Covers Lessons 1–2: IUPAC nomenclature, functional group classes, and structural isomers. Physical-property and bonding questions for Lessons 3–4 are assessed in Checkpoint 2 and the Lesson 4 coverage checklist.
Questions 3 and 5 draw on aldehyde testing and ester naming, which belong to Lessons 13 and 15 and are not part of the Module 7 nomenclature requirement. Answer them for transfer, but a miss there is not a gap in Lessons 1–2.
1. What is the correct IUPAC name for CH₃CH₂CH(OH)CH₂CH₃?
2. Which statement correctly classifies 2-methylbutan-2-ol?
3. A compound has an unbranched carbon chain, molecular formula C₄H₈O, and gives a silver mirror with Tollens' reagent. What is the correct IUPAC name? Extension
4. Which pair represents functional group isomers?
5. What is the correct IUPAC name for the ester CH₃CH₂COOCH₂CH₂CH₃? Extension
MC Answers: 1-B | 2-C | 3-B | 4-C | 5-C
MC Explanations:
1-B: –OH is on C3 of a 5-carbon chain. Numbering from either end gives C3 → pentan-3-ol. Pentan-2-ol would have –OH on C2.
2-C: Structure = CH₃C(OH)(CH₃)CH₂CH₃. The C–OH carbon (C2) is bonded to three other carbons → tertiary.
3-B: A Tollens' silver mirror is given by aldehydes only, so the compound is an aldehyde. Two aldehydes share the formula C₄H₈O: butanal (unbranched) and 2-methylpropanal (branched), so the formula and Tollens' result alone do not decide between them — the stem's unbranched chain does, giving butanal. Butan-2-one (A) is a ketone and gives a negative Tollens' test. Butan-1-ol (C) is C₄H₁₀O, not C₄H₈O.
4-C: Butanal and butan-2-one share formula C₄H₈O but have different functional groups (aldehyde vs ketone) → functional group isomers. Option A = position isomers.
5-C: Cut at the O: left CH₃CH₂COO⁻ = propanoate (3C acid); right –OCH₂CH₂CH₃ = propyl (3C alcohol) → propyl propanoate.
SA1 (4 marks):
(a) 2-methylbutan-1-ol: parent chain butane (4C), methyl at C2, –OH at C1. Full structural: draw all C–C and C–H bonds explicitly with –OH on C1. Condensed: HOCH₂CH(CH₃)CH₂CH₃ [1 + 1 mark]
(b) Pentan-3-one: 5-carbon chain, C=O at C3. Full structural: CH₃–CH₂–C(=O)–CH₂–CH₃ with double bond shown. Condensed: CH₃CH₂COCH₂CH₃ [1 + 1 mark]
SA2 (4 marks):
(a) C₃H₇NO has one degree of unsaturation, so every valid structure must contain one C=O (or a ring). Possibilities: (1) Propanamide (CH₃CH₂CONH₂), amide; (2) 3-aminopropanal (H₂NCH₂CH₂CHO), a primary amine bearing an aldehyde group. [1 mark each structural formula]
(b) Dissolve a little of each in water and test with red litmus paper (or universal indicator). 3-aminopropanal: the free –NH₂ group accepts a proton from water, so the solution is weakly basic → red litmus turns blue. Propanamide: the N lone pair is delocalised into the adjacent C=O, so it is not available to accept a proton and the solution stays essentially neutral → no colour change. [1 mark reagent + observation for each; 1 mark explanation using lone pair delocalisation]
Tick when you have finished all questions.