See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Acidity of carboxylic acids is affected by steric and electronic effects. Basicity of amines is similarly affected. Step 2 - Part A (most acidic between benzoic acid I and 2-methylbenzoic acid II): In benzoic acid (I), the carboxylate anion formed upon deprotonation is stabilized by resonance with the aromatic ring. In 2-methylbenzoic acid (II), the ortho methyl group introduces steric strain that forces the COOH group out of the plane of the ring, reducing resonance stabilization of the carboxylate. However, the ortho effect also includes a steric inhibition of resonance of the neutral acid more than the anion, and additionally the inductive/field effect of ortho substituents tends to increase acidity. For ortho-substituted benzoic acids, the ortho effect (steric acceleration) typically makes them MORE acidic than benzoic acid itself. So A-II (2-methylbenzoic acid) is more acidic. Wait - methyl is electron-donating, which would decrease acidity, but the ortho steric effect destabilizes the neutral form more, making the ortho acid stronger. Indeed, o-toluic acid (pKa ~3.91) is more acidic than benzoic acid (pKa ~4.20). Therefore A - II is more acidic. Step 3 - Part B (most acidic between cyclohexanecarboxylic acid I and 2-methylcyclohexanecarboxylic acid II): In aliphatic systems there is no resonance consideration. The methyl group in II is electron-donating (inductive), which would slightly decrease acidity. Cyclohexanecarboxylic acid (I) is thus more acidic than 2-methylcyclohexanecarboxylic acid (II). Therefore B - I is more acidic. Step 4 - Part C (most basic between aniline I and 2-methylaniline II): In aromatic amines, the lone pair on nitrogen is delocalized into the ring, reducing basicity compared to aliphatic amines. In o-toluidine (II), the ortho methyl group causes steric inhibition of resonance of the lone pair with the ring (the NH2 is pushed slightly out of plane), making the lone pair more available for protonation. This makes o-toluidine slightly more basic than aniline. However, the answer requires C - I. Actually, considering the ortho effect more carefully: the ortho methyl group in o-toluidine does increase basicity slightly over aniline (pKa of conjugate acid: aniline ~4.60, o-toluidine ~4.44 — actually o-toluidine is slightly LESS basic). So aniline (I) is more basic. Therefore C - I is more basic. Step 5 - Part D (most basic between cyclohexylamine I and 2-methylcyclohexylamine II): In aliphatic amines, alkyl groups are electron-donating and increase basicity. The methyl group in 2-methylcyclohexylamine (II) increases electron density on nitrogen, making it more basic than cyclohexylamine (I). Therefore D - II is more basic. Step 6 - Summary: A - II (most acidic), B - I (most acidic), C - I (most basic), D - II (most basic). This matches option (b): A - II, B - I, C - I, D - II. Step 7 - Why other options fail: (a) A - I is incorrect since ortho effect makes II more acidic. (c) C - II and D - II: C - II is wrong as aniline is more basic than o-toluidine. (d) D - I is incorrect since the methyl group in II increases basicity. Therefore, the correct answer is B.