Organic synthesis: A Level Chemistry questions with answers
Free Organic synthesis questions for A Level Chemistry (9701), part of Inorganic & organic. Each question has the answer and the marking points an examiner would look for.
An azo dye is made from benzene in four steps: benzene → nitrobenzene → phenylamine → benzenediazonium chloride → azo dye (coupled with phenol).
(a)
Give the reagents and conditions for the first two steps.
[2]
(b)
Give the reagents for step 3 and explain why the temperature is kept below 10 °C.
[2]
(c)
State the conditions for step 4 and explain why azo compounds are coloured.
[2]
(d)
Explain why phenylamine is a weaker base than ammonia.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
conc HNO₃/H₂SO₄ ≤ 55 °C; Sn/conc HCl then NaOH
concentrated HNO₃ and H₂SO₄, below 55 °C B1 Sn and concentrated HCl, heat, then NaOH(aq) B1
2
(b)
NaNO₂ + HCl; diazonium salt decomposes
NaNO₂ and HCl (HNO₂) B1 the diazonium ion decomposes above 10 °C B1
2
(c)
alkaline phenol; extended delocalisation
phenol in alkaline solution B1 extended delocalised system through the N=N group absorbs visible light B1
2
(d)
N lone pair delocalised into ring
the lone pair on N is delocalised into the benzene ring M1 so it is less available to accept a proton M1
2
Question 2Structured · 10 marksA* challenge
Ethanal can be converted into 2-hydroxypropanoic acid (lactic acid) in two steps.
(a)
Give the reagents and conditions for each step, and name the intermediate.
[4]
(b)
The product of step 1 is a racemic mixture. Explain why, with reference to the mechanism.
[3]
(c)
Lactic acid forms a condensation polymer. Draw one repeat unit and name the type of polymer.
[3]
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Part
Answer and marking points
Marks
(a)
HCN with KCN catalyst (or NaCN/H+), room temperature → 2-hydroxypropanenitrile; then hydrolysis: heat under reflux with dilute HCl(aq)
HCN with CN− catalyst B1 2-hydroxypropanenitrile B1 heat with dilute acid B1 under reflux B1
4
(b)
carbonyl group is planar; CN− attacks equally from either face; equal amounts of both enantiomers formed
planar trigonal carbonyl B1 attack equally likely from either side B1 50 : 50 mixture of enantiomers, optically inactive B1
3
(c)
–O–CH(CH3)–CO– ; polyester
ester link shown B1 correct repeat unit with continuation bonds B1 polyester B1
3
Question 3Structured · 11 marks
This question is about planning organic syntheses.
(a)
Suggest a three-step synthesis of 2-hydroxypropanoic acid from ethanol. Give reagents and conditions for each step and the intermediates.
[6]
(b)
Explain why the product is optically inactive.
[2]
(c)
Identify the functional groups in a compound that reacts with sodium carbonate, gives a positive iodoform test and has a broad IR peak at 3200–3600 cm−1.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
oxidise to ethanal (K₂Cr₂O₇/H⁺, distil); HCN/KCN → 2-hydroxypropanenitrile; hydrolyse with dilute acid
step 1: acidified K₂Cr₂O₇, distil to give ethanal B1 step 2: HCN with KCN catalyst B1 intermediate 2-hydroxypropanenitrile B1 step 3: heat under reflux with dilute HCl B1 product 2-hydroxypropanoic acid B1 correct intermediates named B1
6
(b)
planar carbonyl attacked equally from both sides; racemic mixture
the CN⁻ attacks the planar C=O group equally from either side B1 forming a racemic mixture B1
2
(c)
carboxylic acid; CH₃CH(OH)– group; alcohol O–H
carboxylic acid (reacts with carbonate) B1 CH₃CH(OH)– / CH₃CO– group (iodoform) B1 O–H of alcohol (3200–3600 cm−1) B1
3
Question 4Structured · 11 marks
This question is about converting benzene into useful compounds.
(a)
Suggest a two-step route from benzene to phenylamine, giving reagents and conditions.
[4]
(b)
1-Phenylethanol, C₆H₅CH(OH)CH₃, can be made from benzene in two steps via phenylethanone, C₆H₅COCH₃. Give the reagents and conditions for each step.
[3]
(c)
Explain why the 1-phenylethanol made in this way is a racemic mixture.
[2]
(d)
Describe a chemical test to show that the second step has gone to completion.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
conc HNO₃ + conc H₂SO₄ < 55 °C → nitrobenzene; Sn + conc HCl, heat, then NaOH
concentrated HNO₃ and H₂SO₄ B1 below 55 °C B1 nitrobenzene intermediate B1 Sn and concentrated HCl under reflux, then NaOH(aq) B1
planar carbonyl attacked by H⁻ from both sides equally
NaBH₄ delivers H⁻ to the planar C=O from either face with equal probability B2
2
(d)
2,4-DNPH: no orange precipitate if no ketone left
add 2,4-DNPH: no orange precipitate shows no ketone remains B2
2
Question 5Structured · 3 marks
Chiral drugs are often made as a single enantiomer.
(a)
Explain why it is important to produce only one enantiomer of a drug.
[2]
(b)
Suggest one method of producing a single enantiomer.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
other enantiomer inactive or harmful
only one enantiomer fits the target receptor B1 the other may be inactive or have harmful side effects B1
2
(b)
chiral catalyst / enzyme
use an enzyme or chiral catalyst / start from a natural chiral molecule B1
1
Question 6Structured · 9 marks
2-Hydroxy-2-methylpropanenitrile, (CH₃)₂C(OH)CN, can be made from propene in three steps, as shown in Fig. 6.1.
Fig. 6.1
(a)
Give the reagents and conditions for each step.
[3]
(b)
Name the mechanism of step 3 and describe the first stage.
[2]
(c)
The yields of the three steps are 80%, 60% and 80%. Calculate the overall percentage yield.
[1]
(d)
Use your answer to (c) to calculate the mass of product obtained from 8.40 g of propene. (Mr: propene 42; product 85)
[2]
(e)
State a test that would show that step 2 is complete, and the result.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
steam, H₃PO₄ catalyst, heat and pressure; K₂Cr₂O₇/H₂SO₄, heat; HCN with KCN
step 1: steam with phosphoric acid catalyst (high temperature and pressure) B1 step 2: acidified potassium dichromate(VI), heat B1 step 3: HCN with a little KCN (or KCN then dilute acid) B1
nucleophilic addition B1 CN− attacks the δ+ carbonyl carbon, breaking the C=O π bond B1
2
(c)
38.4 %
0.80 × 0.60 × 0.80 × 100 = 38.4 % B1
1
(d)
6.53 g
n = 8.40/42 = 0.200 mol C1 × 0.384 × 85 = 6.53 g A1
2
(e)
warm with acidified K₂Cr₂O₇: stays orange
warm a sample with acidified potassium dichromate(VI): it stays orange (no green), so no propan-2-ol remains (allow: IR shows no broad O–H absorption at 3200–3600 cm−1) B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in