4 Organic chemistry

Exam analysis

Most tested topics

(a) Introduction

A hydrocarbon is a compound that contains hydrogen and carbon only.

Substance	Hydrocarbon?	Reason
methane, $\ce{CH4}$	yes	it is a compound containing only carbon and hydrogen
ethene, $\ce{C2H4}$	yes	it is a compound containing only carbon and hydrogen
ethanol, $\ce{C2H5OH}$	no	it also contains oxygen
carbon dioxide, $\ce{CO2}$	no	it contains no hydrogen and also contains oxygen

Check every element symbol in the formula. Both C and H must be present, and any additional element means the substance is not a hydrocarbon.

A compound is not a hydrocarbon merely because it contains carbon and hydrogen. The word ‘only’ excludes alcohols, carboxylic acids and every other compound containing an additional element.

Different formula types answer different questions about an organic substance: atom ratio, actual atom count, homologous-series pattern, connectivity, or every bond.

Formula type	What it shows	Example for ethene
empirical	simplest whole-number ratio of atoms	$\ce{CH2}$
molecular	actual number of each atom in one molecule	$\ce{C2H4}$
general	algebraic pattern shared by a homologous series	$\ce{C_nH_{2n}}$ for alkenes
structural	how atoms or groups are connected, with bonds condensed	$\ce{CH2=CH2}$
displayed	every atom and every covalent bond	draw both C atoms, four H atoms and all six bonds

To obtain an empirical formula, divide every molecular subscript by their highest common factor. To obtain a molecular formula from a displayed or structural formula, count each atom; a general formula is not the formula of one particular molecule until a value of $n$ is chosen.

A molecular formula does not show connectivity, so different structures can share it. A structural formula may group bonds such as $\ce{CH3CH2OH}$ ; a displayed formula must show the O–H bond and every other bond explicitly.

Homologous series, functional group and isomerism describe three different relationships: membership of a family, the reactive part of a molecule, and alternative structures for one molecular formula.

Term	Meaning	Diagnostic question
homologous series	a family with the same functional group and general formula, similar chemical reactions, a trend in physical properties, and successive members differing by $\ce{CH2}$	do the compounds follow one family pattern?
functional group	the atom or group of atoms responsible for a compound’s characteristic chemical reactions	which part controls the typical reactions?
isomerism	the existence of compounds with the same molecular formula but different structural formulae	is the atom count unchanged but the connectivity different?

\ce{CH3CH2CH2CH3} \qquad \ce{CH3CH(CH3)CH3}

The two formulae above both have molecular formula $\ce{C4H10}$ but different carbon connectivity, so they are structural isomers.

Isomers must have the same molecular formula, not merely the same empirical or general formula. Members of a homologous series have similar chemical properties and a trend in physical properties; they do not all have identical physical properties.

An IUPAC name combines a carbon-chain stem, any substituent prefixes and a suffix that identifies the principal bond or functional group. Number the chain so the important feature receives the lowest possible locant.

Carbon atoms in the longest relevant chain	1	2	3	4	5	6
stem	meth-	eth-	prop-	but-	pent-	hex-

Feature	Naming pattern	Example
alkane	stem + -ane	$\ce{CH3CH2CH3}$ : propane
alkene	lowest C=C locant + -ene	$\ce{CH2=CHCH2CH3}$ : but-1-ene
alcohol	lowest –OH locant + -ol	$\ce{CH3CH2CH2OH}$ : propan-1-ol
carboxylic acid	stem + -anoic acid; acid carbon is C1	$\ce{CH3CH2COOH}$ : propanoic acid
halogen or branch	numbered prefix before the parent name	$\ce{CH3CHClCH3}$ : 2-chloropropane

Choose the longest chain containing the principal functional group or C=C bond, number it from the nearer important feature, identify substituents, then assemble locants, prefixes, stem and suffix. For an ester, name the alkyl group attached to oxygen first and the acid-derived alkanoate second, for example methyl ethanoate.

Count the parent chain, not every carbon in a branch. Do not omit a locant when more than one position is possible, and keep this specification’s naming work to compounds containing no more than six carbon atoms.

A molecular formula fixes the number of each atom but not their connectivity. Generate possible structures systematically while satisfying normal valencies: carbon forms four bonds, oxygen two, hydrogen and halogens one.

Step	Check
1	write the exact inventory of C, H and any other atoms
2	draw distinct carbon skeletons, beginning with the longest chain and then branching
3	place any multiple bond or functional group in each genuinely different position
4	add H atoms so every atom has its normal valency
5	recount atoms and remove drawings that differ only because they are rotated or reversed
6	convert each valid structural formula to a displayed formula by showing every atom and bond

\ce{CH3CH2CH2CH3} \qquad \ce{CH3CH(CH3)CH3}

For $\ce{C4H10}$ , the only carbon skeletons are a four-carbon chain and a three-carbon chain with one methyl branch. These give butane and methylpropane; redrawing either from the opposite end does not create another isomer.

