Chapter 13: Amines Mock Test
Time: 1 Hour | Maximum Marks: 25
- All questions are compulsory.
- Section A contains Q1 (Multiple Choice) and Q2 (Very Short Answer).
- Section B contains Short Answer Type I questions (2 marks each). Attempt any 4.
- Section C contains Short Answer Type II questions (3 marks each). Attempt any 2.
- Section D contains Long Answer questions (4 marks each). Attempt any 1.
SECTION A
Q1. Select and write the most appropriate answer from the given alternatives: [4 Marks]
-
The correct order of basic strength of methyl substituted amines in an aqueous solution is:
(A) $CH_3NH_2 > (CH_3)_2NH > (CH_3)_3N$(B) $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$(C) $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2$(D) $(CH_3)_2NH > (CH_3)_3N > CH_3NH_2$
-
Hinsberg's reagent, used to distinguish between primary, secondary, and tertiary amines, is:
(A) Benzenesulfonyl chloride(B) Benzoyl chloride(C) p-Toluenesulfonyl chloride(D) Acetyl chloride
-
Which of the following classes of amines gives a foul-smelling compound upon heating with chloroform and alcoholic KOH?
(A) Secondary aliphatic amines(B) Tertiary aliphatic amines(C) Primary aliphatic and aromatic amines(D) Secondary aromatic amines
-
In the Hoffmann bromamide degradation reaction, the primary amine formed has:
(A) One carbon atom more than the amide(B) The same number of carbon atoms(C) One carbon atom less than the amide(D) Two carbon atoms less than the amide
Q2. Answer the following questions in one sentence: [3 Marks]
- Write the IUPAC name of $CH_3-N(CH_3)-CH_2-CH_3$.
- What happens when a primary aliphatic amine reacts with nitrous acid ($HNO_2$) at low temperature?
- Identify the catalyst used in the Sandmeyer reaction to convert benzene diazonium chloride to chlorobenzene.
SECTION B
Attempt any FOUR of the following: [8 Marks]
- Explain why aniline is less basic than ammonia.
- What is the action of bromine water on aniline at room temperature? Write the chemical equation.
- How does Hinsberg's reagent react with a primary amine? Write the reaction.
- Explain Gabriel Phthalimide Synthesis with the overall reaction sequence.
- Write a short note on the coupling reaction to form p-hydroxyazobenzene (orange dye).
SECTION C
Attempt any TWO of the following: [6 Marks]
- Write balanced chemical equations to prepare ethylamine from:
- Nitroethane
- Acetonitrile
- Propionamide
- Explain the Hoffmann bromamide degradation reaction with a suitable example. Why is it called a "degradation" reaction?
- Explain the Carbylamine test with a chemical equation. Why do secondary and tertiary amines fail to give this test?
SECTION D
Attempt any ONE of the following: [4 Marks]
- (a) What is the action of nitrous acid on aniline at low temperature (273 - 278 K)? Name the reaction. [2 Marks]
(b) Why can't primary aromatic amines be prepared by Gabriel Phthalimide synthesis? Explain. [2 Marks] - (a) Explain the relative basicity of primary, secondary, and tertiary methylamines in an aqueous solution. [3 Marks]
(b) Identify product A in the following reaction: $CH_3CN \xrightarrow{Na / C_2H_5OH} \text{A}$. [1 Mark]
Solutions & Marking Scheme
SECTION A [7 Marks]
Q1. Multiple Choice Answers:
1. (B) $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$ [1 Mark. $2^\circ > 1^\circ > 3^\circ$ due to combined inductive, solvation, and steric effects in aqueous state]
2. (A) Benzenesulfonyl chloride [1 Mark for correct option]
3. (C) Primary aliphatic and aromatic amines [1 Mark for correct option. Carbylamine test]
4. (C) One carbon atom less than the amide [1 Mark for correct option. Carbonyl carbon is lost]
Q2. Very Short Answers:
1. IUPAC Name:
N,N-Dimethylethanamine. [1 Mark]
2. Action of $HNO_2$ on $1^\circ$ aliphatic amine:
It forms a highly unstable aliphatic diazonium salt which immediately decomposes into an alcohol with the brisk effervescence of nitrogen gas. [1 Mark]
3. Sandmeyer Reaction Catalyst:
Cuprous chloride ($Cu_2Cl_2$) dissolved in HCl. [1 Mark]
SECTION B [8 Marks]
Q3. Basicity of Aniline vs Ammonia:
In aniline ($C_6H_5NH_2$), the lone pair of electrons on the nitrogen atom is in conjugation with the $\pi$-electrons of the benzene ring. Due to this resonance (+R effect), the lone pair is delocalized over the ring, making it less available for protonation. [1 Mark]
In ammonia ($NH_3$), the lone pair on nitrogen is localized and readily available for donation. Furthermore, unprotonated aniline is highly resonance-stabilized compared to the anilinium ion. Hence, aniline is a weaker base. [1 Mark]
Q4. Bromine water on Aniline:
The $-NH_2$ group is strongly activating. Aniline reacts rapidly with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline. [1 Mark]
$C_6H_5NH_2 + 3Br_2(aq) \rightarrow C_6H_2Br_3NH_2 \downarrow + 3HBr$
(Aniline) (2,4,6-Tribromoaniline)
Q5. Hinsberg's Reagent and $1^\circ$ Amine:
Primary amines react with Benzenesulfonyl chloride ($C_6H_5SO_2Cl$) to form N-alkylbenzenesulfonamide. [1 Mark]
$C_6H_5SO_2Cl + R-NH_2 \rightarrow C_6H_5SO_2NHR + HCl$ [1/2 Mark]
Due to the strong electron-withdrawing sulfonyl group, the hydrogen attached to the nitrogen is highly acidic, making the product soluble in aqueous NaOH. [1/2 Mark]
Q6. Gabriel Phthalimide Synthesis:
It is used to synthesize pure primary aliphatic amines.
