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A Level H2 Chemistry Organic Chemistry Quiz

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Questions

A-Level Chemistry H2 Quiz - Organic Chemistry

Name: __________________________
Class: __________________________
Date: __________________________
Score: ______ / 45

Duration: 60 minutes
Total Marks: 45

Instructions:

Answer all questions.
Write your answers in the spaces provided.
The use of the Data Booklet is relevant to some questions.
Show all working for calculations.

Section A: Multiple Choice & Short Concepts (Questions 1-5)

1. Which of the following compounds will react with 2,4-dinitrophenylhydrazine but not with Tollens' reagent?
A. Propanal
B. Propanone
C. Propanoic acid
D. Propan-1-ol

Answer: ____________________ [1]

2. Which reagent can be used to distinguish between phenol and cyclohexanol?
A. Sodium metal
B. Aqueous bromine
C. Acidified potassium dichromate(VI)
D. Phosphorus(V) chloride

Answer: ____________________ [1]

3. State the type of reaction mechanism involved when ethene reacts with hydrogen bromide to form bromoethane.

Answer: ____________________ [1]

4. Draw the skeletal formula of the organic product formed when but-2-ene reacts with cold, dilute, alkaline potassium manganate(VII).

[1]

5. Explain why phenol is more acidic than ethanol, but less acidic than ethanoic acid.

[2]

Section B: Reactions and Mechanisms (Questions 6-10)

6. Suggest a reagent and condition to convert benzene into nitrobenzene.

Reagent: __________________________
Condition: __________________________ [2]

7. What is the major organic product when 2-methylpropene reacts with hydrogen chloride?

Answer: ____________________ [1]

8. Identify the functional group present in the compound formed when ethanoyl chloride reacts with ammonia.

Answer: ____________________ [1]

9. Compound A has the molecular formula $C_4H_8O$ . It gives a yellow precipitate with 2,4-dinitrophenylhydrazine. When warmed with Tollens' reagent, it forms a silver mirror.

(a) Deduce the structure of A.
 [1]

(b) A is reduced using $NaBH_4$ to form compound B. Draw the structure of B.
 [1]

10. Chlorobenzene and chloroethane undergo hydrolysis under different conditions.

(a) State the reagents and conditions required to hydrolyse chloroethane to ethanol.
 [1]

(b) State the reagents and conditions required to hydrolyse chlorobenzene to phenol.
 [1]

(c) Explain, in terms of bonding, why chlorobenzene is much less reactive towards nucleophilic substitution than chloroethane.
 [3]

Section C: Synthesis and Properties (Questions 11-15)

11. Consider the reaction sequence below:

$\text{Ethene} \xrightarrow{\text{Reagent 1}} \text{Bromoethane} \xrightarrow{\text{Reagent 2}} \text{Ethylamine}$

(a) Identify Reagent 1 and the type of reaction.
 [2]

(b) Identify Reagent 2 and the type of reaction.
 [2]

12. Ethylamine is a stronger base than ammonia. Explain why.

[2]

13. Ethylamine reacts with ethanoyl chloride. Write the equation for this reaction and name the organic product.

[3]

14. An ester C ( $C_4H_8O_2$ ) is hydrolysed by aqueous sodium hydroxide. One of the products is sodium ethanoate.

(a) Deduce the structure of ester C.
 [1]

(b) Write the equation for the hydrolysis of C with aqueous NaOH.
 [2]

15. Compound D is 2-bromopropane.

(a) Draw the mechanism for the reaction of D with aqueous sodium hydroxide to form propan-2-ol. Include all curly arrows, lone pairs, and charges.
 [3]

Section D: Advanced Applications (Questions 16-20)

16. If D (2-bromopropane) is reacted with ethanolic sodium hydroxide under reflux, a different product E is formed.

(a) Name product E.
 [1]

(b) Name the type of reaction.
 [1]

17. Polymer F is formed from the monomer propene.

(a) Draw the repeating unit of polymer F.
 [1]

(b) Explain why polymer F is resistant to biodegradation.
 [2]

18. The following scheme shows the conversion of benzene into phenylamine.

$\text{Benzene} \xrightarrow{\text{Step 1}} \text{Nitrobenzene} \xrightarrow{\text{Step 2}} \text{Phenylamine}$

(a) For Step 1, write the equation for the formation of the electrophile.
 [2]

(b) For Step 2, state the reagents and conditions.
 [1]

19. Phenylamine is converted to a diazonium salt.

(a) State the reagents and temperature required for this diazotisation.
 [2]

(b) Why must the temperature be kept below 10°C during diazotisation?
 [2]

20. Compound B (from Question 9) is heated with concentrated sulfuric acid to form a mixture of alkenes. Name the major alkene formed and explain why it is the major product.

(Note: Assume for this question that Compound B is butan-2-ol to allow for elimination regioselectivity discussion, or explain based on butan-1-ol if strictly following Q9. Given Q9 identified A as Butanal, B is Butan-1-ol. However, to test "major product" logic typically associated with Zaitsev's rule, let us assume the question refers to the dehydration of butan-2-ol which would be formed if A were butanone. Since A is Butanal, B is Butan-1-ol. Dehydration of primary alcohols usually yields the terminal alkene. To maintain syllabus rigor on "major products" in elimination, this question asks about the dehydration of butan-2-ol instead.)

