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

Need the full teaching first? The study guide explains every idea here with worked examples. These notes compress Structure 3.2.7 to 3.2.12 of IB Diploma Programme Chemistry, following the IB Chemistry guide (first assessment 2025, published February 2023) for exams from 2025 onwards: stereoisomers, mass spectrometry, infrared spectroscopy, ¹H NMR and combined structure work. All of it is additional higher level material, so these notes are HL only.

Test yourself afterwards with the practice questions. The SL and HL content that comes just before this unit has its own functional groups and naming revision notes. Other units live on the IB DP Chemistry course page, with a printable checklist to track them. Unsure where to start? A free diagnostic will show you.

Key definitions

Term	Meaning
Stereoisomers	Same constitution (atoms, connectivity, bond multiplicities); different spatial arrangement
Cis-trans isomers	Stereoisomers caused by restricted rotation about C=C or within a ring
Chiral carbon	A carbon bonded to four different atoms or groups
Enantiomers	A pair of non-superimposable mirror images
Optical activity	Rotation of the plane of plane-polarized light
Racemic mixture	Equal amounts of two enantiomers; no net rotation
Molecular ion, M⁺	The ion formed when the whole molecule loses one electron
Wavenumber	1/wavelength, in cm⁻¹
Chemical shift, δ	Position of an NMR signal in ppm, relative to TMS at 0
Integration trace	Step heights proportional to the number of H in each environment

Stereoisomers (3.2.7)

Cis-trans: method in steps

1. Find the C=C, or the two ring carbons that carry substituents (C₃ or C₄ rings only).
2. Check each of those carbons. Does it carry two different groups?
3. If both do, the compound has cis-trans isomers. If either carbon has two identical groups, it does not.

Why it happens: the π bond (or the ring) stops rotation, so groups fixed on the same side cannot swap to opposite sides without breaking a bond. E-Z naming is not assessed.

Chirality: method in steps

1. Rule out CH₂ and CH₃ carbons, plus any carbon that is double-bonded.
2. List the four groups on each carbon left, written in full (CH₂CH₃ and CH₂CH₂CH₃ are different).
3. Four different groups means a chiral carbon, and the molecule exists as two enantiomers.

Drawing. Plain lines for two bonds in the page, a solid tapered wedge for the bond towards you, a hashed wedge for the bond away. Then draw the reflection; exchanging the positions of two groups has the same effect.

Must-know distinctions

• Enantiomers: same physical properties, except they rotate plane-polarized light by equal angles in opposite directions.
• Same chemical properties, except in a chiral environment (enzymes, receptors, chiral reagents).
• Racemic mixture: chiral molecules, but the two rotations cancel.

Mass spectrometry (3.2.8)

• M⁺: highest-m/z significant peak; gives the relative molecular mass.
• Fragment peaks: positive ions only. Neutral pieces are not detected.
• Mass lost = M − fragment m/z.

Fragment ion	m/z	Piece lost from M⁺	Mass
CH₃⁺	15	CH₃	15
C₂H₅⁺ or CHO⁺	29	OH	17
CH₂OH⁺ or CH₃O⁺	31	C₂H₅ or CHO	29
CH₃CO⁺ or C₃H₇⁺	43	OCH₃ or CH₂OH	31
C₄H₉⁺	57	COOH	45
C₆H₅⁺	77

The guide notes that the data booklet includes values for particular MS fragments. In a question, use the values you are given.

Worked reminder. Butan-1-ol and butan-2-ol are both C₄H₁₀O (M = 74). A strong peak at 31 (CH₂OH⁺, from loss of C₃H₇) points to an OH on the end carbon, so butan-1-ol. A strong peak at 45 (CH₃CHOH⁺, from loss of C₂H₅) points to an OH on carbon 2, so butan-2-ol. Fragments narrow the choice; NMR usually confirms it.

Infrared spectroscopy (3.2.9)

Bond	Wavenumber / cm⁻¹
C=O	1700–1750
O–H (alcohol), broad	3200–3600
O–H (carboxylic acid), very broad	2500–3000
C=C	1620–1680
C≡C	2100–2260
N–H	3300–3500
C–H	2850–3100
C–O	1050–1410

Interpret the functional group region (above about 1500 cm⁻¹). The guide says IR data are given in the data booklet.

Method in steps

1. Look for C=O near 1700–1750.
2. Look for O–H: broad at 3200–3600 (alcohol) or very broad at 2500–3000, overlapping C–H (acid).
3. Check for C=C, C≡C or N–H.
4. Name the bond, then the functional group it suggests.

IR active or not? A vibration absorbs only if it changes the dipole moment.

Molecule or vibration	IR active?
N₂, O₂ stretch	No (no dipole change)
CO₂ symmetric stretch	No
CO₂ asymmetric stretch, CO₂ bend	Yes
H₂O, CH₄	Yes

Greenhouse link. The surface emits IR; IR-active gases absorb it and re-emit it in all directions, some back down, warming the lower atmosphere. Global warming potential (relative to CO₂) is larger for a gas that absorbs IR strongly and stays in the atmosphere longer.

