Animated Solution for Chemistry - Organic Chemistry: The correct order of reactivity of the given chlorides with acetate in acetic acid is

Decoding the Reaction Conditions

Welcome to this fantastic organic chemistry problem! We are asked to find the correct order of reactivity for three different cyclic chlorides when they react with acetate in acetic acid.

The very first step in any organic reaction is to decode the environment. Here, acetic acid is a polar protic solvent, and acetate is a relatively weak nucleophile in this medium. This setup screams one thing: the $S_{N}1$ mechanism! In an $S_{N}1$ reaction, the leaving group (the chloride ion) departs first, forming a carbocation intermediate. So, our entire focus must be on the stability of the carbocations formed by these three molecules.

Carbocation Stability

Let's recall the golden rule of $S_{N}1$ reactions. The rate-determining step is the formation of the carbocation. Therefore, the more stable the carbocation intermediate, the lower the activation energy, and the faster the reaction will proceed.

To compare their stabilities, we will look at resonance, inductive effects, and hyperconjugation. Let's ionize each molecule one by one and inspect the resulting carbocations.

The Heavyweight Champion

Let's start with our first molecule, 3-chloro-6-methylcyclohex-1-ene (let's call it Molecule A). When the chloride ion leaves from the top carbon, it leaves behind a positive charge. Notice that this carbon is right next to a double bond, making it an allylic carbocation.

Now, let's count the $\alpha$-hydrogens for hyperconjugation. The adjacent carbon in the ring, which is not part of the double bond, has two hydrogens. The methyl group is at the bottom of the ring, far away from the positive charge, so it doesn't provide any $\alpha$-hydrogens. Therefore, we have exactly $2\alpha$-hydrogens stabilizing this carbocation.

The Middle Ground

Moving on to the second molecule, 3-chloro-4-methylcyclohex-1-ene (Molecule B). Here, the chloride is again on the top carbon, but the methyl group is right next to it.

When the chloride leaves, we get another secondary allylic carbocation. Let's count its $\alpha$-hydrogens. The adjacent carbon in the ring now has a methyl group attached to it, which means it only has one hydrogen left! Thus, we only have $1\alpha$-hydrogen stabilizing this carbocation.

The Underdog

Finally, let's look at the third molecule, 1-(chloromethyl)cyclohex-1-ene (Molecule C). The chloride is on a methyl group outside the ring. When it departs, the positive charge forms on this exocyclic carbon. Because it's adjacent to the ring's double bond, it is a primary allylic carbocation.

Now, how many $\alpha$-hydrogens does it have? The only adjacent carbon is the one in the ring, which is already making four bonds—two in the double bond, one to the rest of the ring, and one to our positively charged carbon. It has $0\alpha$-hydrogens! This means it gets absolutely no stabilization from hyperconjugation.

Now we can easily rank their stabilities. All three are resonance-stabilized allylic carbocations, so hyperconjugation is the ultimate tie-breaker.

Carbocation A has $2\alpha$-hydrogens, making it the most stable. Carbocation B has $1\alpha$-hydrogen, placing it in the middle. Carbocation C has $0\alpha$-hydrogens, making it the least stable.

Since the rate of an $S_{N}1$ reaction is directly proportional to carbocation stability, the reactivity order is A > B > C. This perfectly matches option (a)!

Before we wrap up, think about this: what if we changed the conditions? If we used a strong nucleophile in a polar aprotic solvent like acetone, the mechanism would shift to $S_{N}2$. In an $S_{N}2$ reaction, steric hindrance is the deciding factor, not carbocation stability. In that case, the primary chloride C would be the most reactive, and the tertiary chloride A would be the least reactive, completely reversing our order! Always pay attention to the solvent and the nucleophile.