A valid drawing must match the molecular formula exactly and obey every valency. Different orientations of the same connectivity are not different structures; different atom connectivity is required.

Classify an organic reaction from its overall change: replacement is substitution, joining across an unsaturated bond is addition, and reaction with oxygen that releases energy is combustion.

Type	Overall pattern	Example clue
substitution	one atom or group in an organic molecule is replaced by another	alkane + halogen under ultraviolet light gives a haloalkane and hydrogen halide
addition	atoms add across a C=C bond and two reactants form one organic product	alkene + bromine gives a dibromoalkane
combustion	an organic compound burns in oxygen	oxygen is a reactant and energy is released

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

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

\ce{CH4 + 2O2 -> CO2 + 2H2O}

Use reactants and products to identify the overall pattern. Addition does not replace an atom and normally gives one organic product; substitution retains the carbon framework while replacing a group. Reaction mechanisms are not required.

(b) Crude oil

Crude oil is a mixture of many hydrocarbons: compounds made from hydrogen and carbon only.

The hydrocarbons have different molecule sizes and structures, so they have different physical properties. Because they are mixed rather than chemically bonded into one substance, physical separation can group them into useful fractions.

Crude oil is	Crude oil is not
a mixture containing many hydrocarbon compounds	one pure hydrocarbon
variable in composition	a compound with one fixed formula
separable into fractions by physical properties	separated by breaking covalent bonds during distillation

A fraction obtained from crude oil is still a mixture of hydrocarbons with similar boiling points; it is not normally one pure compound.

Industrial fractional distillation separates crude oil because its hydrocarbons have different boiling-point ranges. It separates physically; no covalent bonds are broken.

Stage	What happens
1	crude oil is heated strongly so most of it vaporises
2	the vapour enters near the bottom of a fractionating column
3	the column is hot at the bottom and becomes cooler toward the top
4	vapours rise, cool and condense at different heights according to boiling point
5	condensed liquids are drawn off as fractions; refinery gases leave at the top and bitumen remains near the bottom

Large, high-boiling molecules condense low in the hot column. Smaller, lower-boiling molecules rise farther before condensing; the lowest-boiling hydrocarbons remain gases and leave from the top.

A fraction condenses over a boiling range because it contains several hydrocarbons. Do not describe cracking or a laboratory flask: the required process is an industrial fractionating column with a temperature gradient.

The main crude-oil fractions are named by their boiling ranges and chosen for uses that fit their physical properties.

Fraction, from top toward bottom	Main use
refinery gases	bottled gases for heating and cooking
gasoline	fuel for cars
kerosene	aircraft fuel
diesel	fuel for diesel engines
fuel oil	fuel for ships and some power stations
bitumen	surfacing roads and roofing

Fractions near the top contain smaller, more easily vaporised molecules and are commonly used as mobile fuels. The very viscous residue at the bottom suits waterproofing and road surfaces rather than vaporising as an engine fuel.

Use the specification names exactly: gasoline is the car-fuel fraction, kerosene is the aircraft-fuel fraction, and fuel oil—not diesel—is the named ship-fuel fraction in this list.

From the top fractions toward the bottom fractions, boiling point and viscosity increase, and colour becomes darker.

Direction through main fractions	Molecules	Boiling point	Viscosity	Colour
top → bottom	generally larger / longer	increases	increases; flows less easily	becomes darker
bottom → top	generally smaller / shorter	decreases	decreases; flows more easily	becomes paler

Larger hydrocarbon molecules have stronger intermolecular attractions overall, so more energy is needed to separate them during boiling. They also move past one another less easily, producing greater viscosity.

Viscosity means resistance to flow, not density. A more viscous fraction flows more slowly; it is not described as having a lower boiling point.

A fuel is a substance that releases heat energy when it burns.

Burning is combustion: the fuel reacts with oxygen and transfers chemical energy to the surroundings as heat. A useful fuel must therefore release energy during combustion, rather than merely being flammable in name.

Gasoline, kerosene and diesel are used as fuels because their hydrocarbons combust in oxygen and the released heat can power engines.

A substance is not defined as a fuel simply because it contains stored chemical energy. The definition requires heat energy to be released when the substance is burned.

Hydrocarbon combustion always forms water from hydrogen. The carbon product depends on the oxygen supply.

Oxygen supply	Type	Possible products
plentiful	complete combustion	carbon dioxide and water
limited	incomplete combustion	carbon monoxide and water, and/or carbon (soot) and water

\ce{CH4 + 2O2 -> CO2 + 2H2O}

\ce{2CH4 + 3O2 -> 2CO + 4H2O}

When balancing, keep the hydrocarbon formula unchanged: balance carbon first, hydrogen second and oxygen last. Limited oxygen does not mean no oxygen; it means there is insufficient oxygen to oxidise all carbon fully to carbon dioxide.

