Question

The major product formed in the following reaction is

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Acid protonates the OH to make it a leaving group; the resulting carbocation rearranges (hydride/alkyl shift) to the most stable, most substituted cation before losing a proton, giving the most substituted (fully conjugated, exocyclic isopropylidene) alkene as major product.

Answer 1

The Correct Option is B

Step 1: Identify the mechanism.
Treating a bicyclic alcohol with \(\mathrm{H^+}\) is acid-catalyzed dehydration (E1). The first step is protonation of the \(\mathrm{-OH}\) to give \(\mathrm{-OH_2^+}\), a good leaving group; water then leaves to give a carbocation.

Step 2: Consider what happens to the initial carbocation.
E1 dehydration goes through the most stable carbocation available, and in polycyclic terpenoid-type skeletons like this one, the cation first formed at the carbon that lost water is not necessarily the most stable one accessible. When a more stable cation can be reached by a 1,2-hydride or alkyl (Wagner-Meerwein type) shift, that shift happens before loss of a proton, since such rearrangements are usually fast compared to the final deprotonation step.

Step 3: Apply Zaitsev's rule to the final elimination.
Once the cation has settled into its most stable form, the proton that leaves in the last step is the one that gives the most substituted, most stable alkene, per Zaitsev's rule for E1 eliminations. Option (B) shows the alkene as a fully substituted exocyclic isopropylidene system conjugated with the ring, the most substituted and thermodynamically most stable double bond among the choices, consistent with an E1 pathway through the most stable accessible cation.

Step 4: Rule out the other options.
Option (A) shows only a disubstituted exocyclic methylene with no gem-dimethyl conjugation, less stable and expected only if no rearrangement occurred. Option (C) shows an extra ring double bond (a diene), not accessible from a simple E1 dehydration of this substrate without an added oxidation step. Option (D) places the double bond fully inside the ring in a way that breaks the conjugation with the isopropylidene unit that a Wagner-Meerwein-directed elimination would favour. Only (B) matches rearrangement to the most stable cation followed by Zaitsev elimination.

Final Answer:
\[ \boxed{\text{Option (B)}} \]