Year 12 Chemistry Module 7 · Organic Chemistry ⏱ ~45 min 5 MC · 3 Short Answer Lesson 2 of 23

IUPAC Nomenclature II, Functional Group Classes & Isomers

Master the naming rules for all eight organic functional group classes, alcohols, aldehydes, ketones, carboxylic acids, esters, amines, amides and haloalkanes, and learn to classify and draw structural isomers systematically.

Today's hook: Ethanol and dimethyl ether share the exact same molecular formula, so why does one get metabolised by your liver while the other is used as an anaesthetic propellant?

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Worksheets

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Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Think First

Ethanol (the alcohol in wine and beer) and dimethyl ether (an industrial solvent and propellant) have exactly the same molecular formula: C₂H₆O. Same atoms, same count, yet one is metabolised by your liver at a predictable rate and the other causes anaesthesia and is used in refrigeration.

Before you read on, try to draw two different structural arrangements using 2 carbons, 6 hydrogens, and 1 oxygen. Write down how you would decide which arrangement is an “alcohol” and which is not. You will return to this at the end of the lesson.

Learning Intentions

goals

Know

The structural signature, suffix/prefix, and a named example for each of the eight functional group classes
The IUPAC suffixes: -ol, -al, -one, -oic acid, -amide, -amine, and the “alkyl alkanoate” pattern for esters
That primary/secondary/tertiary classification applies to alcohols (on C–OH) and amines (on N)
The three types of structural isomers: chain, position, functional group

Understand

Why the position of a carbonyl group (terminal vs internal) determines whether a compound is an aldehyde, ketone, or carboxylic acid
Why amides are neutral while amines are basic (lone pair delocalisation into C=O)
Why the same molecular formula can represent completely different functional groups (functional group isomers)
How IUPAC names encode both the type and position of a functional group unambiguously

Can Do

Identify the functional group class from any structural formula and give the correct IUPAC name
Name esters correctly as “alkyl alkanoate” with the alkyl part first
Classify a pair of structural isomers as chain, position, or functional group isomers with justification
Draw all structural isomers of a given molecular formula systematically

Scan these before reading

vocab

Suffix -olIndicates an alcohol functional group (–OH); position numbered to give lowest locant.

Suffix -alIndicates an aldehyde functional group (–CHO); always at C1, no locant needed.

Suffix -oneIndicates a ketone functional group (C=O between carbons); locant number required.

Suffix -oic acidIndicates a carboxylic acid functional group (–COOH); always at C1.

Structural isomerCompounds with the same molecular formula but different connectivity of atoms.

Functional group isomerA type of structural isomer where compounds have the same formula but different functional groups (e.g., alcohol vs ether).

Key Structural Signatures, Reference Panel

Alcohol

R–OH

Suffix: -ol. Number chain from end closer to –OH. 1°/2°/3° based on carbon neighbours of C–OH. e.g. propan-1-ol, propan-2-ol

Aldehyde

R–CHO

Suffix: -al. Terminal C=O with H on same carbon. Always C1, never needs a locant. e.g. ethanal, propanal

Ketone

R–CO–R′

Suffix: -one. Internal C=O bonded to two carbons. Locant needed for C5+. e.g. butanone (butan-2-one), pentan-3-one

Carboxylic Acid

R–COOH

Suffix: -oic acid. Always at C1, no locant. e.g. methanoic acid, ethanoic acid (vinegar)

Ester

R–COO–R′

Named as alkyl alkanoate. Alkyl (from alcohol, the OR′ part) comes FIRST. e.g. ethyl ethanoate, methyl propanoate

Amine

R–NH₂

Suffix: -amine. 1° = one alkyl on N; 2° = two alkyls; 3° = three alkyls. Weak base (lone pair on N). e.g. ethanamine

Amide

R–CONH₂

Suffix: -amide. Always at C1, no locant. NOT basic (lone pair delocalised into C=O). e.g. ethanamide, propanamide

Haloalkane

R–X  (X = F, Cl, Br, I)

Prefix: fluoro-/ chloro-/ bromo-/ iodo- + locant. e.g. 1-chlorobutane, 2-bromobutane. Can be 1°/2°/3°.

