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CBSE Class 12 Chapter Test: AMINES
General Instructions:
- This question paper consists of 16 questions divided into 5 sections: A, B, C, D, and E.
- Section A comprises 6 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
- Section B comprises 2 Very Short Answer questions of 2 marks each.
- Section C comprises 3 Short Answer questions of 3 marks each.
- Section D comprises 1 Case-Based question of 4 marks.
- Section E comprises 2 Long Answer questions of 5 marks each.
Directions for Q7 & Q8: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Reason (R): In aniline, the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring.
Reason (R): Iron scrap is highly reactive and reduces the nitro group much faster than tin.
(i) Aniline and N-methylaniline
(ii) Ethylamine and Aniline
C6H5NH2 →(NaNO2 + HCl, 273 K) A →(H3PO2 + H2O) B
(i) Primary amines have higher boiling points than tertiary amines.
(ii) Aniline does not undergo Friedel-Crafts reaction.
(iii) Ammonolysis of alkyl halides is not a suitable method for the preparation of pure primary amines.
(i) Carbylamine reaction
(ii) Hoffmann bromamide degradation
(iii) Sandmeyer's reaction
CH3CH2Br →(KCN) A →(LiAlH4) B →(HNO2, 273 K) C
Diazonium salts have the general formula Ar-N2+X-. Primary aliphatic amines form highly unstable alkyl diazonium salts, whereas primary aromatic amines form arene diazonium salts which are stable for a short time in solution at low temperatures (273-278 K). The stability of arene diazonium ions is explained on the basis of resonance. Diazonium salts are very good intermediates for the introduction of -F, -Cl, -Br, -I, -CN, and -OH groups into the aromatic ring. They also act as electrophiles and undergo coupling reactions with highly electron-rich aromatic compounds like phenols and amines to form brightly colored azo dyes.
(a) Why must the diazotization reaction be carried out strictly at low temperatures (0 to 5°C)? [1 Mark]
(b) What happens when benzene diazonium chloride is warmed with water? Write the equation. [1 Mark]
(c) Write the chemical equation for the coupling reaction of benzene diazonium chloride with phenol. Mention the color of the dye formed and the pH medium used. [2 Marks]
(i) Ethanoic acid to Methanamine
(ii) Aniline to p-Bromoaniline
(iii) Benzene to Aniline
(iv) Aniline to Sulphanilic acid
(v) Benzyl chloride to 2-Phenylethanamine
(b) Arrange the following:
(i) In increasing order of basic strength: C6H5NH2, C2H5NH2, (C2H5)2NH, NH3
(ii) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 [2 Marks]
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SOLUTIONS & MARKING SCHEME
1. Answer: (b) Benzenesulphonyl chloride
Reasoning: Hinsberg's reagent is benzenesulphonyl chloride (C6H5SO2Cl). It reacts differently with 1°, 2°, and 3° amines, making it a crucial distinguishing reagent.
2. Answer: (b) (CH3)2NH > CH3NH2 > (CH3)3N > NH3
Reasoning: In an aqueous solution, basicity is determined by a combination of the +I effect, steric hindrance, and solvation (hydration) energy. For the methyl group, this complex interplay makes the secondary amine the strongest base, followed by primary, tertiary, and finally ammonia.
3. Answer: (c) An isocyanide
Reasoning: The reaction is the Carbylamine test. Primary amines react with chloroform and alc. KOH to form isocyanides (carbylamines), which are characterized by their deeply offensive and unbearable odor.
4. Answer: (c) Aniline
Reasoning: Gabriel phthalimide synthesis involves the SN2 nucleophilic attack of the phthalimide anion on an alkyl halide. To prepare aniline, one would need to use chlorobenzene (an aryl halide), which does not undergo nucleophilic substitution under normal conditions due to C-Cl partial double bond character.
5. Answer: (a) Anilinium ion
Reasoning: In a strongly acidic nitrating mixture, the highly basic aniline gets protonated to form the anilinium ion (C6H5NH3+). The -NH3+ group is strongly deactivating and meta-directing, leading to the formation of a significant amount of m-nitroaniline.
6. Answer: (b) Br2 / aq. or alc. NaOH
Reasoning: Hoffmann bromamide degradation "steps down" an amide to a primary amine using bromine in the presence of an aqueous or ethanolic solution of a strong alkali (NaOH or KOH).
7. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: Cyclohexylamine is an aliphatic amine (+I effect increases basicity). Aniline is aromatic; its nitrogen lone pair is delocalized over the benzene ring via resonance, making it significantly less available for protonation. Thus, aniline is the weaker base.
8. Answer: (c) A is true but R is false.
Reasoning: Assertion is true: Fe/HCl is preferred. Reason is false: The actual reason it is preferred is that FeCl2 formed gets hydrolyzed to release HCl continuously during the reaction, requiring only a tiny initial catalytic amount of HCl, making it highly economical.
9. Solution:
(i) Aniline and N-methylaniline: Carbylamine Test. Aniline (1° amine) forms a foul-smelling isocyanide when heated with CHCl3 and alc. KOH. N-methylaniline (2° amine) does not respond to this test.
(ii) Ethylamine and Aniline: Azo Dye Test. Aniline reacts with NaNO2/HCl at 0-5°C followed by addition of alkaline 2-naphthol (or phenol) to form a brilliant orange/red azo dye. Ethylamine forms a highly unstable diazonium salt that decomposes to give brisk effervescence of N2 gas and ethanol, forming no dye.
