IB DP Chemistry Structure 3.2: Isomers and spectroscopic identification of organic compounds

This study guide teaches Structure 3.2.7 to 3.2.12 of IB Diploma Programme Chemistry, following the IB Chemistry guide (published February 2023, first assessment 2025) for exams from 2025 onwards. These understandings are the additional higher level (AHL) half of the guide’s topic “Functional groups: Classification of organic compounds” (Structure 3.2), so this page is HL only.

Shaky on formulas, functional groups or naming? Start with the functional groups and naming study guide. Afterwards, the revision notes condense this unit into tables and steps, and the practice questions give you exam-style problems. Our other units sit on the IB DP Chemistry course page; you can tick them off on the printable checklist, and a free diagnostic spots weak areas.

What this unit covers

Guide reference	Skills you need	Level
Structure 3.2.7	Cis-trans isomers (alkenes, C₃ and C₄ rings); chiral carbons drawn with wedges; enantiomers, optical activity, racemic mixtures	HL only
Structure 3.2.8	Use the molecular ion and fragment peaks in a mass spectrum to deduce structural features	HL only
Structure 3.2.9	Read the functional group region of IR spectra; link IR absorption to greenhouse gases	HL only
Structure 3.2.10	Use the number of ¹H NMR signals, their chemical shifts and integration traces to deduce a structure	HL only
Structure 3.2.11	Use singlets, doublets, triplets and quartets for finer structural detail	HL only
Structure 3.2.12	Combine data from several techniques to identify a molecule	HL only

Stereoisomers (3.2.7)

Stereoisomers share the same constitution (atoms, connectivity and bond multiplicities) but differ in the arrangement of atoms in space. You need two kinds.

Cis-trans isomerism

A compound shows cis-trans isomerism when both of these are true:

• Rotation is restricted. Twisting a C=C bond would destroy the sideways p-orbital overlap of the π bond. In a three- or four-membered ring, the ring itself locks the carbons.
• Each of the two key carbons carries two different groups. If either carbon has two identical groups, only one form exists.

Cis: matching groups on the same side of the double bond (or face of the ring). Trans: opposite sides. Rings are limited to C₃ and C₄, and E-Z naming is not assessed.

Worked example 1: which compounds show cis-trans isomerism?

Compound	What each key carbon carries	Cis-trans?
2-bromobut-2-ene, CH₃CBr=CHCH₃	C2: CH₃ and Br; C3: H and CH₃	Yes
2-methylpent-2-ene, (CH₃)₂C=CHCH₂CH₃	C2: two CH₃ groups	No
1-ethyl-2-methylcyclopropane	C1: H and C₂H₅; C2: H and CH₃	Yes: the ethyl and methyl groups share a face (cis) or not (trans)
1,1-dichlorocyclobutane	C1: two Cl atoms; the other ring carbons: two H atoms	No

Chiral carbons and enantiomers

A chiral carbon is bonded to four different atoms or groups. Because its bonds point to the corners of a tetrahedron, the molecule and its mirror image cannot be superimposed. The two forms are enantiomers. Build both with a model kit or modelling software: no rotation turns one into the other.

Drawing them. Use tapered wedge-dash bonds: two plain lines in the plane of the paper, a solid wedge pointing towards you and a hashed wedge pointing away. Then draw the mirror image beside it.

Worked example 2: 3-bromo-3-methylhexane, CH₃CH₂CBr(CH₃)CH₂CH₂CH₃

1. Carbon 3 carries Br, CH₃, C₂H₅ and C₃H₇. All four differ, so carbon 3 is chiral. A chiral carbon does not need a hydrogen.
2. No other carbon qualifies: each carries at least two H atoms.
3. First enantiomer: C₂H₅ as a plain line up-left, C₃H₇ as a plain line up-right, Br on a solid wedge, CH₃ on a hashed wedge.
4. Mirror image: imagine a vertical mirror. C₂H₅ moves to the right and C₃H₇ to the left, while Br stays on the solid wedge and CH₃ on the hashed one.

Shortcut: exchanging the positions of two groups also produces the partner enantiomer.

Optical activity. Each enantiomer rotates the plane of plane-polarized light by the same angle, one clockwise and one anticlockwise: both are optically active. In a racemic mixture the two enantiomers are present in a 1 : 1 ratio, so the rotations cancel, even though every molecule is chiral.