Carbon monoxide and soot are alternative or simultaneous products of incomplete combustion. Do not list hydrogen gas, and do not omit water simply because the question emphasises the carbon-containing product.

Carbon monoxide is poisonous because it reduces the capacity of blood to transport oxygen.

Incomplete combustion in a limited oxygen supply can produce colourless carbon monoxide. If it is inhaled, less oxygen is delivered by the blood to body tissues, so aerobic respiration cannot be sustained normally.

A faulty or poorly ventilated fuel-burning appliance is dangerous because carbon monoxide can accumulate without visible soot being a reliable warning.

The required explanation is reduced oxygen-carrying capacity of blood. References to haemoglobin are not required, and saying only that carbon monoxide is ‘toxic’ does not explain why.

Inside a car engine, the temperature becomes high enough for nitrogen and oxygen from the air to react, forming oxides of nitrogen.

\ce{N2(g) + O2(g) ->[high\ temperature] 2NO(g)}

Both reactants come from the air drawn into the engine. The high temperature supplies the energy needed for normally unreactive nitrogen and oxygen molecules to react; further oxidation can form other nitrogen oxides.

The nitrogen does not need to be present in the fuel. Oxides of nitrogen form because air is heated strongly in the engine, not because nitrogen is a normal hydrocarbon impurity.

Some hydrocarbon fuels contain sulfur impurities. When the fuel burns, the sulfur also reacts with oxygen and forms sulfur dioxide.

\ce{S(s) + O2(g) -> SO2(g)}

sulfur impurity in fuel → combustion in air → sulfur dioxide released with the exhaust gases

Sulfur dioxide is produced from sulfur impurities, not from the carbon and hydrogen of a pure hydrocarbon. The hydrocarbon’s combustion products and the impurity’s combustion product must be traced separately.

Sulfur dioxide and oxides of nitrogen contribute to acid rain by entering the atmosphere and forming acidic solutions in cloud and rain water.

Pollutant	Main source in this Topic	Route to acid rain
sulfur dioxide, $\ce{SO2}$	burning sulfur impurities in fuels	dissolves and reacts in atmospheric water, producing acidic rain
oxides of nitrogen, $\ce{NO_x}$	high-temperature reaction of nitrogen and oxygen in engines	react with oxygen and water in the atmosphere, producing acidic rain

Acid rain can acidify lakes and soils, harm aquatic life and plants, and react with carbonate stone in buildings and statues.

Acid rain is not the same environmental problem as the greenhouse effect. Carbon dioxide contributes to climate warming, whereas the required pollutants here are sulfur dioxide and oxides of nitrogen.

Catalytic cracking breaks long-chain alkanes into shorter-chain alkanes and alkenes.

Stage	Industrial change
1	vaporise a long-chain alkane fraction
2	pass the vapour over hot silica or alumina catalyst
3	maintain a temperature in the range $600$ – $700\,^{\circ}\mathrm{C}$
4	collect a mixture containing shorter-chain alkanes and alkenes

\ce{C10H22 -> C4H10 + C2H4 + C4H8}

A valid cracking equation conserves every carbon and hydrogen atom. At least one product is an alkene, so its formula follows the alkene pattern rather than the alkane pattern.

Cracking is a chemical reaction that breaks C–C bonds; fractional distillation is a physical separation and does not change molecule size. The required catalysts are silica or alumina, not a generic metal catalyst.

Cracking is necessary because fractional distillation produces fractions in proportions that do not match demand.

Before cracking	What cracking produces	Why this helps
surplus / lower demand for some long-chain fractions	shorter-chain alkanes	helps meet higher demand for useful fuels such as gasoline components
demand for reactive small molecules	alkenes	supplies feedstock for making polymers and other chemicals

Crude oil supply is fixed by its natural composition, but market demand is different. Converting less-demanded long chains into more-demanded short chains and alkenes improves the balance between what refineries obtain and what users need.

Cracking does not create more total carbon or merely separate an existing fraction. It chemically redistributes atoms into smaller molecules, changing the product mix to address supply and demand.

(c) Alkanes

Alkanes form a homologous series with the general formula $\ce{C_nH_{2n+2}}$ , where $n$ is a positive whole number equal to the number of carbon atoms.

$n$	Name	Molecular formula
1	methane	$\ce{CH4}$
2	ethane	$\ce{C2H6}$
3	propane	$\ce{C3H8}$
4	butane	$\ce{C4H10}$
5	pentane	$\ce{C5H12}$

To test a proposed alkane formula, count its carbon atoms, substitute that value for $n$ , and check whether the hydrogen count is $2n+2$ . For example, $n=6$ gives $\ce{C6H14}$ .

