Organic Chemistry: 2025 Syllabus Study Guide

If you are studying Chemistry in 2026, there is an important distinction to get right before you start revising organic chemistry: the Chemistry 11-12 Syllabus (2025) does not begin teaching until 2028. NESA has scheduled the new Year 11 course for Term 1, 2028, the new Year 12 course for Term 4, 2028, and the first HSC examination under the replacement syllabus for 2029. The existing 2017 Chemistry syllabus remains relevant to students completing the current course before that change. This guide explains the new Organic chemistry focus area without mixing it up with the old Module 7 requirements.

## Which Chemistry syllabus applies to you?

The year printed on a syllabus is not necessarily the year you begin studying it. The 2025 syllabus is the replacement document's publication year, not its first teaching year.

The implementation schedule is:

| Period | Chemistry syllabus |
|---|---|
| 2026-2027 | Schools prepare to teach the replacement syllabus. Students continue under the applicable existing course. |
| Term 1, 2028 | New Chemistry 11-12 syllabus begins for Year 11. |
| Term 4, 2028 | NESA schedules the introduction of the new Year 12 course and its school-based assessment requirements. |
| 2029 | First HSC examination for the Chemistry 11-12 Syllabus (2025). |

This creates two distinct groups of students:

- **Students completing the existing course:** Continue to use the Chemistry Stage 6 Syllabus (2017), including its Module 7: Organic Chemistry content.
- **Students entering the replacement course:** Follow the Chemistry 11-12 Syllabus (2025), including its Year 12 Organic chemistry focus area.

Check the [official NESA Chemistry syllabus and implementation information](https://curriculum.nsw.edu.au/learning-areas/science/chemistry-11-12-2025/content) rather than assuming that a resource labelled 2025 applies to every student studying Chemistry in 2025 or 2026.

If you are studying Chemistry in 2026, there is an important distinction to get right before you start revising organic chemistry: the Chemistry 11-12 Syllabus (2025) does not begin teaching until 2028. NESA has scheduled the new Year 11 course for Term 1, 2028, the new Year 12 course for Term 4, 2028, and the first HSC examination under the replacement syllabus for 2029. The existing 2017 Chemistry syllabus remains relevant to students completing the current course before that change. This guide explains the new Organic chemistry focus area without mixing it up with the old Module 7 requirements.

## Which Chemistry syllabus applies to you?

The year printed on a syllabus is not necessarily the year you begin studying it. The 2025 syllabus is the replacement document's publication year, not its first teaching year.

The implementation schedule is:

| Period | Chemistry syllabus |
|---|---|
| 2026-2027 | Schools prepare to teach the replacement syllabus. Students continue under the applicable existing course. |
| Term 1, 2028 | New Chemistry 11-12 syllabus begins for Year 11. |
| Term 4, 2028 | NESA schedules the introduction of the new Year 12 course and its school-based assessment requirements. |
| 2029 | First HSC examination for the Chemistry 11-12 Syllabus (2025). |

This creates two distinct groups of students:

- **Students completing the existing course:** Continue to use the Chemistry Stage 6 Syllabus (2017), including its Module 7: Organic Chemistry content.
- **Students entering the replacement course:** Follow the Chemistry 11-12 Syllabus (2025), including its Year 12 Organic chemistry focus area.

Students starting Year 11 in 2028 will be the first cohort studying the replacement syllabus. Students who started Year 11 in 2027 or earlier remain under the existing course for their HSC.

Check the [official NESA Chemistry syllabus and implementation information](https://curriculum.nsw.edu.au/learning-areas/science/chemistry-11-12-2025/content) rather than assuming that a resource labelled 2025 applies to every student studying Chemistry in 2025 or 2026.

## What does the new Organic chemistry focus area cover?

The new Year 12 course has four focus areas, each allocated 30 indicative hours:

| Year 12 focus area | Indicative hours |
|---|---:|
| Equilibrium | 30 |
| Acid-base reactions | 30 |
| Organic chemistry | 30 |
| Applying chemical ideas | 30 |

Organic chemistry is therefore one quarter of the Year 12 course's indicative teaching time.

The focus area is associated with outcome CH-12-03, which requires students to analyse the structural, chemical, and physical properties of carbon compounds and their uses.

The important point is that this focus area has a defined scope. NESA identifies seven homologous series of simple organic compounds:

1. Alkanes
2. Alkenes
3. Alkynes
4. Alcohols
5. Haloalkanes
6. Carboxylic acids
7. Esters

You'll study these compounds through several connected ideas:

- Molecular and structural representations.
- IUPAC naming conventions, including compounds with up to eight carbon atoms.
- Chain, position, geometric, and functional group isomerism.
- Chemical reactions and reaction pathways.
- Physical properties and intermolecular forces.
- Quantitative calculations involving organic reactions.
- Laboratory and secondary-source investigations.

There is also an important distinction between Organic chemistry and the neighbouring Applying chemical ideas focus area. Some topics that students traditionally associate with organic chemistry, particularly spectroscopy, petroleum, and polymers, are addressed separately.

[IMAGE: A clear navigational diagram of the new Year 12 Chemistry course. Show four equally sized focus areas: Equilibrium, Acid-base reactions, Organic chemistry, and Applying chemical ideas. Highlight the Organic chemistry area and show Working scientifically and depth studies as elements integrated across the course.]

## 1. Organic structures, functional groups, and naming

Before predicting what an organic compound will do, you need to understand what it is made of and how its atoms are connected.

Carbon is particularly useful for building complex molecules because each carbon atom can form four covalent bonds. Carbon atoms can connect into straight chains, branched chains, and rings, while also forming single, double, and triple bonds.

For the new syllabus, you need to recognise several ways of representing these structures.

| Representation | What it tells you | Example |
|---|---|---|
| Molecular formula | Number of each type of atom | \(\ce{C2H6O}\) |
| Condensed structural formula | How atoms and groups are connected | \(\ce{CH3CH2OH}\) |
| Expanded structural formula | Individual atoms and bonds | Every bond is displayed. |
| Skeletal formula | Carbon framework using lines and vertices | Carbon and attached hydrogen labels are omitted where understood. |

A molecular formula alone doesn't always tell you which compound you have.

For example, ethanol and dimethyl ether both have the molecular formula \(\ce{C2H6O}\). However, their atoms are connected differently, giving them different functional groups and chemical properties.

