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Score — click each question to reveal the answer
Q1
The general formula for an open-chain alkane is:
A. CnHn
B. CnH2n
C. CnH2n+2
D. CnH2n−2
✓ Answer: C
Alkane = CnH2n+2. Cycloalkane and alkene both = CnH2n. Alkyne = CnH2n−2.
Slide pp.6,47
Q2
In an alkane, what is the C–C–C bond angle and the hybridisation of each carbon?
A. 120°, sp²
B. 180°, sp
C. 109.5°, sp³
D. 90°, sp³d
✓ Answer: C
Every alkane carbon is sp³ hybridised → tetrahedral, 109.5°. Bond lengths: C–C 154 pm, C–H 110 pm.
Slide pp.8,9
Q3
Name the structure: CH₃CH(CH₃)CH(CH₃)CH₂CH(CH₃)CH₃
Slide p.25
Slide p.25
A. 2,3,5-trimethylhexane
B. 2,4,5-trimethylhexane
C. 2,3,5-trimethylheptane
D. 3,4,6-trimethylhexane
✓ Answer: A
Longest chain = 6 C (hexane). Number from the end that gives the lowest locant set: {2,3,5} vs {2,4,5} — compare at the first difference (3 vs 4) → 2,3,5 wins.
Slide p.25
Q4
Why is 2,2,5-trimethylhexane preferred to "2,5,5-trimethylhexane"?
Slide p.26
Slide p.26
A. Substituents must be in alphabetical order of locant
B. Quaternary carbons must always be C2
C. The chain numbers from whichever end is on the right of the drawing
D. The locant set {2,2,5} is lower than {2,5,5} at the first difference
✓ Answer: D
IUPAC rule: number to give the lowest set of locants at the first point of difference. {2,2,5} vs {2,5,5} — equal at 2, then 2 < 5.
Slide p.26
Q5
When alphabetising substituents in an IUPAC name, the prefix "di" is:
A. Counted only when part of a complex substituent name (e.g. 1,1-dimethylethyl)
B. Always counted as a letter
C. Never counted
D. Treated as "tri" for the purpose of ordering
✓ Answer: A
Multiplier prefixes (di, tri, tetra) attached directly to a substituent (e.g. dimethyl on the parent chain) are NOT counted for alphabetisation: "ethyl" beats "dimethyl" on "e" vs "m". BUT when "di" is embedded inside a complex substituent's own name (e.g. 1,1-dimethylethyl = tert-butyl), it IS counted.
Slide p.28
Q6
Which carbon in (CH₃)₃C–CH₂–CH₃ is tertiary (3°)?
A. Any of the three terminal methyls of the gem-dimethyl group
B. The central C of (CH₃)₃C– (bonded to 4 carbons)
C. The CH₂ in the middle (bonded to 2 carbons)
D. There is no 3° carbon — the molecule has a 4° carbon instead
✓ Answer: D
Carbon types count other carbons attached. The C of (CH₃)₃C– is bonded to 4 carbons → 4° (quaternary). The molecule (2,2-dimethylpropane / neopentane minus one Me = no, this is 2,2-dimethylbutane) has no 3° carbon at all; only 1°, 2°, 4°.
Slide p.23
Q7
In a Newman projection of ethane, the staggered conformation is favoured over the eclipsed by approximately:
A. 1 kJ/mol
B. 84 kJ/mol
C. 12 kJ/mol
D. 230 kJ/mol
✓ Answer: C — 12 kJ/mol
~12 kJ/mol energy gap = the torsional + Van der Waals strain of eclipsed ethane. At room temperature, ~99% of ethane molecules are staggered. The 84 kJ/mol figure on the slide is the thermal kinetic energy — well above the barrier → rotation is fast.
Slide pp.15–17
Q8
For butane, looking down the C2–C3 bond, the MOST stable conformation is:
A. Fully eclipsed (Me/Me at 0°)
B. Gauche (60°)
C. Eclipsed (Me/H at 120°)
D. Anti (180°)
✓ Answer: D — Anti
Anti puts the two methyl groups 180° apart — maximum distance → minimum steric repulsion → lowest energy. Anti < gauche by ~3.8 kJ/mol; anti < eclipsed by ~19 kJ/mol.
