S3.2 Functional groups: classification of organic compounds

Organic compounds are the compounds of carbon, and there are more of them than of all the other elements combined. They range from methane, the main component of natural gas, to the proteins, fats and DNA of living things, plastics and almost every medicine. Sorting them into families by their functional groups makes this variety manageable: compounds with the same functional group react in the same ways, so learning the chemistry of one group covers thousands of compounds.

Carbon and its compounds

Methane, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{4}}}}$, ethanol, ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{2}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}\mathrm{OH}}}$

Why carbon forms so many compounds

Carbon is in group 14, with four valence electrons ($\text{1s}^2\ \text{2s}^2\ \text{2p}^2$). Several features of its bonding combine to give it a chemistry far larger than that of any other element:

• Four bonds per atom. A carbon atom shares its four valence electrons in four covalent bonds, completing an octet. Four bonds allow branching: a carbon atom can join up to four other carbon atoms.

• Strong carbon–carbon bonds. Carbon atoms bond to each other in chains of almost any length and in rings. This ability of an element to form chains of its own atoms is called catenation.

• Multiple bonds. Carbon forms single, double and triple bonds, to other carbon atoms and to oxygen and nitrogen, as in ${{\mathrm{C}{=}\mathrm{C}}}$, ${{\mathrm{C}{\equiv}\mathrm{C}}}$, ${{\mathrm{C}{=}\mathrm{O}}}$ and ${{\mathrm{C}{\equiv}\mathrm{N}}}$

Silicon, the element below carbon in group 14, also forms four bonds, but its chains are short and reactive. The bond enthalpies show why. A C–C bond is strong, and about as strong as a C–O bond, so chains of carbon atoms survive in air at room temperature. An Si–Si bond is much weaker than an Si–O bond, so silicon chains react readily with oxygen, and silicon's natural compounds are almost all built from Si–O bonds, as in quartz and the silicate minerals.

Bond enthalpies from the IB Chemistry data booklet (2025).

Together these features allow an enormous number of structures. The hydrocarbons alone include straight chains, branched chains, rings and chains with double and triple bonds, and every one of them can carry other atoms.

Representing organic molecules

An organic molecule can be represented in several ways. Each type of formula shows some information and hides some, and chemists choose the one that suits the purpose. Butanoic acid, the compound responsible for the smell of rancid butter, is shown in five ways in the panel of formulas.

Empirical and molecular formulas

The molecular formula of butanoic acid is ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{4}}\mathrm{H}{\vphantom{X}}_{\smash[t]{8}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$

Empirical formulas come from experiments. Combustion analysis or percentage composition by mass gives the ratio of atoms, and the relative molecular mass from a mass spectrum then gives the molecular formula.

Full structural formulas

A full structural formula shows double bonds as two lines and triple bonds as three. It shows exactly which atom is joined to which, so it distinguishes between molecules with the same molecular formula. It is usually drawn flat, with bonds at 90° and 180° to each other, so it does not show the true shape of the molecule. The bond angles in butanoic acid are close to 109.5° and 120°, not 90°.

Condensed structural formulas

Butanoic acid is ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{COOH}}}$

• Branches are written in brackets after the carbon atom they are attached to: 2-methylpropane is ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}(\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}})\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

Skeletal formulas

A skeletal formula is the quickest way to draw a large molecule, and it shows the carbon skeleton clearly. Converting a full structural formula of butan-2-ol into a skeletal formula shows how the conventions work.

The rules for drawing a skeletal formula are:

• Draw the carbon chain as a zig-zag; each line is a carbon–carbon bond and each corner or line end is a carbon atom.

• Do not write C atoms or the H atoms bonded to them. Each carbon is assumed to carry enough H atoms to make four bonds in total.

• Write every atom other than carbon and hydrogen (O, N, Cl, Br, …), and write the H atoms bonded to them, as in OH and ${{\mathrm{NH}{\vphantom{X}}_{\smash[t]{2}}}}$.

• Draw double bonds as two lines and triple bonds as three. The carbon atoms on either side of a triple bond are drawn in a straight line, because the bonds around them are linear.

• Draw branches as lines leaving the chain at a corner.

The zig-zag reflects the real shape of the chain. A carbon atom with four single bonds is tetrahedral, with bond angles of about 109.5°, so a chain of such atoms cannot be straight. The skeletal formula draws the angles as 120° on paper, which is close to the real shape.

Stereochemical formulas and three-dimensional models

A stereochemical formula shows the arrangement of the bonds in three dimensions. Bonds in the plane of the paper are drawn as ordinary lines, a bond pointing towards the viewer as a solid wedge that widens towards the front atom, and a bond pointing away as a hashed (dashed) wedge. The drawing of methane with its tetrahedral bond angle of 109.5° is a stereochemical formula. These formulas matter most for molecules containing a chiral carbon atom, where the arrangement in space distinguishes two otherwise identical molecules (HL).

Molecular models make the three-dimensional shape easier to see. Ball-and-stick kits use coloured balls for atoms (by convention black for carbon, white for hydrogen, red for oxygen, blue for nitrogen and green for chlorine) joined by sticks for bonds. Space-filling models show the space each atom's electrons occupy, and molecular modelling software draws molecules that can be rotated on screen and measures their bond lengths and angles.

Choosing a representation

Every representation is a model of the molecule, and each has strengths and limitations.

Converting between formulas

Converting a skeletal formula into a molecular formula is a matter of counting carbon atoms at corners and line ends, then adding the H atoms implied at each one. The skeletal formula of an alcohol is shown in the figure.

Functional groups

In most organic molecules the carbon–hydrogen skeleton is fairly unreactive: C–C and C–H bonds are strong and almost non-polar. The reactions of the molecule happen at a particular atom or group of atoms attached to the skeleton, such as an OH group or a C=C double bond. This part of the molecule is its functional group.

A functional group also affects physical properties such as boiling point and solubility, because it changes the forces between molecules. In general formulas the rest of the molecule is written as R, an alkyl group: a group formed by removing one H atom from an alkane, with the formula ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{n }}\mathrm{H}{\vphantom{X}}_{\smash[t]{2n +1}}}}$

The functional groups each have a name and a structure, and each one defines a family of compounds. The structures of the groups are collected in one figure.

The carbon–carbon double bond of the alkenes and the triple bond of the alkynes also act as functional groups: they are where alkenes and alkynes react.

Telling similar groups apart

Several groups contain the same atoms in different arrangements, and each arrangement has its own name and chemistry. The differences lie in what is attached to the carbon of the C=O group, or to the oxygen atom:

• Carbonyl is a C=O with only carbon or hydrogen attached to the carbon, as in aldehydes (${{{-}\mathrm{CHO}}}$) and ketones (${{\mathrm{C}{-}\mathrm{CO}{-}\mathrm{C}}}$).

• Carboxyl is a C=O and an OH on the same carbon atom, ${{{-}\mathrm{COOH}}}$. It is one group, with properties of its own (it is acidic); it is not a carbonyl group plus a hydroxyl group.

• Ester is a C=O whose carbon is also bonded to an O that is bonded to another carbon, ${{{-}\mathrm{COO}{-}}}$

A systematic check avoids most of these errors: find each oxygen and nitrogen atom in the molecule and look at every atom bonded to it, and to the carbon next to it.

Benzene and the phenyl group

Benzene, ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{6}}\mathrm{H}{\vphantom{X}}_{\smash[t]{6}}}}$, is a ring of six carbon atoms, each bonded to one hydrogen atom. It is drawn as a hexagon with a circle inside, or as a hexagon with alternating double bonds (the Kekulé structure). As in any skeletal formula, each corner of the hexagon is a carbon atom, and the H atoms are not drawn. Compounds that contain a benzene ring are called

When a benzene ring is attached to the rest of a molecule it is called a phenyl group, ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{6}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}}}$–: methylbenzene is ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{6}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

Identifying the functional groups in a larger molecule

Medicines, food additives and natural products often contain several functional groups. Aspartame, a sweetener used in low-sugar soft drinks, contains five.

Saturated and unsaturated compounds

Alkanes are saturated: each carbon atom is bonded to as many hydrogen atoms as possible. Alkenes and alkynes are unsaturated, and so are aromatic compounds such as benzene. Vegetable oils are described as unsaturated because their molecules contain C=C bonds, while animal fats such as butter are mostly saturated. The terms refer to carbon–carbon bonds only: propanone contains a C=O double bond but is described as saturated, because all its carbon–carbon bonds are single.

