Animated Solution for Chemistry - Organic Chemistry: Consider the following bromides, The correct order of $S_{N}1$ reactivity is

The journey to mastering organic chemistry often runs through the heart of reaction mechanisms. Today, we are tackling a classic $S_{N}1$ reactivity problem. This isn't just about memorizing rules; it's about visualizing the invisible dance of electrons and understanding the stabilizing forces that dictate chemical behavior.

When we talk about the $S_{N}1$ (Substitution Nucleophilic Unimolecular) mechanism, we are talking about a two-step process. The first step—the departure of the leaving group—is the slow, rate-determining step.

Imagine the leaving group (in this case, the bromide ion, ${\text{Br}}^{-}$) packing its bags and leaving the molecule. What remains is a positively charged carbon atom, a carbocation. Because this first step is the bottleneck of the reaction, the golden rule emerges: The rate of an $S_{N}1$ reaction is directly proportional to the stability of the intermediate carbocation.

To solve our problem, we must transform our three bromides into their respective carbocations and rank them by stability.

The Power of Resonance

Let's start with molecule (B). When the bromide ion departs, it leaves behind a positive charge on a carbon that is directly adjacent to a carbon-carbon double bond. This is known as an allylic carbocation.

But it gets better. The positively charged carbon is also attached to two methyl groups, making it a tertiary ($3^{\circ }$) allylic carbocation.

Why is this so special? Resonance. The $\pi$ electrons from the adjacent double bond are not static; they can delocalize and shift towards the positive charge, spreading the burden of the electron deficiency across multiple atoms. This delocalization creates a massive stabilizing effect. Because of this powerful resonance stabilization, carbocation (B) is exceptionally stable—the undisputed champion of our trio.

The Role of Hyperconjugation

Next, we examine molecule (C). Removing the bromide ion here yields a secondary ($2^{\circ }$) carbocation.

Unlike (B), there are no double bonds nearby, which means resonance is off the table. Instead, this carbocation must rely on a different stabilizing force: hyperconjugation. Hyperconjugation involves the partial donation of electron density from adjacent $\sigma$ bonds (specifically C-H bonds) into the empty p-orbital of the carbocation.

To quantify this stability, we count the $\alpha$-hydrogens—the hydrogens attached to the carbons directly adjacent to the positive charge. - On the left, we have a $\text{CH}$ group (1 $\alpha$-H). - On the right, we have a ${\text{CH}}_3$ group (3 $\alpha$-H).

This gives us a total of $4\alpha$-hydrogens. While not as powerful as resonance, this hyperconjugative effect provides a solid baseline of stability for our secondary carbocation.

The Struggle of the Primary

Finally, we turn to molecule (A). When its bromide ion leaves, the positive charge lands on the terminal carbon, creating a primary ($1^{\circ }$) carbocation.

Primary carbocations are notoriously unstable. Let's look at its hyperconjugation to see why. The positively charged carbon is adjacent to only one carbon—a ${\text{CH}}_2$ group. This means it has only $2\alpha$-hydrogens available to offer electron density.

With no resonance and minimal hyperconjugation, carbocation (A) is highly unstable and will form very slowly, if at all, under $S_{N}1$ conditions.

Now, we bring it all together. We have evaluated the stabilizing forces for each intermediate: 1. (B) is a tertiary allylic carbocation, massively stabilized by resonance. 2. (C) is a secondary carbocation, moderately stabilized by 4 $\alpha$-hydrogens via hyperconjugation. 3. (A) is a primary carbocation, weakly stabilized by only 2 $\alpha$-hydrogens.

The order of carbocation stability is clearly (B) > (C) > (A). Since $S_{N}1$ reactivity perfectly mirrors this stability, the correct order of reactivity is also (B) > (C) > (A).

By understanding the underlying electronic effects—resonance and hyperconjugation—you can confidently predict the outcome of these reactions. Keep visualizing those electrons, and organic chemistry will become second nature!