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.

 
Step 1 (reduce ketone to alcohol):

NaBH₄, MeOH (or LiAlH₄, then workup)
Gives 2-propanol
Step 2 (dehydrate alcohol to alkene):

conc. H₂SO₄, heat (or POCl₃/pyridine for a cleaner dehydration)
Gives propene

Step 3 (make alkyne from alkene):
1) Br₂ (CCl₄ or CH₂Cl₂) → vicinal dibromide
2) excess NaNH₂, liquid NH₃ (typically 2–3 equiv) → double elimination
3) H₂O / NH₄Cl workup

Step 1: Protonation (activation of the strained σ bond)

H₃O⁺ protonates the bicyclobutane (effectively at/into the very strained central bond).

This generates a bicyclobutonium ion (a nonclassical, bridged cation).
Think of it as the bicyclobutane analogue of a protonated alkene: you've created a cationic "σ-complex".

Step 2: Strain-release ring opening → the more stable carbocation

One of the bicyclobutane C–C bonds cleaves to relieve strain, giving a cyclopropylcarbinyl cation (often in equilibrium with a bicyclobutonium form).

Regioselectivity: the bond breaks in the direction that places the positive charge at the substituted carbon (the carbon bearing the alkyl chain and the methyl group), because that gives the most stabilized (tertiary) carbocation character.

At this point you have exactly the cationic center that, when trapped, becomes the carbon bearing OR in the product.

Step 3: Nucleophilic capture by the alcohol (ROH)

1-propanol attacks the carbocation center (the substituted carbon) to form the new C–O bond.

This gives an oxonium ion (R–O⁺H attached to the carbon skeleton).

Step 4: Deprotonation (regenerate acid)

Water (or another ROH) removes the proton from the oxonium ion.

Product is the ring-opened cyclopropane ether shown, and H₃O⁺ is regenerated (acid-catalyzed).

Notes that help you justify B

The cyclopropane ring in the product is the "residue" of bicyclobutane after selective bond cleavage (strain release).

The alkyl chloride chain is spectator here (it doesn't ionize under these conditions relative to the bicyclobutane activation).

If the nucleophile were H₂O instead of ROH, you'd get the analogous alcohol at the same carbon.