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

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.