How to Identify Missing Reagents in Organic Flow Charts

Your organic reaction flow chart ends at butanone, but the reagent above the arrow from butan-2-ol is missing. What should you write?

The answer is an oxidising agent, such as acidified potassium dichromate, with heating under reflux. Butan-2-ol is a secondary alcohol, and oxidation converts secondary alcohols into ketones.

This gives you the most reliable method for identifying a missing reagent in an organic flow chart: start with the desired product, work backwards to identify the required reaction, then select a reagent and conditions that produce it.

Before moving on, consider one detail. Would the same reagent convert butan-1-ol into butanone?

No. Butan-1-ol is a primary alcohol. Its oxidation produces butanal first and can continue to produce butanoic acid. The position of the hydroxyl group changes the reaction pathway.

The trick is to stop treating reagents as isolated facts to memorise. Each missing reagent must make sense when you compare the starting structure, the desired product, and the reaction conditions.

01Start at the product and work backwards

When you see a reaction flow chart, your first instinct might be to read from left to right and identify the reaction at each arrow.

That works, but there is a more reliable approach when a reagent is missing.

Start at the right-hand side.

Look at the final molecule and ask:

• What is its functional group?
• What type of compound could be converted into this product in one reaction?
• What chemical change must occur?
• Which reagent and conditions cause that change?

For example, suppose the final product is an aldehyde.

An aldehyde contains a carbonyl group, \(\ce{C=O}\), at the end of its carbon chain. A likely precursor is a primary alcohol, which can be oxidised to form the aldehyde.

You can therefore work backwards:

\[ \text{Primary alcohol} \xrightarrow{\text{oxidation}} \text{Aldehyde} \]

Once you recognise the required transformation, choosing the reagent becomes much easier.

A quick reference table for HSC organic reactions

These are useful reaction patterns to recognise when completing an organic synthesis pathway.

These are common HSC reaction patterns, rather than a list of reactions that can be applied without checking the actual structures.

For example, seeing an alcohol does not automatically mean you should choose an oxidising agent. You might need to produce an alkene instead, in which case dehydration is the relevant reaction.

Identify the functional group before choosing the reagent

If you struggle to recognise the product’s functional group, revise that first.

A useful starting point is HSC Chemistry: How to Identify Organic Functional Groups.

You do not need to memorise an entirely separate reaction for every molecule. Instead, learn which transformations apply to each functional group and then check whether the particular structure allows that transformation.

02Check the reaction conditions, not just the reagent

This is where many otherwise reasonable answers go wrong.

Two reaction pathways can use similar reagents but produce different organic products because the conditions are different.

Three distinctions are particularly important.

Aqueous versus ethanolic hydroxide

Consider 1-bromobutane:

\[ \ce{CH3CH2CH2CH2Br} \]

It can undergo two different reactions depending on the solvent used with the hydroxide.

Aqueous sodium hydroxide produces an alcohol.

The reaction is substitution. The hydroxide group replaces the bromine atom.

\[ \ce{CH3CH2CH2CH2Br + NaOH} \]

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

The organic product is butan-1-ol.

Ethanolic potassium hydroxide produces an alkene.

The reaction is elimination. A hydrogen atom and the bromine atom are removed from adjacent carbon atoms, forming a carbon-carbon double bond.

\[ \ce{CH3CH2CH2CH2Br + KOH} \]

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

The organic product is but-1-ene.

Notice that the starting compound is identical.

The difference is the reaction environment.

If a flow chart shows a haloalkane becoming an alcohol, choose aqueous hydroxide. If it shows a haloalkane becoming an alkene, choose ethanolic hydroxide.

Writing simply \(\ce{NaOH}\) or \(\ce{KOH}\) may be insufficient. Include the solvent and heating conditions when they determine the product.

Distillation versus reflux

Oxidation of primary alcohols creates another important distinction.

A primary alcohol can be oxidised to an aldehyde, which can then undergo further oxidation to a carboxylic acid.

