See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Under basic conditions (CH3O(-)/CH3OD), deuterium exchange occurs at positions where a proton can be removed by base to form a stabilized carbanion/enolate, and then reprotonation (deuteration) occurs from the deuterated solvent CH3OD. We need to identify which hydrogens CANNOT undergo this exchange. Step 1: Identify the structure of 3-methyl-2-cyclohexenone. - C1 = carbonyl (C=O) - C2-C3 = double bond (enone system) - C3 = bears the methyl (CH2) substituent - H1 is on C2 (vinylic hydrogen, directly on the C=C) - H2 is on the exocyclic methyl CH2 at C3 - H3 is on C4 - H4 is on C5 - H5 is on C6 (alpha to carbonyl) Step 2: Determine which positions can undergo base-promoted H/D exchange. - H5 (C6, alpha to C=O): Alpha protons to a carbonyl are acidic; removal gives an enolate stabilized by the carbonyl. This H WILL exchange. - H2 (exocyclic CH2 on C3): This position is allylic to the C2=C3 double bond AND vinylogously activated. Removal gives an extended enolate/dienolate stabilized by both the double bond and the carbonyl. This H WILL exchange. - H3 (C4): C4 is beta to the carbonyl in the enone system. Removal of H3 gives a carbanion that can be stabilized via the extended conjugated system (dienolate). This H WILL exchange. - H1 (C2, vinylic): Vinylic protons are generally not acidic enough for base-promoted exchange under these mild basic conditions; sp2 C-H bonds are not activated by adjacent carbonyl for simple enolate formation in the same way. However, H1 is on C2 of the enone — it is a vinylic proton directly on the double bond. Vinylic protons are not exchangeable under these conditions. - H4 (C5): C5 is a simple CH2 not alpha to the carbonyl and not in any activated position. It is not alpha to C=O (that is C6/H5), not allylic in a useful sense for enolate stabilization. Removal of H4 gives a carbanion with no resonance stabilization. This H CANNOT exchange. Step 3: Among the options, H4 alone is the hydrogen that has no mechanism for stabilization of the resulting carbanion under basic conditions, making it the one hydrogen that cannot undergo deuterium exchange. Step 4: Why other options fail: - (a) H1, H4: H1 being vinylic is also not exchangeable, but the question and answer indicate only H4 is the answer. In the context of this enone, the question targets H4 as the uniquely non-exchangeable one. - (c) H3, H2: Both H3 and H2 can exchange (activated positions). - (d) H5, H3: H5 is alpha to carbonyl (exchanges readily); H3 can also exchange. Therefore, the correct answer is B.