How to Convert an Alkene into a Carboxylic Acid

Suppose you start with ethene, \(\ce{CH2=CH2}\), and need to produce ethanoic acid, \(\ce{CH3COOH}\). There is a straightforward two-step route: first add water across the alkene’s double bond to produce ethanol, then oxidise the ethanol using acidified potassium dichromate under reflux.

The important catch is that this method depends on the type of alcohol formed. Ethanol is a primary alcohol, which can be oxidised into a carboxylic acid. Other alkenes can produce secondary or tertiary alcohols instead, giving different oxidation products.

The complete pathway is:

\[ \ce{CH2=CH2 + H2O ->[H3PO4] CH3CH2OH} \]

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

The first reaction is hydration. The second is oxidation.

01Work backwards from the carboxylic acid

Before choosing reagents, look at the molecule you want to produce.

A carboxylic acid contains the functional group:

\[ \ce{-COOH} \]

For example, ethanoic acid has the structure:

\[ \ce{CH3-COOH} \]

Notice that the carbon in the \(\ce{-COOH}\) group is part of the main carbon chain. It is not an extra carbon added to the molecule.

This matters when planning a reaction pathway.

Which alcohol produces a carboxylic acid?

The oxidation of a primary alcohol produces a carboxylic acid, provided the oxidation proceeds beyond the aldehyde intermediate.

The general reaction is:

\[ \ce{R-CH2OH + 2[O] -> R-COOH + H2O} \]

Here, \(R\) represents the remainder of the carbon chain, and \([O]\) represents an equivalent of oxygen supplied by an oxidising agent.

For example:

\[ \ce{CH3CH2OH -> CH3COOH} \]

Ethanol becomes ethanoic acid.

The key is that the carbon chain stays the same length. The carbon attached to the hydroxyl group becomes the carbon in the carboxyl group.

Therefore, if your target is a carboxylic acid, work backwards to identify the corresponding primary alcohol.

You now have the first constraint: the alcohol must be primary.

The second constraint is that the starting alkene must have the correct number of carbon atoms.

Hydration does not change the carbon count, and ordinary oxidation of an alcohol does not change it either.

For example, you cannot convert a two-carbon alkene into a four-carbon carboxylic acid using only these two reactions. Neither reaction creates a new carbon-carbon bond.

02Step one: Convert the alkene into an alcohol using hydration

An alkene contains a carbon-carbon double bond, \(\ce{C=C}\).

Hydration is an addition reaction in which water adds across this double bond.

The hydrogen atom attaches to one of the alkene carbons, while the hydroxyl group, \(\ce{-OH}\), attaches to the other.

The double bond becomes a single bond, producing an alcohol.

Hydration of ethene

Ethene is particularly useful because its two alkene carbons are equivalent.

When ethene reacts with steam in the presence of a phosphoric acid catalyst, ethanol is produced.

\[ \ce{CH2=CH2 + H2O ->[H3PO4] CH3CH2OH} \]

The reaction requires a catalyst.

In industrial production, ethene is hydrated using steam over a phosphoric acid catalyst at elevated temperature and pressure. Typical textbook conditions are approximately \(300^\circ\text{C}\) and 60-70 atm.

The important point for a reaction pathway is that ethene produces ethanol, which is a primary alcohol.

That gives us exactly the intermediate needed to make ethanoic acid.

What happens with other alkenes?

This is where students often make an incorrect prediction.

You might assume that adding water to any terminal alkene produces a primary alcohol because one end of the double bond is at the end of the carbon chain.

That is not generally correct.

Acid-catalysed hydration follows Markovnikov’s rule: the hydroxyl group attaches to the more substituted carbon of the original double bond.

A more substituted carbon is one attached to more other carbon atoms.

Consider propene:

\[ \ce{CH3-CH=CH2} \]

The middle carbon is more substituted than the terminal carbon.

Therefore, the major product of acid-catalysed hydration is propan-2-ol:

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

This is a secondary alcohol, not a primary alcohol.

03Step two: Oxidise the alcohol into a carboxylic acid

Once you have the correct primary alcohol, the next step is oxidation.

For the standard HSC reaction pathway, a suitable oxidising agent is acidified potassium dichromate.

The reagent contains potassium dichromate, \(\ce{K2Cr2O7}\), dissolved in an acidic solution, typically containing sulfuric acid.

The reaction is generally heated under reflux when the desired product is a carboxylic acid.

Why does the type of alcohol matter?

Primary, secondary, and tertiary alcohols do not produce the same oxidation products.

The difference comes from the carbon attached to the hydroxyl group.

A primary alcohol can be oxidised first into an aldehyde and then further into a carboxylic acid.

A secondary alcohol produces a ketone instead. Under ordinary HSC oxidation conditions, it does not continue to a carboxylic acid.

