4 exam-style question sets on Organic reactions (Paper 2), each with a hint and a fully worked answer. The app holds 20 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.
Practise this section in the app →
Question 1
The flow diagram below shows how compound U, a straight-chain ALKANE, is used to prepare other organic compounds. Reaction I is a CRACKING reaction in which W and T are the ONLY products.
- Reaction I (cracking): U → W + T (T has molecular formula C₃H₆)
- Reaction II: W → CH₃CHClCH₃
- Reaction III (elimination): CH₃CHClCH₃ → T
- Reaction IV: T → R (R is a HALOALKANE and the MAJOR product)
- Reaction V: T → S (S is 1,2-dibromopropane)
The following inorganic reagents are available for reactions IV and V:
HBr Br₂ H₂ H₂O NaOH(dilute)
- aDefine the term cracking reaction. (2)
- bIs CH₃CHClCH₃ a PRIMARY, SECONDARY or TERTIARY haloalkane? Give a reason for the answer. (2)
- cWrite down the STRUCTURAL FORMULA of compound W AND the MOLECULAR FORMULA of compound U. (3)
- dFor reaction II, write down the NAME or FORMULA of the inorganic reactant, the type of reaction AND ONE reaction condition. (3)
- eWrite down the TYPE of elimination reaction in reaction III. (1)
- fUsing the correct reagent from the list above, write down a balanced equation, using STRUCTURAL FORMULAE, for reaction IV. (5)
- gWrite down a balanced equation, using STRUCTURAL FORMULAE, for reaction V AND the IUPAC name of the MINOR product of reaction IV. (5)
Hint
Work out W from the product of reaction II; for reaction IV apply Markovnikov's rule to decide the major product.
Worked answer
a) A cracking reaction is the chemical decomposition of a long-chain hydrocarbon (alkane) into smaller, more useful molecules (under high pressure and temperature / with a catalyst). ✓✓
b) SECONDARY ✓ — the C atom bonded to the Cl is attached to TWO other C atoms ✓.
c) W: propane — full structural formula showing a 3-carbon chain with all C–H bonds (condensed: CH₃CH₂CH₃) ✓✓. U: W (C₃H₈) + T (C₃H₆) come only from U, so U = C₆H₁₄ ✓.
d) Inorganic reactant: Cl₂ (chlorine) ✓. Type: substitution (halogenation) ✓. Condition: light/UV (or heat) ✓.
e) Dehydrohalogenation ✓
f) Reagent: HBr ✓. T is propene:
CH₂=CHCH₃ + HBr → CH₃CHBrCH₃
(structural formulae of propene ✓ and 2-bromopropane ✓; correct major product per Markovnikov — H adds to the C with more H atoms ✓; balanced ✓)
g) Reagent for V: Br₂.
CH₂=CHCH₃ + Br₂ → CH₂BrCHBrCH₃ (structural formulae ✓✓; balanced ✓)
Minor product of reaction IV: 1-bromopropane ✓✓
Modelled on Nov 2024 Paper 2, Question 4 — same skills, new scenario.
Question 2
Consider the two-step sequence below.
Reaction 1: Pentan-2-ol is heated with concentrated H₂SO₄ to form alkene X (the MAJOR product) and water.
Reaction 2: X reacts with H₂ over a platinum catalyst to form compound Y.
Which ONE of the following combinations gives the CORRECT IUPAC names of X and Y?
Hint
Zaitsev's rule: the major alkene has the more substituted double bond; hydrogenation then adds H₂ across it.
Worked answer
B. Dehydration of pentan-2-ol can give pent-1-ene or pent-2-ene; by Zaitsev's rule the more substituted alkene, pent-2-ene, is the major product X. Adding H₂ across the double bond (hydrogenation) gives the alkane pentane as Y. A names the minor alkene. C and D are wrong because hydrogenation adds hydrogen, not water — it cannot produce an alcohol. Modelled on Nov 2024 Paper 2, Question 1.3 — same skill, new scenario.
Question 3
Study the reaction sequence below. X, Y and Z are organic compounds.
Reaction I: but-1-ene + HBr → X (major product)
Reaction II: X + NaOH → Y (major product) + NaBr + H₂O
Reaction III: Y + H₂O → Z (major product)
In Reaction II the sodium hydroxide is CONCENTRATED and dissolved in ethanol.
- aWrite down the type of reaction represented by Reaction I. Choose from ADDITION, ELIMINATION or SUBSTITUTION. (1)
- bWrite down the IUPAC name of compound X. Explain why this is the MAJOR product of Reaction I. (4)
- cBesides the concentrated NaOH in ethanol, write down ONE other reaction condition needed for Reaction II. (1)
- dWrite down the IUPAC name of compound Y. Give a reason why it is the major product. (3)
- eWrite down TWO reaction conditions for Reaction III and the IUPAC name of compound Z. (4)
- fWrite down the name of the type of reaction represented by Reaction III. (1)
This question has a diagram, shown in the app.
Hint
Markovnikov for the addition; Zaitsev for the elimination: the major alkene has the double bond between the carbons carrying the fewest hydrogens.
Worked answer
a) ADDITION ✓ (hydrohalogenation).
b) 2-bromobutane ✓✓. In the addition of HBr to but-1-ene, the H atom attaches mainly to the double-bond carbon that already has the LARGER number of H atoms (C1) ✓, so the Br goes to C2 ✓.
c) Strong heating ✓ (hot, concentrated NaOH in ethanol → elimination).
d) But-2-ene ✓✓ — in elimination the major product is the more substituted alkene: the H is removed from the neighbouring C with FEWER H atoms (C3), giving the double bond between C2 and C3 ✓.
e) Conditions: an acid catalyst (dilute H₂SO₄ or H₃PO₄) ✓ and excess water / steam ✓. Z = butan-2-ol ✓✓ (water adds with OH mainly on the more substituted carbon).
f) ADDITION (hydration) ✓.
Modelled on Nov 2025 Paper 2, Question 4 — same skills, new scenario.
Question 4
Ethene reacts with water to form ethanol. Which ONE of the following substances is a suitable catalyst for this hydration reaction?
Hint
Water is added across the double bond — which type of catalyst speeds up hydration?
Worked answer
A. Hydration of an alkene (addition of water across the double bond) is catalysed by an acid — dilute sulfuric acid (or phosphoric acid). B: NaOH is used for hydrolysis or elimination of haloalkanes. C: nickel catalyses hydrogenation. D: copper does not catalyse this addition.
Modelled on Nov 2025 Paper 2, Question 1.2 — same skills, new scenario.