See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the product: The starting material is cyclopentanone. The product is a cyclopentene with an exocyclic =CHD at one carbon and a CD2 group at the adjacent sp3 carbon (C1). More precisely, the product appears to be 3-(deuterio)methylenecyclopent-1-ene with a CHD= on one carbon and the saturated carbon bearing two D atoms (CD2). Re-examining the structure: the product is cyclopent-2-en-1-ylidene type – a cyclopentene ring where C1 has two D substituents (CD2) and C3 has one D on the vinyl carbon (=CHD). This means the carbonyl carbon of cyclopentanone ultimately becomes CD2 (the sp3 carbon) and the alpha carbon becomes part of the double bond bearing one D (=CHD). Step 2 – Retrosynthetic analysis: The product has deuterium at the alpha position (vinylic D, one D) and two D's at what was the carbonyl carbon. To place D's at the alpha carbon via base-catalyzed H/D exchange, we use KOD/D2O first. This exchanges the alpha hydrogens for deuterium. Cyclopentanone has alpha-H's at C2 and C5. KOD/D2O will exchange alpha protons to give alpha-deuterated cyclopentanone (2,2,5,5-tetradeuterocyclopentanone or similar depending on extent). Step 3 – After KOD/D2O: The alpha positions are deuterated. The carbonyl carbon still has C=O. Step 4 – LiAlH4 reduction: LiAlH4 reduces the ketone C=O to CH-OH (or CD-OH if D is delivered, but LiAlH4 delivers H from Al-H). Wait – LiAlH4 delivers H (not D) to the carbonyl carbon. So the carbonyl carbon becomes CHOH (one H from LiAlH4) giving a cyclopentanol with deuterium at alpha positions and H at the carbinol carbon. Product after LiAlH4 and workup: 2,2-dideuterocyclopentanol (with D's at alpha carbons, H at C1 from hydride). Step 5 – H+/Δ (acid-catalyzed dehydration): The alcohol undergoes E1 or E2 elimination under acidic conditions with heat. Dehydration removes H2O. The OH at C1 leaves, and a beta-H (or beta-D) is eliminated. If a D from the alpha carbon is eliminated along with OH, the double bond forms with loss of D2O giving a C=C with remaining D on the ring carbon. The specific regiochemistry and which H/D is eliminated leads to the observed product with =CHD (one D on the vinyl carbon, from the alpha-D remaining after one D is lost during elimination) and CD2 at the sp3 carbon. Step 6 – Why this order (c) works: KOD/D2O first installs D at alpha carbons of the ketone. LiAlH4 then reduces the carbonyl to give –CHDOH or –CH2OH type alcohol (H delivered to carbonyl). H+/Δ finally dehydrates to form the alkene with the observed deuterium pattern. Step 7 – Why other orders fail: - Option (a) KOD/D2O, H+/Δ, LiAlH4: H+/Δ after KOD/D2O would dehydrate but there's no OH yet (still a ketone), so H+/Δ on a ketone just gives aldol or doesn't dehydrate usefully before reduction. - Option (b) H+/Δ first: Acid and heat on cyclopentanone before installing D doesn't help; no OH for dehydration and no D incorporated yet. - Option (d) LiAlH4 first: Reduces ketone to alcohol before D incorporation; then H+/Δ dehydrates without D at alpha, and KOD/D2O last cannot exchange vinylic positions easily to give the right product. Therefore, the correct answer is C.