SECTION A (Answer ALL)
1. Identify the intermolecular forces holding molecules together in:
-
(a) Methyl bromide, CH₃Br
-
(b) Butane, C₄H₁₀
-
© Hexanol, C₆H₁₃OH (5 marks)
2. Write the general formula and molecular formula of each constituent of typical alkane mixtures in petrol (C6 to C10). (5 marks)
3. State whether each compound has cis-trans isomers:
-
(a) 1-bromo-1,2-dichloroethene
-
(b) but-1-ene
-
© 1,1-dichloropropene
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(d) pent-2-ene
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(e) buta-1,3-diene (5 marks)
4. Name the two types of cracking of petroleum fractions and state the conditions required for each. (5 marks)
5. A compound contains 29.3% carbon, 5.7% hydrogen, and 65% bromine. Calculate the empirical formula. [H=1; C=12; Br=80] (5 marks)
6. (a) What is enolization? (1 mark)
(b) Draw the keto-enol isomeric forms of ethanal and propanone. (4 marks)
7. (a) Give the general molecular formula of dialkanoic acids. (1 mark)
(b) Name and draw structures of the first two members of the dialkanoic acid series. (4 marks)
8. (a) What are thermosetting plastics?
(b) Give the structure and IUPAC name of the monomer of:
-
(i) −(CH₂−CH(CH₃))ₙ−
-
(ii) −(CH₂−CCl₂)ₙ− (5 marks)
9. (a) Name the types of isomerism shown by alkenes with four or more carbon atoms. (3 marks)
(b) List two general laboratory methods for preparing alkenes. (2 marks)
10. (a) Name the type of isomerism in disubstituted benzenes. (1 mark)
(b) Draw and name the four aromatic structural isomers of C₈H₁₀. (4 marks)
SECTION B (Attempt ANY TWO)
11. (a) Identify the functional group(s) and series for each compound:
(i) CH₃CH₂CH₂CHO (ii) CH₃CH₂NH₂ (iii) CH₃CH₂CH₂Cl (iv) CH₃CH₂CH₂CH₂OH (v) CH₂=CHCH₂CH₂Br (vi) CH₃CH₂COCH₃ (vii) CH₃COOH (viii) CH₃CH₂OCH₃ (ix) CH₃CH₂COOR (18 marks)
(b) Classify each reaction as addition, substitution, elimination, or hydrolysis:
-
(i) CH₃CH₂CH(OH)CH₃ → CH₃CH=CHCH₃ + H₂O
-
(ii) CH₃CH₂COOCH₃ + H₂O → CH₃CH₂COOH + CH₃OH
-
(iii) CH₃CH=CH−CHCH₃ + H₂ → CH₃CH₂CH₂CH₂CH₃
-
(iv) CH₃CH₂CH(Br)CH₃ + OH⁻ → CH₃CH₂CH(OH)CH₃ + Br⁻ (8 marks)
12. (a) Give structures and names of products of adding to 2-methylbut-2-ene:
(i) Hydrogen (ii) Bromine (iii) Hydrogen chloride (iv) Aqueous KOH (12 marks)
(b) Which class of alkanols is produced by reduction of:
(i) Alkanoic acids (ii) Alkanones (iii) Alkanoates (iv) Alkanal (v) Alkanoic anhydride (10 marks)
© Give the structure of likely products of the reaction between methyl chloride and ammonia. (3 marks)
13. (a) (i) What are isomers?
(ii) Draw and name the alkene isomers with molecular formula C₅H₁₀. (11 marks)
(b) For HCl reacting with 2-methylpropene:
(i) Draw the two possible carbocation intermediates
(ii) Explain which carbocation is more stable
(iii) Give the formula and name of the major product
(iv) Give the formula and name of the minor product (8 marks)
© Which IUPAC names are correct?