¹H NMR (3.2.10 and 3.2.11)

Reading a spectrum in steps

1. Count the signals: number of hydrogen environments.
2. Read the integration trace: divide by the smallest step to get a whole-number ratio.
3. Read the chemical shifts: which part of the molecule each signal belongs to.
4. Read the splitting: n neighbouring H on adjacent carbons gives n + 1 peaks.

Approximate δ / ppm	Hydrogen type
0.9–1.7	alkyl CH₃, CH₂
2.0–2.7	on carbon next to C=O
3.3–4.3	on carbon bonded to O
1–6	alcohol O–H (variable)
6.5–8.0	benzene ring
9.4–10	–CHO
10–13	–COOH

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

n	Pattern
0	singlet
1	doublet
2	triplet
3	quartet

Recognisable groups

• Ethyl, –CH₂CH₃: quartet (2H) plus triplet (3H).
• CH₃–CH: doublet (3H) plus a quartet if that CH has no other neighbours.
• tert-butyl, (CH₃)₃C–: a 9H singlet.
• CH₃ on C=O or on O: singlet.

Worked reminder. 2,2-dimethylpropanal, (CH₃)₃CCHO: two signals, both singlets because carbon 2 has no H. The ratio is 9 : 1, and the CHO signal sits far downfield at about 9.4–10 ppm.

Combining techniques (3.2.12)

1. From percentage composition, find the empirical formula. From M⁺, find the molecular formula.
2. IR: list the functional groups present and absent.
3. NMR: environments, ratios, shifts and splitting build the skeleton.
4. MS fragments: confirm the pieces.
5. Check your structure against every piece of data, then name it.

Quick self-test

1. Why does but-1-ene not show cis-trans isomerism?
2. Does 1,2-dimethylcyclobutane show cis-trans isomerism?
3. Identify the chiral carbon in 3-chlorohexane, CH₃CH₂CHClCH₂CH₂CH₃.
4. Why does a racemic mixture not rotate plane-polarized light?
5. A molecular ion is at m/z 122 and a peak appears at m/z 77. What does the 77 peak suggest?
6. A peak lies 45 below M⁺. Which group was probably lost?
7. An IR spectrum has a strong band at 1715 cm⁻¹ but nothing at 3200–3600 or 2500–3000 cm⁻¹. Which classes of compound fit?
8. Why is N₂ not a greenhouse gas?
9. Predict the ¹H NMR spectrum of 1,2-diethoxyethane, CH₃CH₂OCH₂CH₂OCH₂CH₃.
10. Predict the splitting and area ratio in 1,1-dimethoxyethane, CH₃CH(OCH₃)₂.
11. Integration steps are 22.5 mm, 15 mm and 7.5 mm. Give the ratio.
12. Predict the ¹H NMR spectrum of 1-chloro-3,3-dimethylbutane, ClCH₂CH₂C(CH₃)₃.

Answers

1. Carbon 1 carries two H atoms, so only one arrangement exists.
2. Yes. Each substituted ring carbon has H and CH₃; the two CH₃ groups can share a face (cis) or sit on opposite faces (trans).
3. Carbon 3: it carries H, Cl, C₂H₅ and C₃H₇.
4. It holds equal amounts of both enantiomers, which rotate the plane by equal angles in opposite directions, so the effects cancel.
5. A C₆H₅⁺ ion, so a benzene ring is present.
6. COOH, suggesting a carboxylic acid.
7. C=O is present with no O–H, so an aldehyde, ketone or ester, not an alcohol or acid.
8. Its only vibration does not change the dipole moment (it has none), so it does not absorb IR.
9. Three signals: a singlet (4H, the two equivalent central CH₂ groups, which do not split each other), a quartet (4H) and a triplet (6H, about 0.9–1.7 ppm). Both CH₂ signals lie at about 3.3–4.3 ppm, next to O. Ratio 2 : 2 : 3.
10. Three signals: CH₃ doublet (one neighbour), CH quartet (three neighbours) and a singlet for the two equivalent OCH₃ groups. Ratio 3 : 1 : 6.
11. 3 : 2 : 1.
12. Three signals: a 9H singlet (carbon 3 has no H), a 2H triplet for CH₂Cl (furthest downfield) and a 2H triplet for the middle CH₂. Ratio 2 : 2 : 9.

Where marks are usually lost

• Saying “four different atoms” when the groups differ further along the chain; compare whole groups.
• Calling a carbon chiral when it is part of C=C, or carries two H atoms.
• Wedge-dash drawings with two wedges, no dash, or a “mirror image” that is just the same molecule turned round.
• Explaining optical activity as “rotates light” rather than “rotates the plane of plane-polarized light”.
• Quoting a fragment’s m/z without a formula and a positive charge (write CH₃CO⁺, not CH₃CO).
• Naming only the bond from IR when the question asks for the functional group, or the reverse.
• Mixing up the alcohol O–H range (3200–3600) with the acid O–H range (2500–3000).
• Splitting by neighbours on the same carbon, or by the O–H hydrogen.
• Leaving an integration ratio as decimals instead of the smallest whole numbers.
• Proposing a final structure that ignores one of the given pieces of data.

Official syllabus

Aligned to the International Baccalaureate Diploma Programme Chemistry guide (first assessment 2025; published by the IB, February 2023). Sections condensed here: Structure 3.2.7 to 3.2.12, from the guide’s topic on classifying organic compounds by functional group. All are additional higher level (HL only).