See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the nitrogen type in each compound. x = Piperidine: a saturated secondary amine (sp3 nitrogen). The lone pair on nitrogen is fully available for protonation. This is a typical aliphatic amine, highly basic (pKa of conjugate acid ~11). y = 2-Piperidinone (delta-valerolactam): an amide nitrogen. The lone pair on nitrogen is delocalized into the adjacent carbonyl (C=O) via resonance, making it far less available for protonation. Amides are very weakly basic (pKa of conjugate acid ~0 or negative). z = 1,2,3,4-Tetrahydropyridine (cyclic imine, C=N present): an sp2 imine nitrogen. The lone pair occupies an sp2 orbital (in the plane of the double bond) and is not part of the pi system, so it is available for protonation but the sp2 hybridization means it is held more tightly than an sp3 nitrogen. Imines are moderately basic (pKa of conjugate acid ~5–6), much less than aliphatic amines but much more than amides. Step 2 – Rank basicity. Aliphatic amine (x) >> Imine (z) >> Amide (y) Rationale: x has sp3 N with no resonance delocalization → most basic. z has sp2 N; lone pair is available but less so than sp3 → intermediate. y has amide N; lone pair is strongly delocalized into C=O → least basic. Step 3 – Match to options. Decreasing order: x > z > y, which corresponds to option (b). Step 4 – Why other options fail. (a) x > y > z: wrong because amide (y) is far less basic than imine (z). (c) y > x > z: wrong because amide is least basic, not most. (d) y > z > x: wrong for the same reason. Therefore, the correct answer is B.