Core Content

Misconception to Fix: “The IUPAC name tells you the molecular formula of a compound.”, This is wrong. The IUPAC name describes the carbon skeleton, functional groups, and substituent positions, but different compounds can share the same molecular formula (isomers). IUPAC nomenclature distinguishes isomers by structure, not just atomic composition.

Alcohols, Naming and Primary/Secondary/Tertiary Classification

Hydroxyl group · -ol suffix · 1°/2°/3° based on C–OH neighbours

Alcohols all contain the -OH group, but WHERE on the carbon skeleton it sits, and how many other carbons surround it, determines both the name and the entire chemistry of that alcohol.

The hydroxyl group (–OH) bonded to a carbon is the defining feature of an alcohol. The suffix is -ol, replacing the final -e of the parent alkane name. Number the chain from the end closer to the –OH group to give it the lowest possible locant.

Propan-1-ol: 3C chain, –OH at C1 (numbered from the end closest to –OH)
Propan-2-ol: 3C chain, –OH at C2
Ethanol and methanol: only one possible –OH position, so no locant needed

Alcohols are classified by the number of carbon atoms directly bonded to the carbon bearing the –OH group:

Primary (1°)

C–OH bonded to: 1 other carbon
H on C–OH: 2
Example: Ethanol (CH₃CH₂OH)

Secondary (2°)

C–OH bonded to: 2 other carbons
H on C–OH: 1
Example: Propan-2-ol (CH₃CHOHCH₃)

Tertiary (3°)

C–OH bonded to: 3 other carbons
H on C–OH: 0
Example: 2-methylpropan-2-ol

Method: To classify an alcohol, locate the C–OH carbon and count its CARBON neighbours only (1, 2, or 3). Do not count the –OH oxygen or H atoms attached to it. The number of carbon neighbours gives the classification directly.

Critical Error: Primary/secondary/tertiary classification applies to the C–OH carbon in alcohols and the N atom in amines. Never call an aldehyde “primary” because it sits at C1, that is a misconception. Aldehydes are a separate functional group class; the 1°/2°/3° language applies only where a central atom can have one, two, or three alkyl group attachments.

Which carbon neighbour count correctly identifies propan-2-ol as a secondary alcohol?

Aldehydes, Ketones and Carboxylic Acids

Carbonyl group position · -al / -one / -oic acid · terminal vs internal

Aldehydes, ketones, and carboxylic acids all contain the carbonyl group (C=O), the position of that carbonyl and what else is bonded to it is the only thing separating three completely different functional group classes.

Aldehyde (–CHO): The carbonyl carbon is terminal and also has a hydrogen directly bonded to it. Suffix: -al. The aldehyde carbon is always C1, a locant is never written. Examples: methanal (HCHO), ethanal (CH₃CHO), propanal (CH₃CH₂CHO), butanal.

Ketone (R–CO–R′): The carbonyl carbon is internal bonded to two other carbons. Suffix: -one. A locant is required for chains of five or more carbons. For C4, the only possible internal position is C2, so “butan-2-one” is correct (also written as butanone). Examples: butan-2-one, pentan-2-one, pentan-3-one. The carbonyl cannot be at C1, that would make it an aldehyde.

Carboxylic acid (–COOH): The carbonyl carbon bears both a C=O and an –OH on the same carbon, always terminal. Suffix: -oic acid. Always C1, no locant. Examples: methanoic acid (HCOOH), ethanoic acid (CH₃COOH), propanoic acid, butanoic acid.

