See image — Amines Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the nature of each amine: - Compound 1 (aniline): primary arylamine; lone pair on N is delocalized into the benzene ring, reducing basicity. - Compound 2 (benzylamine): primary alkylamine; the CH2 group insulates N from the ring, so the lone pair is NOT delocalized into the ring; basicity is close to a typical aliphatic amine (pKaH ~9.3). - Compound 3 (2-nitroaniline): primary arylamine with an electron-withdrawing NO2 group ortho to NH2; the nitro group further reduces electron density on N both by induction and resonance (through the ring), making it less basic than aniline (pKaH ~-0.3 vs aniline ~4.6). - Compound 4 (benzamide): the nitrogen lone pair is delocalized into both the carbonyl C=O and the ring; amide nitrogen is far less basic than an arylamine (pKaH ~ -1 or lower). Step 2 - Rank from weakest to strongest base: - Compound 4 (benzamide) is the weakest base because the lone pair is strongly delocalized into the carbonyl group. - Compound 3 (2-nitroaniline) is next because the electron-withdrawing NO2 at the ortho position withdraws electron density from N via resonance and induction more than the plain phenyl ring does. - Compound 1 (aniline) is next; direct conjugation of N lone pair with ring reduces basicity but there is no additional EWG. - Compound 2 (benzylamine) is the strongest base among the four because the methylene spacer prevents conjugation of N lone pair with the ring; basicity is similar to aliphatic amines. Step 3 - Final order (weakest → strongest): 4 < 3 < 1 < 2 Step 4 - Checking options: Option (b) states 4 < 3 < 1 < 2, which matches our analysis. Why other options fail: - (a) places 2 between 4 and 1, which is wrong since benzylamine should be the strongest. - (c) places 3 between 1 and 2, but 2-nitroaniline should be weaker than plain aniline. - (d) ranks 2 as weakest, which contradicts benzylamine being an aliphatic-type amine. Therefore, the correct answer is B.