Unsaturation changes chemical properties. A C=C bond can open up and add other atoms, so alkenes undergo addition reactions that alkanes cannot. Orange bromine water, for example, is decolorized when it is shaken with an alkene but not with an alkane.

Functional groups and chemical reactions

Because reactions take place at the functional group, compounds with the same group react in the same way, whatever the length of the carbon chain. All primary alcohols, from ethanol to hexan-1-ol, can be oxidized to aldehydes and then to carboxylic acids; all alkenes add hydrogen, halogens and water across the C=C bond.

This makes it possible to plan a route from one compound to another as a series of functional group changes. Ethanoic acid can be made from ethene in three steps:

Ethene adds water to give ethanol; ethanol is oxidized to ethanal, and ethanal is oxidized further to ethanoic acid. The reactions of each functional group, and the conditions for them, are described with redox reactions and electron-pair sharing reactions. The amido group also links the amino acids in proteins: two amino acids join in a condensation reaction that forms an amido link and releases water, an HL idea met with condensation polymers.

Homologous series

Methane, ethane, propane and butane all contain only C–C and C–H single bonds, burn in the same way and react with chlorine in the same way. Each molecule has one more carbon atom, and two more hydrogen atoms, than the one before it. They are the first four members of a homologous series, the alkanes.

The members of a homologous series share four features:

• They have the same general formula, such as ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{n }}\mathrm{H}{\vphantom{X}}_{\smash[t]{2n +2}}}}$

General formulas

A general formula follows from the structure. In a straight-chain alkane every carbon atom carries two H atoms, and the two end carbons carry one extra each, so $n$ carbon atoms carry $2n + 2$ hydrogen atoms: ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{n }}\mathrm{H}{\vphantom{X}}_{\smash[t]{2n +2}}}}$

The twelve homologous series in the table are the main families of organic compounds. Their general formulas are not in the data booklet, so they have to be learned.

In the general formulas $n$ is the total number of carbon atoms, except in ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{n }}\mathrm{H}{\vphantom{X}}_{\smash[t]{2n +1}}\mathrm{CONH}{\vphantom{X}}_{\smash[t]{2}}}}$

Trends in physical properties

Within a homologous series the functional group stays the same, so the main difference between successive members is the length of the carbon chain. Physical properties that depend on the forces between molecules, such as boiling point, melting point, viscosity and solubility, change gradually from one member to the next.

Boiling points along a homologous series

The boiling points of the straight-chain alkanes rise steadily from methane to decane. The graph of boiling and melting points against the number of carbon atoms shows the trend.

Alkane molecules are non-polar, so the only forces between them are London (dispersion) forces. These arise from temporary dipoles: at any instant the electrons in a molecule are unevenly spread, and the resulting temporary dipole induces a dipole in a neighbouring molecule. The strength of London forces increases with:

• the number of electrons in the molecule. Each ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}$ unit adds eight electrons. A larger electron cloud is more easily distorted (more polarizable), so the temporary dipoles are larger.

• the area of contact between molecules. Longer chains lie alongside each other over a greater length, so more of each molecule is close to its neighbours.

Boiling requires molecules to separate from each other into the gas, so the energy needed, and the boiling point, rises as the London forces get stronger. Each additional ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}$ causes a smaller rise than the one before: from methane to ethane the boiling point rises by 73 °C, but from nonane to decane by only 23 °C. Adding one ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}$

Melting points also rise with chain length, but less regularly, because melting depends on how well the molecules pack together in the solid as well as on the forces between them. At room temperature the first four alkanes are gases, those from ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{5}}\mathrm{H}{\vphantom{X}}_{\smash[t]{12}}}}$

The same trend is used on a huge scale in oil refineries. Crude oil is a mixture of hydrocarbons, mostly alkanes, with a wide range of chain lengths. In fractional distillation it is vaporized and passed into a tall column that is hot at the bottom and cooler towards the top. The short-chain compounds, with the lowest boiling points, rise furthest before condensing, so the column separates the mixture into fractions of increasing chain length, from refinery gases at the top, through petrol, kerosene and diesel, to the heavy residues used for bitumen at the bottom.

The effect of branching

Isomers with branched chains boil at lower temperatures than their straight-chain isomers. Pentane, 2-methylbutane and 2,2-dimethylpropane all have the formula ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{5}}\mathrm{H}{\vphantom{X}}_{\smash[t]{12}}}}$

With less surface in contact, the temporary dipoles of neighbouring molecules interact less, the London forces are weaker, and less energy is needed to separate the molecules. The more branched the isomer, the lower its boiling point.

The effect of the functional group

Comparing compounds with almost the same molar mass removes the effect of size, because molecules with similar numbers of electrons have London forces of similar strength. Any remaining differences come from the functional group.

• London forces only. Butane is non-polar and has the lowest boiling point.

• Dipole–dipole forces as well. Ethers, esters, aldehydes and ketones contain polar C–O or C=O bonds. Their molecules have permanent dipoles, which attract each other in addition to the London forces, so their boiling points are higher than that of butane. The C=O group of aldehydes and ketones is strongly polar, giving the highest boiling points of this group.

• Hydrogen bonding as well. Alcohols and carboxylic acids have an H atom bonded to oxygen, so their molecules form hydrogen bonds with each other. Hydrogen bonds are the strongest intermolecular forces, and these compounds have by far the highest boiling points. A carboxylic acid boils higher than an alcohol of similar mass because two acid molecules can hold together through two hydrogen bonds at once, using both the C=O and the O–H.

Amines and amides can form hydrogen bonds too, through their N–H bonds. Because nitrogen is less electronegative than oxygen, the hydrogen bonds between amine molecules are weaker than those between alcohol molecules: propylamine, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{NH}{\vphantom{X}}_{\smash[t]{2}}}}$

The effect of a hydrogen-bonding group is largest in small molecules. Methanol boils 227 °C above methane, but hexan-1-ol only 88 °C above hexane. As the chain grows, the London forces between the chains grow while the hydrogen bonding stays the same, so the alcohols become more like the alkanes.

Solubility in water

Small alcohols and carboxylic acids mix with water in all proportions, because their OH groups form hydrogen bonds with water molecules. The hydrocarbon chain cannot hydrogen bond, and as it gets longer it disrupts more of the hydrogen bonding between water molecules than the single OH group can replace. Solubility therefore falls steadily along the series.

Data at 20–25 °C: PubChem and Wikipedia, 2024.

Alkanes are almost insoluble in water, since they cannot form hydrogen bonds with it, but they mix freely with each other and with other non-polar solvents. The same balance between a polar group and a non-polar chain explains why ethanol dissolves both in water and in many oils.

Volatility

A volatile liquid evaporates readily at room temperature, because its intermolecular forces are weak. Volatility follows the reverse of the boiling point trend: it decreases along a homologous series, and branched isomers are more volatile than straight-chain ones. Propanone (boiling point 56 °C) evaporates within seconds from the skin, one reason it suits nail varnish remover, while the small esters are volatile enough to be smelled at low concentrations and give many fruits their scent.

Naming organic compounds

Many organic compounds have traditional names: acetic acid, acetone and wood alcohol. Such names say nothing about structure, and different countries and industries used different ones. The International Union of Pure and Applied Chemistry (IUPAC) publishes rules that give every compound one systematic name from which its structure can be drawn. Acetic acid, the acid in vinegar, is ethanoic acid; acetone, the solvent in many nail varnish removers, is propanone; and the alcohol in alcoholic drinks is ethanol.

The parts of a name

An organic name is built from three parts. The stem gives the number of carbon atoms in the parent chain, the longest continuous chain of carbon atoms that contains the functional group. The suffix (or, for some groups, a prefix) gives the functional group. Prefixes give the substituents, such as alkyl groups and halogen atoms, and locants, numbers that give the positions of the groups on the chain.

For example, 2-methylbutan-1-ol has a four-carbon parent chain (but-), single C–C bonds only (-an-), an OH group on carbon 1 (-1-ol) and a methyl branch on carbon 2 (2-methyl).

The rules

1. Find the longest continuous chain of carbon atoms that contains the functional group (for an alkene, both carbon atoms of the C=C). This chain gives the stem. It does not have to be drawn in a straight line.

2. Identify the functional group and its suffix: -ane, -ene, -yne, -ol, -al, -one or -oic acid. Halogen atoms are named as prefixes (fluoro-, chloro-, bromo-, iodo-).

3. Number the chain from the end that gives the functional group (or the C=C) the lowest possible locant. If there is no suffix group, number from the end that gives the lowest locants to the substituents at the first point of difference.