Group	Carbonyl position	Additional feature	Suffix	Locant?
Aldehyde	Terminal (C1)	H on carbonyl C	-al	Never
Ketone	Internal	Two C neighbours	-one	Yes (C5+)
Carboxylic acid	Terminal (C1)	–OH on carbonyl C	-oic acid	Never

Method, Check C=O First: Before naming any compound with a C=O group: look at what is bonded to the carbonyl carbon. H at chain end → aldehyde (-al). Two carbon neighbours → ketone (-one). –OH at chain end → carboxylic acid (-oic acid). This three-way check takes five seconds and prevents the most common functional group misidentification in Module 7.

Critical Misconception: “Propanal and propanone are the same compound because they both have C=O.” This is wrong. They are functional group isomers (both C₃H₆O), same atoms, completely different arrangements and entirely different chemistry. Propanal has the C=O at C1 with an H; propanone has C=O at C2 flanked by two methyl groups.

Real-World Context: Ethanoic acid = vinegar (5% aqueous solution). Butanoic acid = rancid butter and strong cheese smell. Methanoic acid = bee and ant venom (Latin formica = ant, hence “formic acid”). Ethanal = primary liver metabolite responsible for hangover symptoms after alcohol consumption.

True or False: Propanal and propanone are position isomers because both have the formula C₃H₆O with a C=O group at different positions.

Esters, Amines, Amides and Haloalkanes

Alkyl alkanoate · lone pair basicity · halogen as prefix

Esters smell like fruit, amines smell like fish, amides form the backbone of every protein in your body, and haloalkanes are the building blocks for most organic synthesis reactions, each has a structural signature you can read instantly once you know the pattern.

Esters (R–COO–R′): Named as alkyl alkanoate the alkyl group (from the alcohol, the –OR′ part) comes FIRST, the alkanoate (from the acid, the RCOO– part) comes second.

CH₃COOCH₂CH₃ → ethyl ethanoate (ethanoate from CH₃COO–; ethyl from –OC₂H₅)
CH₃COOCH₃ → methyl ethanoate
CH₃CH₂COOC₄H₉ → butyl propanoate

Amines (R–NH₂): Suffix -amine. Classified by alkyl groups attached to N: primary (1°) = one alkyl (R–NH₂); secondary (2°) = two alkyls (R–NH–R′); tertiary (3°) = three alkyls. Amines are weak bases the lone pair on N accepts H⁺ from acids.

Amides (R–CONH₂): Suffix -amide. The carbonyl carbon is always C1, no locant. Examples: ethanamide (CH₃CONH₂), propanamide. Amides are neutral the nitrogen lone pair is delocalised into the adjacent C=O and is not available for proton acceptance.

Haloalkanes (R–X): Halogen as a prefix, fluoro-, chloro-, bromo-, iodo-, with a locant. Examples: 1-chlorobutane, 2-bromobutane. Classification as 1°/2°/3° follows the same logic as alcohols (count carbon neighbours of the C bonded to X).

Group	Structure	Naming pattern	Key feature
Ester	R–COO–R′	Alkyl alkanoate (alkyl first)	Fruit-like odour; formed from acid + alcohol
Amine	R–NH₂	[chain]-amine	Weak base; lone pair on N available
Amide	R–CONH₂	[chain]-amide	Neutral; lone pair delocalised into C=O
Haloalkane	R–X	[halo]-[alkane] + locant	Polar C–X bond; 1°/2°/3° applies

Ester Naming Trick: Write “alkyl alkanoate” before you start, then fill in: alkyl = name of the –OR′ group (what was the alcohol); alkanoate = name of the RCOO– group (what was the acid). “Ethyl ethanoate”, NOT “ethanoate ethyl.” The alkyl always comes first in the two-word name.

Critical Misconception: “Amides are basic because they contain nitrogen, just like amines.” This is wrong. The nitrogen lone pair in an amide is delocalised by resonance into the adjacent C=O, it is spread across the C–N bond and unavailable to accept a proton. Amines have an undelocalised lone pair that freely accepts H⁺. Never confuse the two: amines are weak bases; amides are neutral.