[IMAGE: A side-by-side comparison of ethanol and dimethyl ether. Show their molecular formulas, condensed structural formulas, expanded structural structures, and skeletal representations where appropriate. Highlight the C-O-H arrangement in ethanol and the C-O-C arrangement in dimethyl ether.]

### Functional groups

A functional group is the specific arrangement of atoms responsible for many of an organic compound's characteristic chemical reactions.

Learn to recognise these groups rather than trying to memorise every molecule individually.

| Homologous series | Important structural feature |
|---|---|
| Alkanes | Carbon-carbon single bonds |
| Alkenes | Carbon-carbon double bond, C=C |
| Alkynes | Carbon-carbon triple bond, C≡C |
| Alcohols | Hydroxyl group, -OH |
| Haloalkanes | Halogen bonded to carbon, C-X |
| Carboxylic acids | Carboxyl group, -COOH |
| Esters | Ester linkage, -COO- |

For example, ethanol contains the hydroxyl group, while ethanoic acid contains the carboxyl group.

They both contain oxygen, but they belong to different homologous series and have different chemical properties.

For more practice identifying these structures, see [HSC Chemistry: How to Identify Organic Functional Groups](https://tutorgum.com/learn/resources/hsc-chemistry-structures-of-organic-acids-and-bases/).

### Homologous series and naming

A homologous series is a family of organic compounds with the same functional group and a consistent structural pattern.

Successive members of a simple homologous series differ by a \(\ce{CH2}\) unit.

For example:

- Methane: \(\ce{CH4}\)
- Ethane: \(\ce{C2H6}\)
- Propane: \(\ce{C3H8}\)
- Butane: \(\ce{C4H10}\)

The naming system tells you how many carbon atoms are in the main chain and which functional group is present.

| Number of carbon atoms | Prefix |
|---|---|
| 1 | Meth- |
| 2 | Eth- |
| 3 | Prop- |
| 4 | But- |
| 5 | Pent- |
| 6 | Hex- |
| 7 | Hept- |
| 8 | Oct- |

Common endings include:

| Homologous series | Naming ending |
|---|---|
| Alkane | -ane |
| Alkene | -ene |
| Alkyne | -yne |
| Alcohol | -ol |
| Carboxylic acid | -oic acid |
| Ester | -oate |

For an ester, the name contains two parts. The first identifies the alkyl group from the alcohol, and the second identifies the carboxylate portion from the acid.

For example, ethyl ethanoate is formed from ethanol and ethanoic acid.

A useful naming procedure is:

1. Identify the longest appropriate carbon chain.
2. Identify the principal functional group and any multiple bonds.
3. Number the chain to give the relevant feature the lowest possible position.
4. Identify branches and substituents.
5. Apply the correct suffix and assemble the complete IUPAC name.

The new syllabus specifically includes straight-chain and branched-chain aliphatic hydrocarbons, haloalkanes, alcohols, organic acids, and esters up to C8.

### Isomerism

Isomers have the same molecular formula but different structures.

This matters because two compounds can contain exactly the same atoms but behave differently.

The important types to recognise are:

| Type | What changes? | Example |
|---|---|---|
| Chain isomerism | Arrangement of the carbon skeleton | Butane and 2-methylpropane |
| Position isomerism | Position of a functional group or multiple bond | But-1-ene and but-2-ene |
| Geometric isomerism | Spatial arrangement around a suitable double bond | Cis-but-2-ene and trans-but-2-ene |
| Functional group isomerism | Functional group itself | Propanoic acid and methyl ethanoate |

Geometric isomerism is particularly worth understanding.

A carbon-carbon double bond restricts rotation. If each carbon in the double bond has two different groups attached, different spatial arrangements can exist.

A triple bond is linear, so it does not produce the usual cis/trans geometric isomerism associated with alkenes.

[IMAGE: Compare the structures of butane and 2-methylpropane, but-1-ene and but-2-ene, and cis-but-2-ene and trans-but-2-ene. Show the carbon skeletons and the spatial arrangement around the double bond accurately.]

### Worked example: Name an alkene from its condensed structural formula

A compound has the condensed structural formula \(\ce{CH3CH2CH=CHCH3}\).

Determine its IUPAC name, molecular formula, homologous series, and whether geometric isomerism is possible.

[IMAGE: Draw the five-carbon structure CH3CH2CH=CHCH3 with all five carbon positions numbered from the end closest to the double bond. Do not display the compound's final name.]

**Step 1**: Count the carbon atoms.

There are five carbon atoms in the longest chain, so the prefix is pent-.

**Step 2**: Locate the double bond.

Number the chain from the end closest to the double bond. The double bond begins at carbon 2.

**Step 3**: Identify the homologous series.

The compound contains a C=C bond, so it belongs to the alkene homologous series. Its suffix is -ene.

**Step 4**: Write the complete name and molecular formula.

The correct name is pent-2-ene.

For an open-chain alkene with one double bond, the general formula is:

\[
\ce{C_nH_{2n}}
\]

Therefore, its molecular formula is \(\ce{C5H10}\).

**Step 5**: Check for geometric isomerism.

Each carbon in the double bond has two different groups attached.

Geometric isomerism is therefore possible. However, the condensed formula alone does not tell us which geometric isomer is present.

## 2. Hydrocarbons and their reactions

Hydrocarbons contain only carbon and hydrogen.

The new syllabus covers three important families.

| Family | Bonding | General formula for open-chain compounds with one relevant bond type |
|---|---|---|
| Alkanes | Single bonds | \(\ce{C_nH_{2n+2}}\) |
| Alkenes | One double bond | \(\ce{C_nH_{2n}}\) |
| Alkynes | One triple bond | \(\ce{C_nH_{2n-2}}\) |

The formulas above apply to acyclic compounds with one double or triple bond, respectively. They do not cover every possible cyclic or multiply unsaturated structure.

### Why do alkanes and alkenes react differently?

Imagine adding bromine water to two separate samples: methane and ethene.

Which one would you predict reacts under ordinary test conditions?

Ethene reacts, while methane does not react appreciably under those same conditions.

The reason is their different bonding.

Alkanes contain only carbon-carbon single bonds. They are relatively unreactive under ordinary conditions and commonly undergo substitution reactions with halogens when exposed to ultraviolet light.

Alkenes contain a carbon-carbon double bond. This consists of one sigma bond and one pi bond. The pi bond provides a reactive site where addition reactions can occur.

During an addition reaction, atoms are added across the double bond, converting it into a single bond.