Consider butan-1-ol:

\[ \ce{CH3CH2CH2CH2OH} \]

The first oxidation produces butanal:

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

\[ \ce{-> CH3CH2CH2CHO + H2O} \]

Further oxidation produces butanoic acid:

\[ \ce{CH3CH2CH2CHO + [O]} \]

\[ \ce{-> CH3CH2CH2COOH} \]

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

How do you control which product forms?

The answer involves the amount of oxidising agent and the apparatus.

In distillation, the aldehyde is removed from the reaction mixture as it forms. This reduces the opportunity for further oxidation.

In reflux, vapour rises into a condenser, cools, and returns to the reaction flask. This keeps volatile substances in contact with the oxidising agent for longer.

For HSC questions, remember:

• Primary alcohol to aldehyde: controlled oxidation and distillation.
• Primary alcohol to carboxylic acid: excess oxidising agent and reflux.
• Secondary alcohol to ketone: oxidation.
• Tertiary alcohol: does not undergo oxidation under ordinary HSC oxidation conditions.

The oxidising reagent might be the same in the first three pathways. The conditions determine the outcome.

Concentrated sulfuric acid can have different roles

Another common source of confusion is concentrated sulfuric acid.

It can act as a catalyst in esterification, but it can also promote dehydration of an alcohol.

These are different reactions.

For dehydration:

\[ \ce{Alcohol ->[conc. H2SO4,\ heat] Alkene + H2O} \]

For esterification:

\[ \ce{Carboxylic\ acid + Alcohol} \]

\[ \ce{<=>[conc. H2SO4,\ heat] Ester + H2O} \]

The other organic reactant matters. Concentrated sulfuric acid does not independently turn an alcohol into an ester.

Always look at the structures on both sides of the arrow before deciding what a reagent is doing.

03Use carbon count and reagent availability as final checks

You can identify a reaction correctly and still construct an impossible pathway.

Two final checks help prevent this.

Check the carbon skeleton

Most of the reactions discussed so far preserve the original carbon skeleton.

For example:

• Substitution changes the functional group but retains the carbon chain.
• Elimination forms a double bond but retains the carbon chain.
• Addition changes the bonding pattern but retains the carbon count.
• Oxidation changes the functional group without normally rearranging the carbon skeleton.

If you begin with a four-carbon molecule and the next product contains five carbon atoms, something is wrong unless another carbon-containing reactant has been introduced.

There is an important exception to the simple one-molecule comparison: esterification.

An ester is produced by combining a carboxylic acid and an alcohol. Both organic reactants contribute carbon atoms.

For example, ethyl propanoate contains five carbon atoms:

\[ \ce{CH3CH2COOCH2CH3} \]

Three come from propanoic acid, and two come from ethanol.

The concentrated sulfuric acid is a catalyst. It does not provide the additional carbon atoms.

This distinction is especially useful when a flow chart contains an ester as its final product.

For a more detailed treatment, see How to Check Atom Conservation in Multi-Step Organic Synthesis.

Check which reagents are actually available

Sometimes an exam question provides a list of available reagents.

Treat this list as a constraint.

Suppose you need to oxidise a primary alcohol, but the available reagents include:

• Aqueous sodium hydroxide
• Concentrated sulfuric acid
• Acidified potassium dichromate
• Ethanolic potassium hydroxide

The appropriate choice is acidified potassium dichromate.

Do not choose concentrated sulfuric acid simply because it contains hydrogen and oxygen. It is not the oxidising agent required for this transformation.

Likewise, if the target requires an alcohol and your reagent list contains both aqueous and ethanolic hydroxide, you must choose the conditions that favour substitution.

One further point: a reaction does not always have a uniquely identifiable reagent. Several oxidising agents, for example, can perform the same general transformation. If the question provides no reagent bank, give a chemically valid option and specify the necessary conditions. Do not assume that the product alone identifies one unique chemical.

04Worked examples

Let’s put the method into practice.