The general formula describes molecular formulae for open-chain alkanes. Do not simplify $\ce{C2H6}$ to the empirical formula $\ce{CH3}$ , and do not use the alkene formula $\ce{C_nH_{2n}}$ .

An alkane is a saturated hydrocarbon: it contains carbon and hydrogen only, and every bond between carbon atoms is a single covalent bond.

Classification	Evidence in an alkane
hydrocarbon	the molecule contains only carbon and hydrogen atoms
saturated	there are no carbon–carbon double or triple bonds; the carbon framework carries the maximum number of hydrogens for that open chain

Because an alkane has no $\ce{C=C}$ bond to open, it does not undergo the addition reactions characteristic of alkenes. Its reaction with a halogen is substitution instead.

Saturated does not mean that every carbon atom is bonded to four hydrogen atoms. Carbon makes four bonds in total; in ethane, for example, each carbon also bonds to the other carbon.

A structural formula shows how atoms are grouped along the carbon chain; a displayed formula shows every atom and every covalent bond.

Carbon atoms	Unbranched name	Molecular formula	Condensed structural formula
1	methane	$\ce{CH4}$	$\ce{CH4}$
2	ethane	$\ce{C2H6}$	$\ce{CH3CH3}$
3	propane	$\ce{C3H8}$	$\ce{CH3CH2CH3}$
4	butane	$\ce{C4H10}$	$\ce{CH3CH2CH2CH3}$
5	pentane	$\ce{C5H12}$	$\ce{CH3CH2CH2CH2CH3}$

For a displayed formula, first join the required carbon skeleton using single bonds. Then add hydrogen atoms until every carbon has four bonds and every hydrogen has one. Count the atoms at the end to confirm the molecular formula.

Structural isomers have the same molecular formula but different carbon connectivity. $\ce{C4H10}$ has straight-chain butane and branched methylpropane; $\ce{C5H12}$ has pentane, 2-methylbutane and 2,2-dimethylpropane.

Rotating or bending the same chain does not create a new isomer. Compare which carbon atoms are connected, and use methane, ethane, propane, butane and pentane for the required unbranched-chain names.

In ultraviolet radiation, an alkane reacts with a halogen by substitution: one halogen atom replaces one hydrogen atom in the alkane.

Reactants	Required condition	Products after one substitution
alkane + chlorine	ultraviolet radiation	chloroalkane + hydrogen chloride
alkane + bromine	ultraviolet radiation	bromoalkane + hydrogen bromide

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

\ce{C2H6 + Br2 ->[UV] C2H5Br + HBr}

Keep the carbon skeleton unchanged, replace exactly one H by Cl or Br, and use the other halogen atom to form $\ce{HCl}$ or $\ce{HBr}$ . Check that every atom is conserved.

For this specification, stop after mono-substitution and state ultraviolet radiation. Further substitutions can occur chemically, but they are outside 4.22; reaction mechanisms are also not required. Do not call this addition.

(d) Alkenes

Every alkene contains the functional group $>\!\ce{C=C}\!<$ : a carbon–carbon double bond with each carbon also joined to the rest of the molecule.

Look for two adjacent carbon atoms connected by two bond lines. In ethene, $\ce{CH2=CH2}$ , the $\ce{C=C}$ bond is the functional group; in propene, $\ce{CH2=CHCH3}$ , the same group is attached to a $\ce{CH3}$ group.

The functional group is specifically $\ce{C=C}$ , not any double bond and not the whole molecule. A $\ce{C=O}$ bond does not make a compound an alkene.

The homologous series of open-chain alkenes with one carbon–carbon double bond has the general formula $\ce{C_nH_{2n}}$ , where $n$ is the number of carbon atoms and $n\ge 2$ .

$n$	Alkene	Molecular formula
2	ethene	$\ce{C2H4}$
3	propene	$\ce{C3H6}$
4	butene isomers	$\ce{C4H8}$

Substitute the carbon count for $n$ and double it to obtain the hydrogen count. A five-carbon member therefore has formula $\ce{C5H10}$ .

Do not use the alkane formula $\ce{C_nH_{2n+2}}$ . The $\ce{C_nH_{2n}}$ pattern here assumes one double bond and no ring; molecules with several double bonds follow a different hydrogen count.

An alkene is an unsaturated hydrocarbon because it contains only carbon and hydrogen, and it has at least one carbon–carbon double bond.

Word	Structural evidence
hydrocarbon	only carbon and hydrogen atoms are present
unsaturated	a $\ce{C=C}$ bond is present, so the molecule can add atoms across that bond

During an addition reaction, the double bond becomes a single bond and each of its carbon atoms forms a new bond. This capacity to add atoms distinguishes an unsaturated alkene from a saturated alkane.

