Question

M and N in the given reaction scheme are (Simmons–Smith on the alkene, then OsO\(_4\)/NMO):

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Simmons–Smith on allylic alcohols is \textbf{directed}: the cyclopropane forms \textbf{syn} to the OH-bearing face.
OsO\(_4\) (with NMO) adds \textbf{syn} to alkenes, giving \textbf{cis}-1,2-diols.
Combine directing/syn rules to set relative configurations in sequences.

Answer 1

The Correct Option is C

Step 1: Simmons–Smith (CH\(_2\)I\(_2\), Zn–Cu). Allylic alcohols direct Simmons–Smith cyclopropanation to the \emph{same} face as the OH (chelation of the carbenoid to the alcohol). Therefore, the double bond of the cyclohexene is converted to a fused cyclopropane (bicyclo[4.1.0]heptane framework) with the newly formed CH\(_2\) group \emph{syn} to the existing OH. This matches the structure of M shown in option (C).
Step 2: OsO\(_4\)/NMO dihydroxylation. Osmium tetroxide adds syn across the C=C to give a vicinal \emph{cis}-diol. Applying this to the substrate furnishes a triol in which the two newly introduced OH groups are on the same face (cis to each other), as depicted for N in option (C).
Step 3: Eliminate other options.

(A) and (D) depict cyclopropanation from the face \emph{anti} to the directing OH (contrary to Simmons–Smith OH-directing effect).
(B) and (D) show triol stereochemistries inconsistent with syn dihydroxylation.
Hence, (C) correctly represents both M and N.