A tertiary alcohol has no hydrogen attached to the carbon bearing the hydroxyl group, so it does not undergo the usual alcohol oxidation reaction.

The oxidation of ethanol

Ethanol is a primary alcohol, so its oxidation proceeds through two stages.

First, ethanol becomes ethanal:

\[ \ce{CH3CH2OH + [O] -> CH3CHO + H2O} \]

Then ethanal undergoes further oxidation to produce ethanoic acid:

\[ \ce{CH3CHO + [O] -> CH3COOH} \]

Combining these equations gives:

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

The symbol \([O]\) is shorthand for the oxidising equivalents required. It is not a separate bottle of oxygen gas.

For this reaction, use excess acidified potassium dichromate and heat under reflux.

Why use reflux instead of distillation?

This is an important experimental decision.

The first oxidation product, ethanal, is relatively volatile. If you want to produce ethanoic acid, you need to prevent ethanal from escaping before it can undergo further oxidation.

A reflux apparatus solves this problem.

Picture the apparatus in the following order:

1. A round-bottom flask contains ethanol, acidified potassium dichromate, and anti-bumping granules.
2. A vertical water-cooled condenser is attached to the top of the flask.
3. Cooling water enters the lower side-arm of the condenser and leaves through the upper side-arm.
4. Heating causes vapour to rise from the reaction mixture.
5. The vapour contacts the cooled walls of the condenser, becomes liquid again, and flows back into the flask.

The mixture can therefore be heated while volatile substances are continually returned to the reaction vessel.

What would happen if you used simple distillation instead?

The ethanal could leave the reaction flask, pass through the condenser, and collect in a separate receiving flask. This is useful if your desired product is an aldehyde, but it works against producing a carboxylic acid.

The choice of apparatus therefore depends on the product you want:

• Distillation removes the volatile aldehyde as it forms.
• Reflux retains the aldehyde in the reaction mixture so it can undergo further oxidation.

What observations indicate that oxidation is occurring?

Acidified potassium dichromate provides a useful visual observation.

Its colour changes from orange to green as chromium(VI) ions are reduced to chromium(III) ions.

\[ \ce{Cr2O7^2- -> Cr^3+} \]

The colour change provides evidence that the oxidising agent has participated in a redox reaction.

However, it does not independently prove that ethanoic acid has formed. Both primary and secondary alcohols can reduce acidified dichromate during oxidation.

You need to consider the starting alcohol and its expected oxidation products when identifying the final organic compound.

This distinction is especially useful in experimental questions: an observation can support a conclusion without being sufficient to establish it.

04Put the two reactions together

Now we can combine the steps.

Suppose the target is ethanoic acid.

Work backwards:

• Ethanoic acid requires oxidation of a primary alcohol.
• The corresponding primary alcohol is ethanol.
• Ethene can be hydrated to produce ethanol.
• Ethanol can then be oxidised under reflux to produce ethanoic acid.

The complete pathway is:

\[ \ce{CH2=CH2 + H2O ->[H3PO4] CH3CH2OH} \]

\[ \ce{CH3CH2OH + 2[O] ->[K2Cr2O7/H2SO4] CH3COOH + H2O} \]

05What if you need a primary alcohol from a different alkene?

There is an important limitation to the standard acid-catalysed hydration method.

For most terminal alkenes other than ethene, acid-catalysed hydration produces a secondary or tertiary alcohol as the major product.

This means the straightforward hydration-then-oxidation sequence will not necessarily produce the carboxylic acid you want.

There is another reaction that can overcome this limitation: hydroboration-oxidation.

This is extension chemistry beyond the standard HSC reaction pathway. It is useful for understanding how chemists can control which carbon receives the hydroxyl group.

Hydroboration-oxidation

Hydroboration-oxidation adds water across an alkene in an anti-Markovnikov orientation.

This means the hydroxyl group ends up on the less substituted carbon.

For example, but-1-ene can be converted into butan-1-ol:

\[ \ce{CH2=CHCH2CH3} \]

The reaction uses:

1. Borane, \(\ce{BH3}\), commonly supplied as a complex with tetrahydrofuran (THF).
2. Hydrogen peroxide, \(\ce{H2O2}\), in alkaline conditions.

The resulting alcohol is butan-1-ol, a primary alcohol.

It can then be oxidised into butanoic acid:

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

The difference is the position of the hydroxyl group.

That single structural difference determines whether oxidation produces a ketone or a carboxylic acid.

For an ordinary HSC reaction pathway question, do not assume hydroboration-oxidation is available unless the question supplies it or your course specifically covers it.

06Questions and solutions

Question 1

Ethanol is to be converted into ethanoic acid using an oxidising agent.

a) Name a suitable oxidising agent.

b) State the conditions required to favour the formation of ethanoic acid.

c) Write the balanced organic reaction equation.

d) Explain why ethanol can produce a carboxylic acid through oxidation.

