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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise 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. All of it is additional higher level material: HL only. Each question supplies the data you need.

New to the content? Begin with the study guide, or refresh with the revision notes. Browse other units via the course page for IB DP Chemistry, track them with the printable checklist, and let a free diagnostic suggest where to begin.

Questions

1. State, with a reason, whether each compound shows cis-trans isomerism.

(a) 1,1-dichloroprop-1-ene, CCl₂=CHCH₃ [1] (b) 4,4-dimethylpent-2-ene, CH₃CH=CHC(CH₃)₃ [1] (c) 1,2-dibromocyclobutane [1]

2. 1-bromo-1-chloroethane, CH₃CHBrCl, exists as a pair of enantiomers.

(a) Identify the chiral carbon and explain why it is chiral. [1] (b) Draw stereochemical formulas of the two enantiomers using wedge-dash bonds. [2] (c) Describe how the two enantiomers differ in their effect on plane-polarized light. [2] (d) A sample of 1-bromo-1-chloroethane has no effect on plane-polarized light. Explain this. [2]

3. Methyl ethanoate, CH₃COOCH₃, and ethyl methanoate, HCOOCH₂CH₃, are isomers. Fragment data: CH₃⁺ 15, C₂H₅⁺ 29, CH₃O⁺ 31, CH₃CO⁺ 43, HCOO⁺ 45. Explain how their mass spectra would allow you to tell them apart. [4]

4. The mass spectrum of 1-phenylethanone, C₆H₅COCH₃, has peaks at m/z 120, 105, 77 and 43. Identify the ions at 105, 77 and 43. [3]

5. Use these IR data: O–H (alcohol) 3200–3600 cm⁻¹; O–H (carboxylic acid) 2500–3000 cm⁻¹; C=O 1700–1750 cm⁻¹; C=C 1620–1680 cm⁻¹.

(a) Compound A absorbs very broadly across 2500–3000 cm⁻¹ and strongly at 1720 cm⁻¹. Identify both bonds and the functional group. [2] (b) Compound B absorbs broadly at 3200–3600 cm⁻¹ and at 1650 cm⁻¹, with nothing at 1700–1750 cm⁻¹. Deduce two features of its structure. [2]

6. These parts are about greenhouse gases.

(a) State which vibrations of a CO₂ molecule absorb infrared radiation, and explain why the symmetric stretch does not. [2] (b) Explain why CH₄ is a greenhouse gas but N₂ is not. [2]

7. Predict the number of ¹H NMR signals, their area ratio and each signal’s splitting.

(a) 2,2-dichlorobutane, CH₃CCl₂CH₂CH₃ [3] (b) diethyl propanedioate, CH₂(COOCH₂CH₃)₂ [3]

8. Ester X, C₅H₁₀O₂, gives four ¹H NMR signals: 1.1 ppm (triplet), 1.3 ppm (triplet), 2.3 ppm (quartet) and 4.1 ppm (quartet), with integration steps of 9.6, 9.6, 6.4 and 6.4 mm. Shifts: H on carbon next to C=O 2.0–2.7 ppm; H on carbon bonded to O 3.3–4.3 ppm.

(a) Deduce the ratio of hydrogen atoms in each environment. [1] (b) Identify X. Explain how the signals support your answer and why X cannot be propyl ethanoate, CH₃COOCH₂CH₂CH₃. [4]

9. Compound W contains 46.1% carbon, 7.8% hydrogen and 46.1% oxygen by mass. Relative atomic masses: C 12.01, H 1.01, O 16.00.

• MS: M⁺ at m/z 104; other peaks at 73 and 45.
• IR: very broad absorption across 2500–3000 cm⁻¹ and strong absorption at 1710 cm⁻¹.
• ¹H NMR: four signals: 2.6 ppm (triplet), 3.4 ppm (singlet), 3.7 ppm (triplet) and 11.5 ppm (singlet), with integration steps of 6.0, 9.0, 6.0 and 3.0 mm.