A compound can contain a $\ce{C=C}$ bond yet fail to be a hydrocarbon if it also contains another element. Both parts of the classification must be justified.

An alkene displayed formula must show every atom and bond, including one $\ce{C=C}$ bond; each carbon must have four bonds in total and each hydrogen one.

Carbon atoms	Unbranched name	Condensed structural formula
2	ethene	$\ce{CH2=CH2}$
3	propene	$\ce{CH2=CHCH3}$
4	but-1-ene	$\ce{CH2=CHCH2CH3}$
4	but-2-ene	$\ce{CH3CH=CHCH3}$

Choose the longest chain containing the double bond. Number from the end that gives the double bond the lowest position, then place that number before “ene”: but-1-ene has its double bond starting at carbon 1; but-2-ene starts at carbon 2.

Structural isomers have the same molecular formula but different connectivity. $\ce{C4H8}$ includes but-1-ene, but-2-ene and branched methylpropene; cyclic alkanes are not alkene answers.

Turning a drawing around does not create a new structural isomer. Cis/trans or E/Z notation is not required, so do not count those labels as extra required names.

Alkenes react with bromine by addition to form dibromoalkanes: the $\ce{C=C}$ double bond becomes a single bond and one bromine atom bonds to each of the two carbon atoms.

\ce{CH2=CH2 + Br2 -> CH2Br-CH2Br}

Ethene forms 1,2-dibromoethane. No atoms are removed: both bromine atoms from $\ce{Br2}$ appear in the one saturated product, while the carbon skeleton is unchanged.

This is addition, not alkane substitution. Ultraviolet radiation is not required, and the product retains no $\ce{C=C}$ bond at the reacted position.

Add bromine water to separate samples and shake: an alkene decolourises the bromine water from orange to colourless, whereas an alkane leaves it orange under these test conditions.

Sample	Observation	Conclusion
alkene	orange bromine water becomes colourless	$\ce{C=C}$ is present and bromine adds across it
alkane	no colour change; bromine water stays orange	no $\ce{C=C}$ is present

Name bromine water, not bromide or bromine alone, and report the colour change—not merely “a reaction occurs”. This comparison is made without ultraviolet radiation; UV would introduce the different alkane substitution reaction.

(e) Alcohols

Alcohols contain the hydroxyl functional group $\ce{-OH}$ , with the oxygen bonded to a carbon atom and to a hydrogen atom.

In methanol, $\ce{CH3OH}$ , and ethanol, $\ce{CH3CH2OH}$ , writing $\ce{OH}$ at the end makes the functional group visible. A displayed formula must show the separate $\ce{C-O}$ and $\ce{O-H}$ bonds.

Do not identify any molecule containing oxygen as an alcohol. The required group is $\ce{-O-H}$ attached to the carbon framework; a carboxylic acid contains $\ce{-COOH}$ and belongs to a different functional group.

The required alcohols have an unbranched carbon chain with one terminal $\ce{-OH}$ group. Their names end in “ol”.

Carbon atoms	Required name	Structural formula
1	methanol	$\ce{CH3OH}$
2	ethanol	$\ce{CH3CH2OH}$
3	propan-1-ol (propanol accepted)	$\ce{CH3CH2CH2OH}$
4	butan-1-ol (butanol accepted)	$\ce{CH3CH2CH2CH2OH}$

To draw a displayed formula, join the carbon atoms with single bonds, attach $\ce{-O-H}$ to the end carbon, then add hydrogens until every carbon has four bonds, oxygen has two and hydrogen has one.

A displayed formula must show the $\ce{O-H}$ bond; writing an unconnected OH label is incomplete. For this objective, propanol means propan-1-ol and butanol means butan-1-ol—not propan-2-ol or butan-2-ol.

Ethanol can be oxidised in three required ways: complete combustion, microbial oxidation in air, and heating with acidified potassium dichromate(VI).

Route	Conditions / oxidant	Main products
complete combustion	burn in air or oxygen	carbon dioxide and water
microbial oxidation	oxygen in air; microorganisms	ethanoic acid
laboratory oxidation	heat with potassium dichromate(VI) in dilute sulfuric acid	ethanoic acid

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

\ce{C2H5OH + O2 -> CH3COOH + H2O}

During oxidation with acidified potassium dichromate(VI), the dichromate colour changes from orange to green. The required acid is dilute sulfuric acid, and the mixture is heated.

Combustion does not produce ethanoic acid: it oxidises ethanol completely to $\ce{CO2}$ and $\ce{H2O}$ . For the dichromate route, do not replace dilute sulfuric acid with phosphoric acid even if a legacy mark scheme once allowed it.

Ethanol is manufactured either by hydrating ethene with steam or by fermenting glucose with enzymes in yeast.

