S3.2 Functional groups: classification of organic compounds | IB DP Chemistry

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🎯What you need to be able to do

Identify and interconvert empirical, molecular, structural (full and condensed) and skeletal formulas, and construct 3D models of organic molecules.
Identify by name and structure the functional groups: halogeno, hydroxyl, carbonyl, carboxyl, alkoxy, amino, amido, ester and phenyl; use the terms saturated and unsaturated.
Identify the homologous series alkanes, alkenes, alkynes, halogenoalkanes, alcohols, aldehydes, ketones, carboxylic acids, ethers, amines, amides and esters.
Describe and explain the trend in melting and boiling points within a homologous series.
Apply IUPAC nomenclature to compounds with up to six carbons in the parent chain containing one type of the halogeno, hydroxyl, carbonyl or carboxyl group, including branched isomers.
Recognise structural isomers: branched, straight-chain, position and functional group isomers, and classify alcohols, halogenoalkanes and amines as primary, secondary or tertiary.
AHL Describe and explain cis–trans isomerism and optical isomerism, and draw stereochemical formulas of enantiomers.
AHL Deduce structural information from mass spectra, infrared spectra and proton NMR spectra, including splitting patterns, and combine the three.

📚The chemistry

Ways of writing an organic molecule

The same compound can be represented five ways, and questions move between them freely. For butan-1-ol:

Empirical
simplest ratio
\( \mathrm{C_4H_{10}O} \) (already simplest here)

Molecular
actual atom count
\( \mathrm{C_4H_{10}O} \)

Condensed structural
shows the sequence
\( \mathrm{CH_3CH_2CH_2CH_2OH} \)

Full structural
every atom and every bond drawn

Skeletal
carbon chain as a zig-zag; carbons and their hydrogens implied; heteroatoms and functional groups shown

Stereochemical
shows 3D arrangement with wedge and dash bonds. Not expected to be drawn except where specifically indicated

Skeletal formulas are worth becoming fluent in early: each vertex and each line end is a carbon, and each carbon carries however many hydrogens bring it to four bonds. They are much faster to draw, and AHL mechanism questions assume them.

Functional groups

A functional group is the atom or group of atoms that gives a compound its characteristic physical and chemical properties. It is the reason organic chemistry is learnable at all: there are millions of compounds but only a handful of behaviours.

halogeno –X
(F, Cl, Br, I)

hydroxyl –OH

carbonyl C=O

carboxyl –COOH

alkoxy –OR

amino –NH₂

amido –CONH₂

ester –COO–

phenyl –C₆H₅

A compound is saturated if it contains only single carbon–carbon bonds, and unsaturated if it contains a double or triple bond. Unsaturation is what makes alkenes react by addition (R3.4) and what allows addition polymerization (S2.4).

Homologous series

A homologous series is a family of compounds in which successive members differ by a common structural unit, usually CH₂. Members share the same functional group, the same general formula and similar chemical properties, and show a gradation in physical properties.

Alkanes \( \mathrm{C}_n\mathrm{H}_{2n+2} \)
–ane

Alkenes \( \mathrm{C}_n\mathrm{H}_{2n} \)
–ene

Alkynes \( \mathrm{C}_n\mathrm{H}_{2n-2} \)
–yne

Halogenoalkanes
chloro–, bromo–, iodo–

Alcohols \( \mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{OH} \)
–ol

Aldehydes RCHO
–al · carbonyl at chain end

Ketones RCOR’
–one · carbonyl within the chain

Carboxylic acids RCOOH
–oic acid

Ethers ROR’
alkoxyalkane

Amines RNH₂
–amine

Amides RCONH₂
–amide

Esters RCOOR’
alkyl –anoate

Trends in physical properties

Melting and boiling points increase as the chain lengthens. Each extra CH₂ adds electrons and surface area, so the London (dispersion) forces between molecules become stronger, and more energy is needed to separate them. Hence the alkanes go gas → liquid → solid as you go down the series, and volatility falls.

Two refinements that get examined:

Branching lowers the boiling point. A branched isomer is more spherical, so the molecules make less surface contact, the London forces are weaker and it boils lower than its straight-chain isomer — even though the molar mass is identical.
The functional group can outweigh the chain. Compare compounds of similar molar mass: an alkane (London only) boils lowest, an aldehyde or ketone (dipole–dipole) higher, an alcohol (hydrogen bonding) higher still, and a carboxylic acid highest of all, because two molecules hydrogen bond to each other twice over as a dimer. Solubility in water follows the same logic and falls off as the hydrocarbon chain lengthens.