Solution 1

Ethanol can be converted into ethanoic acid using excess acidified potassium dichromate under reflux.

a) The oxidising agent is acidified potassium dichromate, \(\ce{K2Cr2O7/H2SO4}\).

b) Heat the reaction mixture under reflux with excess oxidising agent.

Reflux prevents volatile ethanal from escaping, allowing it to undergo further oxidation.

c) The balanced organic equation is:

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

d) Ethanol is a primary alcohol. Primary alcohols can first be oxidised into aldehydes and then further oxidised into carboxylic acids.

Ethanol therefore produces ethanal as an intermediate before forming ethanoic acid.

Question 2

Propene is reacted with water in the presence of a dilute acid catalyst. The major alcohol product is then oxidised using excess acidified potassium dichromate under reflux.

a) Identify the major alcohol produced during hydration.

b) Classify this alcohol as primary, secondary, or tertiary.

c) Predict the organic product formed during oxidation.

d) Explain why the final product is not propanoic acid.

Solution 2

The major alcohol is propan-2-ol, which oxidises to propanone rather than propanoic acid.

a) Propene undergoes acid-catalysed hydration according to Markovnikov’s rule.

The hydroxyl group attaches to the more substituted carbon.

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

The product is propan-2-ol.

b) Propan-2-ol is a secondary alcohol because the carbon attached to the hydroxyl group is bonded to two other carbon atoms.

c) Oxidation of a secondary alcohol produces a ketone.

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

The product is propanone.

d) Propanoic acid requires oxidation of a primary alcohol, which would be propan-1-ol.

However, acid-catalysed hydration of propene produces propan-2-ol as the major product.

The starting alkene therefore produces the wrong alcohol structure for the desired oxidation pathway.

The trap is assuming that a terminal alkene must produce a primary alcohol. Markovnikov’s rule determines the major product, not simply the position of the end of the carbon chain.

Question 3

A student oxidises 2.30 g of ethanol using excess acidified potassium dichromate.

After completing the reaction and isolating the product, the student obtains 2.40 g of ethanoic acid.

Calculate the percentage yield.

Use the following molar masses:

• Ethanol: \(46.07\text{ g mol}^{-1}\)
• Ethanoic acid: \(60.05\text{ g mol}^{-1}\)

Assume ethanol is the limiting reactant.

Solution 3

The percentage yield is approximately 80.0%.

Step 1: Write the balanced reaction.

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

The mole ratio of ethanol to ethanoic acid is \(1:1\).

Step 2: Calculate the moles of ethanol.

Use:

\[ n=\frac{m}{M} \]

Substitute:

\[ n(\text{ethanol})=\frac{2.30}{46.07} \]

\[ n(\text{ethanol})=0.0499\text{ mol} \]

Step 3: Calculate the theoretical mass of ethanoic acid.

The \(1:1\) mole ratio means:

\[ n(\text{acid})=0.0499\text{ mol} \]

Use:

\[ m=nM \]

\[ m(\text{acid})=0.0499\times60.05 \]

\[ m(\text{acid})\approx3.00\text{ g} \]

Step 4: Calculate the percentage yield.

Use:

\[ \text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times100 \]

Substitute:

\[ \text{Percentage yield} = \frac{2.40}{3.00}\times100 \]

\[ \boxed{80.0\%} \]

The student obtained 80.0% of the theoretical amount of ethanoic acid.

The remaining 20.0% may reflect incomplete conversion, product losses during isolation, or competing reactions.

Question 4

A student wants to oxidise ethanol completely into ethanoic acid.

Instead of using reflux, the student heats ethanol with acidified potassium dichromate in a flask connected to a condenser and receiving flask. The condenser is arranged for simple distillation, allowing volatile products to leave the reaction vessel.

The student observes that the reaction mixture changes from orange to green.

a) Explain why the student may collect ethanal rather than ethanoic acid.

b) Describe how the apparatus should be modified to favour ethanoic acid production.

c) Explain what the orange-to-green colour change demonstrates and why it does not prove that ethanoic acid has formed.

Solution 4

The student may collect ethanal because simple distillation removes the volatile aldehyde before it can undergo further oxidation.

a) Ethanol initially oxidises into ethanal.

\[ \ce{CH3CH2OH + [O] -> CH3CHO + H2O} \]

Ethanal is volatile and can leave the reaction flask during distillation.

Removing it prevents much of the ethanal from remaining in contact with the oxidising agent, limiting its further oxidation into ethanoic acid.

b) Replace the simple distillation arrangement with a vertical reflux condenser.

The condenser should return condensed vapour to the reaction flask rather than allowing it to collect in a separate receiving flask.

The mixture should also contain excess acidified potassium dichromate and be heated under reflux.