Method	Reactants	Required conditions
hydration	ethene + steam	phosphoric acid catalyst; about $300\,^{\circ}\mathrm{C}$ ; about $60$ – $70\,\mathrm{atm}$
fermentation	glucose	enzymes in yeast; absence of air; optimum temperature about $30\,^{\circ}\mathrm{C}$

\ce{C2H4 + H2O -> C2H5OH}

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

Hydration produces ethanol as its only product in the equation. Fermentation produces both ethanol and carbon dioxide, so both products and their coefficient 2 must be retained when balancing.

Do not exchange the conditions: phosphoric acid, high temperature and high pressure belong to ethene hydration; yeast enzymes, no air and about $30\,^{\circ}\mathrm{C}$ belong to fermentation.

Fermentation is carried out without air and near the enzymes’ optimum temperature so that glucose is converted to ethanol at a useful rate without losing the desired product.

Condition	Why it is needed
absence of air	prevents ethanol being oxidised to ethanoic acid and prevents aerobic respiration from replacing ethanol production
about $30\,^{\circ}\mathrm{C}$	gives yeast enzymes a fast working rate without denaturing them
not too cold	enzyme-controlled reactions become slow
not too hot	enzymes denature, so their active sites no longer catalyse fermentation

temperature too low → insufficient particle movement and slow enzyme activity; optimum temperature → fastest effective enzyme action; temperature too high → enzyme structure changes → fermentation stops

Saying only that ‘yeast dies’ does not explain the temperature condition. The required explanation concerns enzyme activity and denaturation; absence of air is a separate chemical condition.

(f) Carboxylic acids

Carboxylic acids contain the carboxyl functional group $\ce{-COOH}$ .

Inside this group, the same carbon atom is double-bonded to one oxygen and single-bonded to an $\ce{-O-H}$ group: $\ce{-C(=O)OH}$ . In ethanoic acid, $\ce{CH3COOH}$ , the final carbon is part of the functional group.

Do not circle only $\ce{-OH}$ or only $\ce{C=O}$ : both belong to the one $\ce{-COOH}$ group. An ester has $\ce{-COO-}$ between two carbon groups and no $\ce{O-H}$ in that linkage.

The required unbranched carboxylic acids end in “anoic acid”; the carbon in $\ce{-COOH}$ counts as part of the chain.

Total carbon atoms	Name	Structural formula
1	methanoic acid	$\ce{HCOOH}$
2	ethanoic acid	$\ce{CH3COOH}$
3	propanoic acid	$\ce{CH3CH2COOH}$
4	butanoic acid	$\ce{CH3CH2CH2COOH}$

For a displayed formula, draw the unbranched carbon chain, make the final carbon double-bonded to O and single-bonded to O-H, then add hydrogens so every carbon has four bonds, oxygen two and hydrogen one.

Do not add an extra carbon after naming the alkyl-looking part: $\ce{CH3COOH}$ has two carbons and is ethanoic acid. A displayed formula must show both the $\ce{C=O}$ and $\ce{O-H}$ bonds.

Aqueous carboxylic acids react like other acids: with a suitable metal they form a carboxylate salt and hydrogen; with a metal carbonate they form a carboxylate salt, carbon dioxide and water.

Reactant added	Products	Gas observed
metal	carboxylate salt + hydrogen	$\ce{H2}$ ; effervescence as the metal gets smaller
metal carbonate	carboxylate salt + carbon dioxide + water	$\ce{CO2}$ ; effervescence

\ce{2CH3COOH(aq) + 2Na(s) -> 2CH3COONa(aq) + H2(g)}

\ce{2CH3COOH(aq) + K2CO3(aq) -> 2CH3COOK(aq) + CO2(g) + H2O(l)}

The acid name changes from “anoic acid” to “anoate” in the salt: ethanoic acid forms an ethanoate, such as sodium ethanoate or potassium ethanoate.

Fizzing alone does not identify the gas. A metal produces hydrogen, whereas a carbonate produces carbon dioxide and water; do not omit the carboxylate salt in either equation.

Vinegar is an aqueous solution containing ethanoic acid.

Aqueous means dissolved in water, so vinegar contains water with ethanoic acid molecules in solution. Ethanoic acid gives vinegar its acidic behaviour, including reactions with suitable metals and carbonates.

Vinegar is not pure ethanoic acid and the words are not interchangeable. “Vinegar” names the aqueous mixture; “ethanoic acid” names the acid compound it contains.

(g) Esters

Esters contain the functional group $\ce{-COO-}$ .

The ester linkage is $\ce{-C(=O)-O-}$ : one carbon is double-bonded to oxygen and single-bonded to a second oxygen, which continues to another carbon group.

The complete group includes both oxygen atoms. Unlike $\ce{-COOH}$ in a carboxylic acid, an ester linkage has carbon attached beyond the single-bonded oxygen rather than an $\ce{O-H}$ bond.

Ethanol reacts with ethanoic acid in the presence of an acid catalyst to form the ester ethyl ethanoate and water.

