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Score — click each question to reveal the answer
Q1
Wöhler's 1828 synthesis of urea from ammonium cyanate is historically significant because it:
2019 PP III.1 (paraphrased)
2019 PP III.1 (paraphrased)
A. Showed an organic compound could be made from an inorganic salt without any living tissue
B. Proved the Vital Force Theory was correct
C. First isolated urea from urine
D. Demonstrated that carbon is essential to all life
✓ Answer: A
Wöhler heated NH₄OCN (an inorganic salt) and obtained urea, an unmistakably organic substance — disproving Vital Force Theory.
⚠ A is the exact opposite of the truth. C is a trap — urea had been isolated from urine in 1773; Wöhler's contribution was the lab synthesis, not the isolation.
McMurry 8e Ch 12 · Slide p.7
Q2
Orbital hybridization is best described as:
2019 PP III.2 (paraphrased)
2019 PP III.2 (paraphrased)
A. Sharing of electrons between two different atoms
B. The mixing of atomic orbitals of similar energy on the same atom to form new equivalent orbitals with different shape and direction
C. The combining of orbitals from two adjacent atoms to make a covalent bond
D. Electron promotion from s to p without any orbital reshaping
✓ Answer: B
Hybridization recombines s and p atomic orbitals of the same atom into a new equivalent set with different shape, energy and direction.
⚠ C confuses hybridization with bond formation; in fact hybrid orbitals are formed before bonding.
Slide pp.21–22
Q3
The shape and bond angle around the carbon in methane (CH₄) are:
A. Trigonal planar, 120°
B. Linear, 180°
C. Tetrahedral, 109.5°
D. Square planar, 90°
✓ Answer: C
Carbon in CH₄ is sp³ hybridised → four equivalent hybrid orbitals point to the corners of a tetrahedron at 109.5° (VSEPR minimises repulsion).
Slide pp.22–24
Q4
Which statement about a carbon–carbon double bond (C=C) is TRUE?
2019 PP III.4 inverse
2019 PP III.4 inverse
A. It consists of two equivalent π bonds
B. It consists of one σ + one π bond
C. Both carbons are sp³ hybridised
D. It rotates freely at room temperature
✓ Answer: B
A C=C double bond is 1σ + 1π. Both carbons are sp² hybridised. The π bond locks rotation, which is why cis/trans isomers exist (Ch 3).
⚠ A is the false statement from 2019 PP. Two π bonds belong to the triple bond.
Slide p.30
Q5
Which molecule has polar bonds but is overall NONPOLAR?
A. H₂O
B. CHCl₃
C. HCl
D. CCl₄
✓ Answer: D — CCl₄
CCl₄ is tetrahedral and symmetric — the four C–Cl bond dipoles cancel, giving zero net dipole moment.
⚠ CHCl₃ (B) has the same tetrahedral geometry but the one C–H is much less polar than the three C–Cl, so the dipoles do not cancel → CHCl₃ is polar.
Slide pp.16–17
Q6
Which of the following carbon–carbon bonds is the SHORTEST?
A. C–C in ethane
B. C=C in ethene
C. C≡C in ethyne
D. C–C in cyclohexane
✓ Answer: C — C≡C in ethyne
Triple bond is sp–sp, with 1σ + 2π — the strongest and shortest carbon–carbon bond (~120 pm). Trend: C–C 154 > C=C 134 > C≡C 120 pm.
Slide p.30
Q7
Which is NOT classified as an organic compound, despite containing carbon?
A. Urea, (NH₂)₂C=O
B. Ethanol, CH₃CH₂OH
C. Carbonic acid, H₂CO₃
D. Acetic acid, CH₃COOH
✓ Answer: C — H₂CO₃
CO, CO₂, carbonic acid, carbonates, bicarbonates, cyanides and cyanates contain carbon but are classed as inorganic by convention.
Slide p.8
Q8
In which of the following molecules is the central carbon sp hybridised?
A. Methane, CH₄
B. Ethene, CH₂=CH₂
C. Formaldehyde, HCHO
D. Hydrogen cyanide, HCN
✓ Answer: D — HCN
HCN is linear (H–C≡N); the carbon has 2 σ-bonded atoms and forms a triple bond → sp hybridisation, 180°.
⚠ B and C have C=X double bonds → sp² (120°). A is sp³ (109.5°).
Slide pp.26–27
Q9
Which family of organic compounds has the general formula CnH2n+2?
A. Alkanes
B. Alkenes
C. Alkynes
D. Cycloalkanes
✓ Answer: A — Alkanes
Alkanes: CnH2n+2. Alkenes / cycloalkanes: CnH2n. Alkynes: CnH2n-2.
Slide p.43
Q10
The functional group –COOH defines which family?