CH₃COOCH₂CH₂CH₃ is an ester. What is its correct IUPAC name?

Structural Isomers, Chain, Position and Functional Group

Same molecular formula · different arrangement · three isomer types

Structural isomers prove that molecular formula alone tells you almost nothing useful about a compound, the arrangement of atoms changes everything about physical properties, chemical reactivity, and biological activity.

Structural isomers are compounds with the same molecular formula but different structural arrangements of atoms. There are three types to distinguish:

Chain isomers: Same functional group and molecular formula, but a different carbon skeleton (branching pattern). Example: butane vs 2-methylpropane (both C₄H₁₀). The functional group (alkane) is identical; only the chain architecture changes. Different surface areas produce different boiling points.

Position isomers: Same functional group, same carbon skeleton, but the functional group is at a different position on the chain. Example: propan-1-ol vs propan-2-ol (both C₃H₈O). Both are alcohols with a 3-carbon chain, but –OH at C1 (primary) vs C2 (secondary). This matters chemically: propan-1-ol (primary) is oxidised to propanal then to propanoic acid; propan-2-ol (secondary) is oxidised only to propanone.

Functional group isomers: Same molecular formula but entirely different functional groups. Example: ethanol (C₂H₆O, alcohol) vs methoxymethane (dimethyl ether, C₂H₆O). Another pair: propanal vs propanone (both C₃H₆O, but aldehyde vs ketone).

Chain

Same: Formula, functional group
Different: Carbon skeleton shape
Example pair: butane / 2-methylpropane (C₄H₁₀)

Position

Same: Formula, group, skeleton
Different: Position of group on chain
Example pair: propan-1-ol / propan-2-ol (C₃H₈O)

Functional group

Same: Molecular formula
Different: Functional group itself
Example pair: propanal / propanone (C₃H₆O)

Systematic Method: When asked to draw ALL structural isomers: (1) Vary the carbon skeleton (chain isomers). (2) For each skeleton, vary the position of the functional group (position isomers). (3) Consider different functional groups that give the same molecular formula (functional group isomers). Verify each structure gives the correct molecular formula before naming.

Common Error, Duplicate Structures: Students draw what they believe are two different isomers but actually draw the same compound from a different direction. To confirm structures are genuinely different: assign IUPAC names to both. If the names are identical, the structures are identical regardless of how they are drawn or oriented on the page.

Real-World Context: Ethanol vs methoxymethane (dimethyl ether), same C₂H₆O formula, entirely different biological effects. Ethanol is metabolised by alcohol dehydrogenase in the liver. Methoxymethane (methoxyflurane when halogenated) is the anaesthetic gas in Penthrox emergency inhalers. One molecular formula, two structurally distinct arrangements, completely different pharmacology.

Because the –OH group in ethanol can form hydrogen bonds (donating H to an electronegative acceptor) but the C–O–C ether linkage cannot donate H-bonds, ethanol (bp 78°C) has a dramatically higher boiling point than dimethyl ether (bp −24°C) despite both having the molecular formula C₂H₆O. Structure dictates intermolecular forces, which determine physical properties.

Coming up: In Lesson 3 you will see how chain length and functional group together determine physical properties such as boiling point, the first systematic comparison across homologous series.

Three of these are examples of the SAME type of isomerism. Which is the odd one out?

Worked Examples

Worked Example 1, Naming compounds across functional group classes (straightforward)

Name the following: (a) CH₃CH₂CH₂OH (b) CH₃CH₂CHO (c) CH₃COCH₃ (d) CH₃CH₂COOCH₃ (e) CH₃CH₂NH₂

GIVEN: Five structural formulas. FIND: IUPAC name for each. METHOD: (1) Identify functional group from structural signature. (2) Apply the suffix/prefix rule. (3) Number the chain for lowest locant.