4. Name each substituent as a prefix with its locant. List the prefixes in alphabetical order (bromo, chloro, ethyl, fluoro, iodo, methyl). Use the numeric prefixes di-, tri-, tetra-, penta- and hexa- for two to six of the same substituent, with a locant for each; these multiplying prefixes are ignored when alphabetizing.

5. Write the name as one word: commas between numbers (2,3-), hyphens between numbers and letters (2-methyl).

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Alkanes

Branches on an alkane are alkyl groups. 2-methylbutane, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}(\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}})\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

The most common naming error is choosing a chain that is not the longest. A condensed or skeletal formula is often drawn with the longest chain bent, so every possible path through the carbon skeleton has to be counted.

Halogenoalkanes

Halogen atoms are named as prefixes in the same way as alkyl groups, and the two kinds of prefix are numbered together. Examples are 2-bromobutane, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CHBrCH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

Alkenes and alkynes

The ending -ene shows a C=C double bond, with the locant of its first carbon atom placed before -ene. The chain is numbered so that the double bond has the lowest locant, even if substituents then get higher numbers: ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}{=}\mathrm{CHCH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

Alcohols

The ending -ol shows a hydroxyl group, with its locant: the "e" of the alkane name is dropped before a vowel, so propane becomes propan-1-ol (${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{OH}}}$

Aldehydes and ketones

Aldehydes end in -al. The CHO carbon is always carbon 1, at the end of the chain, so no locant is needed: methanal (${{\mathrm{HCHO}}}$), ethanal (${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CHO}}}$), propanal (${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CHO}}}$

Ketones end in -one, with a locant for the C=O carbon where there is more than one possible position: pentan-2-one (${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{COCH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

Carboxylic acids

Carboxylic acids end in -oic acid. The carbon of the COOH group is counted in the chain and is always carbon 1: methanoic acid (${{\mathrm{HCOOH}}}$, found in ant stings), ethanoic acid (${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{COOH}}}$, about 4–8% of vinegar), propanoic acid, butanoic acid and 2-methylpropanoic acid, ${{(\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}){\vphantom{X}}_{\smash[t]{2}}\mathrm{CHCOOH}}}$

The reverse task, drawing a structure from its name, follows the same logic. For 3-methylpentan-2-ol: draw a five-carbon chain (pent-), number it, put OH on carbon 2 (-2-ol) and a methyl group on carbon 3. The condensed formula is ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}(\mathrm{OH})\mathrm{CH}(\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}})\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

Names of other classes of compound

Esters, ethers, amines and amides are named by patterns of their own. An ester is made from a carboxylic acid and an alcohol, and its name has two words: the alkyl group from the alcohol, then the acid part with the ending -oate.

Naming inorganic compounds

IUPAC rules apply to inorganic compounds too. Ionic compounds are named cation first, then anion, with -ide for a simple anion. Where a metal can form ions with different charges, its oxidation state is given in Roman numerals in brackets: ${{\mathrm{MnO}{\vphantom{X}}_{\smash[t]{2}}}}$ is manganese(IV) oxide, ${{\mathrm{ReCl}{\vphantom{X}}_{\smash[t]{3}}}}$

Binary covalent compounds of two non-metals are named with the same numeric prefixes as organic substituents (mono-, di-, tri-, tetra-, penta-, hexa-) to show the number of each atom: ${{\mathrm{CO}}}$ is carbon monoxide, ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{4}}}}$

Structural isomers

The molecular formula ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{2}}\mathrm{H}{\vphantom{X}}_{\smash[t]{6}}\mathrm{O}}}$ describes two different compounds. Ethanol, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{OH}}}$

Isomers are different compounds with different physical properties, and often different chemical properties. Structural isomers are divided into three kinds: chain, position and functional group isomers. A second kind of isomerism, in which the connectivity is the same but the arrangement in space differs, is stereoisomerism, an HL topic.

Chain isomers

Butane, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

The number of possible chain isomers grows rapidly with the number of carbon atoms.

Data: Wikipedia, "Alkane", 2024.

Octane isomers and petrol

The 18 isomers of ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{8}}\mathrm{H}{\vphantom{X}}_{\smash[t]{18}}}}$ behave very differently in a car engine. In a petrol engine the fuel–air mixture should burn only when the spark plug fires. Straight-chain alkanes tend to ignite too early under compression, an uncontrolled explosion called knocking that wastes energy and can damage the engine. Highly branched alkanes resist knocking much better.

The octane number of a fuel measures its resistance to knocking against two reference compounds. 2,2,4-trimethylpentane, a highly branched isomer of ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{8}}\mathrm{H}{\vphantom{X}}_{\smash[t]{18}}}}$

Position isomers

Propan-1-ol and propan-2-ol (boiling points 97 °C and 82 °C) are position isomers, and so are but-1-ene and but-2-ene, and 1-bromobutane and 2-bromobutane. In each pair the group is on the end carbon of the chain in one isomer and on an inner carbon in the other.

Functional group isomers

Functional group isomers come from the series that share a general formula. They differ more than other isomers, because their chemical reactions are those of different groups.

Boiling points: PubChem, rounded to the nearest °C.

The boiling points follow from the intermolecular forces: the alcohol and the acid form hydrogen bonds between their molecules, but their ether and ester isomers cannot.

Finding every isomer of a formula

Drawing all the structural isomers of a formula needs a system, so that none are missed and none are drawn twice:

1. Draw the longest possible straight chain and place the functional group in each distinct position.

2. Shorten the chain by one carbon, use that carbon as a branch in each distinct position, and place the functional group again.

3. Continue until no further branching is possible, then consider other functional groups that fit the formula.

4. Name each structure; any two with the same name are the same compound.

Primary, secondary and tertiary compounds

Alcohols and halogenoalkanes are classified by the position of the functional group on the carbon skeleton. The classification depends on the number of carbon atoms bonded to the carbon atom that carries the OH or halogen atom.

Butan-1-ol is a primary alcohol, butan-2-ol a secondary alcohol and 2-methylpropan-2-ol a tertiary alcohol. Methanol and chloromethane, in which the carbon atom carries no other carbon atoms, are classed as primary. Halogenoalkanes follow the same rule: 1-bromobutane is primary, 2-bromobutane secondary and 2-bromo-2-methylpropane tertiary.

Amines are classified differently: by the number of carbon atoms (alkyl or aryl groups) bonded to the nitrogen atom. A primary amine is ${{\mathrm{RNH}{\vphantom{X}}_{\smash[t]{2}}}}$, a secondary amine ${{\mathrm{R}{\vphantom{X}}_{\smash[t]{2}}\mathrm{NH}}}$

The classification matters because it decides how a compound reacts. Primary alcohols are oxidized to aldehydes and then carboxylic acids, secondary alcohols to ketones, and tertiary alcohols resist oxidation; and the mechanism by which a halogenoalkane is substituted depends on whether it is primary or tertiary.

Isomers of benzene compounds (HL)

If benzene had the Kekulé structure, with three fixed C=C and three C–C bonds, 1,2-dibromobenzene would exist as two isomers: one with the two Br atoms on carbons joined by a double bond, and one with them on carbons joined by a single bond. Only one 1,2-dibromobenzene has ever been found. In total there are exactly three isomers of ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{6}}\mathrm{H}{\vphantom{X}}_{\smash[t]{4}}\mathrm{Br}{\vphantom{X}}_{\smash[t]{2}}}}$

This fits a ring in which all six carbon–carbon bonds are identical, with the electrons of the double bonds delocalized around the ring. The same is true of the three dimethylbenzenes (1,2-, 1,3- and 1,4-dimethylbenzene), commonly called xylenes. The delocalized model of benzene, and the other physical and chemical evidence for it, are described with the covalent model.

Stereoisomers and cis–trans isomerism (HL)

Structural isomers, such as butan-1-ol and butan-2-ol, have the same molecular formula but their atoms are joined in a different order. A second kind of isomerism is possible even when the order of attachment is the same. Two molecules can contain the same atoms, joined by the same bonds, and still be different compounds because their atoms point in different directions in space.

The word constitution collects the three things a structural formula shows: which atoms are present, which atoms are bonded to which, and whether each bond is single, double or triple. Stereoisomers share all three. They are distinguished only by a three-dimensional model or a stereochemical formula, never by a flat structural formula. The family tree of isomers shows how the two kinds fit together.