Chemical properties. Enantiomers share boiling points and densities and react alike with non-chiral reagents. They behave differently only in a chiral environment, such as an enzyme or a receptor. The two enantiomers of carvone smell different, one of spearmint and one of caraway, because the smell receptors in your nose are chiral.

Mass spectrometry of organic compounds (3.2.8)

In the mass spectrometer, a molecule loses an electron to form the molecular ion, M⁺. The peak with the highest m/z (ignoring small isotope peaks) usually gives the relative molecular mass.

The molecular ion can fragment into a positive ion and a neutral species. Only the positive ion is detected, so the gap between M⁺ and a fragment peak tells you which neutral piece broke off.

Ion	m/z
CH₃CO⁺ or C₃H₇⁺	43
C₆H₅⁺	77
CH₃⁺	15
CHO⁺ or C₂H₅⁺	29
CH₃O⁺ or CH₂OH⁺	31

Neutral piece lost	Mass below M⁺
CH₃	15
OH	17
C₂H₅ or CHO	29
OCH₃ or CH₂OH	31
COOH	45

The guide notes that the data booklet includes values for particular MS fragments; in a question, use the values given.

Worked example 3: methyl 2-methylpropanoate, (CH₃)₂CHCOOCH₃

Account for peaks at m/z 102, 87, 71, 59 and 43.

• 102: M⁺. C₅H₁₀O₂ has M = 5(12) + 10(1) + 2(16) = 102.
• 87: 102 − 15, loss of CH₃.
• 71: 102 − 31, loss of OCH₃, leaving (CH₃)₂CHCO⁺.
• 59: COOCH₃⁺ (12 + 16 + 16 + 15).
• 43: (CH₃)₂CH⁺, a C₃H₇⁺ ion.

Together, the peaks at 71 and 59 point to a methyl ester.

Infrared spectroscopy (3.2.9)

Bonds vibrate at particular frequencies and absorb infrared radiation of matching frequency. Spectra plot transmittance against wavenumber (1/wavelength, in cm⁻¹).

Bond	Wavenumber / cm⁻¹	What it looks like
C=O	1700–1750	strong, sharp
C=C	1620–1680
C≡C	2100–2260
O–H, alcohol	3200–3600	broad
N–H	3300–3500
O–H, carboxylic acid	2500–3000	very broad, overlapping C–H
C–H	2850–3100	in almost all organic molecules
C–O	1050–1410

The guide says IR data are given in the data booklet; use the table the question supplies. Interpret the functional group region, roughly above 1500 cm⁻¹.

Worked example 4. An unknown absorbs very broadly across 2500–3000 cm⁻¹, strongly at 1705 cm⁻¹ and at 1650 cm⁻¹. Interpret the spectrum.

• 2500–3000 cm⁻¹, very broad: O–H of a carboxylic acid.
• 1705 cm⁻¹: C=O, confirming a carboxyl group.
• 1650 cm⁻¹: C=C.

It is an unsaturated carboxylic acid, such as but-2-enoic acid, CH₃CH=CHCOOH, which also has cis and trans forms that the functional group region cannot tell apart.

IR activity and greenhouse gases

A vibration absorbs IR only if it changes the dipole moment. The symmetric stretch of CO₂ keeps the molecule non-polar, so it is IR inactive; the asymmetric stretch and the bend change the dipole and absorb. N₂ and O₂ never gain a dipole as they vibrate, so they do not absorb IR.

Earth’s surface emits infrared radiation. CO₂, H₂O and CH₄ absorb some of it and re-emit it in all directions, partly back towards the surface: they are greenhouse gases. A gas’s global warming potential compares its warming effect with that of the same mass of CO₂ over a set period. It is higher when the gas absorbs IR strongly, especially at wavelengths other gases leave unabsorbed, and when it stays in the atmosphere for a long time.

¹H NMR: signals, shifts and integration (3.2.10)

In a strong magnetic field, hydrogen nuclei absorb radio waves, at slightly different frequencies in different chemical environments.