(a) CH₃CH₂CH₂OH: Terminal –OH → alcohol. 3C chain → prop-. Suffix -ol. –OH at C1 (numbered from the end closest to –OH). ANSWER: propan-1-ol (primary alcohol, C–OH bonded to 1 other carbon).

(b) CH₃CH₂CHO: Terminal C=O with H on same carbon → aldehyde. 3C chain → prop-. Suffix -al. Always C1. ANSWER: propanal.

(c) CH₃COCH₃: Internal C=O flanked by two carbons → ketone. 3C chain → prop-. Suffix -one. C=O at C2 (the only possible internal position in a 3C chain). ANSWER: propanone (locant omitted as unambiguous for 3C).

(d) CH₃CH₂COOCH₃: –COO– linkage → ester. Named as alkyl alkanoate. RCOO– part: CH₃CH₂COO– = propanoate (3C acid). –OR′ part: –OCH₃ = methyl. Alkyl first. ANSWER: methyl propanoate.

(e) CH₃CH₂NH₂: –NH₂ on terminal carbon → primary amine. 2C chain → eth-. Suffix -amine. N bonded to one alkyl group only. ANSWER: ethanamine (primary amine).

Worked Example 2, Drawing all C₄H₈O structural isomers with a carbonyl group (intermediate)

Draw and name all structural isomers of C₄H₈O that contain a carbonyl group. Classify each.

GIVEN: Molecular formula C₄H₈O, carbonyl group (C=O) required. FIND: All structural isomers. METHOD: Consider (a) aldehyde structures (terminal C=O + H), then (b) ketone structures (internal C=O). Vary the carbon skeleton for each.

Aldehyde structures (C=O at C1 with H on carbonyl C):

Straight-chain: CH₃CH₂CH₂CHO = butanal ✔ C₄H₈O ✔

Branched: (CH₃)₂CHCHO = 2-methylpropanal ✔ C₄H₈O ✔

Ketone structures (internal C=O): A 4C ketone has C=O at C2. C=O at C3 in a 4C chain is the same compound numbered from the other end.

CH₃COCH₂CH₃ = butan-2-one ✔ C₄H₈O ✔

ANSWER: Three isomers. (1) Butanal straight-chain aldehyde. (2) 2-methylpropanal branched aldehyde (chain isomer of butanal). (3) Butan-2-one ketone (functional group isomer of both aldehyde isomers, same C₄H₈O formula, different functional group).

Worked Example 3, Evaluate a student’s isomer classification claim (hard, 5 marks)

A student states: “Propan-1-ol and propan-2-ol are functional group isomers because they both have the formula C₃H₈O but the –OH group is in a different place.” Evaluate this statement fully.

Check the molecular formula claim. Propan-1-ol (CH₃CH₂CH₂OH) → C₃H₈O ✔. Propan-2-ol (CH₃CHOHCH₃) → C₃H₈O ✔. The student is correct that both have the formula C₃H₈O and they are structural isomers. [1 mark]

Evaluate the isomer type classification. The student calls them “functional group isomers”, this is WRONG. Functional group isomers have different functional groups. Both propan-1-ol and propan-2-ol contain the –OH (hydroxyl) group, the functional group is identical in both compounds. [1 mark]

State the correct isomer type. These are position isomers: same molecular formula (C₃H₈O), same functional group (–OH, alcohol), same 3-carbon skeleton, but –OH is at C1 in propan-1-ol and at C2 in propan-2-ol. Only the position of the group changes. [1 mark]

Extend with chemical consequences. Propan-1-ol has a primary C–OH (bonded to 1 other carbon); propan-2-ol has a secondary C–OH (bonded to 2 carbons). Primary alcohols are oxidised to aldehydes then carboxylic acids under mild oxidising conditions; secondary alcohols are oxidised only to ketones, oxidation stops at that stage. [1 mark]