There are two kinds of stereoisomerism. Cis–trans isomerism arises when rotation about a bond is restricted, as in a C=C double bond or a ring. Optical isomerism arises when a molecule is not superimposable on its mirror image, most often because it contains a carbon atom bonded to four different groups. In both cases, turning one stereoisomer into the other would mean breaking a bond and re-forming it in a different direction, so at room temperature the two stereoisomers are separate, stable compounds that can be isolated and bottled.

Restricted rotation about a carbon–carbon double bond

A C=C double bond consists of one σ bond and one π bond. The σ bond lies along the axis between the two carbon nuclei. The π bond forms from the side-on overlap of an unhybridized p orbital on each carbon, and its electron density lies above and below the plane that contains the two carbons and the four atoms attached to them. This orbital picture of double bonds is developed with σ and π bonding in the covalent model.

Side-on overlap is only possible while the two p orbitals are parallel. Twisting one carbon relative to the other moves the p orbitals out of line, and a 90° twist removes the overlap completely. Rotation about a C=C bond would therefore require the π bond to break.

The energy needed is large. Using average bond enthalpies from the data booklet, the π bond is worth roughly $614 - 346 = 268\ \text{kJ mol}^{-1}$ (the C=C bond enthalpy minus the C–C bond enthalpy). The kinetic energy of molecular collisions at room temperature is far too small to supply this, so the groups on each carbon of a double bond stay locked in position relative to each other. This is described as restricted rotation about the double bond.

The two conditions for cis–trans isomerism

Restricted rotation alone is not enough. Propene, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}{=}\mathrm{CHCH}{\vphantom{X}}_{\smash[t]{3}}}}$

1. rotation about the bond joining two carbon atoms is restricted (a C=C double bond, or a bond in a small ring);

2. each of those two carbon atoms carries two different atoms or groups.

But-2-ene, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{=}\mathrm{CHCH}{\vphantom{X}}_{\smash[t]{3}}}}$

The names come from Latin: cis means "on this side" and trans means "across". The prefix is written in italics in print and joined to the name with a hyphen: cis-but-2-ene and trans-but-2-ene. The comparison is always made between groups on different carbon atoms, across the double bond. When all four groups on a C=C bond are different, as in ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{=}\mathrm{C}(\mathrm{Br})\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

The test is quick to apply. Each carbon atom of the double bond is checked in turn to see whether its two attached atoms or groups are different. If either carbon carries two identical groups (two H atoms, two ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ groups, two Cl atoms), there are no cis–trans isomers.

Cis–trans isomerism in cycloalkanes

The carbon atoms in a ring cannot rotate freely about the ring bonds, because a full turn would require a ring bond to break. In cyclopropane (${{\mathrm{C}{\vphantom{X}}_{\smash[t]{3}}\mathrm{H}{\vphantom{X}}_{\smash[t]{6}}}}$) and cyclobutane (

The same two conditions apply as for alkenes: two ring carbons must each carry two different groups. 1,2-Dimethylcyclopropane and 1,2-dichlorocyclobutane have cis and trans isomers. So does 1,3-dichlorocyclobutane, where the two substituted carbons are on opposite corners of the ring. A 1,1-disubstituted ring, such as 1,1-dimethylcyclopropane, has no cis–trans isomers because one ring carbon carries both methyl groups and the other ring carbons carry two H atoms each. The trans isomer of 1,2-dimethylcyclopropane happens to be chiral as well, so it also exists as a pair of non-superimposable mirror-image forms (enantiomers).

Properties of cis and trans isomers

Cis and trans isomers have different shapes, so they pack together differently and often have different polarities. Their physical properties therefore differ, sometimes only slightly and sometimes a great deal. 1,2-Dichloroethene shows the effect of polarity most clearly.

Each C–Cl bond is polar, with the chlorine atom partially negative. In cis-1,2-dichloroethene both chlorine atoms are on the same side of the molecule, so the bond dipoles add to give a net dipole and the molecules attract each other through dipole–dipole forces as well as London forces. In the trans isomer the two bond dipoles point in opposite directions and cancel, leaving a non-polar molecule with London forces only. The cis isomer therefore has the higher boiling point.

Melting points depend on packing in the solid as well as on intermolecular forces. Trans isomers are usually more symmetrical and pack more closely, so they often have the higher melting point even when the cis isomer has the higher boiling point.

Data: PubChem compound records (accessed 2026); values rounded. The melting point of the cis acid is quoted between 131 and 135 °C in different sources.

The two but-2-enedioic acids show that the differences can be large. In the cis isomer (maleic acid) the two –COOH groups are close enough to form a hydrogen bond with each other inside the molecule, so fewer hydrogen bonds form between molecules and the solid melts at a much lower temperature. In the trans isomer (fumaric acid) every –COOH group points outwards and hydrogen-bonds to a neighbouring molecule, giving an extended network in the crystal and a melting point of about 287 °C.

Chemical properties can differ too. When cis-but-2-enedioic acid is heated it loses a molecule of water and forms a cyclic anhydride, because its two –COOH groups are held next to each other. The trans isomer cannot do this directly, because its –COOH groups are held on opposite sides of the rigid double bond. The difference matters in living things as well. In the retina, light absorbed by the visual pigment rhodopsin converts its light-sensitive part, 11-cis-retinal, into all-trans-retinal, and this change of shape starts the nerve signal that produces vision.

Chirality and optical isomerism (HL)

Hands are the everyday example of chirality. A left hand and a right hand are mirror images of each other, with the same parts joined in the same order, yet they cannot be placed one on top of the other so that every part matches: a left glove does not fit a right hand. The word chiral comes from the Greek for "hand". Many molecules, including most of the molecules of life, show the same property.

Chiral carbon atoms

A chiral carbon has a tetrahedral arrangement of four different groups, and there are two ways of arranging four different groups around a tetrahedron. The two arrangements are mirror images of each other. Butan-2-ol is the simplest alcohol with a chiral carbon: its C2 carries H, OH, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ and ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

The four groups must be compared as whole groups, not just by the first atom attached. In butan-2-ol, the ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ and ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

Several kinds of carbon atom can be ruled out at once:

• a ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ or ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}$

What remains are ${{\mathrm{CH}}}$ carbons and carbons with no hydrogen at all; each of these must be checked by comparing its four groups. Chiral carbons are common in medicines. Ibuprofen, a painkiller, has one.

Enantiomers: non-superimposable mirror images

A molecule with one chiral carbon is always chiral and exists as a pair of enantiomers. A molecule that can be superimposed on its mirror image is achiral; molecules with no chiral carbon, such as propan-2-ol, are almost always achiral. In a wedge-dash drawing, the two enantiomers of butan-2-ol are reflections of each other in a mirror plane.

Drawing enantiomers with wedge-dash formulas

A flat structural formula cannot show the difference between enantiomers. A stereochemical formula shows the tetrahedral arrangement around the chiral carbon with three kinds of bond:

• a plain line for a bond in the plane of the paper;

• a solid wedge, tapering from narrow at the chiral carbon to wide at the outer atom, for a bond pointing towards the viewer;

• a hashed (dashed) wedge for a bond pointing away from the viewer.

To draw a pair of enantiomers:

1. Draw the chiral carbon with two bonds in the plane at roughly $109^{\circ}$ to each other (for example one up and one down to the left).

2. Add one wedge and one hashed bond on the other side, pointing roughly downwards to the right, so that the four bonds suggest a tetrahedron.

3. Place the four different groups on the four bonds.

4. Draw a vertical dashed line for the mirror and draw the reflection: every bond keeps its type (plane, wedge or hashed) and its height, but left and right are exchanged.

Why enantiomers cannot be superimposed

With a molecular model kit, or a virtual modelling program, the two enantiomers can be built and compared directly. However one model is turned, at most two of its four groups can be made to coincide with the same groups on the other model. Turning enantiomer 2 of butan-2-ol through 180° lines up the ethyl and methyl groups with those of enantiomer 1, but then OH and H are in each other's positions.

The two enantiomers are therefore different molecules. Converting one into the other would require breaking two bonds to the chiral carbon and re-forming them the other way round, which does not happen at room temperature.

Optical activity and plane-polarized light

Enantiomers have identical melting points, boiling points, densities and solubilities, and identical IR and NMR spectra, because every distance and angle within one molecule is the same in its mirror image. The one physical property that tells them apart is their effect on plane-polarized light.

Ordinary light consists of electromagnetic waves whose oscillations occur in every plane containing the direction of travel. A polarizing filter passes only the oscillations in one plane, producing plane-polarized light. When plane-polarized light passes through a solution of one enantiomer, the plane of polarization is rotated.

