A Level Chemistry (A2) chapter 12: Nitrogen compounds MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
96 exam-style questions on Nitrogen 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. Which compound is the weakest base?
A
ethylamine
B
ammonia
C
ethanamide
D
phenylamine
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Answer: C. In ethanamide the nitrogen lone pair is delocalised over the C=O group, making it almost neutral. Phenylamine is a weak base but still stronger than an amide; ethylamine is the strongest of these.
2. Which nitrile gives 2-methylpropan-1-amine, (CH3)2CHCH2NH2, when it is reduced with hydrogen and a nickel catalyst?
A
CH3CH2CH2CN
B
(CH3)3CCN
C
(CH3)2CHCN
D
(CH3)2CHCH2CN
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Answer: C. Reduction turns –C≡N into –CH2NH2 with no change to the carbon skeleton, so the nitrile carbon becomes the CH2 next to N. (CH3)2CHCN gives (CH3)2CHCH2NH2; CH3CH2CH2CN gives butan-1-amine.
3. Why is ethylamine a stronger base than ammonia?
A
the ethyl group withdraws electrons from nitrogen, which weakens the N–H bonds
B
ethylamine has two lone pairs on its nitrogen atom, while ammonia has one
C
the lone pair of ethylamine is delocalised into the ethyl group, stabilising the ion
D
the ethyl group donates electrons to N, making the lone pair more available to accept H+
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Answer: D. Alkyl groups have a positive inductive effect, increasing electron density on the nitrogen. The lone pair accepts H+ more readily, and the ethylammonium ion is stabilised.
4. A solution of benzenediazonium chloride is warmed with water. What are the products?
A
C6H5OH, N2 and HCl
B
C6H5NH2, N2O and HCl
C
C6H5Cl, N2 and H2O
D
C6H5N=NC6H4OH and HCl
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Answer: A. C6H5N2+Cl− + H2O → C6H5OH + N2 + HCl. The azo dye needs a separate coupling partner such as phenol in alkali.
5. What is meant by the isoelectric point of an amino acid?
A
the pH at which the amino acid is fully protonated
B
the temperature at which the zwitterion decomposes
C
the pH at which the amino acid carries a charge of −1
D
the pH at which the amino acid has no overall charge
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Answer: D. At the isoelectric point the amino acid exists mainly as the zwitterion, with no net charge, so it does not move in an electric field.
6. 0.0200 mol of ethanamide is boiled with 50.0 cm3 of 1.00 mol dm−3 NaOH(aq) until hydrolysis is complete and all the ammonia has been driven off. The remaining solution is titrated with 1.00 mol dm−3 HCl(aq) using phenolphthalein, so that only the excess NaOH reacts. What volume of HCl(aq) is needed?
A
10.0 cm3
B
20.0 cm3
C
30.0 cm3
D
50.0 cm3
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Answer: C. CH3CONH2 + OH− → CH3COO− + NH3 uses 0.0200 mol of the 0.0500 mol NaOH, leaving 0.0300 mol: 30.0 cm3 of HCl. 20.0 cm3 is the NaOH used, not the NaOH left.
7. When ethylamine reacts with ethanoyl chloride, two moles of ethylamine are used for each mole of ethanoyl chloride. Why?
A
two ethyl groups become attached to the nitrogen of the amide
B
the second mole reacts with the HCl formed, giving ethylammonium chloride
C
the second mole of ethylamine acts as a catalyst and is regenerated
D
the amide formed reacts with a second mole of ethylamine to give a diamide
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Answer: B. The amide-forming step releases HCl. Ethylamine is a base, so a second mole is used up forming CH3CH2NH3+Cl−, and the product is N-ethylethanamide.
8. 0.930 g of phenylamine (Mr 93.0) reacts completely with excess bromine water at room temperature. What is the maximum mass of organic product?
A
1.72 g
B
2.51 g
C
3.30 g
D
4.79 g
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Answer: C. 0.0100 mol phenylamine gives 0.0100 mol 2,4,6-tribromophenylamine, C6H4Br3N (Mr 329.7): 3.30 g. 1.72 g assumes monosubstitution; 4.79 g is the mass of the 0.0300 mol Br2 used.
9. Ethylamine is prepared by heating bromoethane with ammonia in ethanol under pressure. Why is a large excess of ammonia used?
A
to act as a catalyst for the substitution reaction
B
to make further substitution of the ethylamine less likely
C
to neutralise the ethylamine as soon as it forms
D
to prevent elimination of HBr, which would give ethene
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Answer: B. Ethylamine is itself a nucleophile and can react with more bromoethane. With a large excess of NH3, a bromoethane molecule is much more likely to meet ammonia than ethylamine.
10. Propanamide, CH3CH2CONH2, is reduced with LiAlH4. What is the product?
A
ethylamine, CH3CH2NH2
B
propan-1-ol, CH3CH2CH2OH
C
propan-1-amine, CH3CH2CH2NH2
D
propanenitrile, CH3CH2CN
Show answer
Answer: C. LiAlH4 reduces the C=O group of the amide to CH2, keeping all three carbons and the nitrogen, so a primary amine with the same carbon chain is formed.
96, each with the answer and a short explanation, across 4 topics: Primary and secondary amines, Phenylamine and azo compounds, Amides, Amino acids. Every chapter test draws ten of them, so you can take it many times.
Is the A Level Chemistry (A2) chapter 12 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.