IUPAC nomenclature

The scope is defined tightly: up to six carbons in the parent chain, and one type of the halogeno, hydroxyl, carbonyl or carboxyl groups. Straight-chain and branched isomers are both included.

Find the longest continuous carbon chain that contains the functional group and name it: meth, eth, prop, but, pent, hex.
Number the chain from the end that gives the functional group the lowest number. Only if there is no functional group do you number to give the substituents the lowest numbers.
Name substituents as prefixes with their numbers, using di-, tri-, tetra- for repeats, and place them in alphabetical order.
Add the suffix for the functional group with its position number.

So \( \mathrm{CH_3CH(CH_3)CH_2CH_2OH} \) is 3-methylbutan-1-ol: the longest chain containing the –OH is four carbons (butan), the –OH is on carbon 1 (numbering from that end), and there is a methyl branch on carbon 3.

Structural isomers

Structural isomers have the same molecular formula but different connectivity. Four kinds:

Chain (straight versus branched) — butane and methylpropane.
Position — the same functional group in a different place: propan-1-ol and propan-2-ol.
Functional group — the same formula, a different group entirely: ethanol and methoxymethane (\( \mathrm{C_2H_6O} \)); propanal and propanone (\( \mathrm{C_3H_6O} \)).

Alcohols, halogenoalkanes and amines are classified by how many carbon atoms are attached to the carbon bearing the functional group:

Primary (1°)
one carbon attached (or none)
e.g. propan-1-ol

Secondary (2°)
two carbons attached
e.g. propan-2-ol

Tertiary (3°)
three carbons attached
e.g. 2-methylpropan-2-ol

The classification is not bookkeeping: it decides whether an alcohol can be oxidised (R3.2) and whether a halogenoalkane substitutes by S_N1 or S_N2 (R3.4).

AHL Stereoisomers

Stereoisomers have the same constitution — same atoms, same connectivity, same bond multiplicities — but a different spatial arrangement.

Cis–trans isomerism arises because there is no free rotation about a C=C double bond (the \( \pi \) bond would have to break) or around a ring. Two conditions must both hold: restricted rotation, and two different groups on each of the two carbons. Cis has the two like groups on the same side, trans on opposite sides. It occurs in non-cyclic alkenes and in C₃ and C₄ cycloalkanes. Cis and trans isomers have different physical properties — the cis isomer is usually polar and so boils higher, while the trans isomer packs better and melts higher. E–Z nomenclature is explicitly not assessed.

Optical isomerism arises from a chiral carbon: a carbon bonded to four different groups. Such a molecule and its mirror image are non-superimposable, and the pair are called enantiomers. Draw them with wedge-and-dash bonds, as mirror images across a vertical line.

Enantiomers are identical in every ordinary physical property — same melting point, same solubility, same spectra. They differ in exactly two ways: they rotate plane-polarised light by equal amounts in opposite directions (optical activity), and they behave differently in a chiral environment. That second point is why one enantiomer of a drug can be therapeutic and the other inactive or harmful: enzyme active sites are chiral. A racemic mixture is a 50:50 mixture of the two, and it is optically inactive because the rotations cancel.

AHL Mass spectrometry of compounds

In a mass spectrometer, organic molecules fragment. The peak at the highest \( m/z \) is the molecular ion \( \mathrm{M^{+}} \), which gives the molar mass directly. The other peaks come from fragments, and the difference between the molecular ion and a fragment identifies what was lost. Common losses (in the data booklet): 15 for \( \mathrm{CH_3} \), 17 for \( \mathrm{OH} \), 29 for \( \mathrm{CHO} \) or \( \mathrm{C_2H_5} \), 45 for \( \mathrm{COOH} \).