An enantiomer that rotates the plane clockwise, as seen by an observer looking towards the light source, is labelled (+) (dextrorotatory); its mirror image rotates the plane anticlockwise by the same angle and is labelled (−) (laevorotatory). The size of the rotation depends on the compound, the concentration of the solution, the length of the sample tube, the temperature and the wavelength of the light, but under identical conditions the two enantiomers always give equal and opposite angles. Because enantiomers are told apart in this way, they are also called optical isomers, and their optical activity is the experimental evidence that a sample contains a chiral compound.

The polarimeter

Optical activity is measured with a polarimeter. Light of a single wavelength, traditionally the yellow light of a sodium lamp at 589 nm, passes through a polarizer and then through a tube containing a solution of the sample. A second polarizing filter, the analyser, is then turned until the maximum amount of light passes through it. The angle through which the analyser has been turned, measured from its position with a tube of pure solvent, is the observed rotation $\alpha$.

The two enantiomers of carvone show the effect clearly. One enantiomer is the main flavour compound of spearmint oil and the other of caraway seeds. Under standard conditions they rotate plane-polarized light by $-61^{\circ}$ and $+61^{\circ}$ respectively (their specific rotations).

Racemic mixtures

A racemic mixture forms whenever a chiral centre is created from achiral starting materials without a chiral influence. The reactant cannot favour one arrangement over its mirror image, so the two enantiomers form at the same rate and in equal amounts. Reactions that pass through a planar intermediate, such as the carbocation formed in some nucleophilic substitutions, give racemic products for this reason. Laboratory syntheses of chiral drugs therefore usually give the racemate unless a chiral catalyst or a chiral starting material is used.

A sample that does not rotate plane-polarized light is therefore one of two things. It contains no chiral molecules at all (an achiral compound), or it is a racemic mixture of a chiral compound. The two cases are distinguished by looking at the structure: a compound with no chiral carbon cannot form a racemic mixture.

Enantiomers in chiral environments

Enantiomers react identically with achiral reagents such as water, hydroxide ions or bromine: the rates and products are mirror images of each other, so they are indistinguishable. Their behaviour differs only in a chiral environment, when they interact with another chiral substance.

Living systems are full of chiral molecules. Proteins are built from amino acids that are almost all one enantiomer, so enzymes and the receptor sites on cells are chiral too. A receptor site can fit one enantiomer of a molecule closely, in the way a right hand fits a right glove, while the other enantiomer fits poorly or binds in a different way. As a result the two enantiomers of a drug, a flavour or a scent can have quite different effects in the body. The two carvones are an everyday example: identical in every laboratory test except polarimetry, they smell distinctly different (spearmint and caraway) because the smell receptors in the nose are chiral.

Thalidomide. Thalidomide was introduced in West Germany in 1957 as a sedative, and was soon also prescribed to pregnant women to relieve morning sickness. It has one chiral carbon and was sold as a racemic mixture. By 1961 it had been linked to severe birth defects, especially malformed or missing limbs, and it was withdrawn; more than 10,000 children were affected worldwide. Studies in animals linked the sedative effect mainly to one enantiomer and the damage to the developing embryo mainly to the other.

Selling only the sedative enantiomer would not have prevented the tragedy. The H atom on thalidomide's chiral carbon is easily removed and replaced in the body, so each enantiomer is converted into a mixture of both within hours of being taken. The disaster led to much stricter testing of new medicines, including testing each enantiomer of a chiral drug separately. Thalidomide is still used today, under strict controls that prevent its use in pregnancy, to treat multiple myeloma (a blood cancer) and a painful complication of leprosy.

Ibuprofen gives a milder example. It is usually sold as a racemic mixture, although almost all of its painkilling activity comes from one enantiomer; an enzyme in the body converts much of the other enantiomer into the active one, but not the reverse. Many newer medicines are made and sold as a single enantiomer, which avoids giving patients a second compound that has no benefit and may cause side effects.

Fragmentation in mass spectrometry (HL)

Mass spectrometry of elements separates isotopes and gives relative atomic masses. When an organic compound is analysed, the spectrum contains far more information. The heaviest ion usually gives the relative molecular mass of the compound, and many lighter ions, formed when molecules break apart inside the instrument, show which groups of atoms the molecule contains. Mass spectrometry needs only a tiny sample, and it is used to identify drugs, pollutants and unknown products of synthesis.

The molecular ion

In the most common method the vaporized sample is bombarded with a beam of high-energy electrons. A collision knocks one electron out of a molecule, M, to form a positive ion:

The horizontal axis of a mass spectrum is the mass-to-charge ratio, $m/z$. Almost all the ions formed carry a charge of 1+, so $m/z$ is numerically equal to the mass of the ion. The vertical axis is the relative abundance of each ion, with the tallest peak, the base peak, set to 100 %. The molecular ion lost a single electron, so it has an odd number of electrons and is a radical cation (sometimes written $\mathrm{M}^{\bullet+}$).

The molecular ion peak is the peak at the highest $m/z$, apart from very small peaks one or two units higher that come from heavier isotopes. Its height varies greatly. Aromatic compounds give strong molecular ion peaks, while many alcohols give weak ones because their molecular ions break up so readily: the molecular ion of butan-2-ol is less than 1 % of the base peak.

How molecular ions fragment

Electron bombardment gives the molecular ions much more energy than is needed to remove an electron. The excess energy makes bonds break, and each molecular ion can split into two parts. One part keeps the positive charge and the other is neutral, most often a radical:

Only charged particles are accelerated, separated and detected in the mass spectrometer. The positive ion ${{\mathrm{X}{\vphantom{X}}^{+}}}$ is therefore recorded as a peak, but the neutral fragment $\mathord{\bullet}{{\mathrm{Y}}}$ is not. A fragment ion appears at an $m/z$ value equal to the mass of the molecular ion minus the mass of the neutral fragment it lost. The breaking up of molecular ions in this way is called fragmentation. Ethanol, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{OH}}}$

Each of these ions appears as a peak in the mass spectrum of ethanol. The most abundant ion, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{OH}{\vphantom{X}}^{+}}}$ at $m/z = 31$

Reading the spectrum from right to left gives the structure in pieces. The peak at 46 is the molecular ion, so $M_\text{r} = 46$. The peak at 45 is $\text{M} - 1$: loss of one H atom. The peak at 31 is $\text{M} - 15$, the loss of $\mathord{\bullet}{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

Fragmentation is not random. Bonds break most often where the products are most stable: where a weak bond breaks, or where the positive ion formed is stabilized. Ions in which the positive charge sits next to an oxygen atom, such as ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{OH}{\vphantom{X}}^{+}}}$ and the acylium ion ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CO}{\vphantom{X}}^{+}}}$

Using the table of fragments lost

The data booklet lists the masses of the neutral fragments most commonly lost from a molecular ion. The difference between the $m/z$ of the molecular ion and the $m/z$ of a fragment peak gives the mass lost, and the table suggests what that fragment was.

Data: IB Chemistry data booklet (2025), section 22.

The same masses help to identify the fragment ions themselves. Some ions turn up in the spectra of many compounds:

Fragmentation patterns of homologous series

Compounds in the same homologous series break up in similar ways, so a few patterns go a long way.

Alkanes

Alkanes give clusters of peaks 14 units apart (one ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}$), from the alkyl cations ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}{\vphantom{X}}^{+}}}$

Alcohols

Alcohols often lose water ($\text{M} - 18$), and the C–C bond next to the carbon carrying the –OH group breaks readily. Primary alcohols give a strong peak at 31 (${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{OH}{\vphantom{X}}^{+}}}$

Aldehydes and ketones

Bonds next to the carbonyl group break readily, giving acylium ions (${{\mathrm{RCO}{\vphantom{X}}^{+}}}$), which are stabilized by the oxygen atom. A methyl ketone gives a strong peak at 43 (${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CO}{\vphantom{X}}^{+}}}$

Propanone, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{COCH}{\vphantom{X}}_{\smash[t]{3}}}}$

Carboxylic acids and esters

Carboxylic acids lose OH ($\text{M} - 17$) and COOH ($\text{M} - 45$), and often show ${{\mathrm{COOH}{\vphantom{X}}^{+}}}$ at 45. Ethanoic acid, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{COOH}}}$

Aromatic compounds

A benzene ring is very stable, so aromatic compounds give strong molecular ion peaks. A compound containing a ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{6}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}}}$ group usually gives the phenyl cation,

Isotope peaks: M+1 and M+2

About 1.1 % of carbon atoms are carbon-13, so a small fraction of molecules is one mass unit heavier than the rest. This gives a small M+1 peak just to the right of the molecular ion peak. Its height relative to the molecular ion is roughly 1.1 % for every carbon atom in the molecule, so an M+1 peak about 4.4 % as tall as the molecular ion suggests four carbon atoms.