(i) 2,3-dimethylpentane and 2-methyl-3-methylpentane
(ii) 3-methyl-3-ethylheptane and 3-ethyl-3-methylheptane
(iii) 2,3-dimethyl-4-ethylheptane and 4-ethyl-2,3-dimethylheptane (6 marks)
14. (a) (i) Draw the two straight-chain isomeric alkanols with molecular formula C₄H₁₀O. (2 marks)
(ii) Draw oxidation products when both alkanols are oxidised by acidified potassium heptaoxochromate(VI); write equations using [O]. (6 marks)
(iii) Draw and name the alkanol with formula C₄H₁₀O that is resistant to oxidation by acidified K₂Cr₂O₇. (2 marks)
(b) State possible elimination products from:
(i) 1-bromobutane (ii) 2-bromobutane (iii) 1-bromo-2-methylpropane (iv) 2-bromo-2-methylpropane (8 marks)
© Ethanoic acid can be formed from bromoethane in two steps. Give reagents, conditions, and the mechanism name for step one, and an equation for step two. (8 marks)
IJMBE 2024 Chemistry Paper II: Organic — Full Solutions
SECTION A
Question 1 — Intermolecular Forces
(a) Methyl bromide, CH₃Br
Dipole-dipole forces (and London/van der Waals dispersion forces).
CH₃Br is a polar molecule due to the electronegative Br atom creating a permanent dipole.
(b) Butane, C₄H₁₀
London dispersion forces (van der Waals forces) only.
Butane is a non-polar molecule; only temporary induced dipoles exist.
© Hexanol, C₆H₁₃OH
Hydrogen bonding (plus van der Waals forces).
The −OH group allows H-bonding between O and H of neighbouring molecules.
Question 2 — Alkane Mixtures in Petrol (C6–C10)
General formula of alkanes: CₙH₂ₙ₊₂
| Carbon atoms (n) | Molecular Formula |
|—|---|
| 6 | C₆H₁₄ |
| 7 | C₇H₁₆ |
| 8 | C₈H₁₈ |
| 9 | C₉H₂₀ |
| 10 | C₁₀H₂₂ |
Question 3 — Cis-Trans Isomerism
For cis-trans isomerism to exist, each carbon of the double bond must carry two different groups.
| Compound | Cis-Trans? | Reason |
|—|---|—|
| (a) 1-bromo-1,2-dichloroethene | Yes | C1 has Br and Cl; C2 has Cl and H — all four groups different |
| (b) but-1-ene | No | C1 has two H atoms |
| © 1,1-dichloropropene | No | C1 has two Cl atoms |
| (d) pent-2-ene | Yes | C2 has H & CH₃; C3 has H & C₂H₅ |
| (e) buta-1,3-diene | No | C1 has two H atoms; C4 has two H atoms |
Question 4 — Two Types of Cracking
(i) Thermal Cracking
-
Conditions: High temperature (400–900°C), high pressure (up to 70 atm), no catalyst
-
Long-chain alkanes are broken into shorter alkanes and alkenes
(ii) Catalytic Cracking
-
Conditions: Temperature ~500°C, low/atmospheric pressure, catalyst (zeolite/aluminium silicate)
-
Produces more branched alkanes and aromatic compounds; used for high-octane petrol
Question 5 — Empirical Formula
Given: C = 29.3%, H = 5.7%, Br = 65%
Atomic masses: C=12, H=1, Br=80
| Element | % | ÷ Atomic mass | Mole ratio | Simplest ratio |
|—|---|—|---|—|
| C | 29.3 | 29.3/12 = 2.44 | 2.44/0.81 = 3 | 3 |
| H | 5.7 | 5.7/1 = 5.70 | 5.70/0.81 = 7 | 7 |
| Br | 65.0 | 65/80 = 0.81 | 0.81/0.81 = 1 | 1 |
Empirical formula = C₃H₇Br
Question 6 — Enolization
(a) Enolization is the conversion of a carbonyl compound (keto form) to its enol form, where a hydrogen atom migrates from the α-carbon to the oxygen atom, forming a C=C double bond with an −OH group.