05A common misconception: choosing a reagent by name alone

A particularly tempting mistake is to memorise a reagent without considering the solvent or reaction conditions.

For example:

\[ \text{Haloalkane}+\text{Hydroxide} \]

A student might immediately predict an alcohol.

That is not always correct.

Aqueous hydroxide favours substitution, while hydroxide in ethanol favours elimination under suitable heating conditions.

The same starting material can therefore produce different organic compounds.

When answering an exam question, make sure your reagent description is specific enough to distinguish the possible reactions.

For example, instead of writing:

“Add sodium hydroxide.”

Write:

“Heat with aqueous sodium hydroxide to favour substitution, replacing the bromine atom with a hydroxyl group.”

That explanation identifies the reagent, the conditions, and the chemical reason for choosing them.

06Questions and solutions

Try these questions without looking back at the worked examples. They progress from straightforward reagent identification to more complicated pathways involving competing reactions and structural isomers.

Question 1

A reaction flow chart shows the conversion:

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

The available reagents are:

• Aqueous sodium hydroxide
• Ethanolic potassium hydroxide
• Acidified potassium dichromate

Identify the missing reagent and state the required conditions. Explain the type of reaction occurring.

Solution 1

The missing reagent is aqueous sodium hydroxide, with heating.

The reaction is substitution because the bromine atom in 1-bromopropane is replaced by a hydroxyl group.

The reaction is:

\[ \ce{CH3CH2CH2Br + NaOH} \]

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

The three-carbon chain remains unchanged.

Ethanolic potassium hydroxide would favour elimination, producing an alkene rather than an alcohol. Acidified potassium dichromate is an oxidising agent, so it is unsuitable for replacing bromine with a hydroxyl group.

Question 2

A primary alcohol undergoes two successive oxidation reactions:

\[ \ce{CH3CH2CH2CH2OH} \]

\[ \ce{->[A] CH3CH2CH2CHO} \]

\[ \ce{->[B] CH3CH2CH2COOH} \]

Acidified potassium dichromate is the only available oxidising agent.

Identify the conditions required for A and B.

Explain why the first product must be removed from the reaction mixture if the desired product is butanal.

Solution 2

For A, use limited acidified potassium dichromate, gentle heating, and distillation.

For B, use excess acidified potassium dichromate and heating under reflux.

The starting compound is butan-1-ol, a primary alcohol. Its first oxidation produces butanal.

If the aldehyde remains in contact with the oxidising agent, it can undergo further oxidation to butanoic acid.

Distillation removes the butanal as it forms, limiting this further oxidation.

For the second reaction, the desired product is butanoic acid. Reflux allows prolonged contact between the aldehyde and the excess oxidising agent, favouring complete oxidation.

The two transformations are:

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

\[ \ce{-> CH3CH2CH2CHO + H2O} \]

Then:

\[ \ce{CH3CH2CH2CHO + [O]} \]

\[ \ce{-> CH3CH2CH2COOH} \]

Both reactions retain the four-carbon skeleton.

Question 3

A student needs to convert 1-bromobutane into but-1-ene.

They write the following reagent above the reaction arrow:

\[ \ce{NaOH} \]

Explain why this answer is incomplete.

Identify the correct reagent and conditions. Then predict the organic product if aqueous sodium hydroxide were used instead.

Solution 3

The correct reagent is ethanolic potassium hydroxide, with heating.

The reaction is elimination.

A hydrogen atom and the bromine atom are removed from adjacent carbon atoms, producing a carbon-carbon double bond.

\[ \ce{CH3CH2CH2CH2Br + KOH} \]

\[ \ce{->[ethanol,\ heat] CH2=CHCH2CH3} \]

The organic product is but-1-ene.

Writing only \(\ce{NaOH}\) is incomplete because the solvent determines which reaction is favoured.

If aqueous sodium hydroxide were used instead, substitution would occur, producing butan-1-ol:

\[ \ce{CH3CH2CH2CH2Br + NaOH} \]

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

The key difference is:

• Ethanolic hydroxide favours elimination.
• Aqueous hydroxide favours substitution.