AHL Infrared spectroscopy

Infrared radiation is absorbed by a bond if the vibration changes the dipole moment of the molecule. Each bond absorbs at a characteristic wavenumber (cm⁻¹), and the data booklet lists them, so IR answers the question which functional group is present. The signals worth recognising on sight:

O–H (alcohol) — broad, 3200–3600
O–H (carboxylic acid) — very broad, 2500–3000
N–H (amine, amide) — 3300–3500
C=O — strong and sharp, 1700–1750: the most useful peak in organic chemistry
C=C — 1620–1680

The same physics explains the greenhouse effect: \( \mathrm{CO_2} \), \( \mathrm{H_2O} \) and \( \mathrm{CH_4} \) absorb infrared because their vibrations change the dipole moment, whereas \( \mathrm{N_2} \) and \( \mathrm{O_2} \), being symmetric diatomics, cannot and so are not greenhouse gases.

AHL Proton NMR

1H NMR reports on the different chemical environments of hydrogen atoms. Three pieces of information, and you should read them in this order:

Number of signals = number of different hydrogen environments.
Chemical shift (\( \delta \), in ppm) = what each environment is next to. Values are in the data booklet: roughly 0.9–1.7 for alkyl, 2–3 next to a carbonyl or a benzene ring, 3–4 next to oxygen or a halogen, 9–10 for an aldehyde proton, 10–13 for a carboxylic acid proton.
Integration (relative areas) = the ratio of the numbers of hydrogens in each environment.
Splitting pattern: a signal is split into \( n+1 \) peaks by \( n \) hydrogens on the adjacent carbon. So a singlet means no neighbours, a doublet means one, a triplet two, a quartet three. The classic ethyl group signature is a triplet and a quartet in a 3 : 2 ratio.

Real structure determination combines the techniques: mass spectrometry gives the molar mass and hence the molecular formula, IR gives the functional group, and NMR gives the carbon skeleton and where the group sits on it. Expect a Paper 2 question that supplies all three.

✏️Worked example

(a) Name \( \mathrm{CH_3CH_2CH(CH_3)CH_2Br} \) and \( \mathrm{CH_3CH(OH)CH_2CH_3} \), and classify the second as a primary, secondary or tertiary alcohol.
(b) Draw and name the three structural isomers of \( \mathrm{C_3H_6O} \) that contain a carbonyl group or a C=C, and state the type of isomerism relating any two of them.
(c) AHL A compound X has a molecular ion at \( m/z = 60 \) and a strong fragment at \( m/z = 45 \). Its infrared spectrum shows a very broad absorption at 2500–3000 cm⁻¹ and a strong sharp peak at 1715 cm⁻¹. Its 1H NMR shows two signals, a singlet at \( \delta \) 2.1 with relative area 3 and a singlet at \( \delta \) 11.5 with relative area 1. Deduce the structure of X.

(a) For the first: the longest chain containing the functional group is four carbons, so butane. Number from the end nearest the bromine, giving 1-bromo; there is a methyl on carbon 2. The name is 1-bromo-2-methylbutane.

For the second: four carbons, so butan; the –OH is on carbon 2 whichever end you number from, so it is butan-2-ol. The carbon bearing the –OH is attached to two other carbons, so it is a secondary alcohol.

(b) \( \mathrm{C_3H_6O} \) has one degree of unsaturation.

\( \mathrm{CH_3CH_2CHO} \) — propanal, an aldehyde (carbonyl at the end of the chain).
\( \mathrm{CH_3COCH_3} \) — propanone, a ketone (carbonyl within the chain).
\( \mathrm{CH_2{=}CHCH_2OH} \) — prop-2-en-1-ol, an unsaturated alcohol.

Propanal and propanone contain different functional groups — aldehyde against ketone — so they are functional group isomers. (So is either of them compared with the alkenol.)

(c) Take the three techniques in turn.

Mass spectrum: the molecular ion at 60 gives \( M_{\mathrm{r}} = 60 \). The fragment at 45 is a loss of \( 60 - 45 = 15 \), which is CH₃. So the molecule contains a methyl group.

Infrared: the very broad 2500–3000 cm⁻¹ absorption is characteristic of the O–H of a carboxylic acid (an alcohol O–H would be higher and less broad), and 1715 cm⁻¹ is a C=O. Both together mean a carboxyl group, –COOH.

NMR: two signals means two hydrogen environments, in the ratio 3 : 1 — so four hydrogens in total. The \( \delta \) 11.5 signal is the carboxylic acid proton, and \( \delta \) 2.1 with area 3 is a CH₃ next to a carbonyl. Both are singlets, so neither group has hydrogens on an adjacent carbon — consistent with CH₃ attached directly to the carboxyl carbon, which bears no hydrogen.

