How to Identify an Unknown Intermediate from Its Functional Group

An unknown intermediate, X, has the molecular formula \(\ce{C4H10O}\). It is heated under reflux with acidified potassium dichromate, producing butan-2-one, \(\ce{CH3COCH2CH3}\). Which alcohol was X?

Before choosing, compare butan-1-ol and butan-2-ol. Both have the same molecular formula, and both are alcohols that can be oxidised. However, only butan-2-ol has its hydroxyl group attached to carbon 2, allowing oxidation to produce the ketone butan-2-one.

Therefore, X is butan-2-ol.

This is the central skill in identifying an unknown organic intermediate: work backwards from the functional group in the product, reconstruct the most likely immediate precursor, and then check whether the carbon skeleton, reagent, and reaction conditions support your answer.

01Start with the desired functional group

When an HSC Chemistry question gives you an unknown intermediate, resist the temptation to guess its name from the molecular formula.

Start with the product.

A functional group provides evidence about the type of reaction that has occurred. Once you recognise that group, you can identify likely reactions that produce it and work backwards to the intermediate.

For example:

• A ketone suggests oxidation of a secondary alcohol.
• An aldehyde suggests controlled oxidation of a primary alcohol.
• A carboxylic acid suggests oxidation of a primary alcohol or an aldehyde.
• An alcohol suggests hydration of an alkene or substitution of a haloalkane.
• An alkene suggests dehydration of an alcohol or elimination from a haloalkane.

These are common pathways, not universal rules. More than one reaction can produce the same functional group, so you must also examine the reagents and conditions.

The functional group gives you a starting point

The following table is a useful reference when working backwards from a target molecule.

Here, R and R’ represent carbon-containing groups.

Notice that a carboxylic acid has two possible immediate precursors in the table. Identifying the functional group alone is therefore not enough. You need another piece of evidence to distinguish the possibilities.

This is why reaction conditions matter.

If you need to refresh your understanding of the structural features of each group, see HSC Chemistry: How to Identify Organic Functional Groups.

02Use oxidation to identify the alcohol precursor

Oxidation is particularly useful when identifying unknown alcohol intermediates.

There are three important categories of alcohols:

• Primary alcohols have the carbon containing the hydroxyl group attached to one other carbon.
• Secondary alcohols have that carbon attached to two other carbons.
• Tertiary alcohols have that carbon attached to three other carbons.

This classification predicts the products formed during oxidation.

Primary alcohols

A primary alcohol can first be oxidised to an aldehyde.

With further oxidation, the aldehyde forms a carboxylic acid.

\[ \ce{RCH2OH + [O] -> RCHO + H2O} \]

\[ \ce{RCHO + [O] -> RCOOH} \]

The symbol \([O]\) represents an oxidising equivalent supplied by the oxidising agent. It is shorthand, not a separate oxygen molecule.

For example, propan-1-ol can undergo oxidation to propanal and then propanoic acid.

\[ \ce{CH3CH2CH2OH -> CH3CH2CHO -> CH3CH2COOH} \]

The carbon skeleton stays intact throughout these reactions.

Secondary alcohols

A secondary alcohol oxidises to a ketone.

\[ \ce{RCH(OH)R' + [O] -> RCOR' + H2O} \]

For example:

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

Propan-2-ol becomes propanone.

The carbon attached to the hydroxyl group becomes the carbonyl carbon. Its position does not move along the carbon chain.

Tertiary alcohols

Tertiary alcohols generally do not undergo oxidation with acidified potassium dichromate under standard HSC reaction conditions.

Why?

The carbon attached to the hydroxyl group has no hydrogen atom attached to it. Oxidising this carbon to form a carbonyl group would require breaking carbon-carbon bonds, rather than simply removing hydrogen and forming a stronger carbon-oxygen bond.

This gives you a useful diagnostic comparison.

03The experiment: why reaction conditions matter

Suppose you have an unknown alcohol and want to determine what it produces during oxidation.

You add the alcohol to acidified potassium dichromate and heat the mixture. The original orange solution changes towards green.

What does this observation tell you?

It provides evidence that oxidation has occurred. The dichromate ions, containing chromium(VI), are reduced to chromium(III), which produces the green colour.

However, the colour change does not tell you whether the organic product is an aldehyde, ketone, or carboxylic acid.

To distinguish these products, you need to control the reaction conditions and examine the organic product.

Reflux: allowing oxidation to continue

In a reflux apparatus, the reaction mixture sits in a heated round-bottom flask. A vertical water-cooled condenser is connected directly above the flask.

