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.

Under H₂SO₄/H₂O, start by protonating the terminal alkene (the vinyl group) in a Markovnikov way.
That gives a carbocation on the carbon attached to the ring, and the terminal carbon becomes CH₃ (this is where the "methyl" seen in product A/B ultimately comes from).
Now a second molecule of the alkene acts as the nucleophile. Its terminal alkene attacks the carbocation ("head-to-tail" style) to form a new C–C bond and produce a more substituted (secondary) carbocation in the dimer.
From this secondary carbocation:

Path to A: simply deprotonate (E1) to form the alkene (dimer alkene A).

Path to B: do NOT eliminate. Instead, let a nearby internal double bond (the cyclohexene C=C) act as an intramolecular nucleophile.



 

Get a goog LG  (#AlkylHalide)

From an alkene you can generate OH and Br on vicinal positions (#HalohydrinFormation) using Br2/H2O (OH goes to the more substituted carbon, Br to the less substituted carbon).

Alternatively, you can make a primary bromide using HBr, ROOR (#AntiMarkovnikovAddition) or make an alcohol using BH3·THF then H2O2/NaOH (#HydroborationOxidation) and then convert it if needed.

Make a strong nucleophile (#Alkoxide) using NaH on alcohol   and do SN2

NaH (#StrongBase) deprotonates the alcohol to form an alkoxide (#AcidBaseReaction).

The alkoxide then attacks a primary alkyl bromide (#SN2Mechanism) to form the ether (#EtherSynthesis).

This works best when the halide is primary (#SN2Preferred) because secondary/tertiary halides tend to give elimination instead.

Hello, I am a little stuck on this problem, would appreciate some help. Thank you!

Hi! can i have some help with this question, thank you so much!

Hello, I would appreciate some help with this question, thank you!

Hello, I would appreciate some help with this question, thank you!

Hello, I would appreciate some help with this question, thank you!

Hello, I would appreciate some help with this question, thank you!