(a) Determine the empirical formula of W. [2] (b) Deduce the molecular formula of W. [1] (c) Identify the bonds behind the two IR absorptions and the functional group they show. [2] (d) Deduce the number of hydrogen atoms in each environment and explain the splitting. [3] (e) Deduce the structure of W and identify the ions at m/z 73 and 45. [2]

10. An optically active liquid, V, has the molecular formula C₆H₁₂O.

(a) IR: strong absorption at 1730 cm⁻¹; no broad O–H band. State what this shows. [2] (b) MS: M⁺ at m/z 100; other peaks include 71 and 29. Suggest an ion for each peak. [2] (c) ¹H NMR includes a doublet of area 1 at 9.6 ppm. State what the shift and the splitting show. [2] (d) V is one of: 2-ethylbutanal, 2,3-dimethylbutanal, 3-methylpentanal and 2,2-dimethylbutanal. Identify V, using (c) and its optical activity. [2] (e) A sample of V made in the laboratory from non-chiral starting materials has no optical activity. Suggest why. [1]

Answers

1. (a) No: carbon 1 carries two Cl atoms [1]. (b) Yes: carbon 2 has H and CH₃; carbon 3 has H and C(CH₃)₃ [1]. (c) Yes: carbons 1 and 2 each carry H and Br, and the ring stops rotation [1]. Examiner insight: Name the groups on each carbon; a bare “yes” or “no” gives no reason.

2. (a) Carbon 1: it is bonded to four different groups, H, Br, Cl and CH₃ [1]. (b) The four groups arranged tetrahedrally round carbon 1, using plain lines in the page plus one tapered wedge and one hashed bond [1]; the reflection of that drawing beside it [1]. (c) Each rotates the plane of plane-polarized light [1] by the same angle but in opposite directions [1]. (d) It is a racemic mixture, with equal amounts of the two enantiomers [1]; their equal and opposite rotations cancel, so there is no net rotation [1]. Examiner insight: In (b), make the second drawing a true reflection; a turned copy of the first is the same enantiomer.

3. Both show M⁺ at m/z 74 [1]. Methyl ethanoate gives 43, CH₃CO⁺, by loss of OCH₃ (74 − 31) [1]. Ethyl methanoate gives 45, HCOO⁺, by loss of C₂H₅ (74 − 29) [1], and 29 from C₂H₅⁺ [1]. Examiner insight: Pair each m/z value with a positively charged formula; numbers alone explain nothing.

4. 105: C₆H₅CO⁺, from loss of CH₃ (120 − 15) [1]. 77: C₆H₅⁺ [1]. 43: CH₃CO⁺ [1]. Examiner insight: Write each fragment as a cation; C₆H₅CO without the + sign is incomplete.

5. (a) 2500–3000 cm⁻¹: O–H of a carboxylic acid; 1720 cm⁻¹: C=O [1]. Together they show a carboxyl group, so A is a carboxylic acid [1]. (b) Broad 3200–3600 cm⁻¹: an alcohol O–H group [1]. 1650 cm⁻¹ with nothing at 1700–1750 cm⁻¹: a C=C bond and no C=O, so B is an unsaturated alcohol [1]. Examiner insight: A bond (O–H) is not a functional group (carboxyl); give both when asked.

6. (a) The asymmetric stretch and the bends absorb [1]. In the symmetric stretch, the two C=O bond dipoles change equally and cancel, so the dipole moment does not change [1]. (b) CH₄ vibrations change its dipole moment, so it absorbs infrared emitted by Earth’s surface and re-emits some back down [1]. N₂’s stretch never creates a dipole, so N₂ cannot absorb infrared [1]. Examiner insight: Link absorption to a change in dipole moment; CO₂ itself is non-polar.

7. (a) Three signals [1]; ratio 3 : 2 : 3 (CH₃ on carbon 2, CH₂, end CH₃) [1]; singlet (carbon 2 has no H), quartet, triplet respectively [1]. (b) Three signals [1]; ratio 1 : 2 : 3 (central CH₂, the two equivalent OCH₂, the two equivalent CH₃) [1]; singlet (both neighbouring carbons are C=O with no H), quartet, triplet respectively [1]. Examiner insight: A carbon with no hydrogen blocks splitting; counting hydrogens across it is a common slip.

