Introduction to organic chemistry – IGCSE Chemistry exam questions with mark scheme

1Grade 1–33 marks

The formulae of four compounds are shown.

CH₃Cl C₆H₁₄ C₂H₅OH C₃H₆

(a) State what is meant by the term hydrocarbon. [2]

(b) Which two of the compounds shown are hydrocarbons? [1]

(a) compound / molecule(s) containing carbon and hydrogen (atoms); reject “a mixture of carbon and hydrogen” / “elements” for this mark(1)
(a) only (allow ‘just’, ‘nothing else’); this mark depends on a reference to carbon and hydrogen; ignore references to crude oil or fuels(1)
(b) C₆H₁₄ and C₃H₆ (both needed, no extra answers)(1)
Answer: (a) a compound of hydrogen and carbon only (b) C₆H₁₄ and C₃H₆
Examiner tip: The word 'only' is the second mark. 'Contains carbon and hydrogen' alone also describes ethanol and chloromethane.

2Grade 4–55 marks

The displayed formula of a hydrocarbon, P, is shown.

(a) Give the molecular formula of P. [1]

(b) Give the structural formula of P. [1]

(d) Name P and the homologous series it belongs to. [2]

(a) C₃H₆ (allow H₆C₃; do not allow CH₂CHCH₃ here)(1)
(b) CH₂=CHCH₃ (allow CH₂CHCH₃ or CH₃CH=CH₂; do not allow C₃H₆)(1)
(c) CH₂(1)
(d) propene (allow prop-1-ene; do not allow propane)(1)
(d) alkenes (allow alkene; do not allow alkane)(1)
Answer: (a) C₃H₆ (b) CH₂=CHCH₃ (c) CH₂ (d) propene; alkenes
Examiner tip: Students swap molecular and structural formulae: the molecular formula only counts atoms, the structural formula shows how they are grouped along the chain.

3Grade 4–55 marks

The graph shows the boiling points of the first five straight-chain alkanes.

(a) Describe the trend in boiling point shown by the graph. [1]

(b) Use the graph to estimate the boiling point of hexane, C₆H₁₄. [1]

(c) Use values from the graph to show that the increase in boiling point from one alkane to the next gets smaller as the number of carbon atoms increases. [1]

(d) The alkanes form a homologous series. Give two features, other than a trend in physical properties, that all members of a homologous series have in common. [2]

(a) boiling point increases as the number of carbon atoms increases (allow as the chain gets longer / as the molecules get bigger; ignore 'it goes up' with no reference to number of carbon atoms)(1)
(b) any value in the range 55 °C to 80 °C (unit not needed)(1)
(c) two correct differences read from the graph showing a decrease, e.g. methane → ethane about 73 °C (allow 65–80) and butane → pentane about 37 °C (allow 30–45); any two correct consecutive differences that are shown to decrease(1)
(d) any two from: same general formula; similar chemical properties / react in similar ways; same functional group (if any); each member differs from the next by CH₂(1)
(d) second feature from the list above (ignore 'same physical properties'; do not allow 'same molecular formula')(1)
Answer: (a) increases as the number of carbon atoms increases (b) about 70 °C (55–80 °C) (c) e.g. rise of about 73 °C from C₁ to C₂ but only about 37 °C from C₄ to C₅ (d) e.g. same general formula; differ by CH₂
Examiner tip: In (b) keep the shape of the curve: the rises are getting smaller, so add a bit less than the last rise (about 37 °C) to pentane’s boiling point.

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4Grade 4–55 marks

Four reactions of hydrocarbons are shown.

A C₃H₆ + Br₂ → C₃H₆Br₂
B C₂H₆ + Br₂ → C₂H₅Br + HBr
C C₄H₈ + 6O₂ → 4CO₂ + 4H₂O
D C₂H₄ + H₂O → C₂H₅OH

(a) Classify each of reactions A, B, C and D as addition, substitution or combustion. [3]

(b) Reaction B only happens in the presence of a certain condition. State this condition. [1]

(a) A addition and D addition(1)
(a) B substitution(1)
(a) C combustion(1)
(b) ultraviolet light / UV (light) (allow sunlight; ignore 'heat', 'light' alone)(1)
(c) two molecules / reactants join to form one / a single product (allow 'only one product'; allow 'water adds across the C=C double bond'; ignore 'it is added')(1)
Answer: (a) A addition, B substitution, C combustion, D addition (b) UV light (c) two reactants form a single product
Examiner tip: A substitution swaps an atom (H for Br) and so always gives two products; an addition reaction gives only one.

