IB DP Chemistry Structure 3.2: Functional groups and naming 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.

This set practises the first six understandings of Structure 3.2, Functional groups: Classification of organic compounds, based on the IB Diploma Programme Chemistry guide (published February 2023, examined from 2025). Questions 1 to 4 focus on Structure 3.2.1 to 3.2.3, question 5 covers 3.2.4, questions 6 to 9 cover 3.2.5 and 3.2.6, and the two longer questions at the end combine them. All of this content is shared by SL and HL.

If the content is new to you, read the study guide first; for a quick refresh, the revision notes are enough. The IB DP chemistry hub and the course checklist to print list the rest of the course.

Questions

1. The ethyl ester of the amino acid phenylalanine has the condensed formula C₆H₅CH₂CH(NH₂)COOCH₂CH₃. Identify the three functional groups in this molecule. [3]

2. Hexanedioic acid, HOOC(CH₂)₄COOH, is used to make nylon.

(a) State its molecular formula. [1] (b) Deduce its empirical formula. [1] (c) State how many hydrogen atoms are not drawn in its skeletal formula. [1] (d) State, with a reason, whether hexanedioic acid is saturated or unsaturated. [1]

3. State the homologous series to which each compound belongs. [5]

(a) C₇H₁₆ (b) CH₃CH₂CH₂CONH₂ (c) CH₃(CH₂)₄Br (d) CH₃CH₂COOCH₃ (e) CH₃CH₂CH₂NH₂

4. A chloroalkane, CₙH₂ₙ₊₁Cl, has a molar mass of 106.61 g mol⁻¹. Use Ar values C = 12.01, H = 1.01, Cl = 35.45.

(a) Determine the value of n. [2] (b) Write the condensed structural formula and the IUPAC name of the tertiary isomer with this molecular formula. [2]

5. Approximate boiling points of four straight-chain aldehydes are given. Use these values.

Aldehyde	propanal	butanal	pentanal	hexanal
Boiling point / K	321	348	376	403

(a) Describe the trend and calculate the mean increase in boiling point per CH₂ unit. [2] (b) Explain the trend. [3] (c) Estimate the boiling point of heptanal. [1] (d) 2-methylbutanal is an isomer of pentanal. Predict whether its boiling point is above or below 376 K, and give a reason. [2]

6. Give the IUPAC name of each compound. [5]

(a) (CH₃)₂C=CHCH₂CH₂CH₃ (b) CH₃CH₂C(CH₃)(OH)CH₂CH₃ (c) (CH₃)₂CHCH(CH₃)CHO (d) CH₂=CHCH₂COCH₃ (e) CH₃CCl₂CH₂CH₃

7. A student wrote these names. For each one, state whether it is correct and, if not, give the correct name. [4]

(a) 4,4-dimethylbutanal (b) 2-propylpropanoic acid (c) hex-5-en-3-ol for CH₃CH₂CH(OH)CH₂CH=CH₂ (d) 2-methylbutan-3-one

8. For each pair, state the type of structural isomerism shown, or state that the compounds are not isomers. [4]

(a) pentan-2-ol and pentan-3-ol (b) hexanoic acid and ethyl butanoate (c) 1-chlorohexane and 1-chloro-3-methylpentane (d) hexane and cyclohexane

9. Classify each compound as primary, secondary or tertiary.

(a) CH₃CH₂CH(OH)CH(CH₃)₂ [1] (b) (CH₃)₃CCH₂CH₂Br [1] (c) (CH₃CH₂)₃CCl [1] (d) (CH₃)₂CHNHCH₃ [1] (e) CH₃CH₂CH(NH₂)CH₂CH₃ [1] (f) Explain your answer to (e). [1]

10. Several compounds have the molecular formula C₅H₁₀O₂.

(a) Show that C₅H₁₀O₂ fits the general formula of the carboxylic acids. [1] (b) Pentanoic acid is the straight-chain acid. Write the condensed formulas and IUPAC names of two branched carboxylic acids with this formula. [4] (c) State the type of structural isomerism between pentanoic acid and your answers to (b). [1] (d) Methyl butanoate, CH₃CH₂CH₂COOCH₃, also has this formula. Name its functional group and state the type of isomerism it shows with pentanoic acid. [2] (e) Explain why pentanoic acid has a much higher boiling point than methyl butanoate. [3]

11. Four compounds, A to D, all have the molecular formula C₆H₁₂O.

A: CH₃CH₂CH₂CH₂CH₂CHO B: CH₃CH₂COCH₂CH₂CH₃ C: (CH₃)₃CCH₂CHO D: HOCH₂CH₂CH₂CH=CHCH₃

(a) Give the IUPAC name of each compound. [4] (b) State the type of structural isomerism between A and B, and between A and C. [2] (c) D is unsaturated. State what this means. [1] (d) Predict which of A and C has the higher boiling point. Explain your answer. [2]

Answers

1. Phenyl (C₆H₅–) [1]; amino (–NH₂) [1]; ester (–COO–CH₂CH₃) [1] Examiner insight: Name the functional group, not the class: “amino” and “ester” are what the question asks for, while “amine” or “benzene” answer a different question.

2. (a) C₆H₁₀O₄ [1] (b) 6 : 10 : 4 has a common factor of 2, so C₃H₅O₂ [1] (c) The four CH₂ groups carry 8 hydrogen atoms, which are not drawn; the two O–H hydrogens are shown [1] (d) Saturated: every carbon–carbon bond is single; the C=O bonds are not carbon–carbon bonds [1] Examiner insight: In (c), remember that hydrogen atoms bonded to oxygen stay in a skeletal formula, so counting all ten is a slip.

