How to Use an Organic Reaction Table to Plan a Synthesis

An exam asks you to prepare methyl butanoate using butan-1-ol and methanol. You open your organic reaction table and find an entry for alcohol oxidation and another for esterification. Which one should you use first? More importantly, should you oxidise both alcohols?

The answer becomes clear when you start with the product rather than the starting materials. Methyl butanoate requires a carboxylic acid and an alcohol. The butan-1-ol must be oxidised to butanoic acid, while the methanol must remain an alcohol so it can supply the methyl group.

This is the key to using an organic reaction table without memorising dozens of disconnected equations: work backwards from the desired functional group, then check the carbon skeleton, available reagents, and reaction conditions.

01Use the reaction table backwards

Think of an organic reaction table as a recipe index. Each entry connects three things:

• The starting functional group.
• The chemical change that occurs.
• The product functional group, together with the required reagents and conditions.

You can read the entry in reverse when planning a synthesis. If you want a ketone, for example, you look for a reaction that produces a ketone and identify its required starting material.

A functional group is the part of an organic molecule responsible for its characteristic chemical reactions. Common examples include:

• Alkene: carbon-carbon double bond, \(\ce{C=C}\).
• Alcohol: hydroxyl group, \(\ce{-OH}\).
• Haloalkane: carbon-halogen bond, such as \(\ce{C-Br}\).
• Aldehyde: terminal carbonyl group, \(\ce{-CHO}\).
• Ketone: carbonyl group, \(\ce{C=O}\), within the carbon chain.
• Carboxylic acid: carboxyl group, \(\ce{-COOH}\).
• Ester: \(\ce{-COO-}\) linkage.

The functional group tells you which reaction families are worth considering. The rest of the molecular structure tells you whether a particular reaction will actually produce the molecule you need.

A reverse lookup table

The following table reorganises common HSC organic reactions around the product you want.

This table gives you a starting point, not an automatic answer. You still need to check whether the actual starting molecule can undergo the proposed reaction.

For example, the entry for ketones tells you to look for a secondary alcohol. It does not mean that oxidising any alcohol will produce a ketone.

A practical five-step method

When you receive a synthesis question, use this sequence.

Step 1: Draw the target molecule.

Do not rely on its name alone. Write its structural formula and identify the functional group.

Step 2: Find the reaction that produces that functional group.

Search the product side of your reaction table. Identify the immediate precursor needed to make the target.

Step 3: Check the carbon skeleton.

Count the carbon atoms and check their arrangement. Most of the common reactions in the table preserve the carbon skeleton.

Step 4: Check the reagents and conditions.

Look for details such as aqueous versus ethanolic solutions, concentrated versus dilute acids, excess oxidising agent, and heating under reflux or distillation.

Step 5: Work backwards again if necessary.

If your immediate precursor is not available, find another reaction that produces it. Continue until you reach the starting materials provided in the question.

The important distinction is that you are not memorising every possible equation. You are learning which transformations connect functional groups, then checking whether the particular molecule satisfies the requirements.

If you want to build this into a complete pathway-reading method, see How to Read Organic Reaction Pathways in HSC Chemistry.

02Check the carbon count before accepting a reaction

Imagine trying to turn a three-carbon molecule into a five-carbon molecule using only oxidation. Something is missing.

Oxidation can change the bonding and oxidation state of carbon atoms, but it does not normally create additional carbon atoms in the organic product.

This gives you a useful restriction when reading your table.

Esterification is particularly important because it combines two organic molecules.

Why esterification needs two starting materials

An ester has the general structural arrangement:

\[
\ce{RCOOR’}
\]

The \(\ce{RCOO}\) section comes from a carboxylic acid, while the \(\ce{R’}\) section comes from an alcohol.

Consider methyl butanoate:

\[
\ce{CH3CH2CH2COOCH3}
\]

Its structure contains:

• Four carbon atoms in the butanoate section.
• One carbon atom in the methyl section.

The total is:

\[
4+1=5\text{ carbon atoms}
\]

Therefore, the required reactants are butanoic acid and methanol.

\[
\ce{CH3CH2CH2COOH + CH3OH}
\]

They combine to form methyl butanoate and water.

A common mistake is to count the carbon atoms correctly but assign the wrong starting material to each section of the ester.

The name matters. In methyl butanoate, the methyl group comes from the alcohol, while the butanoate section comes from the acid.

This is why checking the total carbon count is necessary but not always sufficient. You must also check where those carbon atoms end up.

03Reagents and conditions are part of the reaction

Suppose your table says:

Primary alcohol → carboxylic acid.

You might conclude that adding an oxidising agent to a primary alcohol always gives a carboxylic acid.

That is an incomplete prediction.

A primary alcohol can first form an aldehyde. Under further oxidation, the aldehyde can form a carboxylic acid.