10. Solution:
A = C6H5N2+Cl- (Benzene diazonium chloride). This is the diazotization step.
B = C6H6 (Benzene). Hypophosphorous acid (H3PO2) acts as a mild reducing agent, replacing the diazonium group with Hydrogen.
11. Solution:
(i) Primary amines (R-NH2) possess two N-H bonds and can form extensive intermolecular hydrogen bonds. Tertiary amines (R3N) lack N-H bonds and cannot form intermolecular hydrogen bonds. Hence, 1° amines have higher boiling points.
(ii) The catalyst in Friedel-Crafts is Anhydrous AlCl3 (a Lewis acid). Aniline has a lone pair (Lewis base) and immediately reacts with AlCl3 to form a salt. This places a positive charge on nitrogen, strongly deactivating the ring and halting further substitution.
(iii) The primary amine formed initially is nucleophilic. It further reacts with the remaining alkyl halide to form secondary, then tertiary amines, and finally quaternary ammonium salts, yielding an inseparable, complex mixture.
12. Solution:
(i) Carbylamine reaction:
R-NH2 + CHCl3 + 3KOH (alc.) →(Δ) R-NC (Isocyanide) + 3KCl + 3H2O
(ii) Hoffmann bromamide degradation:
R-CONH2 + Br2 + 4NaOH → R-NH2 (1° amine) + Na2CO3 + 2NaBr + 2H2O
(iii) Sandmeyer's reaction:
C6H5N2+Cl- + Cu2Cl2 / HCl → C6H5Cl (Chlorobenzene) + N2
13. Solution:
A = CH3CH2CN (Propanenitrile). Nucleophilic substitution of Br by CN- (steps up the chain).
B = CH3CH2CH2NH2 (Propan-1-amine). Reduction of nitrile yields a primary amine.
C = CH3CH2CH2OH (Propan-1-ol). Aliphatic primary amines react with HNO2 to form highly unstable diazonium salts which immediately decompose to yield alcohols and N2 gas.
14. Solution (Case-Based):
(a) Arene diazonium salts are highly unstable and decompose easily to form phenol and nitrogen gas if the temperature rises above 5°C (278 K). The low temperature prevents their decomposition.
(b) When warmed with water, the diazonium group is replaced by the hydroxyl (-OH) group to form phenol.
Equation: C6H5N2+Cl- + H2O →(Δ) C6H5OH (Phenol) + N2 + HCl.
(c) Coupling with Phenol:
C6H5N2+Cl- + Phenol →(OH-) p-Hydroxyazobenzene + Cl- + H2O.
Color: Orange Dye. Medium: Mildly alkaline (pH 9-10).
15. Solution:
(i) Ethanoic acid to Methanamine (Step-down):
CH3COOH + NH3 →(Δ) CH3CONH2 (Ethanamide) + H2O
CH3CONH2 + Br2 + 4NaOH → CH3NH2 (Methanamine) + Na2CO3 + 2NaBr + 2H2O.
(ii) Aniline to p-Bromoaniline (Protection required):
C6H5NH2 + (CH3CO)2O →(Pyridine) Acetanilide
Acetanilide + Br2/CH3COOH → p-Bromoacetanilide
p-Bromoacetanilide + H2O/H+ → p-Bromoaniline + CH3COOH.
(iii) Benzene to Aniline:
C6H6 + Conc. HNO3 + Conc. H2SO4 (333K) → C6H5NO2 (Nitrobenzene)
C6H5NO2 + Sn/HCl (or Fe/HCl) → C6H5NH2 (Aniline).
(iv) Aniline to Sulphanilic acid:
C6H5NH2 + Conc. H2SO4 → Anilinium hydrogensulphate
Anilinium hydrogensulphate →(Δ 453-473 K) p-Aminobenzenesulphonic acid (Sulphanilic acid) + H2O.
(v) Benzyl chloride to 2-Phenylethanamine (Step-up):
C6H5CH2Cl + ethanolic KCN → C6H5CH2CN (Phenylethanenitrile) + KCl
C6H5CH2CN + LiAlH4 → C6H5CH2CH2NH2 (2-Phenylethanamine).
16. Solution:
(a) Hinsberg's Test (Benzenesulphonyl chloride, C6H5SO2Cl):
1° Amines: React to form N-alkylbenzenesulphonamide.
Equation: C6H5SO2Cl + R-NH2 → C6H5SO2NHR + HCl.
Solubility: It contains a highly acidic hydrogen attached to Nitrogen, making it soluble in aqueous NaOH.
2° Amines: React to form N,N-dialkylbenzenesulphonamide.
Equation: C6H5SO2Cl + R2NH → C6H5SO2NR2 + HCl.
Solubility: It does not contain an acidic hydrogen, making it insoluble in aqueous NaOH.
3° Amines: Do not possess a replaceable hydrogen on nitrogen, hence they do not react with Hinsberg's reagent.
(b) (i) Increasing order of basic strength:
Aniline is the weakest due to resonance. Ethylamine vs Diethylamine in water: 2° is stronger than 1°.
Order: C6H5NH2 < NH3 < C2H5NH2 < (C2H5)2NH.
(b) (ii) Increasing order of boiling point:
Alcohols form stronger H-bonds than amines (O is more electronegative than N). Among isomeric amines, 1° forms more H-bonds than 2°.
Order: (CH3)2NH < C2H5NH2 < C2H5OH.
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