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

Let's begin by carefully examining our starting material, 2-ethylbenzamide. We have a benzene ring with two substituents positioned ortho to each other: an ethyl group ($-C{H}_{2}C{H}_3$) and an amide group ($-CON{H}_2$).

In organic synthesis, whenever you see two reactive functional groups sitting right next to each other on a rigid scaffold like a benzene ring, you should immediately suspect an intramolecular reaction. Their proximity is going to be the key to solving this puzzle.

Radical Bromination

Our first set of reagents is bromine in the presence of light ($Br_2/h u$). This is the classic setup for a free radical substitution. But where will the bromine attack?

The benzylic carbon—the one directly attached to the benzene ring—is the prime target. Why? Because the free radical formed there is highly stabilized by resonance with the aromatic $\pi$-system.

So, the benzylic hydrogen is abstracted, and a bromine atom takes its place. We have now successfully converted our ethyl group into a benzylic bromide. Notice how this benzylic carbon is now attached to an electronegative bromine. It has become highly electrophilic, meaning it's hungry for electrons.

Acid-Base Chemistry

Next, we introduce dilute potassium hydroxide ($KOH$). $KOH$ is a strong base, and it's going to look for the most acidic proton available.

In our molecule, the protons on the amide nitrogen are slightly acidic due to the adjacent electron-withdrawing carbonyl group. The hydroxide ion snatches one of these protons, leaving behind a negatively charged nitrogen (an amidide anion).

Intramolecular Ring Closure

Now, look at what we've created. We have a strong nucleophile (the nitrogen anion) and a great electrophile (the benzylic carbon) right next door in the very same molecule!

This is the perfect setup for an intramolecular $S_{N}2$ reaction. The nitrogen swings around and attacks the benzylic carbon from the back, kicking out the bromide ion as a leaving group.

As the bromide leaves, a new carbon-nitrogen bond is formed, closing the loop. We have created a stable five-membered ring fused to our benzene ring. This type of cyclic amide is called a lactam. Specifically, we've formed 3-methylisoindolin-1-one.

Intramolecular reactions like this are incredibly fast and favorable, especially when they form five or six-membered rings, which have very little ring strain. Comparing our final structure to the given options, it perfectly matches Option (a).