[IMAGE: A molecular-level comparison of methane and ethene reacting with bromine. Show ethene undergoing addition, with one bromine atom attaching to each carbon of the original double bond. Show methane undergoing substitution under UV light, with one hydrogen replaced by bromine. Include the relevant reaction conditions.]

### Important hydrocarbon reactions

The new syllabus specifically requires students to construct equations and models for several reactions.

| Starting compound | Reaction | Main product |
|---|---|---|
| Alkene + hydrogen | Addition | Alkane |
| Alkene + halogen | Addition | Dihaloalkane |
| Alkene + hydrogen halide | Addition | Haloalkane |
| Alkene + water | Addition | Alcohol |
| Alkane + halogen | Substitution | Haloalkane |

For example, ethene reacts with bromine:

\[
\ce{CH2=CH2 + Br2 -> CH2BrCH2Br}
\]

The double bond becomes a single bond, and each carbon forms a new bond with bromine.

In contrast, methane reacts with chlorine under ultraviolet light:

\[
\ce{CH4 + Cl2 -> CH3Cl + HCl}
\]

One hydrogen atom is replaced by chlorine. This is substitution rather than addition.

The distinction is important because students sometimes assume that all unsaturated and saturated hydrocarbons undergo the same reactions.

They don't.

The new syllabus also includes naming haloalkanes, comparing the reactivity of alkanes and alkenes, investigating their properties, and solving quantitative reaction problems.

## 3. Physical properties and intermolecular forces

Organic compounds don't just differ in how they react. Their structures also affect boiling points, solubility, and other physical properties.

This is where you need to connect molecular structure with intermolecular forces.

First, distinguish two types of bonding:

- Intramolecular bonds hold atoms together within an individual molecule.
- Intermolecular forces attract separate molecules to one another.

Boiling involves overcoming intermolecular attractions. It does not normally involve breaking the covalent bonds within the molecules.

### How does carbon chain length affect boiling point?

Consider a series of straight-chain alkanes.

As the carbon chain becomes longer, the boiling point generally increases.

Why?

Larger molecules have more electrons and greater surface contact between neighbouring molecules. This increases the strength of their London dispersion forces.

Think of longer molecules as having more surface area through which they can interact with neighbouring molecules. The analogy isn't a literal sticking force; the actual attraction comes from temporary changes in electron distribution.

More energy is needed to separate the molecules, so the boiling point increases.

[IMAGE: A qualitative graph showing boiling point against carbon chain length for straight-chain alkanes C1-C8. Include molecular illustrations of a short and a longer alkane, showing the greater contact area and London dispersion attractions. Do not invent numerical measurements.]

### Why do alcohols dissolve in water?

Alcohols contain a hydroxyl group, -OH.

This group is polar and can form hydrogen bonds with water molecules.

However, alcohols also contain a non-polar hydrocarbon chain.

This creates an important competition:

- The hydroxyl group encourages interaction with water.
- The hydrocarbon chain resists mixing with water.

For small alcohols, the hydroxyl group has a strong influence, so they are relatively soluble in water.

As the carbon chain becomes longer, the non-polar portion becomes increasingly important. Water solubility decreases.

The same general idea helps explain the solubility trends of carboxylic acids.

### Compare the main intermolecular forces

| Compound | Important intermolecular forces |
|---|---|
| Alkanes | London dispersion forces |
| Alcohols | Hydrogen bonding, dipole-dipole, and London dispersion forces |
| Carboxylic acids | Strong hydrogen bonding, dipole-dipole, and London dispersion forces |
| Esters | Dipole-dipole and London dispersion forces |

Esters are polar, but they cannot donate hydrogen bonds because they do not contain an O-H bond.

Carboxylic acids can form particularly strong hydrogen-bonded associations, which helps explain their relatively high boiling points compared with many esters of similar molecular mass.

The important exam skill is explaining the trend using molecular structure, rather than simply stating which compound has the higher boiling point.

### Enthalpy of combustion

The syllabus also connects alcohol structure with enthalpy of combustion.

Enthalpy of combustion is the heat released when one mole of a substance undergoes complete combustion under specified conditions.

It is usually negative because combustion releases energy.

For a homologous series of alcohols, increasing the carbon chain length generally makes the molar enthalpy of combustion more negative. Larger molecules contain more carbon and hydrogen atoms that can be oxidised during combustion.

You should also understand that oxygen availability affects the products.

Complete combustion produces carbon dioxide and water, while incomplete combustion can produce carbon monoxide and/or carbon, as well as water.

For example, complete combustion of ethanol is represented by:

\[
\ce{C2H5OH + 3O2 -> 2CO2 + 3H2O}
\]

## 4. Alcohols: reactions and production

Alcohols are particularly important because they can undergo several different reactions.

Their general functional group is -OH, but the position of that group and the surrounding carbon structure affect their properties and reactivity.

### Primary, secondary, and tertiary alcohols

Alcohols are classified according to how many carbon atoms are attached to the carbon carrying the hydroxyl group.

| Classification | Carbon atoms attached to the OH-bearing carbon |
|---|---:|
| Primary alcohol | 1 |
| Secondary alcohol | 2 |
| Tertiary alcohol | 3 |

For example:

- Propan-1-ol is primary.
- Propan-2-ol is secondary.
- 2-methylpropan-2-ol is tertiary.

You need to recognise these structures and explain how their properties and reactivity differ.

[IMAGE: Three accurate structural representations of propan-1-ol, propan-2-ol, and 2-methylpropan-2-ol. Highlight the carbon carrying the hydroxyl group and show the number of carbon attachments to that carbon.]

### Important alcohol reactions

The new syllabus includes several reaction types.

**1. Combustion**

Alcohols burn in oxygen.

Complete combustion produces carbon dioxide and water. Incomplete combustion occurs when oxygen is insufficient and can produce carbon monoxide or carbon.

**2. Dehydration**

Alcohols can lose water to form alkenes.

For example, ethanol can undergo dehydration:

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

This reaction uses suitable acidic conditions and heat.

**3. Oxidation**

Oxidation products depend on the type of alcohol.

| Alcohol | Typical oxidation product |
|---|---|
| Primary | Aldehyde, with further oxidation producing a carboxylic acid |
| Secondary | Ketone |
| Tertiary | Not readily oxidised under ordinary alcohol oxidation conditions |

For example, ethanol can be oxidised to ethanal:

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

Further oxidation can produce ethanoic acid:

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

The conditions matter. Controlled oxidation can favour an aldehyde, while further oxidation produces the carboxylic acid.

**4. Substitution**

Alcohols can react with hydrogen halides to form haloalkanes.

