 Solution

Number the epoxide-bearing ring carbon carrying CH${}_2$OH as C1, then proceed through the other epoxide carbon C2 around the ring to C6. Heating with water opens the epoxide at the less substituted C2 by backside attack: the incoming oxygen becomes C2–OH on the front face, while the original epoxide oxygen remains as C1–OH on the back face. Proton transfers complete formation of the triol. Thus the two original epoxide oxygens' roles are distinct: one is retained from B and the new hydroxyl oxygen comes from water.

Iodine next activates the remaining C4=C5 bond as an iodonium ion. The pendant primary alcohol oxygen attacks intramolecularly at C5 to make the five-membered cyclic ether with path O–CH${}_2$–C1–C6–C5, opening the iodonium anti and leaving iodine at C4 on the opposite face. Bicarbonate removes the oxonium proton and neutralizes acid; it is not the source of the ether oxygen. The folded tether selects the face of closure compatible with the bridged structure, accounting for the ether/iodine stereochemistry.

The remaining two hydroxyl groups react with acid-activated 2,2-dimethoxypropane. Successive oxygen attack, methanol departure and intramolecular closure form the isopropylidene ketal of C, protecting that diol without changing its carbon configurations. The terminal primary hydroxyl has already become the cyclic ether and is not one of the two acetonide oxygens. This atom tracking distinguishes the epoxide hydrolysis, iodoetherification and protection stages.