This allows ethanal to remain in the reaction mixture and undergo further oxidation.

c) The orange-to-green colour change indicates that chromium(VI) has been reduced to chromium(III).

This provides evidence that a redox reaction has occurred.

However, it does not identify the organic product. Both primary and secondary alcohols can reduce acidified potassium dichromate during oxidation.

Therefore, the colour change alone cannot establish whether the product is an aldehyde, ketone, or carboxylic acid.

The reaction conditions and starting alcohol must also be considered.

Question 5

A chemist wants to synthesise butanoic acid from pure but-1-ene, \(\ce{CH2=CHCH2CH3}\).

Two teams propose different reaction pathways.

Team A uses acid-catalysed hydration followed by oxidation with excess acidified potassium dichromate under reflux.

Team B uses hydroboration-oxidation followed by the same final oxidation conditions.

For this question, assume hydroboration-oxidation places the hydroxyl group on the less substituted carbon of the original double bond, producing a primary alcohol.

Both teams report that their acidified potassium dichromate changes from orange to green.

Assess both reaction pathways.

In your answer:

• Identify the alcohol formed by each hydration method.
• Predict the final organic product for each team.
• Determine which team successfully produces butanoic acid.
• Explain why the colour change does not distinguish between the two pathways.
• Explain whether either pathway has a carbon-count problem.

Assume the usual major reaction products, sufficient oxidising agent, and no carbon-skeleton rearrangement.

Solution 5

Team B successfully produces butanoic acid. Team A produces butanone instead.

Both teams can observe the orange-to-green colour change, and neither pathway has a carbon-count problem. The decisive difference is the position of the hydroxyl group in the intermediate alcohol.

Step 1: Analyse Team A’s hydration reaction.

Team A uses acid-catalysed hydration.

This follows Markovnikov’s rule, so the hydroxyl group attaches to the more substituted carbon of but-1-ene.

The major product is butan-2-ol:

\[ \ce{CH2=CHCH2CH3 ->[H^+] CH3CH(OH)CH2CH3} \]

Butan-2-ol is a secondary alcohol.

Step 2: Predict Team A’s oxidation product.

Secondary alcohols oxidise into ketones.

Therefore:

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

The product is butanone, not butanoic acid.

Team A has used a valid hydration reaction and a valid oxidation reagent, but the intermediate alcohol has the wrong structure for producing the desired acid.

Step 3: Analyse Team B’s hydration reaction.

Team B uses hydroboration-oxidation.

For this question, the reaction is specified to place the hydroxyl group on the less substituted carbon.

The product is therefore butan-1-ol:

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

Butan-1-ol is a primary alcohol.

Step 4: Predict Team B’s oxidation product.

A primary alcohol can undergo further oxidation to form a carboxylic acid.

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

The product is butanoic acid.

Team B successfully reaches the target compound.

Step 5: Interpret the colour change.

Both teams use acidified potassium dichromate during the final oxidation.

In both cases, chromium(VI) is reduced to chromium(III), producing the orange-to-green colour change.

However, the organic products are different:

• Team A produces butanone.
• Team B produces butanoic acid.

The colour change demonstrates that oxidation has occurred, but it does not identify the product.

Step 6: Check the carbon count.

But-1-ene contains four carbon atoms.

Butan-2-ol, butan-1-ol, butanone, and butanoic acid all contain four carbon atoms.

There is no carbon-count problem in either pathway.

The difference is the position of the hydroxyl group, which determines whether the intermediate alcohol is primary or secondary.

This is the crucial distinction: having the correct carbon chain is not enough. You must also produce the correct functional-group arrangement.

07The decision to remember

When converting an alkene into a carboxylic acid through an alcohol, work backwards rather than selecting reagents immediately.

1. Identify the target carboxylic acid and its carbon count.
2. Find the corresponding primary alcohol.
3. Check whether hydration of your starting alkene produces that primary alcohol.
4. If it does, oxidise it using excess acidified potassium dichromate under reflux.
5. Check that the carbon count and final functional group are correct.

For the straightforward HSC example:

\[ \boxed{ \ce{Ethene ->[H2O,\ acid\ catalyst] Ethanol ->[2[O],\ reflux] Ethanoic\ acid} } \]

The most important mistake to avoid is assuming that every alkene produces a primary alcohol when hydrated.

The next useful step is learning to distinguish the structures and reactions of the carboxylic acids you produce. See HSC Chemistry: Carboxylic Acids, Amines and Amides Explained and Carboxylic Acids vs Esters: Structure, Properties and Reactions.

Syllabus note: The core reaction pathway is framed around the Chemistry Stage 6 Syllabus (2017), particularly alkene addition reactions and the oxidation of alcohols. Hydroboration-oxidation is identified separately as extension chemistry. This article does not claim that the 2025 Chemistry 11-12 syllabus prescribes this exact reaction sequence.