Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

When approaching complex organic reactions, the first step is always to map out the reactive sites. In this problem, our reactant is phenacyl bromide (also known as 2-bromo-1-phenylethan-1-one).

Looking closely at its structure, we can identify two distinct functional groups that are prone to attack: a highly electrophilic ketone ($C=O$) and a primary alkyl bromide ($-C{H}_{2}Br$). The key to solving this problem lies in understanding how our chosen reagent interacts with these specific groups.

We are treating the molecule with sodium borohydride ($NaB{H}_4$) in methanol. $NaB{H}_4$ is a classic, mild, and highly chemoselective reducing agent.

Unlike its aggressive cousin $LiAl{H}_4$, $NaB{H}_4$ is gentle. It serves as a reliable source of nucleophilic hydride ions ($H^{-}$) that specifically target the electrophilic carbon of aldehydes and ketones. Crucially, under standard conditions, $NaB{H}_4$ is not strong enough to reduce isolated carbon-carbon double bonds or displace alkyl halides. Therefore, we can confidently predict that the initial attack will happen exclusively at the ketone.

As the hydride ion ($H^{-}$) attacks the carbonyl carbon, the $\pi$-electrons are pushed up onto the oxygen atom, generating an alkoxide intermediate ($O^{-}$).

Normally, in a simple reduction, this alkoxide would patiently wait to pick up a proton ($H^{+}$) from the methanol solvent to form a stable secondary alcohol. However, organic chemistry is rarely that simple!

Right next door to our newly formed, negatively charged oxygen nucleophile is a carbon atom bonded to a bromine atom. Bromine is an excellent leaving group. Because the nucleophile and the electrophile are tethered together in the same molecule, the conditions are perfect for an intramolecular $S_{N}2$ reaction.

Intramolecular reactions—especially those forming 3, 5, or 6-membered rings—are kinetically extremely fast. The alkoxide oxygen swings around and attacks the adjacent carbon from the backside, kicking out the bromide ion ($B{r}^{-}$).

This rapid cyclization forms a strained but highly favored three-membered cyclic ether, known as an epoxide. The final major product of this elegant sequence is styrene oxide (2-phenyloxirane). This problem is a beautiful reminder to always keep an eye out for hidden intramolecular traps in JEE chemistry!