Cooling water enters at the bottom of the condenser and exits at the top. Vapour from the boiling mixture rises into the condenser, cools, and returns to the flask as liquid.

This prevents volatile substances from escaping and allows the reaction to continue at an elevated temperature.

For a primary alcohol, refluxing with excess oxidising agent allows the aldehyde intermediate to undergo further oxidation to a carboxylic acid.

For a secondary alcohol, the product is a ketone.

Distillation: collecting an aldehyde before further oxidation

Now consider a different experimental objective.

You want to isolate the aldehyde formed from a primary alcohol, rather than allowing it to oxidise further into a carboxylic acid.

Reflux is not the best choice for this purpose. The aldehyde remains in contact with the oxidising agent and can undergo further oxidation.

Instead, use distillation.

The reaction mixture is heated in a flask connected to a sloping water-cooled condenser. Vapour travels from the flask through the condenser, where it cools and becomes liquid. The liquid then flows into a separate collecting vessel.

Because the aldehyde is removed from the reaction mixture as it forms, further oxidation is reduced.

The distinction is important:

• Reflux promotes continued oxidation.
• Distillation allows a volatile product to be collected as it forms.

A student who identifies an aldehyde as the product but ignores the apparatus may incorrectly predict a carboxylic acid.

The experimental setup is part of the chemical evidence, not just laboratory decoration.

One important limitation: an orange-to-green colour change establishes that oxidation has occurred, but it cannot independently identify the exact organic product. You need additional structural or chemical evidence.

04Work backwards using the carbon skeleton

Recognising the functional group gives you a likely precursor. You must now check whether its carbon structure is possible.

Consider butan-2-one:

\[ \ce{CH3COCH2CH3} \]

It contains four carbon atoms. Its carbonyl carbon is at position 2.

If this ketone formed by oxidation of an alcohol, the precursor must have had its hydroxyl group at the same carbon position.

Therefore:

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

This is butan-2-ol.

A useful rule is that oxidation, substitution, hydration, and dehydration generally preserve the carbon skeleton in the standard reaction pathways considered here.

For example:

If the target molecule contains five carbon atoms but your proposed starting material contains only three, you need another carbon-containing reactant or a different pathway.

Do not assume an oxidising agent, acid catalyst, or heat supplies missing carbon atoms.

For a more detailed treatment of carbon accounting across multiple reactions, see How to Check Atom Conservation in Multi-Step Organic Synthesis.

07A reliable method for identifying unknown intermediates

When you encounter a multi-step organic reaction question, use this sequence.

1. Start with the final product.

Draw its structural formula and identify the functional group. Do not rely on the molecular formula alone.

2. Identify the likely final reaction.

Ask what type of reaction could produce that functional group.

For example, a ketone suggests oxidation of a secondary alcohol.

3. Reconstruct the immediate precursor.

Reverse the relevant structural change.

If a carbonyl group formed through oxidation, place the hydroxyl group on the same carbon.

4. Check the carbon skeleton.

Count the carbon atoms and check their positions. Preserve the carbon framework unless the reaction specifically involves joining or breaking carbon-containing fragments.

5. Read the reagents and conditions carefully.

Aqueous and ethanolic sodium hydroxide can produce different products. Distillation and reflux can also lead to different oxidation endpoints.

6. Test your candidate against the entire pathway.

A proposed intermediate must explain every reaction, not just the final product.

If two structures still fit, the available evidence may not be sufficient to identify a unique intermediate. Look for an additional test, product, or reaction condition that distinguishes them.

There is one particularly important complication: working backwards does not always give a unique answer. Some reactions produce the same product from different starting structures.

The dehydration of alcohols provides a useful example.

08Questions and solutions

Question 1

2-Bromopropane is treated with aqueous sodium hydroxide to form an organic intermediate X. X is then heated under reflux with acidified potassium dichromate.

Identify X and the organic product formed during oxidation. Explain the reaction sequence.

Solution 1

X is propan-2-ol, and the oxidation product is propanone.

Aqueous sodium hydroxide promotes substitution of the bromine atom by a hydroxyl group.

The bromine is attached to carbon 2, so the hydroxyl group also becomes attached to carbon 2.

\[ \ce{CH3CHBrCH3 + NaOH -> CH3CH(OH)CH3 + NaBr} \]

The intermediate is propan-2-ol, a secondary alcohol.

Secondary alcohols undergo oxidation to form ketones. Therefore:

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

The organic product is propanone.

