AS Level Chemistry chapter 15: Halogen compounds MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
24 exam-style questions on Halogen compounds, 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. Ethane reacts with chlorine in ultraviolet light. Which product can only be formed in a termination step?
A
hydrogen chloride
B
1,1-dichloroethane
C
chloroethane
D
butane
Show answer
Answer: D. Butane forms only when two ethyl radicals join: CH3CH2• + •CH2CH3 → C4H10. HCl and the chloroethanes are made in propagation steps.
2. Which equation correctly shows the reaction of ethanol with phosphorus(V) chloride?
A
C2H5OH + PCl5 → C2H5Cl + POCl3 + HCl
B
C2H5OH + PCl5 → C2H5Cl + PCl3 + HOCl
C
C2H5OH + PCl5 → C2H4Cl2 + POCl3 + H2
D
2C2H5OH + PCl5 → 2C2H5Cl + POCl3 + H2O
Show answer
Answer: A. The OH is replaced by Cl, and the O and H end up in POCl3 and steamy HCl fumes. All four equations balance, so the test is the products: PCl5 gives POCl3 and HCl, not water, HOCl or hydrogen, and only one Cl goes onto the carbon chain.
3. 1-Bromopropane is heated under reflux with aqueous sodium hydroxide. What is the organic product?
A
propan-1-ol
B
propene
C
propan-2-ol
D
propanal
Show answer
Answer: A. In aqueous solution OH− acts as a nucleophile and replaces Br on the same carbon, giving propan-1-ol. Ethanolic NaOH would favour elimination to propene.
4. In the reaction of 2-bromo-2-methylpropane with NaOH in ethanol, methylpropene is formed. What is the role of the hydroxide ion?
A
a nucleophile that attacks the carbon bonded to bromine
B
a base that removes H+ from a carbon next to the C–Br carbon
C
an electrophile that accepts electrons from the C–Br bond
D
a catalyst that is regenerated at the end
Show answer
Answer: B. In elimination, OH− acts as a base: it takes a proton from a carbon adjacent to the C–Br carbon, a C=C forms, and Br− leaves. Attacking the carbon as a nucleophile is the substitution route, which gives an alcohol.
5. 2.74 g of 1-bromobutane, C4H9Br, is heated under reflux with NaOH(aq). 1.11 g of butan-1-ol is obtained. What is the percentage yield? [Ar: Br 79.9]
A
40.5%
B
54.1%
C
74.9%
D
133%
Show answer
Answer: C. n(C4H9Br) = 2.74 ÷ 136.9 = 0.0200 mol, so the maximum is 0.0200 mol of C4H9OH. Actual = 1.11 ÷ 74.0 = 0.0150 mol; yield = 74.9%. 40.5% compares the masses directly.
6. Equal amounts of four halogenoalkanes are warmed with aqueous silver nitrate in ethanol. Which gives a precipitate fastest?
A
1-chlorobutane
B
1-bromobutane
C
1-iodobutane
D
1-fluorobutane
Show answer
Answer: C. The rate depends on the strength of the C–X bond, not its polarity: C–I is the longest and weakest bond, so it breaks most readily, giving a yellow precipitate of AgI quickly. C–F is the most polar but far too strong.
7. What is the organic product when bromoethane is heated with sodium hydroxide dissolved in ethanol?
A
ethanol
B
ethene
C
ethanal
D
ethane
Show answer
Answer: B. In ethanol, OH− acts as a base and removes a hydrogen from the carbon next to the C–Br, eliminating HBr to give an alkene. In water, OH− acts as a nucleophile and ethanol is formed instead.
8. Bond energies (kJ mol−1): C–Cl 340, C–Br 280, C–I 240. 1-iodobutane gives a precipitate with aqueous silver nitrate in ethanol much faster than 1-chlorobutane. Which statement explains this?
A
The C–I bond is more polar than the C–Cl bond.
B
The C–I bond is weaker, so it breaks more easily.
C
Iodide ions are more soluble in water than chloride ions.
D
Silver iodide is less soluble than silver chloride.
Show answer
Answer: B. Rate depends on how easily the C–X bond breaks, and C–I is the weakest. Polarity would predict the opposite order, since C–Cl is the most polar bond; that is the trap. The solubility of the silver halide decides which precipitate forms, not how fast the halide ion is released.
9. Why does 2-bromo-2-methylpropane react with aqueous sodium hydroxide mainly by an SN1 mechanism?
A
the C–Br bond is weaker in tertiary halogenoalkanes
B
tertiary halogenoalkanes cannot form carbocations
C
hydroxide ions attack tertiary carbon atoms more easily
D
the tertiary carbocation formed is stabilised by the inductive effect of three alkyl groups, and the crowded carbon hinders direct attack by OH−
Show answer
Answer: D. In SN1 the C–Br bond breaks first to give a carbocation, which is only favourable when the carbocation is stabilised by alkyl groups. Steric crowding also blocks the single-step SN2 attack from behind.
10. An unknown halogenoalkane is warmed with aqueous silver nitrate in ethanol. A cream precipitate forms slowly. Which halogen does the compound contain?
A
fluorine
B
chlorine
C
bromine
D
iodine
Show answer
Answer: C. Silver bromide is cream. Silver chloride is white and silver iodide is yellow; a fluoroalkane gives no precipitate because the C–F bond is too strong and AgF is soluble.
24, each with the answer and a short explanation, across 1 topics: Halogenoalkanes. Every chapter test draws ten of them, so you can take it many times.
Is the AS Level Chemistry chapter 15 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.