Slide p.18
Q9
The most stable conformation of cyclohexane is the chair, and a large substituent prefers:
2019 PP III.5 (corrected)
2019 PP III.5 (corrected)
A. The equatorial position (around the ring's equator)
B. The axial position (parallel to ring axis)
C. The boat conformation
D. The flagpole position
✓ Answer: A — Equatorial
Axial substituents suffer 1,3-diaxial repulsions with the axial H's on C3 and C5. Equatorial points outward → no steric clash → lower energy. The bigger the group, the stronger the preference (t-Bu is essentially 100% equatorial).
⚠ This is the most-tested trap of Ch 2. The 2019 paper inverted the rule and asked for T/F. The inverted form is always FALSE.
Slide p.55
Q10
Cyclopropane and propene can be distinguished by which reagent?
2019 PP III.8
2019 PP III.8
A. NaOH(aq)
B. Br₂/CCl₄ (bromine in carbon tetrachloride)
C. AgNO₃
D. Mg(s)
✓ Answer: B
Propene has C=C → rapidly decolourises Br₂ (electrophilic addition). Cyclopropane is saturated → no reaction at room temperature. So Br₂ turns from red-brown to colourless with propene but stays red-brown with cyclopropane.
⚠ True even though both molecules share the same formula C₃H₆ (constitutional isomers): cyclopropane is saturated, propene is unsaturated.
Q11
In the radical chlorination of methane, the role of UV light is to:
A. Excite the methane C–H bonds
B. Convert chloromethane back to methane
C. Homolytically cleave Cl₂ to give 2 Cl· radicals (initiation)
D. Provide heat only — it is not chemically active
✓ Answer: C
UV photons match the Cl–Cl bond dissociation energy and break it homolytically into two Cl radicals. Those radicals then start the propagation chain.
Slide p.44
Q12
The two propagation steps in CH₄ + Cl₂ chlorination are:
A. Cl· + Cl· → Cl₂ ; CH₃· + CH₃· → C₂H₆
B. Cl· + CH₄ → HCl + CH₃· ; CH₃· + Cl₂ → CH₃Cl + Cl·
C. CH₄ + Cl₂ → CH₃Cl + HCl
D. Cl₂ → 2 Cl· ; CH₃· → CH₂ + H·
✓ Answer: B
Propagation = each step consumes one radical and regenerates another, sustaining the chain. A = termination. D = initiation (only the first half).
Slide pp.44–45
Q13
Which alkyl radical is MOST stable?
A. CH₃· (methyl)
B. CH₃CH₂· (1° ethyl)
C. (CH₃)₂CH· (2° isopropyl)
D. (CH₃)₃C· (3° tert-butyl)
✓ Answer: D
3° > 2° > 1° > methyl. More substituted radicals are stabilised by σ–p hyperconjugation (overlap of adjacent C–H sigma orbitals with the radical's p orbital). BDE evidence: 3° C–H ~96.5 kcal/mol < 1° C–H ~101 kcal/mol — 3° forms more easily.
Slide pp.40,42
Q14
An alkyl radical (e.g. CH₃·) has approximately what shape and hybridisation?
A. Tetrahedral, sp³
B. Trigonal planar, sp² (unpaired e⁻ in p orbital)
C. Linear, sp
D. Bent, sp³ with two lone pairs
✓ Answer: B
ESR spectroscopy of CH₃· shows a nearly planar geometry with the unpaired electron in a p orbital perpendicular to the plane — same as a carbocation. Bond angle ~120°.
Slide p.41
Q15
In a homologous series of straight-chain alkanes (CH₄ → C₂H₆ → C₃H₈ → ...), the boiling point:
A. Decreases with chain length
B. Stays constant
C. Increases with chain length (larger London forces)
D. Increases then decreases above C12
✓ Answer: C
Larger alkanes have larger electron clouds → stronger transient-dipole (London dispersion) forces → more energy needed to separate molecules → higher b.p. Even-C members show slightly higher m.p. than the odd-C neighbours because of denser crystal packing.
Slide pp.30–31
Q16
Which intermolecular force dominates between two pentane molecules?
