AS Level Chemistry chapter 21: Organic synthesis MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
22 exam-style questions on Organic synthesis, 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. Which two-step route converts 1-bromopropane into propanal?
A
NaOH in ethanol, heat; then steam with H3PO4
B
KCN in ethanol, heat; then dilute acid, reflux
C
NaOH(aq), heat; then excess acidified K2Cr2O7, heat under reflux
D
NaOH(aq), heat; then acidified K2Cr2O7, heat and distil off the product as it forms
Show answer
Answer: D. Aqueous NaOH substitutes OH for Br to give propan-1-ol. Oxidation with immediate distillation removes propanal before it is oxidised further; refluxing would give propanoic acid.
2. Which two-step route converts ethanol into propanenitrile, CH3CH2CN?
A
NaBr with concentrated H2SO4; then KCN in ethanol, heat under reflux
B
acidified K2Cr2O7, distil; then HCN with KCN
C
NaBr with concentrated H2SO4; then KCN in water, room temperature
D
concentrated H2SO4, heat; then HCN with KCN
Show answer
Answer: A. Ethanol → bromoethane (substitution with HBr made from NaBr and H2SO4), then CN− substitutes Br, adding one carbon. HCN with ethanal would give a hydroxynitrile, not propanenitrile, and KCN must be used in ethanol, heated.
3. 3-Hydroxypropanal, HOCH2CH2CHO, is reduced with NaBH4. What is the organic product?
A
HOCH2CH2CH2OH
B
HOCH2CH2COOH
C
CH3CH2CH2OH
D
HOCH2CH2CH3
Show answer
Answer: A. NaBH4 reduces the aldehyde group to a primary alcohol and leaves the existing –OH unchanged, giving propane-1,3-diol. It does not remove –OH groups, and oxidation to –COOH is the opposite change.
4. 2-Bromopropane is heated with NaOH dissolved in water. Propan-2-ol is the main product. Which by-product forms in a competing reaction, and what type of reaction forms it?
A
propene, by nucleophilic substitution
B
propene, by elimination
C
propane, by reduction
D
propanone, by oxidation
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Answer: B. OH− can act as a base and remove HBr, forming propene by elimination. Substitution is what gives propan-2-ol; NaOH is neither an oxidising nor a reducing agent.
5. Chloroethane is made from ethane and chlorine in ultraviolet light. Which by-product shows that two ethyl radicals have combined in a termination step?
A
ethene
B
butane
C
1,2-dichloroethane
D
hydrogen chloride
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Answer: B. 2CH3CH2• → CH3CH2CH2CH3. HCl is formed in propagation steps, and dichloroethanes come from further substitution, not from radicals combining.
6. But-3-en-1-ol, CH2=CHCH2CH2OH, is heated with an excess of hydrogen bromide. What is the main organic product?
A
BrCH2CH2CH2CH2Br
B
CH3CHBrCH2CH2OH
C
CH2=CHCH2CH2Br
D
CH3CHBrCH2CH2Br
Show answer
Answer: D. With excess HBr both groups react: HBr adds across the C=C (Br goes to the carbon that gives the more stable secondary carbocation) and Br substitutes for the –OH. 1,4-Dibromobutane puts the Br on the wrong carbon of the C=C.
7. In the route propene → 2-bromopropane → propan-2-ol → propanone, what types of reaction occur, in order?
Answer: A. HBr adds across the C=C (electrophilic addition), OH− replaces Br (nucleophilic substitution), and the secondary alcohol is oxidised to a ketone.
8. Butan-2-ol is heated with concentrated H2SO4. How many different alkenes, including stereoisomers, can form?
A
1
B
2
C
3
D
4
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Answer: C. Water can be lost towards C1 (but-1-ene) or C3 (but-2-ene), and but-2-ene exists as cis and trans isomers: 3 alkenes in total. Answering 2 forgets geometrical isomerism.
9. An excess of sodium metal is added to 1 mol of 3-hydroxybutanoic acid, CH3CH(OH)CH2COOH. How many moles of hydrogen gas are formed?
A
0.5
B
1.0
C
1.5
D
2.0
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Answer: B. Sodium reacts with both the alcohol –OH and the acid –OH, and each –OH gives ½H2: 2 × ½ = 1.0 mol. 0.5 counts only the –COOH group; 2.0 forgets that 2 –OH give one H2.
10. Prop-2-en-1-ol, CH2=CHCH2OH, is heated under reflux with an excess of acidified potassium dichromate(VI). What is the organic product?
A
CH2=CHCHO
B
HOCH2CH(OH)CH2OH
C
CH2=CHCOOH
D
CH3CH2COOH
Show answer
Answer: C. Acidified dichromate oxidises the primary alcohol; under reflux with excess oxidant it goes all the way to the carboxylic acid. The C=C is not attacked by dichromate, so it survives. The aldehyde would need distillation as it forms.
22, each with the answer and a short explanation, across 1 topics: Organic synthesis. Every chapter test draws ten of them, so you can take it many times.
Is the AS Level Chemistry chapter 21 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.