For example:

\[
\ce{CH3CH2OH + HBr -> CH3CH2Br + H2O}
\]

The syllabus also requires you to investigate differences in the reactivity of primary, secondary, and tertiary alcohols.

**5. Production from haloalkanes**

Haloalkanes can undergo substitution with aqueous hydroxide ions to produce alcohols.

For example:

\[
\ce{CH3CH2Br + OH^- -> CH3CH2OH + Br^-}
\]

This provides another connection between different homologous series.

[IMAGE: A connected reaction pathway centred on propan-1-ol. Show dehydration to propene, oxidation to propanal and then propanoic acid, substitution with hydrogen halide to form 1-halopropane, and complete combustion to carbon dioxide and water. Include the relevant conditions beside each reaction arrow.]

### Producing alcohols from plants and crude oil

The new syllabus also asks students to evaluate different methods of alcohol production.

Ethanol can be produced from plant material through fermentation.

Glucose is converted into ethanol and carbon dioxide by yeast under suitable anaerobic conditions:

\[
\ce{C6H12O6 -> 2C2H5OH + 2CO2}
\]

Ethanol can also be produced from ethene, which is obtained from petroleum processing:

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

This is an addition reaction involving water.

When comparing these processes, consider:

- Whether the starting material is renewable.
- Energy requirements.
- Land and water use.
- Production costs.
- Environmental impacts.
- Dependence on fossil fuels.

Plant-based ethanol uses renewable biological feedstocks, but this does not automatically make it environmentally harmless. Land use, fertiliser production, processing energy, and transport also matter.

## 5. Organic acids and esters

The final two major areas of the Organic chemistry focus area are carboxylic acids and esters.

These are closely connected because esters can be produced by reacting a carboxylic acid with an alcohol.

### Carboxylic acids

Carboxylic acids contain the -COOH functional group.

The simplest example is methanoic acid, \(\ce{HCOOH}\).

Another familiar example is ethanoic acid, \(\ce{CH3COOH}\).

Carboxylic acids have several important properties:

- They are weak acids in water.
- They can form hydrogen bonds.
- Their boiling points are influenced by strong intermolecular attractions.
- Their water solubility generally decreases as the hydrocarbon chain becomes longer.

The syllabus includes investigating how chain length affects boiling point and solubility.

You should also understand how primary alcohols can be oxidised to produce carboxylic acids.

For example:

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

Propan-1-ol is oxidised to propanoic acid.

You should also be able to identify the presence of a carboxylic acid experimentally.

One common chemical test involves adding a carbonate. Carboxylic acids react with carbonates to release carbon dioxide gas.

For example:

\[
\ce{2CH3COOH + Na2CO3 -> 2CH3COONa + CO2 + H2O}
\]

Effervescence provides evidence that carbon dioxide is being produced.

### Esters

Esters contain the -COO- functional group.

They are commonly associated with fruity smells and are used in fragrances, flavourings, and industrial products.

Their formation is particularly important in the new syllabus.

An ester can be produced through a condensation reaction between an alcohol and a carboxylic acid.

For example, ethanol reacts with ethanoic acid to form ethyl ethanoate and water:

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

This is called esterification.

The reaction is typically performed using an acid catalyst and heat.

The word condensation is important. A small molecule, in this case water, is produced when the larger molecules combine.

[IMAGE: A detailed structural reaction diagram of ethanoic acid reacting with ethanol to form ethyl ethanoate and water. Show the carboxylic acid and alcohol functional groups, the new ester linkage, and the reversible reaction arrow with acid catalyst and heat indicated.]

### Why is esterification important beyond making fragrances?

The new syllabus connects esterification with condensation polymerisation.

A polyester is formed when suitable molecules repeatedly join together through ester linkages.

This means the chemistry of a relatively simple esterification reaction provides a foundation for understanding larger polymer structures.

You will also need to:

- Name and represent esters.
- Identify functional group isomers of carboxylic acids and esters.
- Compare the physical properties of esters, alcohols, and carboxylic acids.
- Solve stoichiometric problems involving ester production.
- Construct flow charts for ester synthesis, including multi-step pathways.
- Investigate the safe production and purification of an ester in the school laboratory.

### Worked example: Calculate the theoretical yield of an ester

A student reacts 0.24 mol of ethanoic acid with 0.18 mol of propan-1-ol.

The reaction produces propyl ethanoate and water.

Determine the limiting reagent and calculate the theoretical maximum mass of ester produced, assuming complete conversion of the limiting reagent.

[IMAGE: Structural reaction equation showing ethanoic acid and propan-1-ol forming propyl ethanoate and water. Clearly distinguish the acid and alcohol reactants and the ester product.]

**Step 1**: Write the balanced reaction equation.

\[
\ce{CH3COOH + CH3CH2CH2OH}
\]

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

The mole ratio between ethanoic acid, propan-1-ol, and propyl ethanoate is 1:1:1.

**Step 2**: Identify the limiting reagent.

The reaction requires equal amounts of acid and alcohol.

However, the student has:

- 0.24 mol of ethanoic acid.
- 0.18 mol of propan-1-ol.

Propan-1-ol is present in the smaller stoichiometric amount, so it is the limiting reagent.

**Step 3**: Determine the maximum amount of ester.

The mole ratio is 1:1.

Therefore:

\[
n(\text{ester})=0.18\ \mathrm{mol}
\]

**Step 4**: Calculate the molar mass of propyl ethanoate.

Its molecular formula is \(\ce{C5H10O2}\).

\[
\begin{aligned}
M &= 5(12.0)+10(1.0)+2(16.0)\\
&=102.0\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

**Step 5**: Calculate the theoretical mass.

Use:

\[
m=nM
\]

Substituting:

\[
\begin{aligned}
m &= 0.18\times102.0\\
&=18.36\ \mathrm{g}
\end{aligned}
\]

The theoretical maximum yield is therefore **18.4 g of propyl ethanoate**.

In an actual experiment, the mass collected will generally be lower because esterification is reversible and purification can result in product losses.

This distinction between theoretical and actual yield is important when interpreting practical results.

## 6. What changed from the 2017 Module 7 syllabus?

If you've been using older HSC Chemistry resources, you may notice that their organic chemistry content looks considerably broader.

That's because the new syllabus is not simply Module 7 with a different name.

The 2017 Chemistry Stage 6 syllabus organised Year 12 into four modules:

- Module 5: Equilibrium and Acid Reactions.
- Module 6: Acid/base Reactions.
- Module 7: Organic Chemistry.
- Module 8: Applying Chemical Ideas.

