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

Let's look at our starting molecule, $3$-hydroxycyclohexanone. It's a cyclohexane ring equipped with two distinct functional groups: a ketone and a secondary alcohol. We are treating it with two reagents sequentially: first with phosphorus tribromide ($PB{r}_3$), and then with alcoholic potassium hydroxide (alc. $KOH$).

Substitution with $PB{r}_3$

Our first reagent is phosphorus tribromide. What does it do? It's a classic reagent for converting alcohols into alkyl bromides. It will selectively attack the $-OH$ group and replace it with a bromine atom via an $S_{N}2$ mechanism.

The ketone group remains completely unaffected by $PB{r}_3$. The resulting intermediate is $3$-bromocyclohexanone.

Elimination with alc. $KOH$

Now for the second step. We add alcoholic $KOH$. This is a strong base, and its primary job is to cause dehydrohalogenation, which is an $E2$ elimination reaction. Specifically, it will pull off a $\beta$-hydrogen and kick out the bromide leaving group to form a carbon-carbon double bond.

But wait, look closely at the intermediate. The carbon with the bromine has two neighboring carbons, $C2$ and $C4$, both of which have hydrogens. This means the base has two choices. It can remove a proton from $C2$, or it can remove a proton from $C4$. This will lead to two different alkene products.

Let's draw both possibilities.

If we remove the hydrogen from $C4$, we get a double bond between $C3$ and $C4$. This is an isolated double bond, separated from the carbonyl group by an $s{p}^3$ hybridized carbon.

But if we remove the hydrogen from $C2$, the new double bond forms between $C2$ and $C3$, right next to the carbonyl group. This creates an $\alpha ,\beta$-unsaturated ketone, which is a conjugated system. In this system, the $\pi$ electrons can delocalize across the $C=C$ and $C=O$ bonds.

Conjugation brings a massive boost in thermodynamic stability due to resonance. Because the conjugated alkene is thermodynamically much more stable, the elimination will overwhelmingly favor this pathway. Therefore, cyclohex-$2$-en-$1$-one is our major product.