See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the compound. The structure shown is vinylamine (ethenamine): H2N–CH=CH2. The question asks for the most stable conformation about the N–C single bond (the bond between nitrogen and the vinyl carbon). Step 2 – Recognize the two limiting conformations about the N–C bond. • Conformation A (pyramidal / sp3 N): The nitrogen lone pair is NOT conjugated with the C=C pi system. N is sp3-hybridized, pyramidal, with H–N–H bond angle ≈107°. The lone pair occupies a tetrahedral orbital pointing away from the pi bond. • Conformation B (planar / sp2 N): The nitrogen lone pair IS conjugated with the C=C pi system (lone pair parallel to the p-orbitals of the double bond). N becomes sp2-hybridized, the entire H2N–CH=CH2 framework is planar, and H–N–H bond angle opens to ≈120°. Step 3 – Determine which is more stable. When the nitrogen lone pair overlaps with the adjacent pi system (as in enamines and vinylamines), resonance stabilization occurs: H2N–CH=CH2 ↔ H2N+=CH–CH2^–. This delocalization lowers the overall energy of the molecule. Therefore the planar sp2 form (Conformation B) is MORE stable due to resonance/conjugation. Step 4 – Describe the Newman projection about the N–C bond for the most stable conformation. Looking down the N→C bond: the front atom (N) has two H's and a lone pair; the back atom (C) has one H and is part of the C=C. In the most stable (planar, conjugated) conformation, the lone pair on N eclipses the pi bond (lone pair is syn/parallel to the pi system), making the molecule planar. The H–N–H angle is ~120° (sp2 N). Step 5 – Why the pyramidal form is less stable. Although sp3 nitrogen is the ground state for simple amines (e.g., NH3, ~107°), the adjacent pi bond provides a driving force for rehybridization to sp2 because the resonance energy gained outweighs the cost of rehybridization. Hence the pyramidal conformation about N–C is the less stable form. Step 6 – Summary. The resonance (non-conjugated) structure depicts sp3 N, pyramidal, H–N–H ≈107°. The pi-complex (conjugated) planar form depicts sp2 N, H–N–H ≈120°, and this is the most stable conformation about the N–C bond. Therefore, the correct answer is Resonance structure has H-N-H bond angle ~107deg (sp3 N, pyramidal); pi-complex (vacant p-orbital) form is planar (sp2 N) with H-N-H ~120deg.