Can you explain to me how to do this kind of question type?

How do I approach this kind of problem?

Hi Mam! I was wondering how to solve this question (with bond enthalpies).

 This is different approach, but the ideas are the same as what we did earlier.

Protonation of the alcohol converts it into a good leaving group.

H₂O leaves, forming a carbocation. This carbocation then undergoes rearrangement to give a more stable carbocation adjacent to the oxygen of the remaining –OH (stabilized by lone pair of oxygen). During this process, ring expansion can occur.

Ring-expansion preference:

If the rearrangement would expand a 6-membered ring to a 7-membered ring, that pathway is generally less favored (more strain / less favorable geometry).

If the rearrangement would expand a 5-membered ring to a 6-membered ring, that pathway is favored because a 6-membered ring is typically more stable than a 5-membered ring.

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).



 

I originally thought one product would be result of formation good LG, then E1 elimination to alkene. When tertiary carbocation forms, could other OH group attack electrophile?

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

Hello! I think my thinking is right, but I would love some clarification for this question. I think major product has two alkenes, but wasn't sure how to go about the other OH, if it is internal nucelophilic attack, for example

To convert the cyclohexene (ring double bond) into the major alcohol shown (anti-Markovnikov on the ring):

1. BH₃·THF
2. H₂O₂, NaOH (hydroboration–oxidation)

This puts OH at the less substituted ring carbon (as drawn).
Step effects

TsCl, pyridine: converts the ring –OH → –OTs (same carbon).

Oxymercuration in water: adds OH Markovnikov to the terminal alkene side chain.

NaBD₄ (demercuration): delivers D to the carbon that originally held Hg → for a terminal alkene this ends up as D on the terminal carbon.
Dess–Martin oxidizes the side-chain secondary alcohol → ketone (tosylate stays unchanged).

Competing pathways
✅ Desired: SN2 (Williamson ether)
Nucleophile:Electrophil,allylic secondary bromide ,Product, allylic ether
Allylic halides are much more reactive toward substitution because the transition state is stabilized (and SN1/SN2 paths are both facilitated). So you can still get a good amount of Williamson ether.
Yes, elimination is possible because it's 2° halide, but allylic activation makes substitution competitive/likely, so the Williamson ether product is still a reasonable intended answer. In a real lab, you might get a mixture (ether + conjugated diene), and you'd optimize solvent/temperature to favor SN2.