The replacement syllabus uses four Year 12 focus areas instead.

The broad differences are:

| Area | 2017 syllabus | 2025 syllabus |
|---|---|---|
| Course organisation | Module 7: Organic Chemistry | Organic chemistry focus area |
| Main organic families | Broader range of organic compounds | Seven specified homologous series |
| Hydrocarbons | Naming, properties, and reactions | Naming, structural representations, isomerism, properties, and specified reactions |
| Alcohols | Production, properties, and reactions | More explicit attention to alcohol classification, oxidation, combustion, dehydration, and reactivity |
| Organic acids | Organic acids and bases | Carboxylic acids, properties, oxidation, and stoichiometry |
| Esters | Included within the broader organic chemistry content | Dedicated ester content, synthesis, purification, and stoichiometry |
| Polymers | Included in Module 7 | Polyester formation in Organic chemistry; other specified polymers in Applying chemical ideas |
| Spectroscopy | Organic analysis in Module 8 | Organic analysis in Applying chemical ideas |

There are some important scope differences.

**First, the new Organic chemistry focus area specifies a narrower set of simple organic homologous series.**

Amines and amides are not included in its specified list. Aldehydes and ketones appear as products in alcohol oxidation, but they are not listed as additional simple homologous series in this focus area.

**Second, some organic content is now explicitly separated into Applying chemical ideas.**

This includes:

- Mass spectrometry.
- Infrared spectroscopy.
- Proton NMR.
- Carbon-13 NMR.
- Petroleum composition and fractional distillation.
- Organic polymers, including polyethylene, polyvinyl chloride, polytetrafluoroethylene, and polypropylene.
- Environmental and health impacts of selected organic substances and polymers.

This is an important distinction. If you only study the Organic chemistry focus area, you will miss organic analysis and polymer content that belongs elsewhere in the Year 12 course.

**Third, the new syllabus makes the required structural and practical skills more explicit.**

You need to work confidently with skeletal structures, isomerism, reaction models, quantitative problems, and investigations.

The underlying chemistry still matters, but the way the content is organised has changed.

### Can you still use old Module 7 resources?

Yes, but use them selectively.

The older tutorgum resources on organic reaction pathways remain useful for understanding how compounds are converted into other compounds.

In particular:

- [How to Construct Organic Synthesis Pathways for HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-organic-synthesis-reaction-pathways/)
- [How to Read Organic Reaction Pathways in HSC Chemistry](https://tutorgum.com/learn/resources/how-to-read-an-organic-reaction-pathway-in-hsc-chemistry/)

These are useful for practising reaction sequences, recognising starting materials and products, and understanding how individual reactions connect.

However, don't treat the old Module 7 content as a complete checklist for the new syllabus.

A reaction appearing in an old HSC question does not automatically mean it is required in the same way under the replacement course.

Similarly, an old paper may not test some of the more explicit structural, practical, and analytical requirements of the new syllabus.

For a closer explanation, read [Can You Use Old HSC Chemistry Papers for the 2025 Syllabus?](https://tutorgum.com/learn/resources/can-you-use-old-hsc-chemistry-papers-for-the-2025-syllabus/).

The safest approach is to use old resources for transferable skills, then check every topic against the [new NESA Chemistry content](https://curriculum.nsw.edu.au/learning-areas/science/chemistry-11-12-2025/content).

## 7. Practical investigations and preparation

Organic chemistry is not just a collection of structures and equations.

The new syllabus explicitly includes practical and secondary-source investigations.

You should expect to encounter investigations involving:

| Investigation | What you need to understand |
|---|---|
| Comparing alkanes and alkenes | Differences in chemical properties and reactivity |
| Measuring alcohol combustion enthalpy | Calorimetry, heat transfer, and experimental limitations |
| Comparing primary, secondary, and tertiary alcohols | How structure affects reactivity |
| Identifying carboxylic acids | Chemical tests and observable evidence |
| Producing and purifying an ester | Condensation reactions, separation, and purification |

There are also secondary-source investigations involving boiling point, solubility, chain length, and the uses of hydrocarbons.

### What should you practise before starting?

You don't need to learn the entire organic chemistry focus area before your teacher introduces it.

However, a few earlier skills will make the material much easier.

**Refresh these areas:**

1. Covalent bonding and valency.
2. Writing molecular and structural formulas.
3. Balancing chemical equations.
4. Mole calculations and stoichiometric ratios.
5. Intermolecular forces and polarity.
6. Basic experimental design and data analysis.

For practical work, become comfortable identifying independent, dependent, and controlled variables.

You should also be able to explain how an experimental result could be affected by heat loss, incomplete reactions, measurement uncertainty, and contamination.

For example, when measuring the enthalpy of combustion of ethanol, some energy heats the surrounding air and apparatus rather than the water. The measured temperature increase therefore underestimates the total heat released by the reaction.

Practical work must be conducted under appropriate school supervision and safety procedures. Hydrocarbons and alcohols can be flammable, and some reagents require careful handling and disposal. Don't attempt these experiments at home.

### What is compulsory, and what can your school decide?

It helps to separate syllabus requirements from the way your school delivers them.

NESA specifies the required content, outcomes, and investigations. Schools are responsible for developing teaching programs, choosing appropriate practical methods and equipment, and organising local assessment schedules within NESA requirements.

The new Year 12 Chemistry course also includes 10 hours of depth studies. At least one depth study must be completed in Year 12, and it must incorporate questioning and predicting alongside content from at least one other focus area.

The precise topic, experimental approach, and timing of your depth study may depend on your school.

This means two schools can teach the same syllabus through slightly different practical programs without either changing the required course content.

For official course assessment and certification requirements, consult the [NESA ACE Rules](https://curriculum.nsw.edu.au/ace-rules).

## 8. Choosing Chemistry: prerequisites, university requirements, and scaling

There are several different decisions involved in choosing Chemistry. They shouldn't be treated as the same thing.

### NESA course requirements

Under the new Chemistry syllabus:

- Year 11 Chemistry is a two-unit elective with no NESA course prerequisite.
- Year 12 Chemistry requires completion of Year 11 Chemistry.

These are NESA course-enrolment rules for the replacement syllabus.

They are not university admission requirements.

### Your school's subject-selection policy

Your school determines which subjects it offers, how they are timetabled, and how students are placed into classes.

It may also provide local subject-selection advice based on previous achievement or readiness.

