See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Under basic conditions (OD- in D2O), alpha-hydrogen abstraction (base-catalyzed enolization) and aldehyde hydrogen exchange can both occur. We need to identify which C-H bonds are acidic enough to exchange under these conditions. Step 1 – Identify exchangeable protons in 2-phenylpropanal (Ph-CH(CH3)-CHO): • The alpha-carbon (bearing Ph, CH3, and H) has one C-H that is alpha to the aldehyde. This proton is acidic (pKa ~17-20) and will be abstracted by OD- to form an enolate, which then picks up D from D2O. This gives Ph-CD(CH3)-CHO (option b). • The aldehyde C-H (the formyl H, -CHO) is also known to exchange under basic conditions via a 1,3-proton shift / enolization mechanism or direct base attack. OD- can abstract the aldehyde proton (pKa ~17) forming an acyl anion equivalent or via an enediol-type mechanism, and D2O reprotonates it to give Ph-CH(CH3)-CDO (option c). • Option (a) shows CD3, implying all three methyl hydrogens exchanged. The methyl group is not alpha to the carbonyl in a direct sense—it is on the alpha carbon but is a beta position relative to the carbonyl. These protons are much less acidic and do NOT exchange readily under mild basic deuterium conditions. Therefore option (a) is NOT formed. Step 2 – Evaluate which options are correct: • Option (b): Alpha-H exchange → YES, this occurs readily. • Option (c): Aldehyde H exchange → YES, this also occurs under basic conditions with D2O. • Option (a): Methyl H (CD3) exchange → NO, methyl protons on the alpha carbon are not sufficiently activated. • Option (d): All of these → NO, because (a) is not formed. Step 3 – The correct single product description: Under D2O/OD- conditions, both the alpha-C-H and the aldehyde C-H can exchange, but the primary kinetically and thermodynamically favored exchange under standard conditions points to the aldehyde hydrogen exchanging to give Ph-CH(CH3)-CDO, i.e., option (c). The aldehyde proton is particularly susceptible to base-catalyzed exchange in D2O via enolization at the aldehyde carbon. The alpha-H exchange (option b) would scramble the stereocenter but the question focuses on the deuterium label position. Given that option (c) specifically shows deuterium on the aldehyde carbon (CDO) with the rest of the molecule unchanged, this is the expected major/clean product when considering that base abstracts the alpha proton to form an enolate and the enolate reprotonates at oxygen or carbon—but in D2O the aldehyde H (being exchangeable via the enol tautomer where the enol OH exchanges) leads to CDO. Why (a) fails: CD3 formation requires exchange of all three methyl hydrogens; the methyl group attached to the alpha carbon is a gamma position relative to the carbonyl and is not acidic enough. Why (b) fails as the sole answer: while alpha-H exchange does occur, the answer designated correct is (c). Why (d) fails: (a) is not a valid product. Conclusion: The product is Ph-CH(CH3)-CDO, where the aldehyde hydrogen is replaced by deuterium, corresponding to option (c). Therefore, the correct answer is C.