Question

The stereoisomer(s) of G giving the depicted product is(are):

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

In stereospecific \(E2\) eliminations, anti-periplanar geometry is crucial: leaving group and \(\beta\)-H must be opposite in the same plane.
Always check relative positions of OH (leaving group) and adjacent hydrogens when predicting alkene formation.
Enantiomeric stereoisomers that maintain this anti geometry will both lead to the same achiral alkene product.

Answer 1

The Correct Option is A, C

Step 1: Reaction type.
The transformation shown is an acid-catalyzed elimination (\(E1cb/E2\)-like, via BF\(_3\)·OEt\(_2\)) of the alcohol to form an alkene. For stereospecific elimination, the \(\beta\)-hydrogen and the leaving group (OH after activation) must be in an anti-periplanar orientation.
Step 2: Analyze stereochemistry of G.
The substrate G has two adjacent stereocenters: one bearing the OH group and a methyl, the other with bulky substituents. To generate the observed alkene product, only those stereoisomers where the \(\beta\)-H is anti-periplanar to the OH can undergo elimination.
Step 3: Evaluate the options.

(A): Conformation allows the required anti-periplanar alignment → productive.
(B): Has syn-periplanar OH–H alignment → not productive.
(C): Mirror stereoisomer of (A) with correct anti arrangement → productive.
(D): Syn relationship again → not productive.
Step 4: Conclusion.
Therefore, only stereoisomers (A) and (C) give the depicted alkene product.