A Level Chemistry (A2) chapter 14: Organic synthesis MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
24 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 reagent could be used to distinguish phenol from phenylmethanol, C6H5CH2OH?
A
bromine water
B
sodium metal
C
ethanoyl chloride
D
dilute hydrochloric acid
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Answer: A. Phenol decolourises bromine water and gives a white precipitate because its ring is activated by the –OH group; the ring in phenylmethanol is not activated in this way. Both compounds fizz with sodium and react with ethanoyl chloride, and neither reacts with dilute HCl.
2. Propan-1-ol is converted in four steps: 1: heat under reflux with acidified K2Cr2O7 2: SOCl2 3: CH3NH2 at room temperature 4: LiAlH4 in dry ether What is the final organic product?
A
CH3CH2CONHCH3
B
CH3CH2CH2NH2
C
CH3CH2CH2NHCH3
D
CH3CH2CH2OH
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Answer: C. Step 1 gives propanoic acid, step 2 propanoyl chloride, step 3 N-methylpropanamide, and step 4 reduces the amide C=O to CH2, giving the secondary amine N-methylpropan-1-amine. CH3CH2CONHCH3 is the product after step 3 only.
3. Methylbenzene is treated with chlorine in ultraviolet light to make (chloromethyl)benzene, C6H5CH2Cl. Which organic by-product is most likely?
A
4-chloromethylbenzene, ClC6H4CH3
B
phenylmethanol, C6H5CH2OH
C
(dichloromethyl)benzene, C6H5CHCl2
D
chlorobenzene, C6H5Cl
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Answer: C. In UV light, free-radical substitution happens on the side chain, and the product can be substituted again to give C6H5CHCl2. Ring substitution (4-chloromethylbenzene) needs a halogen carrier such as AlCl3 in the dark; no water is present to form an alcohol.
4. Bromoethane is converted to propylamine in two steps: 1: heat with KCN in ethanol → CH3CH2CN 2: CH3CH2CN → CH3CH2CH2NH2 What are the types of reaction in step 1 and step 2?
A
nucleophilic addition; hydrolysis
B
nucleophilic substitution; reduction
C
electrophilic substitution; reduction
D
nucleophilic substitution; hydrolysis
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Answer: B. CN− replaces Br− (nucleophilic substitution), then the nitrile is reduced with LiAlH4 or H2/Ni. Hydrolysis of the nitrile would give propanoic acid, not the amine.
5. 4-aminophenol, H2NC6H4OH, is treated with an excess of ethanoyl chloride at room temperature. What is the organic product?
A
CH3CONHC6H4OH
B
CH3CONHC6H4OCOCH3
C
H2NC6H4OCOCH3
D
CH3COC6H3(NH2)OH
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Answer: B. With excess acyl chloride both groups react: the –NH2 forms an amide and the phenolic –OH forms an ester, releasing two HCl. Putting CH3CO– on the ring needs a Friedel–Crafts catalyst.
6. Which sequence converts benzene into phenol?
A
chlorinate with Cl2/AlCl3; then heat under reflux with concentrated NaOH(aq)
B
nitrate; NaNO2 and dilute HCl below 10 °C; then warm the solution with water
C
nitrate; reduce (Sn/HCl, then NaOH); NaNO2/HCl below 10 °C; warm with water
D
nitrate; reduce (Sn/HCl, then NaOH); then warm the product with water at 50 °C
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Answer: C. Phenylamine must be made first and then diazotised; the diazonium ion reacts with water on warming to give phenol. Chlorobenzene is not hydrolysed by NaOH(aq), and nitrobenzene cannot be diazotised.
7. 4-(hydroxymethyl)phenol, HOC6H4CH2OH, has one –OH group on the ring and one on a side-chain carbon. How many moles of NaOH(aq) and of sodium metal react with one mole of it?
A
NaOH: 2 mol; Na: 2 mol
B
NaOH: 1 mol; Na: 2 mol
C
NaOH: 1 mol; Na: 1 mol
D
NaOH: 0 mol; Na: 2 mol
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Answer: B. Only the phenolic –OH is acidic enough to react with NaOH(aq). Sodium reacts with both the phenolic and the alcoholic –OH groups (giving 1 mol of H2 overall).
8. Which sequence of reagents converts benzene into N-phenylethanamide, C6H5NHCOCH3?
A
CH3COCl and AlCl3, heat; then NH3 in ethanol, heated under pressure
B
Cl2 and AlCl3; then NH3 in ethanol under pressure; then CH3COCl
C
nitrate (conc. HNO3/H2SO4); then CH3COCl; then reduce (Sn/conc. HCl)
D
nitrate (conc. HNO3/H2SO4); reduce (Sn/conc. HCl, then NaOH); then CH3COCl
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Answer: D. Nitration and reduction give phenylamine, which reacts with ethanoyl chloride to form the amide. Chlorobenzene does not undergo substitution by ammonia, and Friedel–Crafts acylation gives a ketone, not an amide.
9. In a synthesis, nitrobenzene is converted into phenylamine. Which reagents are used, and what type of reaction is this?
A
conc. HNO3 and conc. H2SO4; electrophilic substitution
B
Sn and conc. HCl, heat, then NaOH(aq); reduction
C
NH3 in ethanol under pressure; nucleophilic substitution
D
acidified KMnO4, heat; oxidation
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Answer: B. The –NO2 group gains hydrogen and loses oxygen to become –NH2, so this is a reduction. The alkali then frees the amine from its salt.
10. Compound X, C8H8O2, gives these results: • an orange precipitate with 2,4-dinitrophenylhydrazine • no silver mirror with Tollens' reagent • dissolves in NaOH(aq) but gives no effervescence with NaHCO3(aq) Which is X?
A
HOC6H4CH2CHO
B
CH3C6H4COOH
C
HOC6H4COCH3
D
CH3COOC6H5
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Answer: C. 2,4-DNPH shows a C=O of an aldehyde or ketone, and the negative Tollens' test rules out the aldehyde. Dissolving in NaOH but not reacting with NaHCO3 shows a phenol rather than a carboxylic acid. An ester C=O does not react with 2,4-DNPH.
24, 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 A Level Chemistry (A2) chapter 14 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.
