JEE Main 2022 — Hydrocarbons Question with Solution

In the first step of hydroboration oxidation, the boron and hydrogen add to the same face of the double bond. In the second step, the OH takes the identical position occupied by the boron. Thus, the OH and H groups that are effectively added to the original double bond are cis. The overall addition takes place in anti Markovnikov's fashion.

The reagent is mercury(II) acetate, which dissociates slightly to form $+\mathrm{Hg}\left(\mathrm{OAc}\right)$. $+\mathrm{Hg}\left(\mathrm{OAc}\right)$ is the electrophile that adds to the pi bond. • The intermediate is a three-membered ring called the mercurinium ion. Overall, the addition of water follows Markovnikov’s rule

 

The reaction does not suffer from rearrangements because there is no carbocation intermediate.