If your school recommends a particular level of achievement in Year 10 Science before choosing Chemistry, ask whether this is a school recommendation or a formal school placement requirement.

Don't confuse it with a statewide NESA prerequisite.

### University prerequisites and assumed knowledge

Universities set their own entry requirements.

There are three important terms:

| Term | Meaning |
|---|---|
| Prerequisite | A subject or result that must be completed to be eligible for admission. |
| Assumed knowledge | Knowledge the university expects you to have, although it may not be required for an offer. |
| Recommended studies | Subjects that may help prepare you for the degree. |

For example, the University of Sydney explains that assumed knowledge is not required to receive an offer, but students without it may find their first-year studies more difficult.

UNSW similarly distinguishes assumed knowledge from admission eligibility and recommends additional preparation where necessary.

See the official explanations from the [University of Sydney](https://www.sydney.edu.au/study/help/advice/prerequisites-and-assumed-knowledge-explained.html) and [UNSW](https://www.unsw.edu.au/study/how-to-apply/undergraduate/entry-requirements).

There is also an important date consideration.

The first HSC examination under the new Chemistry syllabus is in 2029. Those students will generally be considering university entry for 2030.

As at 9 November 2026, the latest published UAC Year 10 subject-selection booklet is the 2026 edition, released in May 2026. It lists courses that participating universities plan to offer in 2029, not 2030.

It provides useful examples of how university requirements differ:

- The University of Sydney's Pharmacy course lists Chemistry and Biology as assumed knowledge, alongside a Mathematics prerequisite.
- UNSW Pharmacy lists Chemistry and Mathematics Advanced as assumed knowledge.
- UNSW Medicine lists Chemistry and Biology as recommended studies, with additional selection criteria including UCAT ANZ and an interview.

These are examples from the 2026 UAC guide for planned 2029 offerings. They are not guarantees of the requirements for 2030 entry.

For the first new-syllabus cohort, check the relevant UAC guide when it is released and the university's own course entry requirements for the year you intend to enrol.

You can access the [UAC 2026 Steps to Uni booklet](https://www.uac.edu.au/assets/documents/year-10/year-10-booklet-2026.pdf) and the [UAC Year 10 and 11 course-selection advice](https://uac.edu.au/future-applicants/year-10-students).

### Should you choose Chemistry because of scaling?

No. Scaling should not be your main reason for selecting a subject.

UAC calculates the NSW ATAR using an aggregate of scaled marks from:

- Your best 2 units of English.
- Your best 8 units from your remaining courses.

Scaling adjusts marks to account for differences in the academic performance of students undertaking different courses. It is not a fixed bonus attached to Chemistry.

The result depends on the relevant cohort and the scaling process for that year.

UAC specifically advises students to consider their interests, abilities, and future plans rather than choosing subjects simply because they appear to scale well.

Read [How Your ATAR Is Calculated](https://uac.edu.au/future-applicants/atar/how-your-atar-is-calculated) for the official explanation.

### What about individual preference?

Your decision should also reflect what you actually enjoy studying.

Chemistry may be a good choice if you enjoy understanding how substances are structured, how reactions occur, and how scientific evidence can be used to explain the world.

It also provides useful preparation for areas such as pharmaceutical science, chemistry, biochemistry, and some health and science degrees.

However, not every student needs Chemistry for their intended university course.

Check actual admission requirements, consider your strengths, and think about whether you are interested in the subject itself.

## 9. A practical way to revise the new Organic chemistry focus area

The most effective way to approach organic chemistry is to connect four things:

1. Structure.
2. Properties.
3. Reactions.
4. Quantities.

For each compound, practise identifying its functional group and naming it correctly.

Then ask how its structure influences its physical properties. Consider its intermolecular forces, boiling point, and solubility.

Next, identify which reactions it can undergo and the conditions required.

Finally, practise writing balanced equations and using mole ratios to solve quantitative problems.

A useful checklist is:

| Skill | What you should be able to do |
|---|---|
| Naming | Name and draw specified organic compounds up to C8 |
| Structures | Interpret molecular, condensed, expanded, and skeletal formulas |
| Isomerism | Identify chain, position, geometric, and functional group isomers |
| Hydrocarbons | Distinguish alkanes, alkenes, and alkynes and predict specified reactions |
| Alcohols | Classify, name, and predict reactions of primary, secondary, and tertiary alcohols |
| Organic acids | Explain properties, oxidation, and relevant reactions |
| Esters | Name, produce, and calculate quantities involved in esterification |
| Physical properties | Explain boiling point and solubility trends using intermolecular forces |
| Practical work | Interpret experimental results and evaluate reliability and safety |
| Quantitative chemistry | Solve mole and stoichiometric problems involving organic reactions |

## Questions and solutions

### Question 1

An organic compound has the condensed structural formula:

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

(a) Give its IUPAC name.

(b) Identify whether it is a primary, secondary, or tertiary alcohol.

(c) Write its molecular formula.

[IMAGE: Draw the expanded and skeletal structures of the supplied four-carbon alcohol. Show the hydroxyl group attached to the second carbon, without displaying the answers.]

### Solution 1

The compound is butan-2-ol, a secondary alcohol with molecular formula \(\ce{C4H10O}\).

The reasoning is:

(a) There are four carbon atoms in the longest chain, giving the prefix but-. The hydroxyl group is attached to carbon 2, so the name is butan-2-ol.

(b) The carbon carrying the hydroxyl group is connected to two other carbon atoms. Therefore, it is a secondary alcohol.

(c) Counting the atoms gives:

- 4 carbon atoms.
- 10 hydrogen atoms.
- 1 oxygen atom.

Therefore, the molecular formula is \(\ce{C4H10O}\).

The important distinction is that the classification depends on the carbon attached directly to the -OH group, not on the total number of carbon atoms in the molecule.

### Question 2

Three open-chain compounds have the molecular formula \(\ce{C4H8}\):

- A: But-1-ene.
- B: But-2-ene.
- C: 2-methylpropene.

(a) Identify which compound can exist as geometric isomers. Explain your reasoning.

(b) Write a balanced equation for the reaction between compound B and bromine.

(c) Explain why the reaction in part (b) is an addition reaction, while the reaction between an alkane and bromine under ultraviolet light is a substitution reaction.

[IMAGE: Show the skeletal structures of but-1-ene, but-2-ene, and 2-methylpropene side by side. Include their carbon-carbon double bonds and branches accurately. Do not show cis/trans arrangements or reaction products.]

