Question

The major product formed in the following reaction is \(\; \mathrm{nBu_3SnH,\ AIBN,\ C_6H_6,\ heat}\)

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Under \(\mathrm{nBu_3SnH/AIBN}\), intramolecular \(\mathrm{C{-}C}\) bond formation usually follows Baldwin’s rules: \(5\)-exo-trig \(>\) \(6\)-endo-trig.
For chair substrates, prefer transition states that place forming bonds pseudo-equatorial to minimize A\(^{1,3}\) strain—this often dictates the trans product.
Tin hydride steps: (i) halogen abstraction to make a radical, (ii) cyclization, (iii) H-atom transfer to cap the radical.

Answer 1

The Correct Option is B

Step 1: Radical generation under \(\mathrm{nBu_3SnH/AIBN}\).
Thermal initiation gives the \(\mathrm{Bu_3Sn\cdot}\) radical, which abstracts iodine from the substrate to produce a carbon radical at the iodo-substituted carbon:
\[ \mathrm{R{-}CH{-}I} \xrightarrow[\text{AIBN, heat}]{\mathrm{Bu_3Sn\cdot}} \mathrm{R{-}CH\cdot} + \mathrm{Bu_3SnI}. \] Step 2: Intramolecular radical addition follows Baldwin’s rules.
The tether bears a terminal alkene; the carbon radical undergoes 5-exo-trig cyclization onto the alkene to form a new \(\mathrm{C{-}C}\) bond and a radical at the adjacent carbon. According to Baldwin’s rules for radical cyclizations, \(5\text{-exo-trig} \gg 6\text{-endo-trig}\); thus formation of a \(\mathrm{THF}\)-like five-membered O-heterocycle is preferred. Step 3: Stereochemical outcome (chairlike transition state).
The cyclohexane adopts a chair; the radical approaches the alkene from the less hindered face, placing the newly formed bond in a pseudo-equatorial orientation. This minimizes A\(^{1,3}\) strain and gives the trans relationship between the substituents depicted in option (B). Step 4: Radical termination (hydrogen transfer).
The newly formed carbon radical is quenched by \(\mathrm{nBu_3SnH}\): \(\mathrm{C\cdot + Bu_3SnH \to C{-}H + Bu_3Sn\cdot}\), delivering the saturated product corresponding to (B).