Hello! I would like some clarification if possible for this question, thank you so much!

Method 1 - exocyclic methylene (═CH₂) → endocyclic alkene + methyl substituent (methylene-cycloalkane → methyl-cycloalkene), while retaining the alcohol.

Step 1 — Protect the alcohol (so acid won't dehydrate it)

Reagents: Ac₂O, pyridine (or DMAP)

Intermediate 1: same skeleton, but –OH → –OAc (acetate ester)

Step 2 — Acid-catalyzed alkene isomerization (double-bond migration)

Reagents: cat. p-TsOH (or cat. H₂SO₄), dry solvent (toluene/benzene/CH₂Cl₂), gentle heat
Intermediate 2: isomerized alkene product skeleton, still –OAc
Step 3 — Deprotect (regenerate the alcohol)

Reagents: NaOMe/MeOH or NaOH (aq/alcohol), then workup

Product: target structure with –OH and the shifted alkene + methyl


Method 2 -"hydrogenation → radical bromination → E2 elimination
H₂/Pt hydrogenation converts the exocyclic =CH₂ into a methyl substituent (–CH₃) on that ring carbon.

Br₂, hν does selective radical bromination at the most substituted (tertiary) C–H, which should be the carbon that now bears the methyl group (gives a tertiary bromide).

E2 with EtO⁻ (heat) eliminates HBr to give an alkene. In this fused system, elimination that would put a double bond at a bridgehead is strongly disfavored (Bredt's rule), so the reaction is biased toward the allowed internal alkene you want (in the 5-member ring).