(b) Keto-Enol forms:
Ethanal (CH₃CHO):
Keto: CH₃−CH=O
Enol: CH₂=CH−OH
Propanone (CH₃COCH₃):
Keto: CH₃−CO−CH₃
Enol: CH₃−C(OH)=CH₂
Question 7 — Dialkanoic Acids
(a) General molecular formula: CₙH₂ₙ(COOH)₂ or CₙH₂ₙ₊₀O₄
More precisely written as: HOOC−(CH₂)ₙ−COOH, general formula CₙH₂ₙ₋₂O₄ (for the diacid chain)
Standard form: CₙH₂ₙ(COOH)₂ where n = 0, 1, 2…
(b) First two members:
1. Ethanedioic acid (Oxalic acid), n=0:
HOOC−COOH
(Molecular formula: C₂H₂O₄)
2. Propanedioic acid (Malonic acid), n=1:
HOOC−CH₂−COOH
(Molecular formula: C₃H₄O₄)
Question 8 — Plastics/Polymers
(a) Plastics that cannot be softened or melted by heat or pressure are called thermosetting plastics (thermosets).
(b) Monomer structures:
(i) −(CH₂−CH(CH₃))ₙ− → Polypropylene
-
Monomer: CH₂=CH−CH₃
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IUPAC name: Propene (propylene)
(ii) −(CH₂−CCl₂)ₙ− → Polyvinylidene chloride
-
Monomer: CH₂=CCl₂
-
IUPAC name: 1,1-dichloroethene (vinylidene chloride)
Question 9 — Isomerism in Alkenes
(a) Types of isomerism in alkenes with 4+ carbons:
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Chain (skeletal) isomerism — different carbon chain arrangements
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Positional isomerism — double bond at different positions
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Geometric (cis-trans) isomerism — different spatial arrangement around C=C
(b) Two laboratory methods of preparing alkenes:
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Dehydration of alcohols — heating alcohol with conc. H₂SO₄ or passing vapour over hot Al₂O₃
-
Dehydrohalogenation of haloalkanes — heating haloalkane with alcoholic KOH
Question 10 — Disubstituted Benzenes
(a) The type of isomerism is positional isomerism (a form of structural isomerism).
(b) Four aromatic structural isomers of C₈H₁₀:
These are the four isomers of dimethylbenzene + ethylbenzene:
1. 1,2-dimethylbenzene (ortho-xylene)
CH₃
/
benzene ring with CH₃ on adjacent carbon
2. 1,3-dimethylbenzene (meta-xylene)
- Two CH₃ groups at positions 1 and 3
3. 1,4-dimethylbenzene (para-xylene)
- Two CH₃ groups at positions 1 and 4 (opposite)
4. Ethylbenzene
- Benzene ring with one −CH₂CH₃ group
SECTION B
Question 11(a) — Functional Groups
| Compound | Functional Group | Series/Family |
|—|---|—|
| (i) CH₃CH₂CH₂CHO | −CHO (aldehyde) | Alkanal |
| (ii) CH₃CH₂NH₂ | −NH₂ (amine) | Alkanamine |
| (iii) CH₃CH₂CH₂Cl | −Cl (halogen) | Haloalkane (chloroalkane) |
| (iv) CH₃CH₂CH₂CH₂OH | −OH (hydroxyl) | Alkanol |
| (v) CH₂=CHCH₂CH₂Br | C=C + −Br | Bromoalkene |
| (vi) CH₃CH₂COCH₃ | C=O (ketone) | Alkanone |
| (vii) CH₃COOH | −COOH (carboxyl) | Alkanoic acid |
| (viii) CH₃CH₂OCH₃ | −O− (ether linkage) | Alkoxyalkane (ether) |