ANSWER SUMMARY: The molecular formula claim is correct (both C₃H₈O, both structural isomers) ✔. The isomer type classification is wrong ✗, the correct type is position isomers, not functional group isomers. The functional group (–OH) is the same in both compounds; only its position on the 3C chain differs. The position difference means propan-1-ol is primary (oxidised → aldehyde → carboxylic acid) while propan-2-ol is secondary (oxidised → ketone only). [1 mark for structured evaluation]

Complete the sentence: Butan-1-ol and butan-2-ol are _______ isomers because they have the same molecular formula C₄H₁₀O, the same functional group (–OH), and the same 4-carbon skeleton, but the –OH group is at a different _______ on the chain.

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The IUPAC Nomenclature tool shows the suffix ‘-ane’ indicates a hydrocarbon that is…

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Place these steps in the correct order for naming an organic compound with a functional group using IUPAC rules:

Count the carbons in the longest continuous chain containing the principal functional group

Identify the highest-priority functional group and assign it the principal characteristic group suffix

Number the chain to give the principal functional group the lowest possible locant

Identify and name substituents (alkyl groups, halogens) as prefixes with locants

Combine prefix(es) + parent chain name + suffix, applying alphabetical order for prefixes

Activities

Activity A, Functional Group Identification and IUPAC Naming

For each compound below: (1) identify the functional group class, (2) write the IUPAC name, and (3) briefly justify your naming decision (which feature of the structure determined the suffix or prefix).

Compound

CH₃CH₂OH

CH₃CH₂CHO

CH₃COCH₂CH₃

CH₃CH₂COOH

CH₃COOCH₂CH₂CH₃

CH₃CH₂CH₂NH₂

Functional group & IUPAC name (with justification)

Activity B, Classify the Isomer Type and Explain the Structural Difference

For each pair of compounds below: (A) confirm they have the same molecular formula, (B) classify the isomer type (chain / position / functional group), and (C) explain the key structural difference that distinguishes them.

1. Butane (CH₃CH₂CH₂CH₃) vs 2-methylpropane (CH₃CH(CH₃)CH₃)

2. Butan-1-ol (CH₃CH₂CH₂CH₂OH) vs Butan-2-ol (CH₃CH₂CHOHCH₃)

3. Butanal (CH₃CH₂CH₂CHO) vs Butan-2-one (CH₃COCH₂CH₃)

Check Your Understanding

Click an option to check your answer. Aim for all 5 before looking at the answers accordion.

1. Which compound is the ester formed from propan-1-ol and ethanoic acid?

2. A compound has the structure CH₃CH₂CH(NH₂)CH₃. What is the IUPAC name and amine classification?

3. Propanal and propanone both have C₃H₆O. What type of isomers are they?

4. Which statement correctly identifies the type of isomers formed by butan-1-ol and butan-2-ol?

5. A student is given an unknown compound with molecular formula C₃H₆O. Which of the following correctly lists ALL possible structural isomers of this formula that contain only one functional group?

Short Answer

6. For each compound below, state the functional group class and give the correct IUPAC name: (a) CH₃CH₂CH₂COOH, (b) CH₃CONH₂, (c) CH₃CH(CH₃)OH, (d) CH₃CH₂COOCH₂CH₃ 4 MARKS

7. Draw and name all structural isomers of C₃H₆O. For each, identify the isomer type relative to the other and explain the key structural difference. 4 MARKS

8. A pharmacologist is investigating two compounds with the molecular formula C₄H₉NO. Compound A reacts with hydrochloric acid to form a salt. Compound B does not react with HCl. (a) Identify the functional groups that could give this difference in reactivity. (b) Name and draw one possible structure for each compound. (c) Identify the type of isomers that Compound A and Compound B represent, and justify your answer. 5 MARKS

Comprehensive Answers

Activity A, Naming Practice

1. CH₃CH₂OH: Alcohol (–OH on terminal carbon). 2C chain. Name: ethanol. Primary alcohol (C–OH bonded to 1 carbon). No locant needed for C2.