A. Hydrogen bonding
B. Dipole–dipole interaction
C. Ionic Coulombic attraction
D. London dispersion (induced dipole–induced dipole)
✓ Answer: D
Pentane is nonpolar (all C–C and C–H bonds essentially nonpolar). The only intermolecular force available is London dispersion.
Slide p.32
Q17
Baeyer strain theory (1885) correctly predicts high strain for which ring?
A. Cyclopropane
B. Cyclohexane
C. Cyclopentane
D. Cycloheptane
✓ Answer: A
Cyclopropane has C–C–C bond angles of 60° — far below the natural 109.5° — giving severe bond-angle strain. All H's are also eclipsed → torsional strain. Baeyer's theory failed for ≥6-membered rings because he wrongly assumed they were planar; cyclohexane is actually strain-free in the chair.
Slide pp.51–53
Q18
cis-1,2-dimethylcyclopropane and trans-1,2-dimethylcyclopropane are best described as:
A. The same compound
B. Stereoisomers (cis/trans)
C. Constitutional (structural) isomers
D. Enantiomers
✓ Answer: B
Same connectivity (both have methyls on adjacent carbons) but different 3-D arrangement — cis has both methyls on the same face of the ring, trans has them on opposite faces. Because the ring blocks rotation, they don't interconvert → isolable stereoisomers.
Slide p.49
Q19
Identify the carbon types in 2-methylpentane: how many 1° / 2° / 3° / 4° carbons are there?
A. 2 × 1°, 3 × 2°, 1 × 3°, 0 × 4°
B. 4 × 1°, 1 × 2°, 0 × 3°, 1 × 4°
C. 3 × 1°, 2 × 2°, 1 × 3°, 0 × 4°
D. 3 × 1°, 1 × 2°, 1 × 3°, 1 × 4°
✓ Answer: C
2-methylpentane = CH₃–CH(CH₃)–CH₂–CH₂–CH₃. C1, C5, and the methyl on C2 are all terminal CH₃ (1° × 3). C3 and C4 are middle CH₂ (2° × 2). C2 is bonded to 3 carbons (3° × 1). No quaternary carbon. Total 6 carbons. ✓
Slide p.23
Q20
In the most stable chair of trans-1,4-dimethylcyclohexane:
A. Both methyls are axial
B. One methyl is axial, one equatorial
C. The boat is more stable than the chair
D. Both methyls are equatorial
✓ Answer: D — Both equatorial
In trans-1,4-disubstituted cyclohexane, both methyl groups can be in the same orientation (both axial OR both equatorial) in either chair. The chair with both methyls EQUATORIAL is favoured (no 1,3-diaxial repulsions).
⚠ Contrast with cis-1,4-: that one is locked into one-axial / one-equatorial in either chair. With cis-1,3- or trans-1,2-: both equatorial possible.
D1Alkane & Cycloalkane+
An alkane is a saturated open-chain hydrocarbon containing only C–C and C–H single bonds; general formula CnH2n+2. A cycloalkane is the corresponding saturated ring; general formula CnH2n (closing the ring "costs" two H's). Every carbon in both is sp³ hybridised → tetrahedral, 109.5°, C–C 154 pm, C–H 110 pm.
Slide pp.6–9, 48
D2Conformation & Newman projection+
A conformation is one of the infinite spatial arrangements of a molecule's atoms produced by rotation about single (σ) bonds. Conformations interconvert rapidly and are not isolable. A Newman projection is a 2-D representation of a molecule viewed along one C–C bond: the front C is a dot (3 bonds radiating to substituents), the back C is a large circle (3 bonds protruding from its edge). Lets you see the torsion angle between front and back substituents.
Slide pp.13–15
D3Eclipsed vs Staggered (ethane); Anti vs Gauche (butane)+
For ethane (Newman down C–C): eclipsed = front and back H's at 0° offset, higher energy by ~12 kJ/mol (torsional + Van der Waals strain); staggered = 60° offset, energy minimum, ~99% population. For butane (Newman down C2–C3): anti = methyls 180° apart, most stable; gauche = methyls 60° apart, ~3.8 kJ/mol higher (steric); eclipsed conformers ~16–19 kJ/mol above anti.