A common mistake is to identify the intermediate as propan-1-ol because sodium hydroxide supplies the hydroxyl group. However, substitution does not move the hydroxyl group to a different carbon. It replaces bromine at its original position.

Question 2

An unknown straight-chain alcohol X has the molecular formula \(\ce{C5H12O}\).

When X is heated under reflux with excess acidified potassium dichromate, it produces pentan-3-one:

\[ \ce{CH3CH2COCH2CH3} \]

Identify X and explain why pentan-1-ol and pentan-2-ol are not suitable candidates.

Solution 2

X is pentan-3-ol, \(\ce{CH3CH2CH(OH)CH2CH3}\).

The product, pentan-3-one, contains a ketone functional group.

A ketone can form through oxidation of a secondary alcohol. The carbon attached to the hydroxyl group becomes the carbonyl carbon.

In pentan-3-one, the carbonyl group is on carbon 3.

Therefore, the starting alcohol must have had its hydroxyl group attached to carbon 3:

\[ \ce{CH3CH2CH(OH)CH2CH3} \]

This is pentan-3-ol.

Now consider the alternative structures.

• Pentan-1-ol is a primary alcohol. Its oxidation would produce pentanal and, with further oxidation, pentanoic acid.
• Pentan-2-ol is a secondary alcohol, but oxidation would produce pentan-2-one, with the carbonyl group on carbon 2.

Neither produces the specified pentan-3-one.

The important evidence is not simply that X is an oxidisable alcohol. Its hydroxyl group must occupy the correct position in the carbon chain.

Question 3

An unknown alcohol X has the molecular formula \(\ce{C4H10O}\).

Two separate portions of X are oxidised using acidified potassium dichromate.

• In the first experiment, the organic product is distilled off as it forms. The product is 2-methylpropanal, \(\ce{(CH3)2CHCHO}\).
• In the second experiment, X is heated under reflux with excess oxidising agent. The product is 2-methylpropanoic acid, \(\ce{(CH3)2CHCOOH}\).

Identify X and explain why the two experiments produce different organic products from the same starting material.

Solution 3

X is 2-methylpropan-1-ol.

Both products contain the same branched four-carbon skeleton. The difference is the oxidation state of the functional group.

In the first experiment, controlled oxidation produces an aldehyde.

Aldehydes can form from primary alcohols, so X must be a primary alcohol.

The aldehyde structure shows that the carbonyl carbon is at the end of the chain. Replacing the carbonyl oxygen arrangement with a hydroxyl group gives:

\[ \ce{(CH3)2CHCH2OH} \]

This is 2-methylpropan-1-ol.

The second experiment provides an independent check. Under reflux with excess acidified dichromate, the primary alcohol undergoes further oxidation to the carboxylic acid.

The two stages are:

\[ \ce{(CH3)2CHCH2OH} \]

\[ \ce{-> (CH3)2CHCHO} \]

\[ \ce{-> (CH3)2CHCOOH} \]

The different apparatus explains the different products.

In the first experiment, distillation removes the aldehyde before it can undergo substantial further oxidation.

In the second experiment, reflux keeps the organic compounds in contact with the oxidising agent, allowing oxidation to continue to the carboxylic acid.

A tempting interpretation is that the two products must have come from different alcohols. That is incorrect. The same primary alcohol can produce both an aldehyde and a carboxylic acid, depending on the reaction conditions.

Question 4

2-Methylpropene undergoes acid-catalysed hydration to produce an alcohol X with the molecular formula \(\ce{C4H10O}\).

When X is heated with acidified potassium dichromate under standard oxidation conditions, the solution does not change from orange to green.

Identify X and explain both observations.

Solution 4

X is 2-methylpropan-2-ol, a tertiary alcohol.

During acid-catalysed hydration, water adds across the carbon-carbon double bond.

For 2-methylpropene, the major product is 2-methylpropan-2-ol:

\[ \ce{CH2=C(CH3)2 + H2O -> (CH3)3COH} \]

The hydroxyl group is attached to the central carbon, which is bonded to three other carbon atoms.

This makes X a tertiary alcohol.

Tertiary alcohols do not undergo oxidation with acidified potassium dichromate under the standard conditions considered here.

Therefore, the dichromate solution remains orange rather than changing towards green.

The misconception is that every alcohol must produce an orange-to-green colour change with acidified dichromate.

That is not correct. The response depends on the alcohol’s structure. Primary and secondary alcohols can be oxidised under these conditions, whereas tertiary alcohols generally cannot.

