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

Let's embark on a fascinating journey through a classic JEE organic chemistry problem. At first glance, the reactant is a six-membered ring: $4$-chlorocyclohex-$5$-en-$1$-one.

Before we even look at the reagents, let's observe the geometry of our starting material. If we assign the carbonyl carbon as position $1$, the chlorine atom sits proudly at position $4$. This means the oxygen and the chlorine are exactly opposite each other, locked in a $1,4$ (para) relationship. Keep this structural fact in your back pocket; it will be our guiding star later on.

Our first reagent is $HBr$. We know that hydrogen bromide loves to undergo electrophilic addition across a $C=C$ double bond. The reaction kicks off with the attack of the $H^{+}$ ion. But organic chemistry is all about choices—where does the proton go?

It will attach itself to the carbon that allows the formation of the most stable carbocation. If the positive charge were to form adjacent to the carbonyl group (at the alpha position), it would be disastrously unstable. Why? Because the carbonyl oxygen is highly electronegative and exerts a strong electron-withdrawing effect ($-I$ and $-M$). Therefore, the carbocation forms further away, and the bromide ion swoops in to attack, giving us a bromo-chloro intermediate.

Next, we introduce alcoholic $KOH$, a notorious agent for dehydrohalogenation ($\alpha ,\beta$-elimination). It hunts for acidic protons to strip away alongside our halogens to forge new double bonds.

Here is where the problem gets incredibly tricky. If alcoholic $KOH$ were to eliminate both $HBr$ and $HCl$ completely, we would end up with a plain cyclohexadienone, which would rapidly tautomerize into a plain phenol. But wait! Look at the options provided in the question. Every single option retains the chlorine atom.

This is a classic JEE anomaly. The question is heavily hinting at an aromatization process where the ring transforms into a stable phenol, but the chlorine must survive the ordeal.

Instead of getting tangled in a mechanistic debate about how the second double bond forms without losing chlorine (which typically requires an oxidant), we use our structural intuition. Remember that $1,4$ relationship we noted at the start? During the aromatization of the ring, the substituents do not magically migrate. The oxygen becomes the $-OH$ group of the phenol, and the chlorine stays exactly where it was.

Since they started in a $1,4$ arrangement, they must end up in a $1,4$ arrangement. This leads us flawlessly to our final answer: $para$-chlorophenol.