How to Choose Between Two Routes to the Same Organic Product

An exam gives you but-2-ene and asks you to produce butan-2-one. You can reach the product by adding water to the alkene, or by adding hydrogen bromide and then replacing the bromine with a hydroxyl group. Both routes appear to work.

Which one should you choose?

The reliable approach is to work backwards from the desired product. Butan-2-one is a ketone, so its immediate precursor should be the secondary alcohol butan-2-ol. You then need to find which available starting materials and reagents can produce that exact alcohol while preserving the four-carbon chain.

If both routes are chemically valid, the shorter route is usually preferable. However, reagent availability, reaction conditions, product selectivity, and purity can make a longer route the better choice.

This is the central skill in planning organic reaction pathways: identify the structure you need, work backwards through suitable functional group transformations, and eliminate routes that violate the carbon-count or reagent constraints.

01Start with the desired organic product

The first mistake students make is looking at the starting material and immediately trying to remember a reaction.

Instead, draw the target product.

Suppose the desired product is butan-2-one:

\[ \ce{CH3COCH2CH3} \]

It has four carbon atoms and a ketone functional group. The carbonyl group, \(\ce{C=O}\), is located on the second carbon.

Now ask: which functional group can be converted into a ketone?

A secondary alcohol can be oxidised to produce a ketone. Therefore, the immediate precursor is butan-2-ol:

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

Notice that the carbon chain has not changed. The hydroxyl group on carbon 2 has simply been converted into a carbonyl group.

This gives us our first useful rule:

Identify the functional group immediately before the target in the reaction pathway.

Some common examples are:

These are useful starting points, not a complete list of every possible organic reaction.

Why the carbon position matters

Suppose you have two possible alcohols:

• Butan-1-ol: the hydroxyl group is attached to carbon 1.
• Butan-2-ol: the hydroxyl group is attached to carbon 2.

Both contain four carbon atoms. However, they do not produce the same oxidation products.

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

A secondary alcohol is oxidised to a ketone.

\[ \begin{aligned} \text{Butan-1-ol} &\rightarrow \text{Butanal}\\ &\rightarrow \text{Butanoic acid} \end{aligned} \]

\[ \text{Butan-2-ol} \xrightarrow{\text{oxidation}} \text{Butan-2-one} \]

The carbon skeleton is the same length, but the functional group occupies a different position.

This is why checking the molecular formula alone is insufficient. You must check the actual structural formula and the location of the functional group.

02Check the carbon count and reaction constraints

Once you know the immediate precursor, work backwards to the available starting materials.

A useful way to picture this is to treat the carbon chain as a row of numbered seats. A reaction might change what is attached to seat 2, but it generally does not move the seats around.

For the reactions commonly used in HSC organic pathways:

• Hydration adds water across a carbon-carbon double bond.
• Substitution replaces a halogen with another functional group.
• Oxidation changes the bonding around a carbon atom.
• Dehydration removes water from an alcohol.

These transformations generally preserve the number and arrangement of carbon atoms in the organic chain.

Esterification is different because it joins two organic fragments. You must count the carbon atoms in both reactants.

Carbon-count rules

For example, butan-2-one contains four carbon atoms. A proposed route through butan-2-ol is sensible because the alcohol also contains four carbon atoms.

If someone proposes making it from a three-carbon compound using only hydration and oxidation, something is wrong. Neither reaction introduces an additional carbon atom.

You would need a separate reaction that forms a carbon-carbon bond, and you should not invent one that is outside the reaction pathways being considered.

Check the reagents, not just the reaction names

Two reactions can look similar on paper but produce different products when their conditions change.

The following transformations are particularly useful when planning HSC organic synthesis pathways.

A few conditions deserve particular attention.

Aqueous NaOH matters. When converting a haloalkane into an alcohol, specify aqueous sodium hydroxide. Simply writing NaOH leaves out an important reaction condition.

The concentration of sulfuric acid matters. Water with an acid catalyst is used for alkene hydration. Concentrated sulfuric acid and heat can dehydrate an alcohol to form an alkene.

The extent of oxidation matters. A primary alcohol can form an aldehyde or undergo further oxidation to a carboxylic acid. Distillation can be used to collect the aldehyde, while reflux with excess oxidant favours further oxidation.

The position of the functional group matters. A secondary alcohol produces a ketone, not an aldehyde or carboxylic acid, under the usual HSC oxidation conditions.