Assemble: \( \mathrm{CH_3COOH} \) has \( M_{\mathrm{r}} = 2(12.01) + 4(1.01) + 2(16.00) = 60.06 \) ✓. X is ethanoic acid, \( \mathrm{CH_3COOH} \).

Check it. Always confirm that your proposed structure reproduces every piece of data, not just the one that suggested it. Ethanoic acid: molar mass 60 ✓; loses CH₃ to give \( \mathrm{COOH^{+}} \) at 45 ✓; has a broad acid O–H and a C=O ✓; has exactly two hydrogen environments in a 3 : 1 ratio ✓; and both are singlets because there is no C–H on either neighbouring atom ✓. A candidate that explains three observations out of four is not the answer. And count the hydrogens: the NMR ratio 3 : 1 gives four hydrogens only if you already know \( M_{\mathrm{r}} \) — integration gives a ratio, never an absolute number.

Reading NMR integration as a count rather than a ratio. A 3 : 1 pattern is equally consistent with 3 and 1 hydrogens, 6 and 2, or 9 and 3 — only the molecular formula, from the mass spectrum, resolves it. The matching IR trap is treating any peak near 3000 cm⁻¹ as an alcohol: the carboxylic acid O–H is much broader and reaches to lower wavenumbers, and it comes with a C=O near 1700 that an alcohol does not have. Read the peaks together, and let the data booklet settle it rather than memory.

📝Practise

Work through these on paper, then reveal the answer.

1. Name each compound and identify the functional group: (a) \( \mathrm{CH_3CH_2CH_2COOH} \), (b) \( \mathrm{CH_3CH_2COCH_3} \), (c) \( \mathrm{CH_3CHClCH_3} \), (d) \( \mathrm{CH_3CH_2CH_2CHO} \).

(a) Four carbons with a –COOH: butanoic acid, carboxyl group. The carboxyl carbon is always carbon 1, so no number is needed. (b) Four carbons with the C=O on carbon 2: butan-2-one, carbonyl group (a ketone). (c) Three carbons with Cl on carbon 2: 2-chloropropane, halogeno group; the carbon bearing the chlorine has two carbons attached, so it is a secondary halogenoalkane. (d) Four carbons with the C=O at the end: butanal, carbonyl group (an aldehyde). Note that (b) and (d) are functional group isomers of one another, both \( \mathrm{C_4H_8O} \).

2. Draw and name all the structural isomers of \( \mathrm{C_4H_9OH} \), and classify each as primary, secondary or tertiary.

There are four. Butan-1-ol, \( \mathrm{CH_3CH_2CH_2CH_2OH} \) — the carbon bearing –OH has one carbon attached, so primary. Butan-2-ol, \( \mathrm{CH_3CH(OH)CH_2CH_3} \) — two carbons attached, secondary. 2-methylpropan-1-ol, \( \mathrm{(CH_3)_2CHCH_2OH} \) — the –OH carbon has one carbon attached, primary. 2-methylpropan-2-ol, \( \mathrm{(CH_3)_3COH} \) — three carbons attached, tertiary. Butan-1-ol and butan-2-ol are position isomers; butan-1-ol and 2-methylpropan-1-ol are chain isomers. The classification matters because primary alcohols oxidise to aldehydes then acids, secondary to ketones, and tertiary are not oxidised at all under the same conditions.

3. Explain why the boiling points are: butane −0.5 °C, 2-methylpropane −12 °C, butan-1-ol 118 °C, butanoic acid 164 °C.

Butane against 2-methylpropane — identical molar mass and identical formula, so the only difference is shape. The branched isomer is more spherical, so its molecules have a smaller surface area of contact with one another and the London (dispersion) forces between them are weaker; less energy is needed to separate them, so it boils lower. Butan-1-ol is far higher because the –OH group lets the molecules hydrogen bond to one another, a much stronger intermolecular force than London forces alone. Butanoic acid is higher again because the carboxyl group has both an O–H and a C=O, so two molecules can hydrogen bond to each other twice, forming a dimer that effectively doubles the mass to be vaporised. The point to carry away is that the functional group can matter far more than molar mass: all four compounds have similar masses and the boiling points span 176 degrees.

4. AHL State the two conditions for cis–trans isomerism, explain why but-2-ene shows it but but-1-ene does not, and state one physical property in which the two isomers of but-2-ene differ.