5Grade 6–75 marks

The structural formulae of five alkanes, A to E, are shown.

A CH₃CH₂CH₂CH₂CH₃
B CH₃CH(CH₃)CH₂CH₃
C CH₃CH₂CH(CH₃)CH₃
D CH₃CH₂CH₂CH₂CH₂CH₃
E C(CH₃)₄

(a) State what is meant by the term isomers. [2]

(b) Two of the formulae represent the same compound. Which two? [1]

(a) (compounds with) the same molecular formula(1)
(a) but different structural formulae / different displayed formulae / different structures / atoms arranged differently (do not allow ‘different molecular formulae’)(1)
(b) B and C(1)
(c) D has a different molecular formula / D is C₆H₁₄ but the others are C₅H₁₂ (allow D has six carbon atoms and the others have five)(1)
(c) B and C are the same compound so they cannot be isomers of each other (allow same structure drawn / written in a different direction)(1)
Answer: (a) same molecular formula, different structural formula (b) B and C (c) D is C₆H₁₄ (not C₅H₁₂); B and C are the same molecule
Examiner tip: Turning a chain round does not make a new isomer: CH₃CH(CH₃)CH₂CH₃ read from the other end is CH₃CH₂CH(CH₃)CH₃.

6Grade 6–7Triple only6 marks

The structural formulae of three organic compounds are shown.

W CH₃CH₂CH₂OH
X CH₃CH₂COOH
Y CH₂=CHCH₂CH₃

(a) State what is meant by the term functional group. [1]

(b) Give the formula of the functional group in W and in X. [2]

(a) an atom or group of atoms that determines the (chemical) reactions / properties of a compound (allow 'responsible for the characteristic reactions of a homologous series'; ignore 'the main part of the molecule')(1)
(b) W: –OH (allow hydroxyl / OH; do not allow OH⁻ or hydroxide)(1)
(b) X: –COOH (allow carboxyl / CO₂H; do not allow –COO– alone)(1)
(c) W: propan-1-ol (allow propanol / 1-propanol)(1)
(c) X: propanoic acid (do not allow propanoic / propanol / ethanoic acid)(1)
(c) Y: but-1-ene (allow 1-butene; do not allow butene alone or but-2-ene)(1)
Answer: (a) atom/group of atoms that determines a compound's reactions (b) W –OH, X –COOH (c) propan-1-ol, propanoic acid, but-1-ene
Examiner tip: Count every carbon including the one in –COOH: CH₃CH₂COOH has three carbons, so it is propanoic acid, not ethanoic acid.

7Grade 6–75 marks

A hydrocarbon Q contains 83.3% carbon by mass. The relative formula mass (M_r) of Q is 72.

[A_r: H = 1, C = 12]

(a) Show that the empirical formula of Q is C₅H₁₂. [3]

(b) Deduce the molecular formula of Q. [1]

(a) % hydrogen = 100 − 83.3 = 16.7(1)
(a) moles: C = 83.3 ÷ 12 = 6.94 and H = 16.7 ÷ 1 = 16.7 (allow 6.9)(1)
(a) ratio 16.7 ÷ 6.94 = 2.4 so C : H = 1 : 2.4 = 5 : 12 (must show the step to a whole-number ratio, e.g. ×5; do not award if 2.4 is rounded to 2)(1)
(b) C₅H₁₂, because the empirical formula mass is (5 × 12) + 12 = 72, equal to the M_r (working must be shown; allow 72 ÷ 72 = 1)(1)
(c) fits the general formula C_nH_2n+2 (with n = 5, 2n + 2 = 12) (allow it has the maximum number of hydrogen atoms / it is saturated; ignore 'only single bonds' unless linked to the formula)(1)
Answer: (a) C 6.94 : H 16.7 = 1 : 2.4 = 5 : 12 (b) C₅H₁₂ (c) fits CₙH₂ₙ₊₂ (n = 5)
Examiner tip: A ratio of 1 : 2.4 must not be rounded to 1 : 2; multiply both numbers by 5 to reach whole numbers.

8Grade 8–95 marks

Alkane R has a relative formula mass (M_r) of 86.