For unsymmetrical alkenes, also consider Markovnikov’s rule. In the addition of water or a hydrogen halide, the hydrogen attaches to the carbon that already has more hydrogen atoms. This helps predict the major product and therefore the position of the new functional group.

04Work backwards from esters differently

Ketones and carboxylic acids can often be traced backwards through oxidation reactions.

Esters require a slightly different approach because they contain two carbon-based fragments.

Consider ethyl ethanoate:

\[ \ce{CH3COOCH2CH3} \]

An ester forms through the reaction between a carboxylic acid and an alcohol.

\[ \text{Carboxylic acid} + \text{Alcohol} \rightleftharpoons \text{Ester} + \text{Water} \]

To work backwards, identify the two parts of the ester:

• The ethanoate portion comes from ethanoic acid.
• The ethyl portion comes from ethanol.

This is an important naming rule: the first part of an ester name identifies the alcohol-derived group, while the second part identifies the acid-derived group.

05A practical method for comparing two possible routes

When an exam question gives you multiple possible pathways, use the following decision process.

The order matters.

There is no point comparing the number of steps in two pathways if one of them produces the wrong structural isomer.

Similarly, a reaction that produces the correct major product may still produce unwanted minor products. If the question requires a pure product, consider whether an additional separation step is needed.

When writing an HSC response, show every intermediate, write the reagent above or beside each reaction arrow, and state the conditions where they affect the outcome.

Avoid writing vague descriptions such as:

\[ \text{Alcohol} \rightarrow \text{Ketone} \]

Instead, identify the actual compounds and the oxidising agent.

06Questions and solutions

Question 1

Propene is the only organic starting material available. A chemist wants to produce propan-2-ol.

Two possible pathways are being considered:

• Direct hydration of propene.
• Addition of hydrogen bromide followed by substitution using aqueous sodium hydroxide.

Outline both pathways, including the intermediates and reagents. Which pathway requires fewer reaction steps, and why is propan-2-ol the expected major product?

Solution 1

Direct hydration is the shorter pathway, while both routes can produce propan-2-ol as the major product.

Step 1: Identify the target.

Propan-2-ol is a secondary alcohol. It can be produced directly by hydration of propene.

Step 2: Write the direct hydration reaction.

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

Use water with an acid catalyst.

Because propene is unsymmetrical, Markovnikov’s rule predicts that the hydroxyl group attaches to carbon 2, producing propan-2-ol as the major product.

Step 3: Construct the alternative pathway.

First, add hydrogen bromide to propene.

\[ \ce{CH3CH=CH2 + HBr -> CH3CHBrCH3} \]

The major product is 2-bromopropane.

Next, use aqueous sodium hydroxide to replace bromine with a hydroxyl group.

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

Step 4: Compare the routes.

The direct hydration pathway involves one reaction.

The haloalkane pathway involves two reactions: addition of HBr, followed by substitution with aqueous NaOH.

Therefore, direct hydration is preferable when both pathways are available because it requires fewer steps and fewer reagents.

Question 2

A student wants to produce butan-2-one using the following sequence:

1. Convert a haloalkane into an alcohol using aqueous sodium hydroxide.
2. Oxidise the alcohol using excess acidified potassium dichromate under reflux.

Two possible starting materials are available:

• 1-bromobutane
• 2-bromobutane

Both contain four carbon atoms.

Which starting material will produce butan-2-one through the specified sequence? Show the reaction pathway for both starting materials and explain why the other pathway fails.

Solution 2

2-bromobutane is the correct starting material because it produces butan-2-ol, which can be oxidised to butan-2-one.

Step 1: Consider the substitution reaction of 1-bromobutane.

Aqueous sodium hydroxide replaces the bromine atom with a hydroxyl group.

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

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

Step 2: Determine its oxidation products.

Under further oxidation with excess acidified dichromate and reflux, butan-1-ol produces butanoic acid through an aldehyde intermediate.

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

The intermediate is butanal.

Therefore, this pathway does not produce the required ketone.

Step 3: Consider the substitution reaction of 2-bromobutane.

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

The product is butan-2-ol, a secondary alcohol.

Step 4: Oxidise the secondary alcohol.

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

The product is butan-2-one.

The distinction is the position of the hydroxyl group in the intermediate alcohol. Both starting materials contain four carbon atoms, but only 2-bromobutane places the hydroxyl group on carbon 2 after substitution.

Question 3

A chemist has 0.80 mol of ethanol as the only organic feedstock and wants to produce 0.30 mol of ethyl ethanoate.