2. CH₃CH₂CHO: Aldehyde (terminal C=O with H on same carbon). 3C chain. Suffix -al. Name: propanal. Always at C1; no locant written.

3. CH₃COCH₂CH₃: Ketone (internal C=O flanked by two carbons). 4C chain. Suffix -one. C=O at C2. Name: butan-2-one (also accepted: butanone).

4. CH₃CH₂COOH: Carboxylic acid (terminal –COOH). 3C chain. Suffix -oic acid. Name: propanoic acid. Always C1; no locant.

5. CH₃COOCH₂CH₂CH₃: Ester (–COO–). Alkanoate part: CH₃COO– = ethanoate. Alkyl part: –OCH₂CH₂CH₃ = propyl. Name: propyl ethanoate.

6. CH₃CH₂CH₂NH₂: Amine (–NH₂). 3C chain. Suffix -amine. –NH₂ at C1. Name: propan-1-amine (or propylamine). Primary amine, N bonded to one alkyl group. Basic because the lone pair on N is available to accept H⁺.

Activity B, Isomer Identification

1. Butane vs 2-methylpropane: Both C₄H₁₀. Chain isomers same functional group class (alkane), same molecular formula, but different carbon skeleton: butane has a straight 4-carbon chain; 2-methylpropane has a branched 3-carbon chain with a methyl group at C2.

2. Butan-1-ol vs Butan-2-ol: Both C₄H₁₀O. Position isomers same formula, same functional group (–OH, alcohol), same 4-carbon skeleton, but –OH is at C1 (primary alcohol) in butan-1-ol and at C2 (secondary alcohol) in butan-2-ol.

3. Butanal vs Butan-2-one: Both C₄H₈O. Functional group isomers same formula, but different functional groups. Butanal has a terminal C=O with H (aldehyde, suffix -al); butan-2-one has an internal C=O between two carbons (ketone, suffix -one).

Multiple Choice Explanations

1. B, Propyl ethanoate. Ester from propan-1-ol + ethanoic acid. Alkyl part (from propan-1-ol) = propyl. Alkanoate part (from ethanoic acid) = ethanoate. Alkyl comes first → propyl ethanoate. Option A (ethyl propanoate) would come from ethanol + propanoic acid. Option D misspells the suffix (-propanate is not correct IUPAC).

2. C, Butan-2-amine; primary. CH₃CH₂CH(NH₂)CH₃: longest chain = 4C (butane). –NH₂ at C2 from the end that gives the lower locant (C2 vs C3 from the other end). Name: butan-2-amine. Primary amine because N is bonded to only one alkyl group (–NH₂ on C2; N has H₂ + one carbon neighbour). Option A has correct name but wrong classification; secondary amine requires two alkyl groups on N.

3. B, Functional group isomers. Propanal has a terminal C=O (aldehyde); propanone has an internal C=O (ketone). These are entirely different functional groups, the hallmark of functional group isomers. Option A incorrectly describes position isomers (same functional group, different position), but here the group itself differs.

4. C, Position isomers. Butan-1-ol and butan-2-ol: both C₄H₁₀O ✔, same –OH group ✔, same 4-carbon skeleton ✔, –OH at C1 vs C2. This matches the definition of position isomers exactly. Options A and D incorrectly treat 1° vs 2° as different functional group classes, they are classifications of the same class (alcohol), not different classes.

5. D, Propanal and propanone. C₃H₆O: propanal (CH₃CH₂CHO, aldehyde) and propanone (CH₃COCH₃, ketone) both satisfy the formula. Propan-1-ol and propan-2-ol are C₃H₈O (two extra H), a different molecular formula, so they are not C₃H₆O isomers. Options A and B each list only one of the two correct isomers.