3. (a) alkane [1] (b) amide [1] (c) halogenoalkane [1] (d) ester [1] (e) amine [1] Examiner insight: For (b) and (e), check whether the nitrogen is bonded to a C=O carbon: CONH₂ is an amide and NH₂ alone is an amine.

4. (a) 14.03n + 1.01 + 35.45 = 106.61 [1]; 14.03n = 70.15, so n = 5 (C₅H₁₁Cl) [1] (b) CH₃CH₂C(CH₃)₂Cl [1]; 2-chloro-2-methylbutane [1] Examiner insight: Write the equation from the general formula before solving it; a bare “n = 5” gives the reader no method to follow.

5. (a) Boiling point increases as the chain gets longer [1]; differences 27, 28 and 27 K, so the mean is about 27 K per CH₂ [1] (b) Each successive member has one more CH₂, so more electrons and a larger surface area [1]. London (dispersion) forces between molecules become stronger [1]. More energy is needed to separate the molecules, so the boiling point is higher [1] (c) 403 + 27 = about 430 K [1] (d) Below 376 K [1]; the branched molecule is more compact, so there is less surface contact and the London forces are weaker [1] Examiner insight: In (b), the forces being overcome are between molecules; writing that covalent bonds break on boiling contradicts the explanation.

6. (a) 2-methylhex-2-ene [1] (b) 3-methylpentan-3-ol [1] (c) 2,3-dimethylbutanal [1] (d) pent-4-en-2-one [1] (e) 2,2-dichlorobutane [1] Examiner insight: In (d), the ketone suffix takes the lowest locant before the C=C does, so numbering from the C=C end gives the wrong name.

7. (a) Incorrect: the chain extends into a methyl group, so it is 4-methylpentanal [1] (b) Incorrect: the longest chain containing COOH has five carbons, so it is 2-methylpentanoic acid [1] (c) Correct: the -ol gets the lowest locant (3, not 4) before the C=C is considered [1] (d) Incorrect: the carbonyl must have the lower number, so it is 3-methylbutan-2-one [1] Examiner insight: A name with a substituent on the last chain carbon, or an alkyl prefix longer than the chain beyond it, is a sign that the longest chain has been missed.

8. (a) Position isomers [1] (b) Functional group isomers, both C₆H₁₂O₂ [1] (c) Branched and straight-chain (chain) isomers, both C₆H₁₃Cl [1] (d) Not isomers: hexane is C₆H₁₄ and cyclohexane is C₆H₁₂ [1] Examiner insight: Write out both molecular formulas before naming an isomer type; in (d), that check alone gives the answer.

9. (a) Secondary: the C–OH carbon is bonded to two carbons [1] (b) Primary: the C–Br carbon is bonded to one carbon [1] (c) Tertiary: the C–Cl carbon is bonded to three carbons [1] (d) Secondary amine: the nitrogen is bonded to two carbons [1] (e) Primary amine [1] (f) For amines you count the carbon atoms bonded to the nitrogen, and here the N is bonded to only one carbon [1] Examiner insight: In (e), counting the carbons around the C–N carbon, as you would for an alcohol, leads to “secondary”, which is the wrong rule for amines.

10. (a) CₙH₂ₙO₂ with n = 5 gives 2 × 5 = 10 hydrogen atoms, so C₅H₁₀O₂ fits [1] (b) Any two of: CH₃CH₂CH(CH₃)COOH [1], 2-methylbutanoic acid [1]; (CH₃)₂CHCH₂COOH [1], 3-methylbutanoic acid [1]; or (CH₃)₃CCOOH, 2,2-dimethylpropanoic acid (c) Branched and straight-chain (chain) isomerism [1] (d) Ester [1]; functional group isomerism [1] (e) Pentanoic acid has an O–H in its carboxyl group, so its molecules form hydrogen bonds with each other [1]. Methyl butanoate has no O–H, so only dipole–dipole and London forces act between its molecules [1]. Breaking the stronger hydrogen bonds between acid molecules takes more energy, so the acid boils higher [1] Examiner insight: In (b), each formula and its name should match; a correct name attached to a different structure does not show the skill.

11. (a) A hexanal [1]; B hexan-3-one [1]; C 3,3-dimethylbutanal [1]; D hex-4-en-1-ol [1] (b) A and B: functional group isomers (aldehyde and ketone) [1]; A and C: branched and straight-chain (chain) isomers [1] (c) The molecule contains at least one carbon–carbon multiple bond (here a C=C) [1] (d) A [1]; its straight chain gives more surface contact than the compact branched C, so its London forces are stronger [1] Examiner insight: In (a), aldehyde names carry no number for the CHO group, but a ketone such as B always needs one to show where the C=O sits.

Where marks are usually lost

• Naming the class (amine, ester compound) when the question asks for the functional group.
• Leaving a common factor in an empirical formula.
• Forgetting the hydrogen atom of –OH when counting atoms from a skeletal formula.
• Treating a C=O bond as making a compound unsaturated.
• Choosing the longest chain without checking that it contains the functional group or the C=C.
• Numbering so a C=C or methyl gets the lowest number when an -ol, -one or -oic acid is present.
• Explaining boiling points by covalent bonds breaking.
• Classifying amines by the carbon atom instead of the nitrogen atom.
• Calling compounds isomers without first checking their molecular formulas.

Next steps

• Structure 3.2 revision notes for a final read-through
• Structure 3.2 study guide for the full explanations
• The IB DP chemistry course hub and the printable checklist
• Try all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Source: the International Baccalaureate’s Diploma Programme Chemistry guide, issued in February 2023 for courses examined from 2025. This page follows Structure 3.2, Functional groups: Classification of organic compounds, understandings 3.2.1 to 3.2.6, which are common to SL and HL.