromobutane with aqueous sodium hydroxide.

\[ \ce{CH3CH2CH2CH2Br + NaOH -> CH3CH2CH2CH2OH + NaBr} \]

The bromine atom is replaced by a hydroxyl group.

The product is butan-1-ol.

Stage 2: Oxidation

Heat the butan-1-ol with excess acidified potassium dichromate under reflux.

\[ \ce{CH3CH2CH2CH2OH + 2[O] -> CH3CH2CH2COOH + H2O} \]

The primary alcohol is oxidised to a carboxylic acid.

The final product is butanoic acid.

Why is ethanolic sodium hydroxide unsuitable for Stage 1?

Ethanolic sodium hydroxide with heat favours elimination.

Instead of producing butan-1-ol, the reaction would favour the formation of but-1-ene:

\[ \ce{CH3CH2CH2CH2Br + NaOH -> CH2=CHCH2CH3 + NaBr + H2O} \]

This produces an alkene rather than an alcohol.

The next step in the proposed pathway requires an alcohol to undergo oxidation to a carboxylic acid. The alkene does not provide that required intermediate.

The key planning decision is to work backwards from the target.

You need a primary alcohol immediately before the carboxylic acid, so the first reaction must be substitution rather than elimination.

The carbon count also checks out: all the organic intermediates and the final product contain four carbon atoms.

Question 4

2-bromobutane, \(\ce{CH3CHBrCH2CH3}\), is heated with ethanolic potassium hydroxide.

The organic product has the molecular formula \(\ce{C4H8}\).

A student claims that the product must be but-1-ene because it is an alkene containing four carbon atoms.

Evaluate this claim.

Identify the possible constitutional alkene products and explain how their formation depends on the structure of the starting material.

Solution 4

The student’s conclusion is incorrect. Both but-1-ene and but-2-ene can form through elimination of 2-bromobutane.

The molecular formula alone does not identify the position of the double bond.

Start with:

\[ \ce{CH3CHBrCH2CH3} \]

Bromine is attached to carbon 2.

For elimination to occur, a hydrogen must also be removed from a carbon adjacent to carbon 2.

There are two possible neighbouring carbons.

Possibility 1: Hydrogen removed from carbon 1

Bromine leaves carbon 2, and a hydrogen is removed from carbon 1.

The double bond forms between carbons 1 and 2.

\[ \ce{CH2=CHCH2CH3} \]

The product is but-1-ene.

Possibility 2: Hydrogen removed from carbon 3

Bromine leaves carbon 2, and a hydrogen is removed from carbon 3.

The double bond forms between carbons 2 and 3.

\[ \ce{CH3CH=CHCH3} \]

The product is but-2-ene.

But-2-ene can also exist as cis and trans geometric isomers. These are not additional constitutional isomers.

Both possible alkene structures contain four carbon atoms and have the molecular formula \(\ce{C4H8}\).

The important structural constraint is the position of the leaving group and the availability of hydrogen atoms on neighbouring carbons.

The formula tells you the number of atoms. It does not tell you where the double bond must be.

A question asking for a single major product needs sufficient information about the reaction conditions or product distribution. Do not assume that a molecular formula identifies a unique structure.

Question 5

Two straight-chain bromobutane isomers, A and B, have the molecular formula \(\ce{C4H9Br}\).

Each is separately treated with aqueous sodium hydroxide. The resulting alcohols are then oxidised using excess acidified potassium dichromate under reflux.

The following products are identified:

A separate portion of B is treated with ethanolic sodium hydroxide and heat. Two constitutional alkene products are identified:

• But-1-ene.
• But-2-ene.

A student makes two claims:

1. A and B must produce the same alcohol because they have the same molecular formula.
2. The reaction of B with ethanolic sodium hydroxide is oxidation because hydrogen is removed.

Determine the structures of A and B, classify every reaction, and evaluate both claims.

Your explanation must account for why the different reaction pathways produce different functional groups despite preserving the four-carbon skeleton.

Solution 5

A is 1-bromobutane, and B is 2-bromobutane.

Both student claims are incorrect.

The key is that the two starting compounds are structural isomers. They have the same molecular formula but different arrangements of atoms.

That difference determines the alcohol formed during substitution, which then determines the product of oxidation.

Part 1: Identify A

A produces butanoic acid after substitution and oxidation.

A carboxylic acid is produced through oxidation of a primary alcohol in this reaction pathway.

Therefore, the intermediate must be butan-1-ol.

The starting halogenoalkane must be 1-bromobutane.