### Solution 2

Only compound B, but-2-ene, can exist as geometric isomers, and its reaction with bromine produces 2,3-dibromobutane.

**(a) Geometric isomerism**

For geometric isomerism to occur, each carbon atom in the double bond must have two different groups attached.

- But-1-ene has two hydrogen atoms attached to its terminal double-bond carbon, so it cannot have geometric isomers.
- But-2-ene has a hydrogen atom and a methyl group attached to each double-bond carbon, allowing two different spatial arrangements.
- 2-methylpropene has two identical methyl groups attached to one double-bond carbon and two hydrogen atoms attached to the other, so it cannot have geometric isomers.

Therefore, only but-2-ene has geometric isomers.

**(b) Reaction with bromine**

The balanced equation is:

\[
\ce{CH3CH=CHCH3 + Br2 -> CH3CHBrCHBrCH3}
\]

The product is 2,3-dibromobutane.

**(c) Addition versus substitution**

In the reaction with bromine, the carbon-carbon double bond in but-2-ene is converted into a single bond.

Each carbon forms a new bond with a bromine atom. Nothing is replaced by bromine, so this is an addition reaction.

An alkane behaves differently. Under ultraviolet light, a hydrogen atom can be replaced by a halogen atom.

For example:

\[
\ce{CH4 + Cl2 -> CH3Cl + HCl}
\]

This is substitution because one hydrogen atom is replaced by chlorine.

The trap is assuming that both reactions are simply examples of a hydrocarbon reacting with a halogen. The type of bonding in the starting compound determines the reaction pattern.

### Question 3

A student performs a combustion experiment using ethanol.

The following hypothetical measurements are recorded:

| Measurement | Value |
|---|---:|
| Mass of water | 150.0 g |
| Initial water temperature | 20.0°C |
| Final water temperature | 33.5°C |
| Mass of ethanol consumed | 0.460 g |
| Specific heat capacity of water | \(4.18\ \mathrm{J\,g^{-1}\,^\circ C^{-1}}\) |

Assume the heat absorbed by the apparatus is negligible for the calculation.

(a) Calculate the heat absorbed by the water.

(b) Calculate the experimental molar enthalpy of combustion of ethanol.

(c) Explain two reasons why the experimental value may differ from the actual enthalpy of combustion.

[IMAGE: Draw a school combustion calorimetry setup. Show an ethanol spirit burner below a metal container holding water, a thermometer immersed in the water, and the measured separation between the burner and container. Include the initial and final mass measurement context without displaying calculated results.]

### Solution 3

The experimental molar enthalpy of combustion is approximately \(-846\ \mathrm{kJ\,mol^{-1}}\).

**(a) Heat absorbed by the water**

Use:

\[
q=mc\Delta T
\]

Where:

- \(q\) is heat absorbed in joules.
- \(m\) is the mass of water in grams.
- \(c\) is the specific heat capacity.
- \(\Delta T\) is the temperature change in degrees Celsius.

First calculate the temperature change:

\[
\Delta T=33.5-20.0=13.5^\circ\mathrm{C}
\]

Substitute:

\[
\begin{aligned}
q&=150.0\times4.18\times13.5\\
&=8464.5\ \mathrm{J}
\end{aligned}
\]

Therefore:

\[
q=8.46\ \mathrm{kJ}
\]

**(b) Molar enthalpy of combustion**

The molar mass of ethanol is:

\[
M=46.0\ \mathrm{g\,mol^{-1}}
\]

Calculate the amount of ethanol consumed:

\[
\begin{aligned}
n&=\frac{m}{M}\\
&=\frac{0.460}{46.0}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

The heat released by combustion is taken as negative:

\[
\Delta H_c=-\frac{q}{n}
\]

Substitute:

\[
\begin{aligned}
\Delta H_c&=-\frac{8.46}{0.0100}\\
&=-846\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

**(c) Sources of experimental error**

Two important sources of error are:

1. Heat loss to the surroundings: Some energy heats the surrounding air and apparatus rather than the water. The measured temperature increase is therefore smaller than it would be in an ideal calorimeter.

2. Incomplete combustion: If ethanol burns with insufficient oxygen, some carbon monoxide or soot may form. Less energy is released than during complete combustion.

Both effects can make the experimental enthalpy less negative than the actual enthalpy of complete combustion.

### Question 4

An unknown organic compound X has the molecular formula \(\ce{C3H6O2}\).

A hypothetical analysis gives the following results:

- Its infrared spectrum contains a strong absorption near \(1740\ \mathrm{cm^{-1}}\).
- There is no broad absorption associated with a carboxylic acid O-H group.
- Its proton NMR spectrum contains two singlets, each integrating to three hydrogen atoms.
- The two signals occur near 2.1 ppm and 3.7 ppm.

(a) Identify the functional group and determine the most likely structure of X.

(b) Explain why propanoic acid is inconsistent with the evidence.

(c) Identify an alcohol and a carboxylic acid that could react to produce X through esterification.

[IMAGE: Draw a schematic infrared spectrum with a strong carbonyl absorption near 1740 cm-1 and no broad carboxylic acid O-H region. Beside it, draw a schematic proton NMR spectrum containing two singlets with equal three-proton integrations near 2.1 ppm and 3.7 ppm. Do not identify the compound.]

### Solution 4

Compound X is methyl ethanoate, an ester formed from methanol and ethanoic acid.

**(a) Identify the compound**

The strong absorption near \(1740\ \mathrm{cm^{-1}}\) indicates a carbonyl group, C=O.

The absence of a broad carboxylic acid O-H absorption suggests that X is not a carboxylic acid.

The proton NMR spectrum contains two distinct environments:

- A three-hydrogen signal near 2.1 ppm, corresponding to \(\ce{CH3CO-}\).
- A three-hydrogen signal near 3.7 ppm, corresponding to \(\ce{-OCH3}\).

These features match the structure:

\[
\ce{CH3COOCH3}
\]

The compound is methyl ethanoate.

**(b) Why isn't it propanoic acid?**

Propanoic acid has the same molecular formula, \(\ce{C3H6O2}\), but a different functional group.

It contains a carboxyl group and would show a broad carboxylic acid O-H absorption.

Its proton NMR spectrum would also have a different pattern of hydrogen environments.

The absence of the acid O-H feature makes propanoic acid inconsistent with the supplied evidence.

**(c) Identify the reactants**

Methyl ethanoate is formed from:

- Methanol, \(\ce{CH3OH}\).
- Ethanoic acid, \(\ce{CH3COOH}\).