[A_r: H = 1, C = 12]

(a) Use the general formula of the alkanes to deduce the molecular formula of R. Show your working. [2]

(b) Give the empirical formula of R. [1]

(c) R has several structural isomers. Give the structural formulae of two different isomers of R that each have a main chain of five carbon atoms. [2]

(a) 12n + 2n + 2 = 86 / 14n + 2 = 86 (allow trial of C_nH_2n+2 masses, e.g. C₅H₁₂ = 72, C₆H₁₄ = 86)(1)
(a) n = 6 so C₆H₁₄ (correct answer with no working scores 1 only)(1)
(b) C₃H₇ (allow ecf from an incorrect molecular formula in (a) if correctly simplified)(1)
(c) CH₃CH(CH₃)CH₂CH₂CH₃ (allow CH₃CH₂CH₂CH(CH₃)CH₃ or a correct displayed formula)(1)
(c) CH₃CH₂CH(CH₃)CH₂CH₃ (allow a correct displayed formula); names not required; no mark for a second formula that is the first one written from the other end(1)
Answer: (a) 14n + 2 = 86, n = 6, C₆H₁₄ (b) C₃H₇ (c) CH₃CH(CH₃)CH₂CH₂CH₃ and CH₃CH₂CH(CH₃)CH₂CH₃
Examiner tip: Moving the branch to the other end of the chain does not make a new isomer: CH₃CH₂CH₂CH(CH₃)CH₃ is the same molecule as CH₃CH(CH₃)CH₂CH₂CH₃.

9Grade 8–96 marks

Butane, C₄H₁₀, and but-1-ene, C₄H₈, are both hydrocarbons, but they belong to different homologous series.

Compare the structures of butane and but-1-ene and the types of reaction that each one undergoes.

In your answer, refer to addition, substitution and combustion reactions. [6]

butane has only (C–C) single bonds / is saturated, whereas but-1-ene has a C=C double bond / is unsaturated(1)
butane is an alkane (general formula C_nH_2n+2) and but-1-ene is an alkene (general formula C_nH_2n); allow but-1-ene has two fewer hydrogen atoms(1)
butane reacts with a halogen / bromine by substitution, only in ultraviolet light(1)
in substitution a hydrogen atom is replaced by a bromine atom, giving two products, e.g. C₄H₁₀ + Br₂ → C₄H₉Br + HBr (allow bromobutane and hydrogen bromide)(1)
but-1-ene reacts with bromine by addition across the C=C (no UV needed), giving one product, C₄H₈Br₂ / dibromobutane; allow but-1-ene decolourises bromine water(1)
both undergo combustion: complete combustion gives carbon dioxide and water, e.g. 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O / C₄H₈ + 6O₂ → 4CO₂ + 4H₂O; allow incomplete combustion gives carbon monoxide / carbon(1)
Answer: Butane: saturated alkane (single bonds only, CₙH₂ₙ₊₂), substitution with bromine in UV light (C₄H₉Br + HBr). But-1-ene: unsaturated alkene with C=C (CₙH₂ₙ), addition of bromine giving one product (C₄H₈Br₂). Both burn to CO₂ and H₂O.
Examiner tip: Give the reaction type and the condition or product for each compound — 'butane reacts with bromine' scores nothing until you say substitution in UV light.

10Grade 4–54 marks

Dodecane is an alkane found in kerosene. Each molecule of dodecane contains 12 carbon atoms.

(a) Give the molecular formula of dodecane. [1]

(b) Give the empirical formula of dodecane. [1]

Explain why all alkenes have the same empirical formula. [2]

(a) C₁₂H₂₆(1)
(b) C₆H₁₃ (allow ecf from (a) if correctly simplified)(1)
(c) (alkenes have) the general formula C_nH_2n / the number of H atoms is always twice the number of C atoms(1)
(c) so the simplest ratio of C : H is always 1 : 2 / empirical formula is CH₂ (do not allow CH₂ stated with no link to the ratio or general formula for both marks)(1)
Answer: (a) C₁₂H₂₆ (b) C₆H₁₃ (c) general formula CₙH₂ₙ, so C : H always 1 : 2, giving CH₂
Examiner tip: An empirical formula is the simplest whole-number ratio — divide C₁₂H₂₆ by 2, not by 12.

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