Slide pp.13–18
D4Carbon types: 1° / 2° / 3° / 4°+
A carbon's "degree" is the number of other carbons it is bonded to: primary (1°) = bonded to 1 C (terminal CH₃); secondary (2°) = bonded to 2 C (chain CH₂); tertiary (3°) = bonded to 3 C (CH on a branch point); quaternary (4°) = bonded to 4 C (no H). This classification predicts radical stability (3° > 2° > 1° > methyl), governs the Lucas test in Ch 4 (alcohols), and the rate of SN1/SN2 in haloalkanes.
Slide p.23
D5Radical halogenation & chain mechanism+
A reaction in which a C–H bond of an alkane is replaced by a C–X bond (X = Cl, Br) under UV light or 300 °C. Three-step radical chain mechanism: (1) Initiation — UV/heat homolyses X–X to give 2 X·. (2) Propagation — X· abstracts H from R–H giving HX + R·; then R· abstracts X from X₂ giving R–X + X· (regenerates X· → chain continues). (3) Termination — two radicals combine (X·+X·, X·+R·, R·+R·) and the chain stops. Selectivity follows radical stability: 3° > 2° > 1° > methyl (hyperconjugation).
Slide pp.36–45
D6Chair conformation, axial vs equatorial, 1,3-diaxial strain+
The chair is the most stable conformation of cyclohexane: all C–C–C angles ≈ 111° (essentially strain-free), all adjacent C–H bonds staggered (no torsional strain). Each ring carbon carries one axial H (perpendicular to ring plane, parallel to C₃ symmetry axis) and one equatorial H (around the ring's equator, ~109.5° from axial). Substituents prefer the equatorial position because axial substituents suffer 1,3-diaxial repulsion with the axial H's on C3 and C5. The bigger the substituent, the stronger the preference (Me < Et < iPr < tBu).
Slide pp.54–55
E1
State the six IUPAC rules for naming alkanes. Apply them to name (CH₃)₂CH–CH₂–CH(CH₂CH₃)–CH₂–CH₂–CH₃.
10 marksThe six rules
- Find the longest continuous chain of carbons — this is the parent.
- Name the parent as (C-count prefix + "-ane").
- Number the chain to give substituents the lowest set of locants.
- Write each substituent with its locant in front: e.g. "2-methyl".
- List substituents alphabetically. Multiplier prefixes di/tri/tetra are NOT counted for alphabetisation when directly attached.
- For a ring, prefix "cyclo" to the parent name.
Applied to the given structure
Longest chain = 7 carbons (heptane). Number from the end that gives lowest locants. The isopropyl group sits on C2 and the ethyl on C4 with this numbering. Substituents alphabetised: "ethyl" before "(1-methylethyl)" (i.e. isopropyl) on "e" vs "m" — or use the common "isopropyl" name with i for alphabetising. Final name: 4-ethyl-2-(1-methylethyl)heptane — equivalently 4-ethyl-2-isopropylheptane.
Marking (10): rules 1–6 stated correctly (6) · longest-chain identified (1) · correct numbering / lowest locants (1) · alphabetisation rule applied (1) · correct final name (1).
E2
Using a Newman projection looking down the C2–C3 bond of butane, describe the four limiting conformations (0°, 60°, 120°, 180°). Rank them in order of increasing stability and explain why.
8 marksThe four conformations
- 0° — fully eclipsed (Me/Me): the two methyls superimpose. Highest energy (~19 kJ/mol above anti) due to both torsional strain (eclipsed C–H bonds) and severe steric (Van der Waals) clash between methyls.
- 60° — gauche: methyls staggered but only 60° apart. ~3.8 kJ/mol above anti due to residual steric repulsion between gauche methyls.
- 120° — eclipsed (Me/H): a methyl eclipses a hydrogen on the back carbon. ~16 kJ/mol above anti — torsional + some steric.
- 180° — anti: methyls 180° apart, maximum distance. Most stable: staggered + no Me/Me clash.
Stability order (least → most stable)
Fully eclipsed (0°) < Me/H eclipsed (120°) < gauche (60°) < anti (180°).
Why
Two factors govern: (1) torsional strain — eclipsed C–H bonds repel; staggered relieves this. (2) steric (Van der Waals) strain between the methyl groups — minimised when methyls are anti, worst when fully eclipsed.
