Chemistry · Year 12 · Module 7 · Lesson 2
HSC Exam Practice
IUPAC Nomenclature II: Functional Group Classes & Isomers
Short answer
1.Short answer — IUPAC naming and functional groups
Define the term structural isomers.
Identify the functional group class and give the correct IUPAC name for each of the following compounds.
(a) CH3CH2CH2OH
(b) CH3CH2CHO
(c) CH3COCH2CH3
(d) CH3COOC2H5
Distinguish between a primary, secondary, and tertiary alcohol. In each case, state the criterion used for classification and give one named example.
Explain why amines are classified as weak bases while amides are neutral, with reference to the nitrogen lone pair in each functional group.
Describe the systematic method for drawing all structural isomers of a given molecular formula. Your description must identify the three categories of structural isomer and the order in which they should be considered.
Name the ester formed when propan-1-ol reacts with propanoic acid, and write its condensed structural formula. Identify which part of the name comes from the alcohol and which from the acid.
Data response
2.Data response — boiling points of C4H10 isomers
The table and figure below show boiling points and molecular shapes of the two structural isomers with the molecular formula C4H10.
| Compound | Condensed formula | Molecular shape | Boiling point (°C) |
|---|---|---|---|
| Butane | CH3CH2CH2CH3 | Extended linear chain | –1 |
| 2-methylpropane | CH3CH(CH3)CH3 | Compact branched | –12 |
(a) Classify the type of structural isomers that butane and 2-methylpropane represent, and justify your answer. (2 marks)
(b) Using the data in the table and your knowledge of intermolecular forces, account for the difference in boiling point between butane and 2-methylpropane. (3 marks)
3.Data response — isomers of C3H6O
The molecular formula C3H6O can represent two different functional group classes that each contain a C=O group.
(a) Name both structural isomers of C3H6O that contain a C=O group, and write the condensed structural formula for each. (2 marks)
(b) Classify the type of structural isomers these two compounds represent, and explain the structural feature that distinguishes them. (2 marks)
(c) Identify why an IUPAC name can encode the difference between propanal and propanone even though both have the formula C3H6O. In your answer, state the structural feature detected by the suffix –al versus the suffix –one, and explain what a student would look for in a structural formula to determine which suffix applies. (3 marks)
Extended response
4.Extended response — isomers and functional groups
Evaluate the claim: “Knowing the molecular formula of an organic compound is sufficient to determine its IUPAC name, its functional group class, and its chemical properties.”
In your response, refer to specific examples of structural isomers from Module 7 Lesson 2 to support your evaluation. Marking criteria are in the answer key.
Model answers
Q1.1 — Definition of structural isomers (2 marks)
Structural isomers are compounds that share the same molecular formula but have different structural arrangements of atoms (different connectivity). [1] They may differ in the carbon skeleton shape, the position of a functional group, or the type of functional group. [1]
Q1.2 — IUPAC naming (4 marks, 1 each)
(a) CH3CH2CH2OH: Alcohol; 3C chain, –OH at C1; propan-1-ol. Primary alcohol.
(b) CH3CH2CHO: Aldehyde; terminal C=O with H; 3C chain; suffix –al; propanal.
(c) CH3COCH2CH3: Ketone; internal C=O at C2 in a 4C chain; suffix –one; butan-2-one (also: butanone accepted).
(d) CH3COOC2H5: Ester; alkanoate = ethanoate (from CH3COO–); alkyl = ethyl (from –OC2H5); alkyl first; ethyl ethanoate.