- Phthalimide reacts with ethanolic KOH to form potassium phthalimide.
- This salt is heated with an alkyl halide ($R-X$) to form N-alkyl phthalimide (via $S_N2$ reaction).
- Alkaline hydrolysis of N-alkyl phthalimide yields the primary amine ($R-NH_2$) and phthalic acid salt. [2 Marks for explanation of the three steps]
Q7. Coupling Reaction (Azo Dye):
Benzene diazonium chloride reacts with phenol in a mildly alkaline medium (pH 9-10) to form a highly colored orange azo dye (p-hydroxyazobenzene). The diazonium ion acts as an electrophile and attacks the para position of the phenol ring. [1 Mark]
$C_6H_5-N_2^+Cl^- + H-C_6H_4-OH \xrightarrow{OH^-} C_6H_5-N=N-C_6H_4-OH + HCl$
[1 Mark for reaction]SECTION C [6 Marks]
Q8. Preparation of Ethylamine ($C_2H_5NH_2$):
- (i) From Nitroethane (Reduction):
$CH_3-CH_2-NO_2 + 6[H] \xrightarrow{Sn/HCl} CH_3-CH_2-NH_2 + 2H_2O$ [1 Mark] - (ii) From Acetonitrile (Mendius or Catalytic reduction):
$CH_3-C \equiv N + 4[H] \xrightarrow{Na/C_2H_5OH \text{ or } LiAlH_4} CH_3-CH_2-NH_2$ [1 Mark] - (iii) From Propionamide (Hoffmann degradation):
$CH_3-CH_2-CONH_2 + Br_2 + 4NaOH \xrightarrow{\Delta} CH_3-CH_2-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$ [1 Mark]
Q9. Hoffmann Bromamide Degradation:
When a primary amide is heated with bromine and an aqueous/ethanolic solution of NaOH or KOH, it yields a primary amine. [1 Mark]
Example (Acetamide to Methylamine):
$CH_3CONH_2 + Br_2 + 4NaOH \xrightarrow{\Delta} CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
[1 Mark]Why "Degradation"? It is called a degradation (step-down) reaction because the resulting primary amine has one carbon atom less than the parent amide. The carbonyl carbon of the amide is removed as a carbonate ion. [1 Mark]
Q10. Carbylamine Test:
When a primary amine (aliphatic or aromatic) is heated with chloroform and alcoholic KOH, it forms an isocyanide (carbylamine) which has an intolerable, foul smell. [1 Mark]
$R-NH_2 + CHCl_3 + 3KOH(alc) \xrightarrow{\Delta} R-N \equiv C + 3KCl + 3H_2O$ [1 Mark]
Why $2^\circ$ and $3^\circ$ fail: The mechanism of the carbylamine reaction requires the loss of two hydrogen atoms from the nitrogen of the amine to form the $-N \equiv C$ triple bond. Since secondary amines have only one hydrogen on nitrogen, and tertiary amines have none, they cannot form the isocyanide product. [1 Mark]
SECTION D [4 Marks]
Q11. (a) Diazotization [2 Marks] (b) Gabriel Synthesis limitation [2 Marks]
(a) Action of $HNO_2$ on Aniline (Diazotization):
Aniline reacts with nitrous acid (generated in situ from $NaNO_2$ and $HCl$) at a low temperature of 273 - 278 K (0°C to 5°C) to form a stable aromatic diazonium salt called benzene diazonium chloride. [1 Mark]
$C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273-278 \text{ K}} C_6H_5N_2^+Cl^- + 2H_2O$
Name of reaction: Diazotization. [1 Mark]
(b) Why Gabriel Phthalimide Synthesis fails for Aromatic Amines:
To prepare a primary aromatic amine (like aniline), potassium phthalimide would need to react with an aryl halide (like chlorobenzene) via nucleophilic substitution ($S_N2$). [1 Mark]
However, aryl halides do not undergo nucleophilic substitution readily because the C-X bond has a partial double bond character due to resonance. Hence, the phthalimide anion cannot displace the halogen from the benzene ring. [1 Mark]
Q12. (a) Basicity in Aqueous Solution [3 Marks] (b) Identify Product [1 Mark]
(a) Relative basicity of methylamines in aqueous solution:
The order is: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ ($2^\circ > 1^\circ > 3^\circ > \text{Ammonia}$). [1 Mark]
In the gaseous phase, the order is strictly based on the +I inductive effect ($3^\circ > 2^\circ > 1^\circ$). However, in an aqueous solution, basicity is determined by a combination of three competing factors: [1 Mark]
- +I Effect: Alkyl groups push electron density onto nitrogen, increasing basicity ($3^\circ$ should be best).
- Solvation (Hydration) Effect: The protonated amine is stabilized by hydrogen bonding with water. Primary amines have more N-H bonds to form H-bonds, making them highly stabilized ($1^\circ$ should be best).
- Steric Hindrance: Bulky methyl groups block the approach of a proton ($3^\circ$ suffers the most).
The delicate balance of these three factors makes the secondary ($2^\circ$) amine the strongest base. [1 Mark]
(b) Identify Product A:
$CH_3CN \xrightarrow{Na / C_2H_5OH} CH_3-CH_2-NH_2$
Product A is Ethylamine (Ethanamine). [1 Mark]
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