AS Level Chemistry chapter 19: Nitrogen compounds MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
44 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. Ethylamine (Mr 45.0) is made from bromoethane (Mr 109) and excess ammonia with a 60.0% yield. What mass of bromoethane is needed to obtain 4.50 g of ethylamine?
A
6.54 g
B
10.9 g
C
18.2 g
D
30.3 g
Show answer
Answer: C. 4.50 g = 0.100 mol of amine; at 60.0% yield, 0.100 / 0.600 = 0.167 mol of bromoethane = 18.2 g. 10.9 g ignores the yield and 6.54 g multiplies by 0.600 instead of dividing.
2. 2-Bromobutane is heated with excess ammonia in ethanol in a sealed tube. What is the main organic product?
A
CH3CH(NH2)CH2CH3
B
CH3CH2CH2CH2NH2
C
CH3CH=CHCH3
D
CH3CH(OH)CH2CH3
Show answer
Answer: A. Substitution keeps the position of the functional group: NH2 replaces Br on carbon 2. The amine does not move to the end of the chain, and the alcohol would need aqueous hydroxide.
3. In the reaction of bromoethane with ammonia to form ethylamine, how does ammonia act?
A
as a nucleophile, using its lone pair to attack the δ+ carbon atom
B
as an electrophile, accepting electrons from bromine
C
as a free radical
D
as an acid, donating a proton to bromoethane
Show answer
Answer: A. The C–Br bond is polar, leaving the carbon δ+. The lone pair on nitrogen attacks this carbon and bromide leaves: nucleophilic substitution.
4. 3.27 g of bromoethane (Mr 109) is heated with excess ammonia, and 0.90 g of ethylamine (Mr 45.0) is obtained. What is the percentage yield?
A
27.5%
B
50.0%
C
66.7%
D
75.0%
Show answer
Answer: C. 3.27 ÷ 109 = 0.0300 mol, so the maximum is 0.0300 mol of ethylamine = 1.35 g. Yield = 0.90 ÷ 1.35 × 100 = 66.7%. 27.5% wrongly compares the two masses directly.
5. Why is a large excess of ammonia used when preparing ethylamine from bromoethane?
A
to make the reaction exothermic
B
to reduce further substitution, in which the ethylamine formed reacts with more bromoethane
C
to neutralise the ethylamine
D
because ammonia is the catalyst
Show answer
Answer: B. Ethylamine is itself a nucleophile and can attack another bromoethane molecule, giving secondary and tertiary amines. Excess ammonia makes it more likely that bromoethane meets NH3 rather than the amine.
6. In the reaction of bromoethane with ammonia, the first species formed after the C–Br bond breaks is an ion. Which ion is it?
A
CH3CH2+
B
NH4+
C
NH2−
D
CH3CH2NH3+
Show answer
Answer: D. The nitrogen forms a fourth bond to carbon, giving the ethylammonium ion; a second NH3 then removes H+ from it to give ethylamine and NH4+. That is why two NH3 are used per bromoethane.
7. Bromoethane is heated with ammonia in ethanol, but the ammonia is not in excess. Which additional product is likely to form?
A
ethene
B
ethanol
C
ethanenitrile
D
diethylamine, (CH3CH2)2NH
Show answer
Answer: D. Ethylamine also has a lone pair on nitrogen, so it can attack more bromoethane, giving diethylamine and further products. A large excess of ammonia makes this less likely.
8. Which statements about the reaction of 1-bromobutane with excess ammonia in ethanol are correct? 1 The C–Br bond breaks heterolytically. 2 Ammonia uses its lone pair to form a new C–N bond. 3 It is an addition reaction.
A
1 and 2 only
B
1 and 3 only
C
2 and 3 only
D
1, 2 and 3
Show answer
Answer: A. The bromine leaves as Br− taking both bonding electrons (heterolytic), and the nitrogen lone pair forms the C–N bond. One group replaces another, so it is substitution, not addition.
9. Propanal, CH3CH2CHO, is warmed with HCN and a little KCN. What is the name of the product?
A
1-hydroxybutanenitrile
B
2-hydroxypropanenitrile
C
3-hydroxybutanenitrile
D
2-hydroxybutanenitrile
Show answer
Answer: D. The product CH3CH2CH(OH)CN has four carbons, counting the nitrile carbon as C1, with OH on C2. 2-Hydroxypropanenitrile forgets that the CN group adds a carbon atom.
10. 1-Bromopropane and 2-bromopropane are each heated with excess ammonia in ethanol under pressure. Which statement about the two amines formed is correct?
A
They are identical, because both contain three carbon atoms.
B
They are stereoisomers with the molecular formula C3H9N.
C
They are positional isomers with the molecular formula C3H7N.
D
They are positional isomers with the molecular formula C3H9N.
Show answer
Answer: D. Propylamine, CH3CH2CH2NH2, and CH3CH(NH2)CH3 differ only in where the NH2 group is, so they are structural (positional) isomers, not stereoisomers. Each has formula C3H9N.
44, each with the answer and a short explanation, across 2 topics: Primary amines, Nitriles and hydroxynitriles. Every chapter test draws ten of them, so you can take it many times.
Is the AS Level Chemistry chapter 19 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.