Chlorine and bromine produce much larger isotope peaks. Chlorine is about 75 % chlorine-35 and 25 % chlorine-37, so a compound with one chlorine atom gives two molecular ion peaks two units apart in a 3 : 1 ratio. Bromine is roughly half bromine-79 and half bromine-81, so a compound with one bromine atom gives two peaks of almost equal height. The mass spectrum of bromoethane shows this pattern at 108 and 110.

An M and M+2 pair in a 3 : 1 ratio therefore points to one chlorine atom, and a pair in a 1 : 1 ratio to one bromine atom. Fragments that keep the halogen atom show the same pairs, while fragments that have lost it, such as ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{2}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}{\vphantom{X}}^{+}}}$

Interpreting a mass spectrum

Interpreting a spectrum follows the same steps each time: find the molecular ion, account for the large peaks as losses from it, then fit the pieces into a structure consistent with any other information given.

Mass spectrometry is one of the main techniques in drug testing. In sports anti-doping laboratories and forensic laboratories, the substances in a urine or blood sample are first separated by gas or liquid chromatography, and each one is passed straight into a mass spectrometer. The fragmentation pattern of each substance is compared with a library of reference spectra, so a banned substance can be identified, and its amount measured, even when only a few nanograms are present.

Infrared spectroscopy (HL)

Chemists rarely identify a new compound by its chemical reactions alone. Instruments that measure how molecules interact with electromagnetic radiation give faster and more certain evidence. Infrared spectroscopy (IR) shows which bonds are present, and so which functional groups a molecule contains. Proton nuclear magnetic resonance spectroscopy ($^{1}\text{H}$ NMR) shows how the hydrogen atoms are arranged, and mass spectrometry gives the relative molecular mass and the pieces a molecule breaks into. Used together, these techniques can establish the full structure of an unknown compound.

How molecules absorb infrared radiation

The atoms in a molecule are never still. Each covalent bond behaves like a spring joining two masses: it can lengthen and shorten (a stretching vibration), and the angle between two bonds can open and close (a bending vibration). Every vibration has a natural frequency that depends on the strength of the bond and on the masses of the atoms at each end.

Infrared radiation has frequencies in the same range as these vibrations. When the frequency of the radiation matches the natural frequency of a vibration, the molecule absorbs a photon and vibrates with a larger amplitude. Radiation of all other frequencies passes through. A spectrometer passes infrared radiation through a sample and records which frequencies are absorbed.

Wavenumber

Infrared spectra are plotted against wavenumber, the number of wavelengths per centimetre, rather than against frequency or wavelength. Wavenumber is the reciprocal of the wavelength in centimetres:

Since $c = f\lambda$ and $E = hf$, wavenumber is proportional to both the frequency and the energy of the radiation. A higher wavenumber therefore means a more energetic photon and a stiffer or lighter vibration. Organic spectra usually run from $4000\ \text{cm}^{-1}$ (a wavelength of $2.5\ \mu\text{m}$

What sets the wavenumber of a bond

Two factors decide where a bond absorbs, as they would for a spring with masses on it:

• Bond strength. A stronger, stiffer bond vibrates at a higher frequency. Multiple bonds between the same atoms absorb at higher wavenumbers than single bonds: C≡C at $2100\text{–}2260\ \text{cm}^{-1}$, C=C at $1620\text{–}1680\ \text{cm}^{-1}$, and C–C much lower, in the fingerprint region.

• Mass of the atoms. Heavier atoms vibrate more slowly. Bonds to hydrogen, the lightest atom, absorb at the highest wavenumbers (O–H, N–H and C–H between about 2500 and $3600\ \text{cm}^{-1}$

Bonds to hydrogen therefore absorb at the left of a spectrum, multiple bonds in the middle and single bonds between heavier atoms at the right, as the chart of characteristic ranges shows.

Which molecules absorb: IR activity

A vibration absorbs infrared radiation only if it changes the dipole moment of the molecule. The oscillating electric field of the radiation can push and pull only on separated charges, so a vibration that moves partial charges relative to each other couples to the radiation, while one that leaves the charge distribution symmetrical does not.

This rule follows from bond polarity. A polar bond such as C=O, O–H or C–Cl has a large dipole that changes strongly as the bond stretches, so these bonds give strong absorptions. Diatomic molecules of one element (${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$

A molecule with no net dipole can still absorb if some of its vibrations make it temporarily polar. Carbon dioxide is linear and non-polar, but two of its three kinds of vibration distort the molecule unsymmetrically, as the three vibrations of carbon dioxide show.

In the symmetric stretch the two bond dipoles stay equal and opposite, so the dipole moment remains zero. In the asymmetric stretch one C=O bond is longer than the other at any instant, and in the bend the molecule is no longer linear; both create a changing dipole and absorb. Water and methane are IR active in the same way: water is polar and every one of its vibrations changes its dipole, and methane has vibrations that distort its tetrahedral symmetry.

Greenhouse gases and the absorption of infrared radiation

The Sun's radiation reaching the Earth is mostly visible and near-infrared, with some ultraviolet, because the Sun's surface is very hot (about 5800 K). The atmosphere absorbs little of it, so most reaches and warms the ground and oceans. The much cooler Earth (average surface temperature about 288 K, or 15 °C) emits its energy as longer-wavelength infrared radiation, most of it between about $200$ and $2000\ \text{cm}^{-1}$ (5 to 50 μm).

A greenhouse gas is a gas whose molecules absorb this outgoing infrared radiation: water vapour, carbon dioxide, methane and dinitrogen monoxide (${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$) are the main ones. Nitrogen and oxygen, which make up 99% of dry air, are IR inactive and let it pass. After absorbing a photon, a greenhouse gas molecule passes the energy to other molecules in collisions or re-emits infrared radiation in a random direction, so part of the energy is sent back towards the surface. This raises the temperature of the lower atmosphere and the surface, an effect called the greenhouse effect.

Each gas absorbs only in the bands set by its IR-active vibrations. Between about $800$ and $1250\ \text{cm}^{-1}$ none of the main gases absorbs strongly, and much of the Earth's radiation escapes through this atmospheric window. A gas that absorbs inside the window has a particularly large effect, because it blocks radiation that would otherwise escape.

The GWP of a gas depends on three properties: how strongly each molecule absorbs infrared radiation (a large change in dipole moment gives intense absorption), whether it absorbs at wavenumbers where the Earth emits strongly and other gases do not, and how long it stays in the atmosphere before it is removed by chemical reactions.

Data: IPCC Sixth Assessment Report, Working Group I (2021), Table 7.SM.7 (GWP of methane rounded from 27.9); absorption wavenumbers from the NIST Chemistry WebBook. Some recent measurements suggest a shorter lifetime for ${{\mathrm{SF}{\vphantom{X}}_{\smash[t]{6}}}}$, nearer 1000 years.

Over a century, a kilogram of methane causes about 28 times as much warming as a kilogram of carbon dioxide, although each methane molecule lasts only about 12 years. Sulfur hexafluoride, used as an insulating gas in high-voltage electrical switchgear, combines very strong absorption by its six polar S–F bonds, an absorption band inside the atmospheric window and a lifetime of a thousand years or more. Carbon dioxide has the lowest GWP of these gases, but it is emitted in far larger amounts than any other, so it causes most of the human contribution to global warming. Water vapour is the most abundant greenhouse gas; its concentration is controlled by temperature rather than by emissions, so it amplifies warming caused by the other gases.

Reading an infrared spectrum

An IR spectrum plots percentage transmittance (the percentage of the radiation that passes through the sample) against wavenumber. Two conventions make spectra look unlike most graphs. The wavenumber axis runs from high values on the left ($4000\ \text{cm}^{-1}$) to low values on the right, and absorptions appear as dips pointing down from a baseline near 100% transmittance, because they are frequencies at which less radiation reaches the detector. Spectroscopists still call these dips peaks or bands.