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

Reactant	Contribution to the ester name
ethanol	ethyl
ethanoic acid	ethanoate

The acid acts as a catalyst: it increases the reaction rate without being used up. Concentrated sulfuric acid is a suitable catalyst in the preparation.

Water is the other product; this is not the hydration of ethene. The alcohol supplies the first part of the ester name and the acid supplies the “anoate” part.

Ethyl ethanoate has structural formula $\ce{CH3COOCH2CH3}$ and connectivity $\ce{CH3-C(=O)-O-CH2-CH3}$ .

Part	Atoms in ethyl ethanoate
ethanoate side	$\ce{CH3-C(=O)-}$
ester linkage	$\ce{-C(=O)-O-}$
ethyl side	$\ce{-O-CH2-CH3}$

For a displayed formula, show every atom and bond: draw $\ce{C=O}$ , attach that carbon to $\ce{CH3}$ and to O, then attach the O to $\ce{CH2CH3}$ . Check carbon has four bonds, oxygen two and hydrogen one.

Do not write $\ce{CH3CH2COOCH3}$ : that connectivity is methyl propanoate. In ethyl ethanoate, the ethyl group is attached after the single-bonded oxygen.

An ester name has two parts: the alcohol supplies the first alkyl name, and the carboxylic acid supplies the second name ending in “anoate”.

Alcohol	Acid	Ester	Structural formula
methanol	methanoic acid	methyl methanoate	$\ce{HCOOCH3}$
methanol	propanoic acid	methyl propanoate	$\ce{CH3CH2COOCH3}$
ethanol	methanoic acid	ethyl methanoate	$\ce{HCOOCH2CH3}$
ethanol	propanoic acid	ethyl propanoate	$\ce{CH3CH2COOCH2CH3}$

To work backwards from $\ce{RCOOR'}$ , split the structure at the single $\ce{C-O}$ bond: $\ce{RCOO-}$ identifies the acid and $\ce{-R'}$ identifies the alcohol. Restore $\ce{-OH}$ to each reactant: $\ce{RCOOH}$ and $\ce{R'OH}$ .

Do not count all carbons as one chain. The group after oxygen is named first, while the chain containing the carbonyl carbon becomes the “anoate” part.

Esters are volatile compounds with distinctive smells, and they are used as food flavourings and in perfumes.

Property	Consequence	Use
volatile	molecules evaporate readily and reach the nose	perfumes
distinctive smells	particular esters can reproduce recognisable aromas	food flavourings and perfumes

A distinctive sweet or fruity smell can provide evidence that an ester formed in a small-scale preparation. In a laboratory, detect odours only by the instructed safe wafting method, never by inhaling directly.

Volatile means evaporates readily; it does not mean chemically reactive. Do not claim every ester smells pleasant—the syllabus requires distinctive smells and the named uses.

A small sample of ethyl ethanoate can be prepared by warming ethanol with ethanoic acid and an acid catalyst in a water bath.

Stage	Action and purpose
1	place small quantities of ethanol and ethanoic acid in a test tube
2	carefully add the instructed acid catalyst
3	stand the tube in a hot water bath to warm the mixture and increase the reaction rate
4	after heating, allow the mixture to cool and identify ester formation by its distinctive smell using the instructed safe wafting method

Use a water bath rather than a direct Bunsen flame because ethanol, ethyl ethanoate and the mixture are flammable and could ignite.

The acid is a catalyst, not the ester reactant. Do not heat the flammable mixture directly, smell it closely, or invent unverified separation stages beyond the supplied procedure.

(h) Synthetic polymers

An addition polymer forms when many small molecules called monomers join to make one long-chain molecule.

For an alkene monomer, the carbon-carbon double bond opens: the two carbon atoms form a single bond in the polymer backbone and each can bond to the next repeat unit.

n\ce{CH2=CH2 -> [-CH2-CH2-]_{n}}

Monomer	Addition polymer
small molecule	long-chain molecule
contains a reactive $\ce{C=C}$ bond	backbone contains $\ce{C-C}$ single bonds
many molecules join	no small molecule is eliminated

Addition polymerisation does not produce water or another small-molecule by-product. Do not describe the monomer as a repeat unit until its double bond has been changed to the polymer backbone.

To draw an addition-polymer repeat unit, change the monomer's $\ce{C=C}$ bond to $\ce{C-C}$ , keep every group attached to the same carbon, and show continuation bonds through brackets with $n$ outside.