A. Aldehydes
B. Carboxylic acids
C. Amides
D. Esters
✓ Answer: B — Carboxylic acids
–COOH = carboxyl group = carboxylic acid (R–COOH). Aldehyde = –CHO; amide = –CONH₂; ester = –COO–R′.
Slide p.43 Table 1.4
Q11
How many structural isomers does C₄H₁₀ have?
A. 1
B. 3
C. 2
D. 4
✓ Answer: C — 2
n-Butane (CH₃CH₂CH₂CH₃) and isobutane / 2-methylpropane ((CH₃)₃CH). Two chain isomers.
Slide p.45 (textbook practice)
Q12
Ethanol (CH₃CH₂OH) and dimethyl ether (CH₃OCH₃) are an example of:
A. Chain isomers
B. Positional isomers
C. Functional-group isomers
D. Geometric isomers
✓ Answer: C — Functional-group isomers
Same molecular formula C₂H₆O, but different functional groups — one is an alcohol (–OH), the other an ether (–O–).
Slide p.45
Q13
Why is single (σ) bond rotation free, while a double (π) bond's rotation is restricted?
A. σ bond overlap is end-to-end and symmetric about the bond axis; π overlap is sideways and would break on rotation
B. σ bonds are weaker than π bonds, so they break and re-form
C. σ bonds have lower bond energy
D. Single bonds are longer, so atoms move more freely
✓ Answer: A
σ overlap is cylindrically symmetric about the internuclear axis, so rotation doesn't reduce it. π overlap is above and below the plane — rotating one C 90° would tear the π bond apart (energy cost ~270 kJ/mol).
Slide p.33
Q14
The "like-dissolves-like" rule predicts that:
A. Ethanol (polar) dissolves well in hexane (nonpolar)
B. Glucose (polar) dissolves well in water (polar)
C. NaCl (ionic) dissolves well in benzene (nonpolar)
D. Cholesterol (nonpolar) dissolves well in water (polar)
✓ Answer: B — Glucose in water
Polar dissolves polar, nonpolar dissolves nonpolar. Glucose has 5 –OH groups; it hydrogen-bonds extensively with water.
Slide p.11
Q15
In a bond-line drawing, an unmarked vertex represents:
A. A hydrogen atom
B. Any heteroatom
C. A carbon atom with implicit hydrogens to fill the 4-bond valence
D. A lone pair
✓ Answer: C
In bond-line (skeletal) structures: each vertex/end-point is a carbon; hydrogens attached to carbon are not drawn (count: 4 minus the number of bonds drawn at that vertex); heteroatoms (O, N, X) are written explicitly.
Slide p.38
Q16
The most stable conformation of ethane is:
A. Eclipsed
B. Gauche
C. Anti-periplanar (does not apply to ethane)
D. Staggered
✓ Answer: D — Staggered
Staggered: H's on adjacent carbons offset by 60° → minimum torsional strain. Eclipsed: H's aligned → higher energy by ~12 kJ/mol.
Slide p.33
Q17
A homologous series is a series of compounds that:
A. Share the same functional group and successive members differ by CH₂
B. Are all geometric isomers of one another
C. Have the same molecular formula but different connectivity
D. Are stereoisomers with identical physical properties
✓ Answer: A
Homologs share a functional group + general formula; each next member adds one CH₂. e.g. methanol → ethanol → 1-propanol.
Slide p.44
Q18
Which atom has the HIGHEST electronegativity on the Pauling scale (relevant to organic chemistry)?
A. Carbon (2.5)
B. Nitrogen (3.0)
C. Oxygen (3.5)
D. Fluorine (4.0)
✓ Answer: D — Fluorine, 4.0
Pauling-scale order: F (4.0) > O (3.5) > N, Cl (3.0) > Br (2.8) > C, S (2.5) > H (2.1). This trend governs every polar-bond prediction in OC.
McMurry 8e Ch 12
Q19
Which carbon valence rule is correct for stable, neutral organic compounds?
A. C makes 2 bonds, O makes 1, N makes 3
B. C makes 4 bonds, O makes 2, N makes 3, H/X make 1
C. C makes 3 bonds, O makes 3, N makes 1
D. C makes 4 bonds, O makes 4, N makes 4
✓ Answer: B
Standard valences in neutral organics: C = 4, O/S = 2, N/P = 3, H/X = 1. This is the rule for completing condensed and bond-line structures.
Slide p.37
Q20
Which functional group accounts for the peptide bond between two amino acids?
A. Ester
B. Anhydride
C. Carboxylic acid
D. Amide
✓ Answer: D — Amide
The peptide bond is an amide (–CO–NH–) formed between the α-COOH of one amino acid and the α-NH₂ of the next, with loss of H₂O. Same functional group is the β-lactam in penicillin.