Correction for Clarity: Let us assume Compound G is butan-2-ol. Name the major alkene formed when butan-2-ol is dehydrated and explain why.

[2]

Answers

A-Level Chemistry H2 Quiz - Organic Chemistry (Answer Key)

1. B
Reasoning: Propanone is a ketone. Ketones react with 2,4-DNPH (condensation) but do not reduce Tollens' reagent (only aldehydes do). Propanoic acid and propan-1-ol do not react with 2,4-DNPH. [1]

2. B
Reasoning: Phenol undergoes electrophilic substitution with aqueous bromine to form a white precipitate of 2,4,6-tribromophenol. Cyclohexanol does not react with aqueous bromine under these conditions. [1]

3. Electrophilic addition [1]

4.
Structure: Butane-2,3-diol.
Skeletal formula showing OH groups on carbons 2 and 3.
$CH_3-CH(OH)-CH(OH)-CH_3$
[1]

Phenol vs Ethanol: The lone pair on the oxygen in phenol delocalises into the benzene ring. This stabilises the phenoxide ion formed after deprotonation, making the O-H bond more polar and $H^+$ easier to lose. In ethanol, the ethyl group releases electrons (inductive effect), destabilising the ethoxide ion. [1]
Phenol vs Ethanoic Acid: In ethanoic acid, the negative charge on the carboxylate ion is delocalised over two electronegative oxygen atoms, providing greater stability than the delocalisation into the carbon ring in phenoxide. Thus, ethanoic acid is a stronger acid. [1]

Reagent: Concentrated $HNO_3$ and concentrated $H_2SO_4$ [1]
Condition: Temperature below 55°C (or warm at 50-55°C) [1]

7. 2-chloro-2-methylpropane [1]
Reasoning: Markovnikov's rule applies. The major product forms via the more stable tertiary carbocation intermediate.

8. Amide [1]

9.
(a) Butanal ( $CH_3CH_2CH_2CHO$ ) [1]
Reasoning: Aldehyde (positive Tollens') with 4 carbons.
(b) Butan-1-ol ( $CH_3CH_2CH_2CH_2OH$ ) [1]

10.
(a) Aqueous NaOH (or KOH), heat under reflux. [1]
(b) Aqueous NaOH, high temperature (300°C) and high pressure (60 atm). [1]
(c)

In chlorobenzene, the lone pair on the chlorine atom overlaps with the $\pi$ -system of the benzene ring (delocalisation). [1]
This gives the C-Cl bond partial double bond character, making it shorter and stronger. [1]
Therefore, it is difficult to break the C-Cl bond via nucleophilic attack. [1]

11.
(a) Reagent 1: HBr (gas or conc.). Type: Electrophilic Addition. [2]
(b) Reagent 2: Excess ethanolic ammonia, sealed tube/heat. Type: Nucleophilic Substitution. [2]

12. The ethyl group is electron-releasing (positive inductive effect). This increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton ( $H^+$ ). [2]

13.

Equation: $CH_3CH_2NH_2 + CH_3COCl \rightarrow CH_3CONHCH_2CH_3 + HCl$ [1]
Product Name: N-ethylethanamide [1]
(1 mark for correct equation balance/formulae). [1]

14.
(a) Ethyl ethanoate ( $CH_3COOCH_2CH_3$ ) [1]
Reasoning: Hydrolysis gives ethanoate ion, so the acid part is ethanoic. The alcohol part must be ethanol to sum to $C_4$ .
(b) $CH_3COOCH_2CH_3 + NaOH \rightarrow CH_3COO^-Na^+ + CH_3CH_2OH$ [2]

15.
(a) $S_N2$ Mechanism:

Arrow from lone pair on $OH^-$ to the central carbon (C2). [1]
Arrow from C-Br bond to Br atom. [1]
Product: Propan-2-ol + $Br^-$ . [1]

16.
(a) Propene [1]
(b) Elimination [1]

17.
(a)
$\begin{bmatrix} -CH(CH_3)-CH_2- \end{bmatrix}_n$
(Show bonds extending outside brackets). [1]
(b)

The polymer backbone consists of strong non-polar C-C and C-H bonds. [1]
These bonds are chemically inert and resistant to attack by enzymes or microorganisms found in nature. [1]

18.
(a)

$HNO_3 + H_2SO_4 \rightarrow NO_2^+ + HSO_4^- + H_2O$ [2]
(1 for correct reactants/products, 1 for correct charges/stoichiometry).
(b) Sn / conc HCl, heat under reflux (followed by NaOH). [1]

19.
(a) $NaNO_2$ and dilute $HCl$ (or $H_2SO_4$ ). Temperature: 0-10°C. [2]
(b) Diazonium salts are unstable and decompose explosively or form phenols randomly at higher temperatures. Keeping it cold stabilises the $N_2^+$ group for controlled substitution. [2]

20.

Major Product: But-2-ene [1]
Reason: According to Zaitsev's rule, the major product in elimination reactions is the more substituted alkene (the one with more alkyl groups attached to the double-bonded carbons), which is more stable due to hyperconjugation. [1]