The proposed pathway involves:

1. Oxidising some ethanol to ethanoic acid.
2. Reacting the ethanoic acid with unoxidised ethanol.

Assume complete oxidation and 100% conversion during esterification for the theoretical calculation.

Determine:

a) The amount of ethanol that must be oxidised.

b) The amount of ethanol that must remain unoxidised.

c) Whether the original 0.80 mol of ethanol is sufficient.

Explain why oxidising the entire ethanol supply before esterification would be a mistake.

Solution 3

Yes, the synthesis is theoretically possible. The chemist must oxidise 0.30 mol of ethanol and reserve another 0.30 mol as the alcohol reactant.

Step 1: Identify the two reactants required for esterification.

Ethyl ethanoate forms from ethanoic acid and ethanol.

\[ \ce{CH3COOH + CH3CH2OH <=> CH3COOCH2CH3 + H2O} \]

The mole ratio of ethanoic acid to ethanol to ethyl ethanoate is:

\[ 1:1:1 \]

Step 2: Calculate the required amount of ethanoic acid.

To produce 0.30 mol of ethyl ethanoate, the reaction requires:

\[ n(\text{ethanoic acid})=0.30\text{ mol} \]

Step 3: Calculate the ethanol required for oxidation.

One mole of ethanol produces one mole of ethanoic acid under complete oxidation.

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

Therefore:

\[ n(\text{ethanol oxidised})=0.30\text{ mol} \]

Step 4: Calculate the ethanol that must remain unoxidised.

The esterification reaction also requires 0.30 mol of ethanol.

\[ n(\text{ethanol unoxidised})=0.30\text{ mol} \]

The total ethanol required is:

\[ n_{\text{total}}=0.30+0.30 \]

\[ n_{\text{total}}=0.60\text{ mol} \]

Step 5: Check the available supply.

The chemist has 0.80 mol available.

\[ n_{\text{remaining}}=0.80-0.60 \]

\[ \boxed{n_{\text{remaining}}=0.20\text{ mol}} \]

The supply is sufficient, with 0.20 mol remaining.

The theoretical maximum ester production from 0.80 mol ethanol is 0.40 mol because two ethanol molecules are required per ester molecule: one becomes the carboxylic acid, and the other remains the alcohol.

Oxidising all the ethanol would produce ethanoic acid but leave no ethanol to act as the second reactant. The synthesis would then be unable to proceed through the proposed esterification pathway.

In practice, esterification is reversible, so the actual yield may be lower than this theoretical maximum.

Question 4

A chemist hydrates but-1-ene to produce an alcohol mixture.

For this question, assume the isolated mixture contains the following illustrative amounts:

• 0.75 mol butan-2-ol
• 0.25 mol butan-1-ol

The entire mixture is then heated under reflux with excess acidified potassium dichromate.

Predict the organic products and their amounts, assuming complete oxidation.

Explain whether this pathway produces pure butan-2-one directly.

Solution 4

The oxidation produces 0.75 mol of butan-2-one and 0.25 mol of butanoic acid. Therefore, the reaction mixture does not contain pure butan-2-one.

Step 1: Identify the two different alcohols.

Butan-2-ol is a secondary alcohol.

Butan-1-ol is a primary alcohol.

Although they have the same molecular formula, they undergo different oxidation reactions.

Step 2: Determine the oxidation product of butan-2-ol.

A secondary alcohol produces a ketone.

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

The mole ratio is 1:1.

Therefore:

\[ 0.75\text{ mol butan-2-ol} \rightarrow 0.75\text{ mol butan-2-one} \]

Step 3: Determine the oxidation product of butan-1-ol.

A primary alcohol first forms an aldehyde and can undergo further oxidation to produce a carboxylic acid.

Because excess oxidising agent and reflux are specified, the expected final product is butanoic acid.

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

The mole ratio is also 1:1.

Therefore:

\[ 0.25\text{ mol butan-1-ol} \rightarrow 0.25\text{ mol butanoic acid} \]

Step 4: Evaluate the purity of the product.

The organic products are:

The reaction produces the desired ketone, but it also produces an unwanted carboxylic acid.

A separation step would therefore be needed to isolate pure butan-2-one.

The important distinction is between producing the desired compound as one of the products and producing it as the only organic product.

The supplied 75:25 composition is illustrative data for this question, not a measured yield for but-1-ene hydration.

Question 5

A chemist wants to synthesise ethyl propanoate:

\[ \ce{CH3CH2COOCH2CH3} \]

The available organic starting materials are:

• 0.50 mol propan-1-ol
• 0.40 mol ethanol

Sufficient acidified potassium dichromate and concentrated sulfuric acid are available.