| (ix) CH₃CH₂COOR | −COO− (ester linkage) | Alkanoate (ester) |
Question 11(b) — Reaction Classification
(i) CH₃CH₂CH(OH)CH₃ → CH₃CH=CHCH₃ + H₂O
→ Elimination (loss of H₂O from adjacent carbons)
(ii) CH₃CH₂COOCH₃ + H₂O → CH₃CH₂COOH + CH₃OH
→ Hydrolysis (ester broken down by water)
(iii) CH₃CH=CH−CHCH₃ + H₂ → CH₃CH₂CH₂CH₂CH₃
→ Addition (H₂ added across double bond)
(iv) CH₃CH₂CH(Br)CH₃ + OH⁻ → CH₃CH₂CH(OH)CH₃ + Br⁻
→ Substitution (Br replaced by OH)
Question 12(a) — Addition to 2-Methylbut-2-ene
2-methylbut-2-ene: CH₃−C(CH₃)=CH−CH₃
(i) + Hydrogen (H₂): [catalyst: Ni, heat]
CH₃−C(CH₃)=CH−CH₃ + H₂ → CH₃−CH(CH₃)−CH₂−CH₃
Product: 2-methylbutane
(ii) + Bromine (Br₂):
CH₃−C(CH₃)=CH−CH₃ + Br₂ → CH₃−CBr(CH₃)−CHBr−CH₃
Product: 2,3-dibromo-2-methylbutane
(iii) + Hydrogen Chloride (HCl): [Markovnikov’s rule — H adds to C with more H]
CH₃−C(CH₃)=CH−CH₃ + HCl → CH₃−CCl(CH₃)−CH₂−CH₃
Product: 2-chloro-2-methylbutane (major)
(iv) + Aqueous KOH: (hydration — acts as H₂O in this context)
CH₃−C(CH₃)=CH−CH₃ + H₂O → CH₃−C(OH)(CH₃)−CH₂−CH₃
Product: 2-methylbutan-2-ol
Question 12(b) — Reduction Products
| Reduced compound | Class of Alkanol produced |
|—|---|
| (i) Alkanoic acids (RCOOH) | Primary alkanol |
| (ii) Alkanones (RCOR’) | Secondary alkanol |
| (iii) Alkanoates (RCOOR’) | Primary alkanol |
| (iv) Alkanal (RCHO) | Primary alkanol |
| (v) Alkanoic anhydride | Primary alkanol |
Question 12© — Methyl Chloride + Ammonia
Reaction: CH₃Cl + NH₃ → stepwise substitution
Step 1: CH₃Cl + NH₃ → CH₃NH₂ + HCl
(methylamine — primary amine)
Step 2: CH₃Cl + CH₃NH₂ → (CH₃)₂NH + HCl
(dimethylamine — secondary amine)
Step 3: CH₃Cl + (CH₃)₂NH → (CH₃)₃N + HCl
(trimethylamine — tertiary amine)
Step 4: CH₃Cl + (CH₃)₃N → (CH₃)₄N⁺Cl⁻
(tetramethylammonium chloride — quaternary salt)
Likely products: CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, (CH₃)₄N⁺Cl⁻
Question 13(a) — Alkene Isomers of C₅H₁₀
(i) Definition: Isomers are compounds with the same molecular formula but different structural arrangements of atoms.
(ii) Alkene isomers of C₅H₁₀:
-
Pent-1-ene: CH₂=CH−CH₂−CH₂−CH₃
-
Pent-2-ene: CH₃−CH=CH−CH₂−CH₃
-
2-methylbut-1-ene: CH₂=C(CH₃)−CH₂−CH₃
-
3-methylbut-1-ene: CH₂=CH−CH(CH₃)−CH₃
-
2-methylbut-2-ene: CH₃−C(CH₃)=CH−CH₃
-
Cyclopentane (same formula, not an alkene — may be excluded)
Question 13(b) — HCl + 2-Methylpropene
2-methylpropene: CH₂=C(CH₃)₂
(i) Two possible carbocation intermediates:
Carbocation 1 (secondary/tertiary):
CH₃−C⁺(CH₃)−CH₃ ← H adds to CH₂=
(tertiary carbocation)
Carbocation 2 (primary):
⁺CH₂−CH(CH₃)₂ ← H adds to C(CH₃)₂=
(primary carbocation)
(ii) More stable carbocation:
Tertiary carbocation [CH₃−C⁺(CH₃)−CH₃] is more stable because three alkyl groups donate electrons inductively, dispersing the positive charge better than in a primary carbocation.