McMurry 8e Ch 12 + Ch 18
D1Organic compound+
A carbon-containing compound, with conventional exceptions: CO, CO₂, carbonic acid (H₂CO₃), carbonates (MCO₃), bicarbonates (MHCO₃), cyanides (MCN) and cyanates (MOCN) — all of which contain C but are classed as inorganic. Typical organics are built from C, H, O, N, S, P and halogens; held together by covalent bonds; have low m.p./b.p., low density, low water solubility and good solubility in organic solvents.
McMurry & Ballantine 8e Ch 12 · Slide pp.8–11
D2Covalent bond & electronegativity+
A covalent bond is a chemical bond formed by the sharing of one or more electron pairs between two atoms. Electronegativity (χ) is the tendency of an atom to attract the bonding electrons toward itself. When Δχ ≈ 0, the bond is nonpolar covalent (C–C, C–H). When Δχ is appreciable (~0.5–1.9), the bond is polar covalent with a dipole moment (C–O, C–Cl, O–H). When Δχ > 1.9, the bond is essentially ionic (Na–Cl).
Slide pp.13–15
D3Orbital hybridization (sp³, sp², sp)+
The mathematical recombination of atomic orbitals of similar energy on the same atom to give a new set of equivalent hybrid orbitals with different shape, energy and direction. Total orbital count is conserved. For carbon: sp³ (1s + 3p → 4 hybrids, tetrahedral, 109.5°, e.g. CH₄); sp² (1s + 2p → 3 hybrids + 1 leftover p, trigonal planar, 120°, e.g. C=C, C=O); sp (1s + 1p → 2 hybrids + 2 leftover p, linear, 180°, e.g. C≡C, HCN).
Slide pp.21–27 · Pauling 1931
D4Functional group+
A specific atom or small group of atoms embedded in an organic molecule that undergoes characteristic chemical reactions and therefore determines the molecule's main chemical behaviour. The carbon skeleton is the scaffolding; the functional group is the reactive site. Compounds with the same functional group belong to the same family and share similar reactivity (e.g. all alcohols R–OH react with Na to give H₂).
Slide p.41 · Table 1.4
D5Homologous series+
A family of compounds with the same functional group and the same general formula, in which successive members differ by one CH₂ unit. Adjacent members are homologs. Homologs have similar chemical reactivity (same functional group) but their physical properties (b.p., m.p., density) change gradually with chain length. Example: CH₃OH → C₂H₅OH → C₃H₇OH → C₄H₉OH.
Slide p.44
D6Structural (constitutional) isomerism+
The existence of two or more compounds with the same molecular formula but different connectivity of atoms (different order in which atoms are bonded). Three subtypes: (1) Chain — different carbon skeletons (n-butane vs isobutane, both C₄H₁₀). (2) Positional — same skeleton, functional group at a different position (1-propanol vs 2-propanol). (3) Functional-group — same atoms but different functional groups (ethanol vs dimethyl ether, both C₂H₆O). Stereoisomers (different 3-D arrangement, same connectivity) are introduced in Chapter 3.
Slide p.45 · McMurry 8e Ch 12
E1
State the differences between organic and inorganic compounds with respect to elemental composition, bonding type, melting point, solubility and reactivity. Explain why these differences arise.
8 marksFive point-of-difference table
- Elements: organics built mainly from C, H, plus O, N, S, P, halogens; inorganics involve most periodic-table elements.
- Bonding: organics are covalent (electron sharing); inorganics are often ionic (electron transfer, e.g. NaCl).
- Melting / boiling points: organics low (most < 300 °C); inorganics often very high (NaCl 801 °C).
- Solubility: organics low solubility in water, good in organic solvents (like-dissolves-like); ionic inorganics often water-soluble.
- Reactivity: organics are combustible, react slowly, give side products and low yields; ionic inorganic reactions are usually fast and clean.
Why these differences arise
All five differences trace back to bond type. Covalent bonds give discrete molecules held together internally by strong bonds but to other molecules only by weak intermolecular forces → low m.p./b.p. Ionic compounds form 3-D lattices of charged species → high m.p./b.p. and water solubility. Organic reactions involve breaking strong covalent bonds, so they are slower; ionic reactions are simply ion exchange.
Marking (8): 5 differences listed correctly (5) · bonding-type-as-root-cause explanation (2) · named example (1).
E2
Describe the three hybridization states of carbon (sp³, sp², sp). For each state, state the geometry, bond angle, number of σ and π bonds at the carbon, and give one example.
9 markssp³ hybridization
- Mixing: 1 × 2s + 3 × 2p → 4 equivalent hybrid orbitals.