8. (a) Dividing by 3.2: 9.6 : 9.6 : 6.4 : 6.4 = 3 : 3 : 2 : 2 [1]. (b) X is ethyl propanoate, CH₃CH₂COOCH₂CH₃ [1]. Two quartet (2H) plus triplet (3H) pairs show two ethyl groups [1]. The 2.3 ppm CH₂ is next to C=O and the 4.1 ppm CH₂ is bonded to O, so an ethyl sits on each side of the ester link [1]. Propyl ethanoate would show a 3H singlet near 2.0 ppm, and its middle CH₂ (five neighbouring H) would be neither quartet nor triplet [1]. Examiner insight: Say why the rejected isomer fails, not only why your choice fits.

9. (a) Moles in 100 g: C 46.1/12.01 = 3.84; H 7.8/1.01 = 7.72; O 46.1/16.00 = 2.88 [1]. Dividing by 2.88 gives 1.33 : 2.68 : 1; multiplying by 3 gives 4 : 8 : 3, so C₄H₈O₃ [1]. (b) M(C₄H₈O₃) = 104, equal to M⁺, so the molecular formula is also C₄H₈O₃ [1]. (c) 2500–3000 cm⁻¹: O–H; 1710 cm⁻¹: C=O [1]. Together they show a carboxyl group, so W is a carboxylic acid [1]. (d) Dividing by 3.0 gives 2 : 3 : 2 : 1, matching 8 H [1]. The 3.4 ppm singlet (3H) is CH₃ bonded to O with no neighbouring H; the 11.5 ppm singlet (1H) is –COOH [1]. The two 2H triplets are adjacent CH₂ groups splitting each other; the 3.7 ppm one is bonded to O, the 2.6 ppm one is next to C=O [1]. (e) W is 3-methoxypropanoic acid, CH₃OCH₂CH₂COOH [1]. 73 is CH₂CH₂COOH⁺, from loss of OCH₃ (104 − 31); 45 is CH₃OCH₂⁺ (or COOH⁺) [1]. Examiner insight: In (a), scale 1.33 : 2.68 : 1 to whole numbers; rounding 1.33 down to 1 gives a wrong formula.

10. (a) A C=O group is present [1]; there is no O–H, so V is not an alcohol or a carboxylic acid [1]. (b) 71: C₅H₁₁⁺, from loss of CHO (100 − 29) [1]. 29: CHO⁺ (or C₂H₅⁺) [1]. (c) 9.6 ppm: an aldehyde –CHO hydrogen [1]; doublet: the next carbon carries exactly one H [1]. (d) The doublet rules out 3-methylpentanal (triplet) and 2,2-dimethylbutanal (singlet) [1]. Of the two left, 2-ethylbutanal is not chiral (its carbon 2 carries two ethyl groups), so V is 2,3-dimethylbutanal, (CH₃)₂CHCH(CH₃)CHO, whose carbon 2 carries H, CH₃, CH(CH₃)₂ and CHO [1]. (e) The synthesis produces equal amounts of both enantiomers, a racemic mixture, so the rotations cancel [1]. Examiner insight: In (d), name the four groups on the chiral carbon rather than just saying “it is chiral”.

Where marks are usually lost

• Writing “rotates light” instead of “rotates the plane of plane-polarized light”.
• Explaining a racemic mixture as containing non-chiral molecules.
• Fragment ions written without a positive charge, or masses lost confused with fragment masses.
• Confusing the alcohol O–H band with the much broader acid O–H band.
• Counting hydrogens on the same carbon, or across a carbon with no H, when applying the n + 1 rule.
• Treating 1.33 in a mole ratio as 1 instead of multiplying through by 3.

Next steps

• Revisit the study guide.
• Try all free 10-minute diagnostics.
• Earlier Structure 3.2 work: functional groups and naming practice.
• Keep the revision notes.
• Book a free trial class.
• Other units: IB DP Chemistry course page and printable checklist.

Official syllabus

Questions written against the International Baccalaureate Diploma Programme Chemistry guide, first assessment 2025 (published by the IB in February 2023), Structure 3.2.7 to 3.2.12; all additional higher level, HL only.