• Number of signals = number of different hydrogen environments. Equivalent hydrogens, such as the three in a CH₃ group or two CH₃ groups on the same carbon, give one signal.
• Chemical shift, δ in ppm, is measured relative to tetramethylsilane (TMS) at 0 ppm. Hydrogens near electronegative atoms or a C=O group are deshielded and appear further from 0.
• Integration trace: the step heights over the signals are in the ratio of hydrogen atoms in each environment.

Type of hydrogen	Approximate δ / ppm
CH₃ or CH₂ in an alkyl chain	0.9–1.7
H on carbon next to C=O	2.0–2.7
H on carbon bonded to O (alcohol, ether, ester)	3.3–4.3
O–H in an alcohol	1–6, variable
H on a benzene ring	6.5–8.0
–CHO	9.4–10
–COOH	10–13

The guide says ¹H NMR data are given in the data booklet; these ranges are approximate.

Worked example 5: 3-methylpentan-3-ol, CH₃C(OH)(CH₂CH₃)₂

Integration steps of 9, 12, 18 and 3 mm divided by 3 give 3 : 4 : 6 : 1: the CH₃ on carbon 3 (3 H), the two CH₂ groups (4 H), the two end CH₃ groups (6 H) and the OH (1 H).

Splitting patterns (3.2.11)

A signal is split by hydrogen atoms on adjacent carbons. With n such neighbours, it splits into n + 1 peaks.

Neighbouring H (n)	Pattern	Peak ratio
0	singlet	1
1	doublet	1 : 1
2	triplet	1 : 2 : 1
3	quartet	1 : 3 : 3 : 1

Equivalent hydrogens do not split each other, and an alcohol’s O–H signal usually appears as a singlet.

Worked example 6. In 3-methylpentan-3-ol, carbon 3 has no hydrogen, so its CH₃ is a singlet. Each CH₂ has three neighbours: a quartet. Each end CH₃ has two: a triplet. The OH is a singlet. A 2H quartet with a 3H triplet signals an ethyl group.

Combining techniques (3.2.12)

MS gives the molar mass and fragments, IR the functional groups, ¹H NMR the carbon-hydrogen framework.

Worked example 7. Compound Z contains 54.5% C, 9.2% H and 36.3% O by mass. Its mass spectrum shows M⁺ at 88 with peaks at 73, 45 and 43. IR: broad 3200–3600 cm⁻¹ and strong 1715 cm⁻¹. ¹H NMR: 1.4 ppm (doublet, 3H), 2.2 ppm (singlet, 3H), 3.6 ppm (singlet, 1H), 4.2 ppm (quartet, 1H). Identify Z.

1. Moles in 100 g: C 54.5/12.01 = 4.54; H 9.2/1.01 = 9.11; O 36.3/16.00 = 2.27. The ratio is 2 : 4 : 1, so the empirical formula is C₂H₄O (M = 44).
2. 88/44 = 2, so the molecular formula is C₄H₈O₂.
3. IR: alcohol O–H and C=O. With no 2500–3000 band and no signal at 10–13 ppm, Z is not an acid.
4. NMR: the 2.2 ppm singlet (3 H) is CH₃ on C=O with no neighbours, so CH₃CO–. The 4.2 ppm quartet (1 H) is a CH on a carbon bonded to O, next to a CH₃. The 1.4 ppm doublet (3 H) is that CH₃. The 3.6 ppm singlet is the OH.
5. MS: 43 is CH₃CO⁺; 45 is CH₃CHOH⁺; 73 is 88 − 15.

Z is 3-hydroxybutan-2-one, CH₃COCH(OH)CH₃. Carbon 3 carries four different groups, so Z is also chiral.

Common errors

• Calling an alkene cis-trans when one carbon of the C=C has two identical groups.
• Drawing two wedges, or two dashes, on a chiral carbon instead of one of each.
• Saying a racemic mixture is inactive because its molecules are not chiral; the rotations cancel instead.
• Counting hydrogens on the same carbon, or the O–H hydrogen, as splitting neighbours.
• Treating a fragment’s m/z as the mass lost; subtract it from M⁺.

Official syllabus

Aligned to the International Baccalaureate Diploma Programme Chemistry guide, first assessment 2025, published by the IB in February 2023. Sections taught here: Structure 3.2.7 to 3.2.12, all additional higher level, so HL only.