Question

The major product formed in the following reaction is:

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

\(\mathrm{SmI_2}\) reduces the ketone to a ketyl radical that cyclizes onto the tethered \(\alpha,\beta\)-unsaturated ester through carbon, not oxygen; count the chain length to see it is a 5-exo-trig cyclization, giving a fused bicyclic alcohol with a pendant \(\mathrm{CH_2CO_2Me}\) group after the \(\mathrm{MeOH}\) quench.

Answer 1

The Correct Option is B

Step 1: Identify the substrate and the reagent.
The starting material is a cyclopentanone bearing, on the ring carbon next to the carbonyl, a side chain \(\mathrm{{-}CH_2{-}CH_2{-}CH{=}CH{-}CO_2Me}\), an \(\alpha,\beta\)-unsaturated ester (a Michael acceptor) tethered to the ketone through two \(\mathrm{CH_2}\) groups. \(\mathrm{SmI_2}\) in \(\mathrm{THF}\) is a single-electron reducing agent well known for generating ketyl radicals from ketones and aldehydes, which can then cyclize onto a pendant alkene, a ketyl-olefin radical cyclization. \(\mathrm{MeOH}\) is added afterward as a proton source to quench the resulting samarium species.

Step 2: Generate the ketyl radical.
\(\mathrm{SmI_2}\) transfers one electron to the ketone carbonyl, converting it into a ketyl radical anion. The carbonyl carbon, now radical in character, is tethered through the ring's alpha carbon and the two-carbon \(\mathrm{CH_2CH_2}\) chain to the alkene of the unsaturated ester.

Step 3: Count atoms to find the ring size formed on cyclization.
Counting from the ketyl carbon (radical center) to the nearer alkene carbon: ketyl carbon (1), ring alpha-carbon bearing the chain (2), first \(\mathrm{CH_2}\) (3), second \(\mathrm{CH_2}\) (4), and the nearer alkene carbon (5). Cyclization of the ketyl radical onto this nearer alkene carbon closes a new five-membered ring, a 5-exo-trig radical cyclization, strongly favored kinetically over larger ring sizes. This forms a new carbon-carbon bond and fuses a second five-membered ring onto the original cyclopentane, giving a bicyclic, hydrindane-type, fused 5,5-bicyclic skeleton.

Step 4: Follow the radical onto the ester and quench with MeOH.
After the new \(\mathrm{C{-}C}\) bond forms, the radical, or after a second electron transfer the samarium enolate, sits on the carbon alpha to the ester carbonyl, stabilized by conjugation with the \(\mathrm{CO_2Me}\) group. \(\mathrm{MeOH}\), added at the end, protonates this samarium enolate at the alpha carbon, restoring a simple \(\mathrm{{-}CH_2{-}CO_2Me}\) group, the carbon-carbon double bond of the original unsaturated ester is gone, consumed in forming the new ring bond. Meanwhile, the original ketone oxygen, now on the ring-fusion carbon, is protonated to a tertiary alcohol once the ketyl electron pair is used to make the new \(\mathrm{C{-}C}\) bond.

Step 5: Assemble the product.
The product is a bicyclic, fused 5,5, hydrocarbon skeleton carrying a tertiary \(\mathrm{OH}\) at the ring-fusion carbon, from the reduced ketone, and a pendant \(\mathrm{{-}CH_2{-}CO_2Me}\) group on the new ring, exactly as drawn in option (B).

Why the other options are wrong:
Option (A) shows an oxygen-containing six-membered ring (a pyran), which would only form if the ketyl oxygen itself, rather than the ketyl carbon, made the new bond to the alkene, that is not how \(\mathrm{SmI_2}\) ketyl-olefin cyclizations work, the new bond forms at carbon.
Option (D) likewise shows an oxygen in the new ring (an oxepane), for the same reason, this does not match a carbon-radical cyclization.
Option (C) shows a six-membered carbocycle fused to the cyclopentane, with the ester carbon directly attached to the ring, corresponding to a 6-exo cyclization pathway with no \(\mathrm{CH_2}\) linker left over, which does not match the atom count of the actual tether, the tether length gives 5-exo, forming a five-membered ring with a pendant \(\mathrm{CH_2CO_2Me}\), not a direct ring-fused ester.

Final Answer:
The product of the \(\mathrm{SmI_2}\)-mediated ketyl-olefin radical cyclization, quenched with \(\mathrm{MeOH}\), is the fused bicyclic alcohol with a pendant \(\mathrm{{-}CH_2CO_2Me}\) group shown in option (B). \[ \boxed{\text{(B)}} \]