The first reaction is substitution:

\[ \ce{CH3CH2CH2CH2Br + NaOH -> CH3CH2CH2CH2OH + NaBr} \]

The second reaction is oxidation:

\[ \ce{CH3CH2CH2CH2OH + 2[O] -> CH3CH2CH2COOH + H2O} \]

Therefore, A is 1-bromobutane.

Part 2: Identify B

B produces butan-2-one after substitution and oxidation.

Butan-2-one is a ketone.

A secondary alcohol is oxidised to a ketone, so the intermediate must be butan-2-ol.

The starting halogenoalkane must therefore be 2-bromobutane.

The first reaction is substitution:

\[ \ce{CH3CHBrCH2CH3 + NaOH -> CH3CH(OH)CH2CH3 + NaBr} \]

The second reaction is oxidation:

\[ \ce{CH3CH(OH)CH2CH3 + [O] -> CH3COCH2CH3 + H2O} \]

Therefore, B is 2-bromobutane.

Part 3: Check the elimination evidence

The separate sample of B produces but-1-ene and but-2-ene.

This independently supports the identification of B as 2-bromobutane.

Its bromine atom is attached to carbon 2, so elimination can remove a hydrogen from either adjacent carbon.

Removing hydrogen from carbon 1 produces but-1-ene.

Removing hydrogen from carbon 3 produces but-2-ene.

If the starting material had been 1-bromobutane, only one adjacent carbon position would be available for this simple elimination pathway, producing but-1-ene.

The two observed constitutional products therefore provide structural evidence about the starting material.

Part 4: Evaluate the first student claim

The claim that A and B must produce the same alcohol is incorrect.

Both compounds have the molecular formula \(\ce{C4H9Br}\), but their bromine atoms occupy different positions.

This positional difference is preserved during substitution.

• 1-bromobutane produces a primary alcohol.
• 2-bromobutane produces a secondary alcohol.

These alcohols have different oxidation pathways.

The primary alcohol produces a carboxylic acid, while the secondary alcohol produces a ketone.

The molecular formula alone cannot tell you which product will form. You need the structural formula.

Part 5: Evaluate the second student claim

The reaction of B with ethanolic sodium hydroxide is elimination, not oxidation.

The organic reactant loses bromine and a hydrogen atom from an adjacent carbon, forming a carbon-carbon double bond.

The reaction can be represented as:

\[ \ce{CH3CHBrCH2CH3 + NaOH -> CH3CH=CHCH3 + NaBr + H2O} \]

or

\[ \ce{CH3CHBrCH2CH3 + NaOH -> CH2=CHCH2CH3 + NaBr + H2O} \]

To check the student’s claim, consider the total oxidation numbers of carbon.

For 2-bromobutane, \(\ce{C4H9Br}\), hydrogen has an oxidation number of \(+1\), and bromine has an oxidation number of \(-1\).

The sum of the carbon oxidation numbers is therefore:

\[ x+9(+1)-1=0 \]

\[ x=-8 \]

For either butene, \(\ce{C4H8}\):

\[ x+8(+1)=0 \]

\[ x=-8 \]

The total carbon oxidation number remains unchanged.

Although individual carbon oxidation numbers can redistribute as the double bond forms, there is no overall increase in the carbon oxidation-number sum.

The reaction is classified as elimination because hydrogen and bromine are removed and a carbon-carbon double bond forms.

The two claims fail for different reasons:

• The first ignores the connectivity of structural isomers.
• The second confuses the removal of hydrogen with oxidation, without checking the overall structural change and oxidation numbers.

This is why a reaction map must be read as a sequence of structural transformations rather than a collection of reagent names.

07What to do next in an organic reaction map

Once you can distinguish oxidation, substitution, and elimination, the next challenge is planning an entire synthesis pathway.

Start at the target functional group. Identify the intermediate needed immediately before it, select a reaction that produces that intermediate, and work backwards until you reach the starting material.

At every arrow, check three things:

1. Has the functional group changed in the expected way?
2. Has the carbon skeleton been preserved?
3. Are the reagent and reaction conditions appropriate for that transformation?

For broader coverage of organic reaction pathways, see HSC Chemistry Module 7: Organic Reaction Map.

For more practice with oxidation states and redox reasoning, revisit Oxidation numbers in redox reactions: HSC Chemistry Guide.

Syllabus scope: This explanation is framed around the organic reaction pathways in the Chemistry Stage 6 (2017) Module 7 context. It does not claim that every pathway discussed is an assessed requirement of the Chemistry 11-12 Syllabus (2025).