Monomer	Polymer repeat unit	Polymer
$\ce{CH2=CH2}$	$\ce{[-CH2-CH2-]_{n}}$	poly(ethene)
$\ce{CH2=CHCH3}$	$\ce{[-CH2-CH(CH3)-]_{n}}$	poly(propene)
$\ce{CH2=CHCl}$	$\ce{[-CH2-CHCl-]_{n}}$	poly(chloroethene), PVC
$\ce{CF2=CF2}$	$\ce{[-CF2-CF2-]_{n}}$	poly(tetrafluoroethene), PTFE

Check four features: no $\ce{C=C}$ remains; both backbone carbons are present; all H atoms and substituents are retained; and each end has a continuation bond crossing the bracket.

Brackets alone are not enough. A repeat unit with a double bond, a missing substituent, or no continuation bonds does not represent the addition polymer correctly.

The two carbon atoms joined in an alkene monomer become the two-carbon backbone of its addition-polymer repeat unit. The conversion is reversible on paper.

Direction	Method
monomer $\rightarrow$ repeat unit	change $\ce{C=C}$ to $\ce{C-C}$ , retain attached groups, add two continuation bonds and brackets
repeat unit $\rightarrow$ monomer	remove brackets and continuation bonds, then change the backbone $\ce{C-C}$ between the two repeat carbons to $\ce{C=C}$

\ce{CH2=CHCH3 <=> [-CH2-CH(CH3)-]_{n}}

After restoring the double bond, each carbon must still have four bonds. Groups such as $\ce{CH3}$ , $\ce{Cl}$ or $\ce{F}$ stay on the carbon where they appeared in the repeat unit.

Do not insert a double bond between repeat units or remove side groups. The monomer is one alkene molecule, so it has no brackets, $n$ , or continuation bonds.

Many addition polymers are chemically inert and cannot be broken down by microorganisms, so they are non-biodegradable and persist after disposal.

Disposal route	Environmental problem	Explanation
landfill	limited space is occupied for a long time	inert, non-biodegradable polymers decompose extremely slowly
burning	harmful gases may be released	combustion can produce toxic gases; burning poly(chloroethene) can produce hydrogen chloride

A complete explanation links a property to its consequence: inert $\rightarrow$ not biodegraded $\rightarrow$ long-term landfill accumulation; burning $\rightarrow$ toxic gases $\rightarrow$ harm to organisms or air quality.

Do not say that inert polymers react with soil or that burning always makes them harmless. Name the disposal method and its distinct problem rather than giving the same vague pollution claim twice.

In condensation polymerisation, a dicarboxylic acid reacts with a diol to form a polyester and water.

Substance	Two functional groups	Role
dicarboxylic acid	two $\ce{-COOH}$ groups	supplies carbonyl-containing parts of ester links
diol	two $\ce{-OH}$ groups	supplies oxygen-containing parts of ester links
polyester	many $\ce{-C(=O)-O-}$ links	long-chain condensation polymer

n\ce{HOOC-R-COOH} + n\ce{HO-R'-OH -> [-(C(=O)-R-C(=O)-O-R'-O)-]_{n}} + \ce{H2O}

Each new ester linkage forms when the reacting groups lose the elements of water. Because both monomers have a functional group at each end, the reaction can continue to build a chain.

A diol plus a dicarboxylic acid is not addition polymerisation: a small molecule is eliminated and the backbone contains ester linkages. The precise water coefficient depends on how polymer end groups are represented.

To construct a polyester repeat unit, remove $\ce{OH}$ from each carboxyl group and $\ce{H}$ from each alcohol group, then join the remaining fragments through $\ce{-C(=O)-O-}$ ester links.

Reactant	Structural formula	Retained chain fragment
ethanedioic acid	$\ce{HOOC-COOH}$	$\ce{-C(=O)-C(=O)-}$
ethanediol	$\ce{HO-CH2-CH2-OH}$	$\ce{-O-CH2-CH2-O-}$

n\ce{HOOC-COOH} + n\ce{HO-CH2-CH2-OH -> [-C(=O)-C(=O)-O-CH2-CH2-O-]_{n}} + 2n\ce{H2O}

Draw both retained fragments, join them through an ester linkage, place brackets around one complete acid-plus-diol repeat, and put continuation bonds through the bracket. Finally check that the repeat contains both carbonyl carbons and both $\ce{CH2}$ groups.

Do not change a carbonyl $\ce{C=O}$ to $\ce{C-O}$ , omit an oxygen, or copy the addition-polymer method. A polyester repeat unit must contain the $\ce{-C(=O)-O-}$ linkage and fragments from both monomers.

A biopolyester is a polyester that is biodegradable: microorganisms can break it down into simpler substances.

Polymer	After disposal
biodegradable polyester (biopolyester)	can be decomposed by microorganisms
non-biodegradable polymer	persists because microorganisms cannot break it down

Biodegradability can reduce long-term persistence and accumulation compared with a non-biodegradable polymer when suitable biological conditions are present.

Some polyesters are biodegradable; this does not mean that every polyester or every polymer is biodegradable. 'Bio' here identifies the ability to undergo biological decomposition.