Short Answer Model Answers

Q6 (4 marks, 1 mark each):

(a) CH₃CH₂CH₂COOH: Carboxylic acid (–COOH at terminal carbon). 4C chain. Name: butanoic acid [1].

(b) CH₃CONH₂: Amide (–CONH₂). 2C chain. Suffix -amide. Name: ethanamide [1].

(c) CH₃CH(CH₃)OH: The carbon with –OH: CH₃–CH(OH)–CH₃ → 3-carbon chain with –OH at C2 → propan-2-ol. Secondary alcohol (C–OH bonded to 2 other carbons). Name: propan-2-ol [1].

(d) CH₃CH₂COOCH₂CH₃: Ester. Alkanoate part: CH₃CH₂COO– = propanoate (3C acid). Alkyl part: –OCH₂CH₃ = ethyl. Name: ethyl propanoate [1].

Q7 (4 marks): C₃H₆O has two structural isomers containing a single C=O functional group:

Isomer 1: Propanal (CH₃CH₂CHO), aldehyde. Terminal C=O with H on carbonyl C [1].

Isomer 2: Propanone (CH₃COCH₃), ketone. Internal C=O flanked by two methyl groups [1].

These are functional group isomers [1]: same molecular formula C₃H₆O, but different functional groups. Key structural difference: in propanal, C=O is at C1 with an H (aldehyde); in propanone, C=O is at C2 between two carbon groups (ketone) [1].

Q8 (5 marks):

(a) Compound A reacts with HCl → must contain an amine (–NH₂): the undelocalised lone pair on N accepts H⁺ from HCl, forming an ammonium salt (R–NH₃⁺Cl⁻) [1]. Compound B does not react with HCl → must contain an amide (–CONH₂): the N lone pair is delocalised into the adjacent C=O by resonance and cannot accept a proton, amides are neutral [1].

(b) Compound A (amine): 4-aminobutan-2-one (CH₃COCH₂CH₂NH₂) contains a primary amine and a ketone, formula C₄H₉NO ✔ [1]. Compound B (amide): butanamide (CH₃CH₂CH₂CONH₂) contains an amide at C1, formula C₄H₉NO ✔ [1].

(c) Compound A and Compound B are functional group isomers [1]: same molecular formula (C₄H₉NO) but completely different functional groups, Compound A contains an amine (–NH₂) and Compound B contains an amide (–CONH₂). Because the functional groups differ (not just their positions), these are functional group isomers, not position isomers.

Review Drills

What is the IUPAC name for the compound formed when ethanol reacts with propanoic acid to form an ester?

How do you distinguish between a 1°, 2°, and 3° alcohol? Give one example of each.

Why are amines weak bases but amides are neutral? Use lone pair location to explain.

Classify these three isomer pairs: (A) butane / 2-methylpropane, (B) propan-1-ol / propan-2-ol, (C) propanal / propanone.

What structural feature must be present to confirm a compound is an aldehyde rather than a ketone? How does this affect the IUPAC suffix?

A student writes the name “ethanoate methyl” for an ester. What is wrong and what is the correct name?

Revisit Your Initial Thinking

Back at the start, you were asked why ethanol gets metabolised by your liver while dimethyl ether is used as an anaesthetic propellant. Now you know: the –OH group in ethanol can be oxidised by alcohol dehydrogenase in the liver, the enzyme recognises and acts on the hydroxyl group. The C–O–C ether linkage in dimethyl ether is chemically inert under physiological conditions: no enzyme can oxidise it, so it passes through the body without being broken down. The difference comes down entirely to the functional group. Same molecular formula (C₂H₆O), different connectivity, completely different biochemistry. These are functional group isomers: ethanol has O bonded to C and H (C–O–H = hydroxyl/alcohol), while dimethyl ether has O bonded to two carbons (C–O–C = ether).

Mark lesson as complete

Tick when you’ve finished all activities and checked your answers.

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