- Geometry: tetrahedral. Bond angle: 109.5°.
- σ : 4. π : 0.
- Example: methane CH₄, all alkanes, R–OH carbon, R–NH₂ carbon.
sp² hybridization
- Mixing: 1 × 2s + 2 × 2p → 3 hybrid orbitals + 1 unhybridised p.
- Geometry: trigonal planar. Bond angle: 120°.
- σ : 3. π : 1 (formed by the leftover p sideways overlap).
- Example: ethene CH₂=CH₂, the carbonyl carbon of aldehydes/ketones, benzene.
sp hybridization
- Mixing: 1 × 2s + 1 × 2p → 2 hybrid orbitals + 2 unhybridised p.
- Geometry: linear. Bond angle: 180°.
- σ : 2. π : 2 (two perpendicular π bonds from the two leftover p's).
- Example: ethyne HC≡CH, HCN, CO₂.
Why VSEPR predicts these geometries
Electron domains repel one another, so the most stable arrangement maximises the angle between them: 4 domains → tetrahedron; 3 → trigonal planar; 2 → linear.
Marking (9): three correct geometries + angles (3) · σ/π count for each (3) · one valid example each (1.5) · VSEPR reasoning (1.5).
E3
Explain the difference between a polar bond and a polar molecule, using CCl₄, CHCl₃ and H₂O as examples.
6 marksPolar bond
A covalent bond between two atoms of different electronegativities. The more electronegative atom acquires partial negative charge (δ−), the other partial positive (δ+). The bond has a dipole moment (a vector pointing from δ+ to δ−).
Polar molecule
A molecule whose net dipole moment is non-zero. Polar bonds are necessary but not sufficient — if all bond dipoles cancel by symmetry, the molecule is nonpolar.
Three worked examples
- CCl₄ — tetrahedral, four equal C–Cl bond dipoles pointing to the corners of a tetrahedron; vector sum is zero → polar bonds, nonpolar molecule.
- CHCl₃ — tetrahedral but unsymmetric (3 Cl + 1 H); the C–H bond is much less polar than C–Cl, so the three Cl dipoles do not cancel the lone C–H direction → net dipole → polar molecule.
- H₂O — bent (104.5°); two O–H dipoles add to give a strong net dipole → polar molecule.
Marking (6): definition of polar bond + electronegativity (1.5) · definition of polar molecule + role of geometry/symmetry (1.5) · three correct examples with reasoning (3).
E4
Fill in the following functional-group table for the named families: alkane, alkene, alkyne, alcohol, aldehyde, ketone, carboxylic acid, amine, amide, ester. For each, state the functional group and one example.
10 marksAnswer table
- Alkane — no functional group (only C–C and C–H single bonds) — ethane CH₃CH₃.
- Alkene — C=C — ethene CH₂=CH₂.
- Alkyne — C≡C — ethyne HC≡CH.
- Alcohol — –OH on sp³ C — ethanol CH₃CH₂OH.
- Aldehyde — –CHO (terminal carbonyl) — acetaldehyde CH₃CHO.
- Ketone — >C=O (carbonyl flanked by two C's) — acetone CH₃COCH₃.
- Carboxylic acid — –COOH — acetic acid CH₃COOH.
- Amine — –NH₂ (1°), >NH (2°), >N– (3°) — methylamine CH₃NH₂.
- Amide — –CO–NH₂ — acetamide CH₃CONH₂.
- Ester — –COO–R′ — ethyl acetate CH₃COOCH₂CH₃.
Marking (10): 1 mark per family (correct functional group symbol + valid example).
E5
Define isomerism. Distinguish chain, positional and functional-group isomerism, and give a worked example of each.
7 marksDefinition
Isomers are compounds with the same molecular formula but a different arrangement of atoms. Isomerism is the phenomenon. Structural (constitutional) isomers differ in which atom is connected to which; stereoisomers (covered in Chapter 3+) differ only in the 3-D arrangement.
Chain (skeletal) isomerism
Carbon skeleton is connected differently. Example: C₄H₁₀ — n-butane CH₃CH₂CH₂CH₃ (straight) and isobutane / 2-methylpropane (CH₃)₃CH (branched).
Positional isomerism
Functional group at a different position on the same skeleton. Example: C₃H₈O — 1-propanol CH₃CH₂CH₂OH (–OH on C1) and 2-propanol CH₃CH(OH)CH₃ (–OH on C2).
Functional-group isomerism
Same atoms, different functional group entirely. Example: C₂H₆O — ethanol CH₃CH₂OH (alcohol) and dimethyl ether CH₃OCH₃ (ether).
Marking (7): definition of isomers (1) · chain isomerism + example (2) · positional + example (2) · functional-group + example (2).