For example:

\[
\ce{CH3CH2CH2OH}
\]

Propan-1-ol can undergo two stages of oxidation:

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

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

The final product depends on how the reaction is carried out.

Experimental explanation: Why distillation and reflux give different products

Imagine an investigation designed to answer this question:

Can we produce propanal from propan-1-ol without allowing it to oxidise further into propanoic acid?

The reaction uses acidified potassium dichromate as the oxidising agent.

The solution initially appears orange because of the dichromate ions. As oxidation occurs, the dichromate is reduced to green chromium(III) ions.

However, the apparatus and reaction conditions determine whether the aldehyde is removed or remains in contact with the oxidising agent.

The first arrangement: heating under reflux

Place propan-1-ol and acidified potassium dichromate in a round-bottom flask. Heat the flask with a vertical water-cooled condenser attached above it.

Vapour rises from the heated mixture, enters the condenser, cools, and returns as liquid to the same flask.

This arrangement is called reflux.

If excess oxidising agent is present, the propanal produced remains in contact with the oxidising agent. It can undergo further oxidation to propanoic acid.

The prediction: Reflux with excess oxidising agent favours the carboxylic acid rather than allowing the aldehyde to be collected.

What does the colour change tell us?

The orange-to-green colour change indicates that dichromate ions have been reduced during the oxidation reaction.

It provides evidence that an oxidation reaction has occurred. However, it does not establish whether the organic product is propanal or propanoic acid.

The colour change identifies the behaviour of the oxidising agent, not the exact identity of the organic product.

The revised arrangement: distillation

To favour the aldehyde, arrange the apparatus for distillation instead.

As propanal forms, it can vaporise and travel through the condenser into a separate receiving flask. It is removed from the reaction mixture rather than continuously returning to the oxidising agent.

This reduces the opportunity for further oxidation.

The reaction table can therefore distinguish between:

The key experimental lesson is that the reagent alone does not tell you the complete outcome. The apparatus, reaction conditions, and extent of oxidation all matter.

This is exactly why you should never ignore the conditions column in an organic reaction table.

04Worked examples

05Questions and solutions

The following questions progress from straightforward reaction-table lookup to more demanding synthesis problems. For each one, use the product-first method before selecting a reaction.

Question 1

2-bromopropane is converted into propan-2-ol.

Using an organic reaction table:

a) Identify the reaction type.

b) State a suitable reagent and conditions.

c) Write a balanced chemical equation.

d) Explain whether the number of carbon atoms changes.

Solution 1

2-bromopropane undergoes substitution with aqueous sodium hydroxide when heated, producing propan-2-ol.

a) Reaction type

The reaction is nucleophilic substitution at the level required for this reaction pathway.

The bromine atom is replaced by a hydroxyl group.

b) Reagent and conditions

Use aqueous sodium hydroxide and heat.

The aqueous condition is important because it favours substitution to form the alcohol.

c) Balanced equation

\[
\ce{CH3CHBrCH3 + NaOH}
\]

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

d) Carbon count

2-bromopropane contains three carbon atoms.

Propan-2-ol also contains three carbon atoms.

The carbon skeleton is preserved because substitution changes the attached functional group rather than adding or removing carbon atoms.

Question 2

A student wants to produce propanal from propan-1-ol.

They consult an organic reaction table and identify controlled oxidation as the appropriate reaction. However, they heat propan-1-ol with excess acidified potassium dichromate under reflux.

The solution changes from orange to green.

a) Explain why the chosen conditions are unsuitable for selectively producing propanal.

b) Identify the likely final organic product under these conditions.

c) Describe how the apparatus and reaction conditions should be changed to favour propanal.

d) Explain why the observed colour change does not prove that propanal has formed.

Solution 2

The student’s reflux arrangement with excess oxidising agent favours propanoic acid, not the selective production of propanal.

a) Why the conditions are unsuitable

Propan-1-ol is a primary alcohol.

Its oxidation can occur in two stages:

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

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

The first product is propanal. Further oxidation produces propanoic acid.

Under reflux, the volatile aldehyde remains in contact with the oxidising agent because its vapour condenses and returns to the reaction flask.

With excess oxidising agent, further oxidation can occur.

b) Likely final product

The expected product is propanoic acid.

\[
\ce{CH3CH2CH2OH + 2[O]}
\]

\[
\ce{-> CH3CH2COOH + H2O}
\]

c) How to favour propanal

Use controlled oxidation and simple distillation.

The reaction mixture is heated, and the propanal produced travels through a condenser into a separate receiving flask.

Removing the aldehyde as it forms reduces its contact with the oxidising agent, limiting further oxidation.

The amount of oxidising agent must also be controlled.

d) Why the colour change is insufficient

Acidified potassium dichromate changes from orange to green when dichromate ions are reduced to chromium(III) ions.

This confirms that an oxidation reaction has occurred.

It does not distinguish between propanal and propanoic acid. Both can be produced through oxidation of propan-1-ol, so the product must be identified separately.

The trap is confusing evidence that a reaction occurred with evidence of exactly which product formed.

Question 3

Two different alcohols, A and B, each have the molecular formula \(\ce{C4H10O}\).

Their oxidation products are investigated.

• Alcohol A produces butanone.
• Alcohol B produces butanoic acid when heated under reflux with excess acidified potassium dichromate.

a) Identify the structural formula and alcohol classification of A.

b) Identify the structural formula and alcohol classification of B.

c) A student suggests that B could be 2-methylpropan-1-ol because it also contains four carbon atoms. Evaluate this suggestion.

d) Explain why the molecular formula alone is insufficient to select the correct reaction-table entry.

Solution 3

A is butan-2-ol, a secondary alcohol, and B is butan-1-ol, a primary alcohol.

The different oxidation products arise from the position of the hydroxyl group and the arrangement of the carbon skeleton.

a) Identify alcohol A

Butanone is a ketone.

Ketones are produced by oxidising secondary alcohols.

The required structure is:

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

This is butan-2-ol.

Its hydroxyl group is attached to carbon 2, which is bonded to two other carbon atoms. Therefore, it is a secondary alcohol.

Its oxidation is:

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

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

b) Identify alcohol B

Butanoic acid is produced by further oxidation of a primary alcohol.

The required structure is:

\[
\ce{CH3CH2CH2CH2OH}
\]

This is butan-1-ol.

Its hydroxyl group is attached to a terminal carbon, which is bonded to only one other carbon atom. Therefore, it is a primary alcohol.

Its oxidation is:

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

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

c) Evaluate the suggested alternative

The suggestion is incorrect.

2-methylpropan-1-ol has the structure:

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

It is a primary alcohol, but its carbon skeleton is branched.

Oxidation produces 2-methylpropanoic acid:

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

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

This is not butanoic acid.

Both acids contain four carbon atoms, but their structures are different.

d) Why the molecular formula is insufficient

The molecular formula \(\ce{C4H10O}\) does not tell us where the hydroxyl group is located or whether the carbon chain is branched.

Different structural isomers can therefore belong to different reaction pathways or produce different oxidation products.

You must identify the actual structure before choosing the reaction-table entry.

Question 4

A chemist has butan-1-ol available and wants to produce butan-2-one.

The reaction table contains the following transformations:

• Alcohols can undergo dehydration to form alkenes.
• Alkenes can undergo hydration to form alcohols.
• Secondary alcohols can undergo oxidation to form ketones.

For this question, assume acid-catalysed hydration of but-1-ene produces butan-2-ol as the major product.

a) Construct the shortest pathway from butan-1-ol to butan-2-one using the listed transformations.

b) State the reagents and conditions for each step.

c) Write the chemical equations.

d) Explain why directly oxidising butan-1-ol does not produce the desired ketone.

Solution 4

The required pathway is dehydration, followed by hydration, followed by oxidation.

The crucial step is changing the position of the hydroxyl group before oxidation.

Step 1: Dehydrate butan-1-ol

Butan-1-ol is a primary alcohol.

Heating it with concentrated sulfuric acid produces but-1-ene through dehydration.

\[
\ce{CH3CH2CH2CH2OH}
\]

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

This removes water and creates a carbon-carbon double bond.

Step 2: Hydrate but-1-ene

The target is butan-2-one, which requires a secondary alcohol as its immediate precursor.

The specified hydration conditions produce butan-2-ol as the major product.

\[
\ce{CH2=CHCH2CH3 + H2O}
\]

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

Use water or steam with an appropriate acid catalyst.

Notice that the hydroxyl group is now attached to carbon 2 rather than carbon 1.

Step 3: Oxidise butan-2-ol

Butan-2-ol is a secondary alcohol.

Oxidation with acidified potassium dichromate and heating produces butan-2-one.

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

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

Step 4: Explain why direct oxidation fails

Direct oxidation of butan-1-ol follows the primary alcohol pathway.

It produces butanal initially and can undergo further oxidation to butanoic acid.

It does not directly produce butan-2-one because the hydroxyl group is attached to the wrong carbon.

The dehydration and hydration steps provide a way to change the position of the hydroxyl group before oxidation.

Step 5: Check the carbon skeleton

Every transformation preserves the four-carbon skeleton.

The pathway is therefore consistent with the available reaction types and the required product structure.

Question 5

A chemist wants to prepare butyl ethanoate.

The only available carbon-containing starting materials are:

• 1-bromobutane, \(\ce{CH3CH2CH2CH2Br}\).
• Ethene, \(\ce{CH2=CH2}\).

The reaction table provides these transformations:

A student proposes the following route:

• Convert 1-bromobutane into butan-1-ol.
• Oxidise butan-1-ol into butanoic acid.
• Convert ethene into ethanol.
• React butanoic acid with ethanol to produce an ester.

The student argues that the proposed ester must be butyl ethanoate because the total number of carbon atoms is six.

a) Determine whether the student’s proposed route produces butyl ethanoate.

b) Construct a valid route to butyl ethanoate using the available starting materials.

c) Write the structural formula of the final product and the balanced esterification equation.

d) Explain why checking only the total number of carbon atoms gives an incorrect conclusion.

Solution 5

The student’s proposed route produces ethyl butanoate, not butyl ethanoate. A valid route must produce ethanoic acid from ethene and butan-1-ol from 1-bromobutane.

This is a more demanding problem because the two starting materials must supply the correct sections of the ester.

Step 1: Identify the structure of the target ester

Butyl ethanoate has the structural formula:

\[
\ce{CH3COOCH2CH2CH2CH3}
\]

Its two sections are:

• Ethanoate: two carbon atoms, supplied by ethanoic acid.
• Butyl: four carbon atoms, supplied by butan-1-ol.

Therefore, the required esterification reactants are:

\[
\ce{CH3COOH}
\]

and

\[
\ce{CH3CH2CH2CH2OH}
\]

We need to produce these two compounds separately.

Step 2: Produce the four-carbon alcohol

The starting material is 1-bromobutane.

Aqueous sodium hydroxide and heat cause substitution, replacing bromine with a hydroxyl group.

\[
\ce{CH3CH2CH2CH2Br + NaOH}
\]

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

The product is butan-1-ol.

This supplies the four-carbon alcohol section of the target ester.

Step 3: Produce the two-carbon carboxylic acid

The second starting material is ethene.

Hydration produces ethanol:

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

Next, oxidise the ethanol using excess acidified potassium dichromate under reflux.

\[
\ce{CH3CH2OH + 2[O]}
\]

\[
\ce{-> CH3COOH + H2O}
\]

The product is ethanoic acid.

This supplies the two-carbon acid section of the target ester.

Step 4: Combine the two correct fragments

We now have:

• Butan-1-ol from 1-bromobutane.
• Ethanoic acid from ethene.

Carry out esterification using concentrated sulfuric acid as a catalyst and heat.

\[
\ce{CH3COOH + CH3CH2CH2CH2OH}
\]

\[
\ce{<=> CH3COOCH2CH2CH2CH3 + H2O}
\]

The product is butyl ethanoate.

Step 5: Identify the error in the student’s proposed route

The student’s route produces butanoic acid from butan-1-ol:

\[
\ce{CH3CH2CH2CH2OH}
\]

\[
\ce{-> CH3CH2CH2COOH}
\]

The ethene is converted into ethanol:

\[
\ce{CH2=CH2 -> CH3CH2OH}
\]

Combining these two compounds through esterification gives:

\[
\ce{CH3CH2CH2COOH + CH3CH2OH}
\]

\[
\ce{<=> CH3CH2CH2COOCH2CH3 + H2O}
\]

This product is ethyl butanoate.

It has a different arrangement of carbon atoms around the ester linkage.

Step 6: Why the carbon count is not enough

Both esters contain six carbon atoms.

Both have the molecular formula:

\[
\ce{C6H12O2}
\]

However, they are structural isomers. They have the same molecular formula but different arrangements of atoms.

The student’s mistake was checking that the total carbon count was correct without checking which starting material supplied each section of the ester.

This is the important final check: a valid synthesis must produce the correct molecular structure, not merely the correct molecular formula.

06What to do after choosing a reaction

An organic reaction table becomes much more useful when you stop treating it as a list of equations.

Start with the desired product. Identify its functional group, find the immediate precursor, and work backwards until you reach the available starting materials.

Then check three things before accepting your pathway:

1. Structure: Does each reaction produce the correct functional group in the correct position?
2. Carbon count: Are all carbon atoms accounted for, including those supplied by a second organic reactant?
3. Conditions: Do the reagents, solvent, temperature, and reaction setup actually favour the desired product?

Once you have a proposed pathway, the next job is to check atom conservation across every reaction arrow. This is particularly important in multi-step synthesis, where an intermediate can look reasonable but still lead to the wrong final structure.

For that next step, see How to Check Atom Conservation in Multi-Step Organic Synthesis.

Course scope: This guide uses common reaction families from the Chemistry Stage 6 Syllabus (2017), which remains the taught NSW Chemistry course through 2027. The Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with the first HSC examination under the new syllabus in 2029. Check your school’s prescribed reaction table and course requirements for the relevant syllabus edition. See NESA’s Chemistry Stage 6 course information for the implementation timeline.