Two proposed pathways are:

Pathway A: Oxidise 0.35 mol of propan-1-ol to propanoic acid, then react it with 0.35 mol of ethanol.

Pathway B: Oxidise 0.35 mol of ethanol to ethanoic acid, then react it with 0.35 mol of propan-1-ol.

A student argues that both pathways produce ethyl propanoate because both resulting esters contain five carbon atoms and have the molecular formula \(\ce{C5H10O2}\).

Evaluate the two pathways.

Determine:

a) Which pathway produces the required ester.

b) The ester produced by the alternative pathway.

c) Whether 0.35 mol of the required ester is theoretically achievable from the available starting materials.

Assume complete oxidation and 100% esterification when calculating the theoretical maximum. Explain why matching the molecular formula and carbon count does not establish that the correct product has been made.

Solution 5

Pathway A produces the required ethyl propanoate and is stoichiometrically feasible. Pathway B produces propyl ethanoate, which is a different structural isomer.

Step 1: Work backwards from the target ester.

The target is:

\[ \ce{CH3CH2COOCH2CH3} \]

Split the ester at the bond between the carbonyl carbon and the ester oxygen.

This identifies two required components:

• Propanoic acid: \(\ce{CH3CH2COOH}\), containing three carbon atoms.
• Ethanol: \(\ce{CH3CH2OH}\), containing two carbon atoms.

The carbon count is:

\[ 3+2=5 \]

The acid-derived portion must contain three carbons, while the alcohol-derived portion must contain two.

Step 2: Evaluate Pathway A.

Propan-1-ol is a primary alcohol. It can be oxidised to propanoic acid.

\[ \ce{CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O} \]

The propanoic acid can then react with ethanol.

\[ \ce{CH3CH2COOH + CH3CH2OH <=> CH3CH2COOCH2CH3 + H2O} \]

This produces ethyl propanoate, which is the required product.

Step 3: Check whether the quantities are sufficient.

The esterification reaction has a 1:1 mole ratio between propanoic acid, ethanol, and ethyl propanoate.

For 0.35 mol of ester, the required amounts are:

\[ n(\text{propanoic acid})=0.35\text{ mol} \]

\[ n(\text{ethanol})=0.35\text{ mol} \]

To produce 0.35 mol of propanoic acid, oxidise 0.35 mol of propan-1-ol.

The available propan-1-ol is 0.50 mol, so:

\[ 0.50-0.35=0.15\text{ mol} \]

There is sufficient propan-1-ol.

The available ethanol is 0.40 mol, so:

\[ 0.40-0.35=0.05\text{ mol} \]

There is also sufficient ethanol.

Therefore, 0.35 mol of ethyl propanoate is theoretically achievable under the stated assumptions.

Step 4: Evaluate Pathway B.

In this pathway, ethanol is oxidised to ethanoic acid:

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

The ethanoic acid then reacts with propan-1-ol:

\[ \ce{CH3COOH + CH3CH2CH2OH <=> CH3COOCH2CH2CH3 + H2O} \]

The product is propyl ethanoate, not ethyl propanoate.

Step 5: Explain why the two esters are different.

The two products have the same molecular formula:

\[ \ce{C5H10O2} \]

However, their carbon chains are connected differently.

These compounds are structural isomers.

The tempting mistake is to count the total carbon atoms and assume that any combination producing five carbons will give the desired ester.

But esterification does not rearrange the carbon chains. The acid-derived and alcohol-derived fragments retain their respective positions in the product.

Therefore, the structure of the target ester determines which starting material must be oxidised and which must remain as the alcohol.

This is the most important check when comparing possible ester synthesis pathways.

06What to do after choosing a route

A chemically valid reaction pathway is only part of a successful synthesis.

Once the target compound has been produced, you may still need to separate it from unreacted starting materials, water, catalysts, and unwanted organic products.

That is a different decision from choosing the reaction pathway. Instead of examining functional groups and reaction conditions, you compare physical properties such as boiling point, solubility, and volatility.

For the next step, see How to Choose Separation Techniques in HSC Chemistry.

Syllabus scope: This article follows the Chemistry Stage 6 Syllabus (2017), which remains the Year 12 course in 2026. The Chemistry 11-12 Syllabus (2025) is scheduled to begin implementation in 2028, with its first HSC examination in 2029. See NESA’s syllabus transition information.