The reagent label must therefore include the solvent.

Question 4

A flow chart converts but-2-ene into butanone through butan-2-ol.

\[ \ce{CH3CH=CHCH3} \]

\[ \ce{->[A] CH3CH(OH)CH2CH3} \]

\[ \ce{->[B] CH3COCH2CH3} \]

The available reagents are:

• Hydrogen with a nickel catalyst
• Steam with a phosphoric acid catalyst
• Acidified potassium dichromate
• Concentrated sulfuric acid

Identify A and B, including their conditions.

Explain why hydrogen with a nickel catalyst is not suitable for the first reaction, even though it reacts with alkenes.

Solution 4

A is steam with a phosphoric acid catalyst, and B is acidified potassium dichromate with heating under reflux.

The first reaction is hydration.

Water is added across the carbon-carbon double bond, producing an alcohol.

\[ \ce{CH3CH=CHCH3 + H2O} \]

\[ \ce{->[H3PO4] CH3CH(OH)CH2CH3} \]

The second reaction is oxidation.

Butan-2-ol is a secondary alcohol, so oxidation produces butanone.

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

\[ \ce{-> CH3COCH2CH3 + H2O} \]

Acidified potassium dichromate is therefore the correct oxidising agent.

Hydrogen with a nickel catalyst is a tempting alternative because it also reacts with the double bond in an alkene.

However, hydrogenation adds hydrogen across the double bond, producing an alkane:

\[ \ce{CH3CH=CHCH3 + H2} \]

\[ \ce{->[Ni] CH3CH2CH2CH3} \]

This produces butane, not butan-2-ol.

The crucial difference is that hydration introduces the elements of water, allowing the hydroxyl group to form.

The carbon count remains four in both the hydration and oxidation reactions.

Question 5

A chemist wants to produce 2-methylpropanoic acid using a two-step reaction pathway:

\[ \ce{P ->[A] X ->[B] (CH3)2CHCOOH} \]

The starting haloalkane, P, is one of three structural isomers with molecular formula \(\ce{C4H9Br}\).

The possible starting compounds are:

Compound I: 1-bromobutane

\[ \ce{CH3CH2CH2CH2Br} \]

Compound II: 1-bromo-2-methylpropane

\[ \ce{(CH3)2CHCH2Br} \]

Compound III: 2-bromo-2-methylpropane

\[ \ce{(CH3)3CBr} \]

The available conditions are:

• A: Aqueous sodium hydroxide with heating, or ethanolic potassium hydroxide with heating.
• B: Acidified potassium dichromate, with either distillation or reflux using excess oxidising agent.

The intended pathway must contain exactly one substitution reaction followed by one oxidation reaction. No carbon-skeleton rearrangement is allowed.

Determine:

1. Which starting compound must be selected.
2. The correct reagent and conditions for A.
3. The correct reagent and conditions for B.
4. Why the other two starting compounds cannot produce the specified target through the required pathway.

Solution 5

The correct starting compound is compound II, 1-bromo-2-methylpropane.

Use aqueous sodium hydroxide with heating for A, followed by excess acidified potassium dichromate under reflux for B.

This question requires you to consider more than the molecular formula. You must track the carbon skeleton, the position of the bromine atom, and the type of alcohol formed.

Step 1: Work backwards from the target.

The target is 2-methylpropanoic acid:

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

It contains a carboxylic acid functional group.

Under the specified reaction pathway, it can be produced by oxidising a primary alcohol.

The required intermediate is therefore 2-methylpropan-1-ol:

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

Notice that the hydroxyl group must be attached to the terminal carbon.

Step 2: Identify which haloalkane produces this alcohol.

A substitution reaction replaces the bromine atom with a hydroxyl group without rearranging the carbon skeleton.

For compound II:

\[ \ce{(CH3)2CHCH2Br} \]

Replacing bromine with hydroxyl gives:

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

This is exactly the required intermediate.