The infrared spectrum of ethanol has the features of most spectra. The region from $4000$ to about $1500\ \text{cm}^{-1}$ is the functional group region. Its absorptions come mostly from the stretching of a single bond (O–H, C–H, C=O), so each can be assigned to a functional group using the data booklet. The region below about $1500\ \text{cm}^{-1}$ is the fingerprint region: it contains many overlapping absorptions from C–C and C–O stretches and from bending vibrations of the whole molecule.

No two compounds have the same fingerprint region, even when their functional groups are identical. A spectrum that matches a reference spectrum from a database in every absorption, including the fingerprint region, identifies the compound, and extra absorptions show that an impurity is present.

Characteristic absorptions in the data booklet

Section 20 of the data booklet lists the wavenumber ranges at which bonds absorb, with the intensity of each absorption.

Data: IB Chemistry data booklet (2025), section 20.

The shape of an absorption carries information as well as its position. The O–H stretch of an alcohol is broad because hydrogen bonding between molecules gives each O–H bond a slightly different environment and strength, so the sample absorbs over a range of wavenumbers rather than at one. In carboxylic acids the molecules form strongly hydrogen-bonded pairs (dimers), which weakens the O–H bond further and spreads its absorption from about $2500$ to $3300\ \text{cm}^{-1}$, often with the sharper C–H absorptions sitting on top of it. The C=O stretch is strong and fairly sharp because the bond is very polar.

Identifying functional groups from IR spectra

Most organic molecules contain C–H bonds, so an absorption at $2850\text{–}3090\ \text{cm}^{-1}$ tells little. The useful evidence comes from the presence or absence of a few strong absorptions, checked in a fixed order:

1. Is there a strong C=O absorption at $1700\text{–}1750\ \text{cm}^{-1}$? If so, the compound is an aldehyde, ketone, carboxylic acid or ester.

2. Is there an O–H absorption? A broad band at $3200\text{–}3600\ \text{cm}^{-1}$ indicates an alcohol or phenol; a very broad band at $2500\text{–}3000\ \text{cm}^{-1}$

The infrared spectra of ethanoic acid and ethyl ethanoate show how the presence or absence of a single band separates two classes of compound that both contain C=O.

The same reasoning applies to every class of compound in the data booklet, summarized in the table of key absorptions.

IR spectroscopy separates classes of compound well, but compounds in the same class often give very similar functional group regions. Propanal and propanone both show C=O at $1700\text{–}1750\ \text{cm}^{-1}$ and no O–H, and butan-1-ol and butan-2-ol differ only in the fingerprint region. $^{1}\text{H}$ NMR and mass spectrometry are needed to tell such isomers apart.

Following a reaction by IR

Because IR shows which functional groups are present, it is used to follow reactions that convert one group into another. When ethanol is oxidized by acidified potassium dichromate(VI), the broad alcohol O–H band at $3200\text{–}3600\ \text{cm}^{-1}$ disappears. If the product is distilled off as it forms, a C=O band appears at about $1730\ \text{cm}^{-1}$ (ethanal, with no O–H). If the mixture is heated under reflux, the product is ethanoic acid, with a C=O band and the very broad acid O–H band at $2500\text{–}3000\ \text{cm}^{-1}$

Breath analysis

Evidential breath analysers used by police forces measure the concentration of ethanol in exhaled air by infrared absorption. A beam of infrared radiation passes through a sample chamber filled with breath, and the fraction absorbed at a wavenumber characteristic of ethanol increases with its concentration. Instruments measure at about $3.4\ \mu\text{m}$ ($2940\ \text{cm}^{-1}$, the C–H stretch), at about $9.5\ \mu\text{m}$ ($1050\ \text{cm}^{-1}$

The O–H absorption of ethanol is never used, because exhaled breath is saturated with water vapour, which absorbs strongly over the same range. The C–H band has a weakness of its own: almost every organic compound absorbs near $2940\ \text{cm}^{-1}$, including the propanone found in the breath of people with poorly controlled diabetes, so a reading at that wavenumber alone could overstate the ethanol concentration. Measuring the C–O band as well, or comparing absorbances at more than one wavenumber, lets the instrument detect such interfering substances.

$^{1}\text{H}$ NMR spectroscopy: environments, chemical shift and integration (HL)

How NMR spectroscopy works

The nucleus of a hydrogen atom (a single proton) has a property called spin, which makes it behave like a tiny bar magnet. With no external field, all orientations of this magnet have the same energy. In a strong magnetic field the proton can take only two orientations: aligned with the field (lower energy) or opposed to it (higher energy).

A proton in the lower state can absorb electromagnetic radiation whose photon energy exactly matches the gap $\Delta E$ and flip to the higher state. This absorption is called nuclear magnetic resonance. In the fields used in spectrometers the gap is very small, so the radiation absorbed is in the radio-frequency region of the electromagnetic spectrum, typically 300 to 900 MHz.

Different hydrogen atoms absorb at different frequencies because electrons shield the nucleus. The electrons around a proton circulate in the applied field and set up a small magnetic field of their own that opposes it, so the nucleus experiences a slightly weaker field than the one applied. The more electron density around a hydrogen atom, the more it is shielded, and the lower the frequency at which it absorbs.

An electronegative atom nearby (O, N, a halogen) pulls electron density away from the hydrogen atom, which is then deshielded: it experiences more of the applied field and absorbs at a higher frequency. The π electrons of C=O, C=C and benzene rings circulate in a way that also deshields hydrogen atoms attached to or near them. Each hydrogen atom therefore absorbs at a frequency that reflects its position in the molecule.

In practice the sample, a few milligrams of compound, is dissolved in a solvent such as ${{\mathrm{CDCl}{\vphantom{X}}_{\smash[t]{3}}}}$, in which deuterium (${{{\vphantom{X}}^{\hphantom{2}}_{\hphantom{}}{\vphantom{X}}^{\smash[t]{\vphantom{2}}\mathllap{2}}_{\vphantom{2}\mathllap{\smash[t]{}}}\mathrm{H}}}$

The same physics is the basis of magnetic resonance imaging (MRI) in hospitals. A patient lies in a magnetic field of 1.5 or 3 tesla, and the radio-frequency signals of the hydrogen nuclei in water and fat molecules are used to build cross-sectional images of soft tissue. MRI uses no ionizing radiation, unlike X-ray imaging.

Chemical environments and the number of signals

Hydrogen atoms in the same chemical environment absorb at the same frequency and give one signal. Two hydrogen atoms are in the same environment, or are equivalent hydrogen atoms, when they are bonded to the same kind of atom and are related by the symmetry of the molecule or by free rotation about a single bond. The three hydrogen atoms of a ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ group are always equivalent, because the group rotates rapidly.

To count environments, draw the full structural formula, then work along the molecule asking whether each hydrogen atom could be swapped for another by a mirror plane or a rotation of the whole molecule. The displayed formulas of propan-1-ol and propan-2-ol, isomers with the same molecular formula, show how symmetry changes the count.

In propan-1-ol the two ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}$ groups are different, because one is next to the OH group and one is not, so the molecule has four environments. Propan-2-ol has a mirror plane through the central carbon, so its two ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$

The two methyl groups of methyl ethanoate both contain three hydrogen atoms, but they are in different environments: one is attached to the C=O carbon and the other to the oxygen atom of the ester group. In nitrobenzene the ring hydrogen atoms fall into three sets: two next to the ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ group, two one carbon further away, and one opposite it.

Chemical shift

The frequency at which a proton absorbs depends on the strength of the magnet, so absolute frequencies would differ from one spectrometer to another. Instead, the position of each signal is measured relative to a reference compound, tetramethylsilane (TMS), ${{\mathrm{Si}(\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}){\vphantom{X}}_{\smash[t]{4}}}}$

TMS is chosen as the reference for several reasons:

• its twelve hydrogen atoms are all equivalent, so it gives a single, strong signal;

• silicon is less electronegative than carbon, so the hydrogen atoms in TMS are more shielded than those in almost all organic compounds, and its signal lies to the right of theirs, away from the signals being studied;

• it is chemically inert, non-toxic and volatile (boiling point 27 °C), so it can be removed easily from the sample afterwards.

Spectra are plotted with $\delta$ increasing from right to left, so TMS appears at the right-hand edge. Most organic hydrogen atoms appear between 0 and 13 ppm. The more deshielded a hydrogen atom, the larger its chemical shift, as the chart of chemical shift regions shows.

Section 21 of the data booklet gives typical chemical shifts. In each structure the bold H is the hydrogen atom whose shift is given, R is an alkyl group and Hal is F, Cl, Br or I.

Data: IB Chemistry data booklet (2025), section 21. The values may vary with the solvent and conditions.