The reaction is:

\[
\ce{CH3OH + CH3COOH <=> CH3COOCH3 + H2O}
\]

This is esterification.

The important reasoning is that molecular formula alone cannot distinguish a carboxylic acid from an ester. The functional group and spectral evidence are needed.

### Question 5

A chemist investigates the esterification of ethanoic acid and ethanol.

The reaction is:

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

For this idealised problem, treat all four species as dissolved in a homogeneous liquid mixture.

At a fixed temperature, the equilibrium constant is:

\[
K_c=4.00
\]

Initially, a 1.00 L reaction vessel contains:

- 1.00 mol of ethanoic acid.
- 1.00 mol of ethanol.
- No ester.
- No water.

An acid catalyst is present.

The mixture is allowed to reach equilibrium.

The chemist then selectively removes 0.500 mol of water without removing any other substance or changing the volume or temperature. The mixture is allowed to reach a new equilibrium.

(a) Calculate the equilibrium amounts of all four substances before water is removed.

(b) Calculate the new equilibrium amount of ester after water removal.

(c) A student suggests adding more catalyst after the second equilibrium is reached to increase the amount of ester. Evaluate this suggestion.

(d) Explain why removing water changes the equilibrium composition but does not change the equilibrium constant.

[IMAGE: Draw an idealised 1.00 L esterification reaction vessel containing ethanoic acid and ethanol. Show ester and water being produced, followed by a selective water-removal connection introduced after the first equilibrium. Indicate that the vessel volume and temperature remain constant. Do not display equilibrium concentrations or calculated answers.]

### Solution 5

The initial equilibrium contains approximately 0.667 mol of ester. After removing 0.500 mol of water, the new equilibrium contains approximately 0.771 mol of ester. Adding more catalyst will not increase the equilibrium amount.

This question combines stoichiometry, equilibrium, and esterification. The important detail is that removing a product changes the reaction quotient, but adding a catalyst does not change the equilibrium constant.

**(a) Initial equilibrium**

The reaction has a 1:1:1:1 mole ratio.

Initially, the concentration of ethanoic acid and ethanol is:

\[
[\text{acid}]_0=[\text{alcohol}]_0=1.00\ \mathrm{mol\,L^{-1}}
\]

Let \(x\) be the concentration of ester formed at equilibrium.

The equilibrium concentrations are:

| Substance | Initial concentration | Equilibrium concentration |
|---|---:|---:|
| Ethanoic acid | 1.00 | \(1.00-x\) |
| Ethanol | 1.00 | \(1.00-x\) |
| Ester | 0 | \(x\) |
| Water | 0 | \(x\) |

The equilibrium expression is:

\[
K_c=\frac{[\text{ester}][\text{water}]}{[\text{acid}][\text{alcohol}]}
\]

Substitute the equilibrium concentrations:

\[
4.00=\frac{x^2}{(1.00-x)^2}
\]

Taking the positive square root:

\[
2=\frac{x}{1.00-x}
\]

Rearranging:

\[
2-2x=x
\]

\[
x=\frac{2}{3}=0.667
\]

Therefore, the initial equilibrium concentrations are:

| Substance | Equilibrium amount |
|---|---:|
| Ethanoic acid | 0.333 mol |
| Ethanol | 0.333 mol |
| Ester | 0.667 mol |
| Water | 0.667 mol |

Because the volume is 1.00 L, the numerical values of concentration in mol L⁻¹ and amount in mol are the same.

**(b) New equilibrium after removing water**

The chemist removes 0.500 mol of water.

The amount of water decreases from 0.667 mol to:

\[
n(\ce{H2O})=0.667-0.500
\]

\[
n(\ce{H2O})=0.167\ \mathrm{mol}
\]

The other substances remain unchanged immediately after removal.

The reaction quotient is now:

\[
Q_c=\frac{(0.667)(0.167)}{(0.333)(0.333)}
\]

\[
Q_c\approx1.00
\]

Since:

\[
Q_c<K_c
\]

the forward reaction occurs until equilibrium is restored.

This produces more ester and water while consuming ethanoic acid and ethanol.

Let \(y\) be the additional concentration of ester formed.

The new equilibrium concentrations are:

\[
\begin{aligned}
[\text{acid}]&=\frac13-y\\
[\text{alcohol}]&=\frac13-y\\
[\text{ester}]&=\frac23+y\\
[\text{water}]&=\frac16+y
\end{aligned}
\]

Substitute into the equilibrium expression:

\[
4=\frac{(2/3+y)(1/6+y)}{(1/3-y)^2}
\]

Expanding and simplifying gives:

\[
18y^2-21y+2=0
\]

Using the quadratic formula:

\[
y=\frac{21\pm\sqrt{297}}{36}
\]

The physically valid solution is:

\[
y=0.1046\ \mathrm{mol\,L^{-1}}
\]

The other root is impossible because it would produce negative reactant concentrations.

Therefore, the new ester concentration is:

\[
[\text{ester}]=0.667+0.1046
\]

\[
[\text{ester}]=0.7713\ \mathrm{mol\,L^{-1}}
\]

In the 1.00 L vessel:

\[
\boxed{n(\text{ester})=0.771\ \mathrm{mol}}
\]

Removing water has increased the equilibrium amount of ester from 0.667 mol to approximately 0.771 mol.

**(c) Will adding more catalyst increase the amount of ester?**

No. Adding more catalyst will not change the equilibrium composition at the same temperature.

A catalyst lowers the activation energy for both the forward and reverse reactions.

It allows equilibrium to be reached more quickly, but it does not change the equilibrium constant or favour one direction over the other.

The student has confused reaction rate with equilibrium position.

**(d) Why does water removal change the composition but not the equilibrium constant?**

Removing water decreases the reaction quotient, making it smaller than the equilibrium constant.

The forward reaction then produces additional ester and water until the system reaches equilibrium again.

However, the equilibrium constant depends on temperature, not on the initial quantities of reactants and products.

Because the temperature remains constant, \(K_c\) remains 4.00 throughout the process.

This is the important distinction: removing a product can increase ester production, but adding a catalyst cannot increase the final equilibrium yield.

---

The next useful connection is between reaction pathways and analytical chemistry. Reaction pathways help you predict what a compound can become, while infrared spectroscopy, mass spectrometry, and NMR help you work backwards from experimental evidence to determine what a compound actually is. Both sides of that problem are important in the new Chemistry 11-12 course.