Question 5

An unknown pure compound X is known to be an alcohol with the molecular formula \(\ce{C4H10O}\).

Three experiments produce the following results.

Experiment A: X is dehydrated using concentrated sulfuric acid and heat. The only alkene identified is 2-methylpropene, \(\ce{CH2=C(CH3)2}\).

Experiment B: A fresh portion of X is oxidised using acidified potassium dichromate. The organic product is distilled off as it forms. The isolated product is 2-methylpropanal, \(\ce{(CH3)2CHCHO}\).

Experiment C: Another fresh portion of X is heated under reflux with excess acidified potassium dichromate. The final organic product is 2-methylpropanoic acid, \(\ce{(CH3)2CHCOOH}\).

Two possible structures appear consistent with the dehydration result:

• 2-methylpropan-1-ol, \(\ce{(CH3)2CHCH2OH}\)
• 2-methylpropan-2-ol, \(\ce{(CH3)3COH}\)

Both can undergo dehydration to produce 2-methylpropene.

Which structure is X?

Explain why the dehydration result alone cannot distinguish the two candidates, why the oxidation results can, and why it would be incorrect to assume that hydration of the alkene must reproduce the original alcohol.

Assume the compound is pure, no carbon-skeleton rearrangement occurs, and all reported organic products have been correctly identified.

Solution 5

X is 2-methylpropan-1-ol. The oxidation experiments establish that it is a primary alcohol, while the dehydration experiment alone cannot distinguish it from the tertiary alcohol.

This question requires you to compare three independent pieces of evidence rather than rely on one reaction.

Step 1: Examine the dehydration result.

The identified alkene is 2-methylpropene:

\[ \ce{CH2=C(CH3)2} \]

Both proposed alcohols can form this product.

For 2-methylpropan-1-ol:

\[ \ce{(CH3)2CHCH2OH -> (CH3)2C=CH2 + H2O} \]

For 2-methylpropan-2-ol:

\[ \ce{(CH3)3COH -> (CH3)2C=CH2 + H2O} \]

In the first reaction, the hydroxyl group leaves from the terminal carbon, and a hydrogen is removed from the neighbouring carbon.

In the second reaction, the hydroxyl group leaves from the central carbon, and a hydrogen is removed from one of its neighbouring methyl groups.

Both reactions produce the same carbon-carbon double bond.

Therefore, Experiment A does not uniquely identify X.

Step 2: Use the controlled oxidation result.

Experiment B produces 2-methylpropanal.

An aldehyde forms through controlled oxidation of a primary alcohol.

Of the two proposed candidates:

• 2-methylpropan-1-ol is primary.
• 2-methylpropan-2-ol is tertiary.

The aldehyde result therefore supports 2-methylpropan-1-ol and rules out 2-methylpropan-2-ol.

The carbon skeleton also matches. Both the alcohol and aldehyde contain four carbon atoms, with the same branching arrangement.

Step 3: Check the result against further oxidation.

Experiment C produces 2-methylpropanoic acid.

This is exactly what you would expect if a primary alcohol underwent further oxidation after forming an aldehyde.

\[ \ce{(CH3)2CHCH2OH} \]

\[ \ce{-> (CH3)2CHCHO} \]

\[ \ce{-> (CH3)2CHCOOH} \]

The two oxidation experiments are consistent with the same starting compound.

By contrast, 2-methylpropan-2-ol is a tertiary alcohol. It would not normally produce either the aldehyde or the carboxylic acid under these oxidation conditions.

Step 4: Resolve the apparent contradiction involving hydration.

This is the most important conceptual part of the question.

You might reason that if 2-methylpropan-2-ol dehydrates to 2-methylpropene, then adding water back to 2-methylpropene must regenerate 2-methylpropan-2-ol.

That prediction is tempting, but it assumes that dehydration and hydration are perfectly reversible at the level of individual molecular structures.

They are not.

Dehydration can produce the same alkene from two different alcohol structures. It is therefore a many-to-one transformation: two different starting compounds produce the same product.

Under acid-catalysed hydration conditions, 2-methylpropene produces mainly 2-methylpropan-2-ol because the hydroxyl group ends up on the more substituted carbon.

That does not prove that the original alcohol, before dehydration, had its hydroxyl group at that position.

Step 5: Combine all three experiments.

Only 2-methylpropan-1-ol satisfies all three observations.

The key lesson is that a reaction product does not always uniquely reveal its starting structure. When several pathways are possible, use independent evidence to eliminate competing structures rather than simply reversing the last reaction.