Conditions: (i) restricted rotation, provided by a C=C double bond or a ring — rotation about a double bond would require the \( \pi \) bond to break; and (ii) two different groups attached to each of the two carbons involved. But-2-ene, \( \mathrm{CH_3CH{=}CHCH_3} \), satisfies both: each double-bonded carbon carries a CH₃ and an H, which are different, so the two methyls can be on the same side (cis) or opposite sides (trans). But-1-ene, \( \mathrm{CH_2{=}CHCH_2CH_3} \), fails the second condition: carbon 1 carries two hydrogens, which are identical, so swapping them gives the same molecule. Physical difference: cis-but-2-ene is polar (the two methyl groups are on the same side, so the small bond dipoles do not cancel) and therefore has the higher boiling point; trans-but-2-ene is symmetrical and non-polar, but packs more efficiently into a lattice and so has the higher melting point.

5. AHL Define a chiral carbon and explain how enantiomers differ. Identify the chiral carbon in 2-hydroxypropanoic acid (lactic acid), \( \mathrm{CH_3CH(OH)COOH} \).

A chiral carbon is a carbon atom bonded to four different groups. A molecule containing one is non-superimposable on its mirror image, and the two mirror-image forms are enantiomers. Enantiomers have identical ordinary physical properties — the same melting point, boiling point, density, solubility and spectra — and identical chemical behaviour in an achiral environment. They differ in exactly two respects: they rotate the plane of plane-polarised light by equal amounts in opposite directions, and they behave differently in a chiral environment, such as an enzyme active site, which is why one enantiomer of a drug may be active and the other useless or harmful. A racemic mixture, equal amounts of both, is optically inactive because the rotations cancel. In lactic acid, the chiral centre is carbon 2: it carries –CH₃, –OH, –COOH and –H, four different groups. Represent the two enantiomers with wedge-and-dash bonds, drawn as mirror images.

6. AHL A compound Y has \( M^{+} \) at \( m/z = 88 \) and a strong fragment at \( m/z = 43 \). Its IR spectrum shows a strong peak at 1740 cm⁻¹ and no absorption above 3100 cm⁻¹. Its 1H NMR has three signals: a triplet at \( \delta \) 1.3 (area 3), a singlet at \( \delta \) 2.0 (area 3), and a quartet at \( \delta \) 4.1 (area 2). Deduce the structure.

IR first. A strong peak at 1740 cm⁻¹ is a C=O. The absence of any broad absorption above 3100 rules out both an alcohol O–H and a carboxylic acid O–H. A carbonyl with no O–H, at that rather high wavenumber, indicates an ester. NMR next. The triplet (3H) with quartet (2H) pair is the classic ethyl group signature: the CH₃ is split into a triplet by the two hydrogens on the adjacent CH₂ (\( n+1 = 3 \)), and the CH₂ into a quartet by the three hydrogens on the CH₃ (\( n+1 = 4 \)). The quartet sits at \( \delta \) 4.1, well downfield, so that CH₂ is attached to an oxygen — an –OCH₂CH₃ group. The singlet at \( \delta \) 2.0 with area 3 is a CH₃ whose neighbour bears no hydrogens, at a shift typical of a methyl next to a carbonyl: a CH₃CO– group. Assemble: \( \mathrm{CH_3COOCH_2CH_3} \), ethyl ethanoate. Check every datum. \( M_{\mathrm{r}} = 4(12.01) + 8(1.01) + 2(16.00) = 88.1 \) ✓, matching \( M^{+} = 88 \). The fragment at 43 is a loss of 45, which is \( \mathrm{OC_2H_5} \), leaving \( \mathrm{CH_3CO^{+}} \) ✓. Total hydrogens from the integration ratio 3 : 3 : 2 is eight ✓. Every observation is accounted for, which is what a complete deduction looks like — a structure that explains three observations out of four is not the answer.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

The SDBS spectral database (AIST, Japan) — free real mass, IR and NMR spectra for thousands of compounds. Look up a molecule you know and check you can read its spectra before you try to deduce an unknown.
MolView — build a molecule in 3D, rotate it, and see immediately why a chiral centre gives non-superimposable mirror images.
Your data booklet — the IR characteristic frequencies table, the 1H NMR chemical shift table and the mass-spectrometry fragment list are all in it. None of them should be memorised, and all of them should be findable in seconds.