(iii) Major product:
CH₃−CCl(CH₃)−CH₃
2-chloro-2-methylpropane
(iv) Minor product:
ClCH₂−CH(CH₃)₂
1-chloro-2-methylpropane
Question 13© — Correct IUPAC Names
(i) 2,3-dimethylpentane vs 2-methyl-3-methylpentane
→ 2,3-dimethylpentane is correct — all substituents cited together with locants in one name
(ii) 3-methyl-3-ethylheptane vs 3-ethyl-3-methylheptane
→ 3-ethyl-3-methylheptane is correct — substituents listed alphabetically (ethyl before methyl)
(iii) 2,3-dimethyl-4-ethylheptane vs 4-ethyl-2,3-dimethylheptane
→ 4-ethyl-2,3-dimethylheptane is correct — alphabetical order (ethyl before dimethyl)
Question 14(a)(i) — Straight-chain C₄H₁₀O Isomers
Molecular formula C₄H₁₀O — two straight-chain isomeric alkanols:
1. Butan-1-ol:
CH₃−CH₂−CH₂−CH₂OH
(primary alcohol)
2. Butan-2-ol:
CH₃−CH₂−CH(OH)−CH₃
(secondary alcohol)
Question 14(a)(ii) — Oxidation with Acidified K₂Cr₂O₇
Butan-1-ol (primary) — oxidised to aldehyde, then acid:
CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O
(butanal)
CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH
(butanoic acid)
Butan-2-ol (secondary) — oxidised to ketone only:
CH₃CH₂CH(OH)CH₃ + [O] → CH₃CH₂COCH₃ + H₂O
(butanone / methyl ethyl ketone)
Question 14(a)(iii) — Alkanol Resistant to Oxidation
An alkanol resistant to oxidation by acidified K₂Cr₂O₇ is a tertiary alcohol.
2-methylpropan-2-ol:
CH₃
|
CH₃−C−OH
|
CH₃
Molecular formula: C₄H₁₀O
Tertiary alcohols have no H on the carbon bearing −OH, so oxidation cannot proceed.
Question 14(b) — Elimination Products from Haloalkanes
(i) 1-bromobutane (with alcoholic KOH):
CH₃CH₂CH₂CH₂Br → CH₃CH₂CH=CH₂
Only one product: but-1-ene
(ii) 2-bromobutane:
CH₃CH₂CHBrCH₃ →
Major: CH₃CH=CHCH₃ (but-2-ene) — more substituted, Zaitsev's rule
Minor: CH₃CH₂CH=CH₂ (but-1-ene)
Two products possible (cis and trans but-2-ene also)
(iii) 1-bromo-2-methylpropane:
(CH₃)₂CHCH₂Br →
Only: (CH₃)₂C=CH₂ — 2-methylpropene (one possible product)
(iv) 2-bromo-2-methylpropane:
(CH₃)₃CBr →
Only: CH₂=C(CH₃)₂ — 2-methylpropene
Only one product possible
Question 14© — Bromoethane → Ethanoic Acid (2 Steps)
Step 1: Bromoethane → Ethanol
CH₃CH₂Br + NaOH(aq) → CH₃CH₂OH + NaBr
-
Reagent: Aqueous NaOH (or KOH)
-
Condition: Heat/reflux
-
Mechanism: Nucleophilic substitution (SN2)
Step 2: Ethanol → Ethanoic Acid
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
-
Reagent: Acidified potassium dichromate(VI) / K₂Cr₂O₇ + H₂SO₄
-
Condition: Heat under reflux (to prevent stopping at aldehyde stage)
Note: For drawing structural formulae in your exam, always show each bond clearly. For Section B, show all working steps to earn full marks.