Q1.3 — Primary, secondary, tertiary alcohol (3 marks)
Classification is based on the number of carbon atoms directly bonded to the carbon bearing the –OH group (count C neighbours only). [1]
Primary (1°): C–OH bonded to 1 other carbon. Example: ethanol (CH3CH2OH). [0.5]
Secondary (2°): C–OH bonded to 2 other carbons. Example: propan-2-ol (CH3CHOHCH3). [0.5]
Tertiary (3°): C–OH bonded to 3 other carbons. Example: 2-methylpropan-2-ol. [1 for definition + example pair]
Q1.4 — Amines weak base; amides neutral (3 marks)
In an amine (R–NH2), the lone pair on the nitrogen atom is not involved in any resonance and is freely available to accept a proton (H+) from an acid → amine acts as a weak base. [1]
In an amide (R–CONH2), the lone pair on nitrogen is delocalised by resonance into the adjacent C=O π system. [1] This spreads the electron density across the C–N bond and makes the lone pair unavailable to accept H+. Amides therefore do not react with acids and are neutral. [1]
Q1.5 — Systematic method for drawing isomers (3 marks)
Step 1 — Chain isomers: vary the carbon skeleton shape (draw all possible chain lengths and branching patterns) while keeping the functional group the same. [1]
Step 2 — Position isomers: for each skeleton from Step 1, move the functional group to every possible position along the chain. [1]
Step 3 — Functional group isomers: consider entirely different functional groups that can be formed from the same molecular formula. Verify each structure gives the correct molecular formula before naming. [1]
Q1.6 — Ester from propan-1-ol + propanoic acid (3 marks)
Ester name: propyl propanoate. [1] Condensed structural formula: CH3CH2COOCH2CH2CH3. [1] The propanoate part (CH3CH2COO–) comes from propanoic acid (the acid component); the propyl part (–OC3H7) comes from propan-1-ol (the alcohol component). The alkyl group (propyl) is written first in the name. [1]
Q2.1(a) — Isomer type: butane vs 2-methylpropane (2 marks)
Chain isomers. [1] Both have the molecular formula C4H10 and are both alkanes (same functional group class). They differ only in the carbon skeleton: butane is unbranched (CH3CH2CH2CH3); 2-methylpropane is branched (CH3CH(CH3)CH3). [1]
Q2.1(b) — Boiling point difference (3 marks)
Both compounds are non-polar alkanes, so the only intermolecular forces present are London (dispersion) forces. [1] Butane has an extended linear chain with greater surface area available for intermolecular contact → stronger London forces. [1] 2-methylpropane’s branched, more compact shape has reduced surface area → fewer and weaker London interactions → less energy required to separate molecules → lower boiling point (–12°C vs –1°C). [1]
Q3.1(a) — Two C3H6O isomers with C=O (2 marks)
Propanal (CH3CH2CHO): aldehyde, terminal C=O with H at C1. [1]
Propanone (CH3COCH3): ketone, internal C=O at C2 between two methyl groups. [1]
Q3.1(b) — Isomer type (2 marks)
Functional group isomers. [1] Same molecular formula C3H6O, but entirely different functional groups: propanal has an aldehyde (–CHO) with a terminal C=O carrying a hydrogen, while propanone has a ketone (C=O) with an internal carbonyl flanked by two carbon groups. [1]
Q3.1(c) — IUPAC suffix encodes structural difference (3 marks)
The suffix –al indicates that the compound is an aldehyde: the C=O group is terminal (at C1), and the carbonyl carbon also bears one hydrogen atom directly bonded to it (–CHO). In propanal (CH3CH2CHO), a student looks for a C=O at the end of the chain with an H on the same carbon. [1]
The suffix –one indicates that the compound is a ketone: the C=O group is internal, flanked by two other carbon atoms. In propanone (CH3COCH3), the carbonyl carbon has no H atoms — it is bonded to two methyl groups. [1]
A student distinguishes them by checking what is bonded to the C=O carbon: an H at chain end → aldehyde (–al); two carbon neighbours, no H → ketone (–one). The IUPAC suffix therefore encodes the position and environment of the C=O group unambiguously. [1]
Q4.1 — Extended response marking criteria (7 marks)
Band 5–6 features expected:
[1] States that molecular formula alone is insufficient to determine IUPAC name or functional group class, because structural isomers can share the same formula.
[1] Correctly defines structural isomers as same formula / different structural arrangement.
[1] Provides a specific named example of chain isomers (e.g. butane / 2-methylpropane, both C4H10) and explains the naming difference.
[1] Provides a specific named example of functional group isomers (e.g. propanal / propanone, both C3H6O) and explains how the functional group is different.
[1] Links different functional groups to different structural features and naming: e.g. propanal (aldehyde, –al, terminal C=O with H) vs propanone (ketone, –one, internal C=O between two carbons); OR amine reacts with HCl to form a salt (basic) while amide does not (neutral). At least one specific structural/naming difference stated that is drawn directly from lesson content.
[1] Notes that position isomers (same formula, same functional group, different position) also have different properties and names, using a specific example (e.g. propan-1-ol vs propan-2-ol, both C3H8O; different 1°/2° classification).
[1] Reaches a clear evaluative judgement — the claim is incorrect — with a concise summary statement that explains why molecular formula alone cannot identify structure, name, or properties without knowing the structural arrangement.