Therefore, compound II is the correct starting material.

Step 3: Identify reagent A.

The conversion of a haloalkane into an alcohol is substitution.

Use aqueous sodium hydroxide with heating:

\[ \ce{(CH3)2CHCH2Br + NaOH} \]

\[ \ce{-> (CH3)2CHCH2OH + NaBr} \]

Ethanolic potassium hydroxide would favour elimination, producing an alkene rather than the required primary alcohol.

Step 4: Identify reagent B.

The intermediate is 2-methylpropan-1-ol, a primary alcohol.

Its oxidation first produces 2-methylpropanal and can continue to produce 2-methylpropanoic acid.

To obtain the carboxylic acid, use excess acidified potassium dichromate and heat under reflux.

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

\[ \ce{-> (CH3)2CHCOOH + H2O} \]

Therefore, B requires excess acidified \(\ce{K2Cr2O7}\) under reflux.

Step 5: Explain why the other starting compounds fail.

Compound I, 1-bromobutane, initially looks reasonable.

It also contains four carbon atoms and undergoes substitution to form an alcohol.

However, the resulting alcohol is butan-1-ol:

\[ \ce{CH3CH2CH2CH2OH} \]

Oxidation produces butanoic acid:

\[ \ce{CH3CH2CH2COOH} \]

This is a straight-chain carboxylic acid, not the branched 2-methylpropanoic acid.

The carbon skeleton cannot simply rearrange during the specified substitution and oxidation reactions.

Compound III, 2-bromo-2-methylpropane, presents a different problem.

Substitution would produce 2-methylpropan-2-ol:

\[ \ce{(CH3)3COH} \]

This is a tertiary alcohol.

Tertiary alcohols do not undergo oxidation to carboxylic acids under ordinary HSC oxidation conditions. Producing the target would require a different reaction pathway.

Step 6: Check the complete answer.

The correct pathway is:

\[ \ce{(CH3)2CHCH2Br} \]

\[ \ce{->[NaOH(aq),\ heat] (CH3)2CHCH2OH} \]

\[ \ce{->[K2Cr2O7/H+,\ excess,\ reflux] (CH3)2CHCOOH} \]

The four-carbon branched skeleton is retained throughout.

The important distinction is that all three starting compounds have the same molecular formula and carbon count, but their structures produce different intermediates and reaction possibilities.

A correct reagent must work with the actual structure, not just its molecular formula.

07A reliable method to remember

When you encounter an incomplete organic flow chart, follow this sequence:

1. Start at the target. Identify its functional group and structural arrangement.
2. Find the immediate precursor. Work backwards one reaction at a time.
3. Name the transformation. Decide whether it involves substitution, elimination, addition, dehydration, hydration, oxidation, or esterification.
4. Select the reagent. Match the reagent to the chemical change.
5. Specify the conditions. Check the solvent, temperature, concentration, and whether distillation or reflux is required.
6. Check the constraints. Confirm that the carbon skeleton is consistent and every required organic reactant is available.
7. Read forwards again. Make sure every reaction produces the next compound shown.

The last step is particularly useful. A pathway can look convincing when you work backwards but still contain an incorrect intermediate or an incompatible reaction condition.

Once you can identify the missing reagents, the next challenge is checking whether the entire synthesis is chemically possible. That means accounting for every carbon atom, recognising which reactants contribute atoms, and checking that each intermediate can genuinely undergo the next reaction.

That is the skill developed in How to Check Atom Conservation in Multi-Step Organic Synthesis.

Syllabus note: As at November 2026, Year 12 students are still following the Chemistry Stage 6 Syllabus (2017). The Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028, with Year 12 implementation in Term 4 of 2028 and the first HSC examination in 2029. The reaction examples in this resource are framed around familiar 2017 Module 7 organic chemistry pathways. Students entering the new syllabus should check its specific content requirements rather than assume every example is examinable. 【turn468656view0】【turn587123view1】