A Level Chemistry (A2) chapter 8: Hydrocarbons MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
24 exam-style questions on Hydrocarbons, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. In the bromination of benzene using Br2 and AlBr3, what happens in the final step of the mechanism?
A
the intermediate loses H+, which reacts with AlBr4− to form HBr and regenerate AlBr3
B
the intermediate gains a Br− ion from AlBr4− to give 1,2-dibromocyclohexa-3,5-diene
C
AlBr3 reacts with the intermediate and is used up, forming AlBr4− and HBr
D
the intermediate loses Br+ to AlBr4−, regenerating benzene and bromine
Show answer
Answer: A. The positive intermediate loses a proton to restore aromatic stabilisation: H+ + AlBr4− → HBr + AlBr3. The catalyst is regenerated, so it is not used up.
2. A student wants to make 1-bromo-3-nitrobenzene from benzene in two steps. Which route gives it as the main product?
A
nitrate with concentrated HNO3 and H2SO4, then react with Br2 and AlBr3
B
react with Br2 and AlBr3, then nitrate with concentrated HNO3 and H2SO4
C
react with Br2 in ultraviolet light, then nitrate with concentrated HNO3 and H2SO4
D
nitrate with concentrated HNO3 and H2SO4, then shake with bromine water at room temperature
Show answer
Answer: A. The –NO2 group directs to the 3-position, so it must go on first, and bromination of the deactivated ring still needs AlBr3. Brominating first gives mainly the 2- and 4-nitro products, and bromine water alone does not react with nitrobenzene.
3. Which route converts methylbenzene into 3-nitrobenzoic acid as the main product?
A
nitrate with concentrated HNO3 and H2SO4; then oxidise with hot alkaline KMnO4 then dilute acid
B
nitrate with dilute HNO3 at room temperature; then oxidise with hot alkaline KMnO4 then dilute acid
C
oxidise with hot alkaline KMnO4 then dilute acid; then nitrate with concentrated HNO3 and H2SO4
D
oxidise with hot alkaline KMnO4 then dilute acid; then nitrate with dilute HNO3 at room temperature
Show answer
Answer: C. The –COOH group directs to the 3-position, so the side chain must be oxidised first. Nitrating first would put –NO2 at the 2- and 4-positions, because –CH3 directs there. Dilute HNO3 at room temperature only nitrates very reactive rings such as phenol.
4. The enthalpy change of hydrogenation of cyclohexene, C6H10, is −120 kJ mol−1. The enthalpy change of hydrogenation of benzene to cyclohexane is −208 kJ mol−1. By how much is benzene stabilised compared with a hypothetical cyclohexa-1,3,5-triene that has three separate C=C bonds?
A
88 kJ mol−1
B
152 kJ mol−1
C
208 kJ mol−1
D
360 kJ mol−1
Show answer
Answer: B. Three isolated C=C bonds would give 3 × (−120) = −360 kJ mol−1; benzene releases only 208, so it is 360 − 208 = 152 kJ mol−1 more stable. 88 kJ mol−1 wrongly compares 208 with 120 rather than with 360.
5. 4-methylbenzoic acid (–COOH on C1, –CH3 on C4) is nitrated with a mixture of concentrated nitric and sulfuric acids, and one nitro group is substituted. What is the main product?
A
4-methyl-2-nitrobenzoic acid
B
4-(nitromethyl)benzoic acid
C
4-methyl-3-nitrobenzoic acid
D
4-nitrobenzoic acid
Show answer
Answer: C. –CH3 directs to the ring positions next to it (C3 and C5), and –COOH directs to the 3-positions relative to itself, which are again C3 and C5. Both groups favour C3 (equivalent to C5), while neither favours C2. The nitrating mixture substitutes into the ring, not into the side chain, and does not replace the –CH3 group.
6. Methylbenzene is warmed with ethanoyl chloride and anhydrous AlCl3, and one acyl group is substituted. Which compounds are the main products?
A
C6H5CH2COCH3 only
B
1-(3-methylphenyl)ethanone only
C
1-(2-methylphenyl)ethanone and 1-(4-methylphenyl)ethanone
D
2-methylbenzoic acid and 4-methylbenzoic acid
Show answer
Answer: C. The CH3CO+ electrophile substitutes into the ring, and –CH3 directs to the 2- and 4-positions, giving CH3C6H4COCH3 isomers. Side-chain substitution does not occur under these conditions, and no carboxylic acid forms.
7. Methylbenzene reacts with chlorine under two different sets of conditions. Which row gives the correct main products?
A
ultraviolet light: 2- and 4-chloromethylbenzene; AlCl3 in the dark: C6H5CH2Cl
B
ultraviolet light: C6H5CH2Cl; AlCl3 in the dark: 3-chloromethylbenzene
C
ultraviolet light: C6H5CH2Cl; AlCl3 in the dark: 2- and 4-chloromethylbenzene
D
ultraviolet light: 3-chloromethylbenzene; AlCl3 in the dark: C6H5CH2Cl
Show answer
Answer: C. UV light produces chlorine free radicals, which substitute in the methyl side chain, as in alkanes. A halogen carrier produces an electrophile that attacks the ring, and the methyl group directs it to the 2- and 4-positions.
8. Nitrobenzene is heated with a mixture of concentrated nitric and sulfuric acids. What is the main organic product?
A
1,3-dinitrobenzene
B
1,2-dinitrobenzene
C
1,4-dinitrobenzene
D
a mixture of 1,2- and 1,4-dinitrobenzene
Show answer
Answer: A. The –NO2 group directs a second electrophile to the 3-position, so 1,3-dinitrobenzene is formed. The 2- and 4-directing groups in the syllabus are –NH2, –OH and –R.
9. Phenylethene, C6H5CH=CH2, is heated with excess hydrogen and a nickel catalyst until no more hydrogen reacts. How many moles of H2 react with one mole of phenylethene?
A
1
B
3
C
4
D
5
Show answer
Answer: C. One mole adds across the side-chain C=C and three moles hydrogenate the benzene ring, giving ethylcyclohexane. Answering 1 forgets that the ring is hydrogenated under these conditions.
10. Why does benzene usually undergo substitution rather than addition with electrophiles such as bromine?
A
substitution keeps the delocalised π system, but addition would destroy its extra stability
B
benzene contains no π electrons at all, so it cannot take part in any addition reactions
C
addition products would be too crowded, with extra atoms on every carbon of the ring
D
the bromine molecule is too large to add across two neighbouring carbons of the ring
Show answer
Answer: A. The delocalised system gives benzene about 150 kJ mol−1 of extra stability. Substitution restores the ring and keeps this stabilisation; addition would lose it, so it is energetically unfavourable.
24, each with the answer and a short explanation, across 1 topics: Arenes. Every chapter test draws ten of them, so you can take it many times.
Is the A Level Chemistry (A2) chapter 8 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