The ranges follow the shielding argument. Alkyl hydrogen atoms, surrounded only by carbon and hydrogen, are the most shielded (about 1 ppm). A neighbouring C=O group moves a signal to about 2–2.7 ppm, and an oxygen or halogen atom on the same carbon to 3.3–4.8 ppm. Hydrogen atoms on C=C and on benzene rings are deshielded by π electrons (4.5–9 ppm). The aldehyde hydrogen, attached directly to the C=O carbon, and the acid O–H hydrogen are the most deshielded of all (9–13 ppm). The alcohol O–H signal can appear anywhere from 1 to 6 ppm, because its position depends on the extent of hydrogen bonding, which changes with concentration and solvent.

Integration: the relative number of hydrogen atoms

The area under each signal is proportional to the number of hydrogen atoms in that environment. Spectrometers measure the areas electronically and draw an integration trace, a line that rises in a step at each signal. The height of each step is proportional to the area of the signal beneath it.

The integration trace of ethanol rises in steps of 20 mm, 10 mm and 30 mm, a ratio of $2 : 1 : 3$. Ethanol has six hydrogen atoms, so the signals contain two, one and three hydrogen atoms: the ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{\mathrm{OH}}}$

Of equal amounts of different compounds, the one with the most equivalent hydrogen atoms in one environment gives the largest signal. TMS, with twelve equivalent hydrogen atoms, gives a larger single signal than equimolar 2-methylpropane, benzene or octane.

Splitting patterns in $^{1}\text{H}$ NMR spectra (HL)

A low-resolution spectrum, recorded on a simple instrument, shows one peak for each environment. A high-resolution spectrum, recorded with a stronger and more uniform magnetic field, shows that many of these signals are clusters of closely spaced lines. The pattern of lines in each cluster is caused by the hydrogen atoms on the neighbouring carbon atoms.

Spin–spin coupling and the n + 1 rule

Each hydrogen nucleus is a small magnet that is aligned with or opposed to the applied field. A proton on a neighbouring carbon atom therefore feels a field that is slightly increased or slightly decreased by the neighbours, depending on how their spins are arranged. Each different arrangement of neighbouring spins shifts the absorption by a small, fixed amount, so one signal becomes several lines. This interaction between the nuclei on adjacent atoms is called spin–spin coupling.

With $n$ equivalent neighbouring hydrogen atoms there are $n + 1$ possible values of their combined spin, so the signal is split into $n + 1$ lines. The heights of the lines follow the number of arrangements that give each combination: one way for all three spins aligned, three ways for two aligned and one opposed, and so on. These ratios are the numbers in Pascal's triangle.

The splitting describes the neighbours of a group, not the group itself. A ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ group next to a ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}$

Rules for predicting splitting

• Only hydrogen atoms on adjacent carbon atoms split a signal. Hydrogen atoms further away have a negligible effect, so a ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ attached to a C=O carbon, to the O atom of an ester, or to a carbon with no hydrogen atoms is a singlet.

• Equivalent hydrogen atoms do not split each other. The three hydrogen atoms of a methyl group do not split their own signal, and ethane, with six equivalent hydrogen atoms, gives a singlet.

• Coupling is mutual. If group A splits group B, group B splits group A, and the lines in both clusters are the same distance apart. A 3H triplet and a 2H quartet with the same spacing are the signature of an ethyl group, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}{\vphantom{X}}^{-}}}$

The high-resolution spectrum of ethyl ethanoate shows the three most common patterns together.

The signal at 4.12 ppm (2H) is in the ${{\mathrm{R}{-}\mathrm{C}({=}\mathrm{O}){-}\mathrm{O}{-}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}{\vphantom{X}}^{-}}}$

Recognizing common groups from their patterns

The 2-propyl pattern appears in the spectrum of propan-2-ol, where one hydrogen atom splits the six methyl hydrogen atoms into a doublet and is itself split by six neighbours.

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Combining evidence to determine a structure (HL)

No single technique gives a complete structure. Mass spectrometry gives the relative molecular mass and pieces of the molecule but says little about how they are joined; IR shows functional groups but not where they are; $^{1}\text{H}$ NMR shows how the hydrogen atoms are arranged but does not directly detect atoms such as oxygen or bromine. Each technique tests a different feature, so a proposed structure that fits all the evidence is very likely to be correct.

In a combined analysis the mass spectrum is used mainly for the relative molecular mass, from the molecular ion. The fragment peaks, read as losses from ${{\mathrm{M}{\vphantom{X}}^{+}}}$ in the way described for fragmentation in mass spectrometry, then confirm or rule out a structure suggested by the other data.

A strategy for an unknown compound

The molecular formula itself contains structural information. An open-chain alkane ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{n }}\mathrm{H}{\vphantom{X}}_{\smash[t]{2n +2}}}}$

Worked analysis of an unknown ester

Compound X is a colourless liquid with a fruity smell. Combustion analysis shows that it contains 54.5% carbon, 9.2% hydrogen and 36.3% oxygen by mass. Its mass spectrum is the first piece of spectroscopic evidence.

The infrared spectrum of X has no absorption above $3100\ \text{cm}^{-1}$.

The $^{1}\text{H}$ NMR spectrum of X, with the integration, completes the data.

Choosing a technique to distinguish two compounds

Many problems ask which technique distinguishes two given compounds. The approach is to predict what each technique would show for each compound and look for a difference that the data booklet supports.

For isomers in the same class, IR is usually the weakest technique. Propanal and propanone, both ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{3}}\mathrm{H}{\vphantom{X}}_{\smash[t]{6}}\mathrm{O}}}$

Spectra are now collected in public databases such as the Spectral Database for Organic Compounds (SDBS), run by Japan's National Institute of Advanced Industrial Science and Technology, and the NIST Chemistry WebBook. Matching the IR, mass and NMR spectra of an unknown against tens of thousands of reference spectra makes the identification of known compounds fast and reliable. A completely new compound has no reference spectrum, and its structure must be deduced from first principles, as for compound X; for complex molecules chemists add techniques such as X-ray crystallography, which locates every atom in a crystal.

Applying the ideas in exam questions

Paper 1A: naming a branched alkane

Paper 1A: members of a homologous series

Paper 1A: functional group isomers

Paper 1B: boiling points of the alcohols

A student measured the boiling points of five straight-chain alcohols, each with its OH group on carbon 1, by distilling a sample of each and reading a thermometer placed level with the side arm of the flask.

Paper 2: two isomers with the formula C₄H₈O₂

Paper 1A: conditions for cis–trans isomerism (HL)

Paper 1A: counting chiral carbon atoms (HL)

Paper 1A: the species responsible for a peak (HL)

Paper 2: the enantiomers of lactic acid (HL)

Paper 2: two isomeric alcohols in the mass spectrometer (HL)

Paper 1A: distinguishing propanal from propanone (HL)

Paper 1A: number of signals and their ratio (HL)

Paper 1B: following the oxidation of propan-1-ol by IR (HL)

A student oxidized propan-1-ol with acidified potassium dichromate(VI) in two ways: product P was distilled off as soon as it formed, and product Q was obtained after heating under reflux for 30 minutes. The table gives the main absorptions in the infrared spectra.

Paper 2: identifying an unknown compound ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{4}}\mathrm{H}{\vphantom{X}}_{\smash[t]{8}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$

Paper 2: deducing a structure from a high-resolution spectrum (HL)

Self-check

• To do:

  I can explain why carbon forms so many compounds.

• To do:

  I can write empirical, molecular, full structural, condensed structural and skeletal formulas, and convert between them.

• To do:

  I can state what each type of formula shows and what it does not show.

• To do:

  I can identify the halogeno, hydroxyl, carbonyl, carboxyl, alkoxy, amino, amido, ester and phenyl groups by name and structure, including in large molecules.

• To do:

  I can distinguish carbonyl from carboxyl, ester from ether, and amino from amido groups.

• To do:

  I can define saturated and unsaturated compounds.

• To do:

  I can state the features of a homologous series and identify the twelve homologous series from their structures and general formulas.

• To do:

  I can derive and use general formulas, including the next member of a series.

• To do:

  I can explain the trend in boiling points along a homologous series in terms of London forces.

• To do:

  I can explain the effects of branching and of the functional group on boiling point, volatility and solubility in water.

• To do:

  I can name straight-chain and branched compounds with up to six carbon atoms in the parent chain containing a halogeno, hydroxyl, carbonyl or carboxyl group, and draw them from their names.

• To do: