Question 1Level 2 · ExamSingle correct
Amines are classified as primary, secondary or tertiary by the number of carbon atoms bonded to nitrogen, and as aryl (aromatic) amines when nitrogen is bonded directly to a benzene ring. Which one of the following is a secondary aryl amine? Exactly one option is correct.
A2-Methylpropan-2-amine, (CH3)3C–NH2
BN-Methyl-1-phenylmethanamine, C6H5CH2NHCH3
CN,N-Dimethylaniline, C6H5N(CH3)2
DN-Methylaniline, C6H5NHCH3
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Answer: D
Count the carbon atoms on nitrogen and check whether one of them is a ring carbon.
(A) N carries one carbon (and two H): primary aliphatic amine. The carbon is tertiary, but the amine is classified by nitrogen, not by carbon.
(B) N carries two carbons (CH2 of benzyl and CH3): secondary, but neither carbon is a ring carbon, so it is an aliphatic (aralkyl) amine.
(C) N carries three carbons, one a ring carbon: tertiary aryl amine.
(D) N carries two carbons, one a ring carbon, plus one H: secondary aryl amine.
Answer: (D).
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
Consider (I) C6H5NH2, (II) (C2H5)2NH, (III) (C2H5)3N, (IV) C2H5NH2 and (V) NH3. The correct order of increasing pKb in water at 298 K is ______. Exactly one option is correct.
AII < IV < III < V < I
BII < III < IV < V < I
CIII < II < IV < V < I
DIV < II < III < V < I
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Answer: B
Lower pKb means a stronger base. In water, basicity depends on the +I effect of alkyl groups, on solvation (H-bonding) of the protonated amine, and on steric hindrance to solvation.
For ethyl groups the order of basicity is (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 (pKb = 3.00, 3.25, 3.29, 4.75). The tertiary amine is less basic than the secondary one because its cation is poorly solvated and crowded, but the larger ethyl groups still leave it slightly more basic than ethanamine. (For methyl amines the order is different: (CH3)2NH > CH3NH2 > (CH3)3N.)
Aniline is far weaker (pKb = 9.38) because its nitrogen lone pair is delocalised into the ring.
Increasing pKb: II < III < IV < V < I.
Answer: (B).
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
Gabriel synthesis is carried out by heating potassium phthalimide with an organic halide and then hydrolysing the product with aqueous NaOH. Which one of the following amines can be prepared in good yield by this method? Exactly one option is correct.
AAniline, C6H5NH2
BN-Methylethanamine, CH3CH2NHCH3
C2-Phenylethan-1-amine, C6H5CH2CH2NH2
D2-Methylpropan-2-amine, (CH3)3CNH2
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Answer: C
The phthalimide anion displaces halide by nucleophilic substitution (SN2) to give an N-alkylphthalimide; hydrolysis then frees a primary amine.
(A) Aniline would need bromobenzene; aryl halides do not undergo this substitution, so aniline cannot be made this way.
(B) The method gives only primary amines; a secondary amine is not formed.
(C) 2-Bromo-1-phenylethane, C6H5CH2CH2Br, is a primary halide; it reacts cleanly and hydrolysis gives C6H5CH2CH2NH2.
(D) This would need a tertiary halide, (CH3)3CBr, which does not undergo SN2 with the bulky phthalimide anion (elimination dominates).
Answer: (C).
Difficulty: Very hard