Marking (8): four conformations correctly described with angles (4) · correct stability ranking (2) · explanation invoking torsional + steric strain (2).
E3
Write a complete mechanism for the chlorination of methane (CH₄ + Cl₂ → CH₃Cl + HCl) under UV light. Label initiation, propagation and termination steps.
8 marksConditions
Methane + Cl₂ mixed and either heated to 300 °C or irradiated with UV (hν).
1. Initiation
UV homolyses the Cl–Cl bond: Cl₂ —hν→ 2 Cl·. Each fragment leaves with one of the two bonding electrons (homolytic cleavage).
2. Propagation (two steps form the chain)
- Step 2a (H abstraction): Cl· + CH₄ → HCl + CH₃·
- Step 2b (Cl abstraction): CH₃· + Cl₂ → CH₃Cl + Cl·
The regenerated Cl· restarts step 2a → the chain propagates many cycles before termination. Net: CH₄ + Cl₂ → CH₃Cl + HCl.
3. Termination — any radical–radical recombination stops the chain
- Cl· + Cl· → Cl₂
- CH₃· + Cl· → CH₃Cl
- CH₃· + CH₃· → CH₃CH₃ (ethane)
Marking (8): initiation with UV homolysis labelled (2) · both propagation steps balanced with the chain-carrier shown (3) · at least two termination steps (2) · net equation (1).
E4
Why is the chair the most stable conformation of cyclohexane, and why does a large substituent prefer the equatorial position over axial? Illustrate with tert-butylcyclohexane.
7 marksWhy chair is most stable
- All six C–C–C bond angles ≈ 111°, essentially the strain-free sp³ value (109.5°) → no bond-angle strain.
- All neighbouring C–H bonds are staggered (when viewed along any C–C bond) → no torsional strain.
- The alternative boat is ~30 kJ/mol less stable: flagpole H's eclipsed and forced into Van der Waals contact → torsional + steric strain.
Axial vs equatorial
In the chair, each carbon carries one axial and one equatorial substituent. An axial substituent points up (or down) on the same face as the axial H's at C3 and C5 — producing 1,3-diaxial repulsion. An equatorial substituent points away from the ring → no steric clash. Therefore equatorial is favoured.
tert-Butylcyclohexane
The two chairs (t-Bu axial vs t-Bu equatorial) interconvert by ring-flip. The energy difference is so large (~22 kJ/mol; t-Bu has three methyls clashing with C3-axial and C5-axial Hs) that more than 99.99% of the molecules adopt the chair with t-Bu equatorial. tert-Butyl effectively "locks" the chair.
Marking (7): chair stability from angle + staggered (2) · boat instability (1) · 1,3-diaxial concept (2) · t-Bu equatorial illustration (2).
E5
Compare and explain the relative stabilities of methyl, primary, secondary and tertiary alkyl radicals. Reference experimental BDE data and the concept of hyperconjugation.
7 marksStability order
3° > 2° > 1° > methyl. The more substituted the radical, the more stable.
Evidence from bond-dissociation energy (BDE)
- 1° C–H BDE ≈ 101 kcal/mol (forms 1° radical).
- 2° C–H BDE ≈ 98.5 kcal/mol.
- 3° C–H BDE ≈ 96.5 kcal/mol.
Less energy is needed to remove an H from a more substituted carbon → that resulting radical is more stable.
Hyperconjugation explains it
An alkyl radical is sp² hybridised with the unpaired electron in a p orbital perpendicular to the molecular plane. Filled sp³ C–H σ orbitals on neighbouring carbons can donate electron density into this partly-filled p orbital (σ–p hyperconjugation). The more C–H bonds adjacent to the radical centre, the more donors are available → greater stabilisation. A tertiary radical has 9 C–H σ bonds available; a methyl radical has 0.
Consequence for halogenation selectivity
In radical chlorination/bromination of a branched alkane, the H most easily removed is the one whose loss gives the most stable radical — usually the 3° H. This is why tert-butyl chloride forms preferentially when 2-methylpropane is brominated.
Marking (7): correct stability order (1) · BDE values stated and interpreted (3) · hyperconjugation mechanism (2) · consequence for selectivity (1).