Animated Solution for Chemistry - Organic Chemistry: The increasing order of reactivity of the following halides for the $S_{N}1$ reaction is I. $C{H}_{3}CH(Cl)C{H}_{2}C{H}_3$ II. $C{H}_{3}C{H}_{2}C{H}_{2}Cl$ III. $p-H_{3}CO-C_6{H}_4-C{H}_{2}Cl$

When we dive into the world of nucleophilic substitution, the $S_{N}1$ (Substitution Nucleophilic Unimolecular) reaction stands out as a fascinating two-step dance. Unlike its cousin, the $S_{N}2$ reaction, which happens in a single concerted step, the $S_{N}1$ mechanism requires patience.

The first step is the slow, agonizing departure of the leaving group (in our case, a chloride ion, $C{l}^{-}$). This heterolytic bond cleavage leaves behind a carbon atom that is desperately short of electrons—a carbocation. Because this first step is the hardest and slowest, it acts as the rate-determining step (RDS).

Therefore, the golden rule of $S_{N}1$ reactions is simple: The faster you can form the carbocation, the faster the overall reaction. And how do you form a carbocation quickly? By making sure it is as stable as possible once it forms.

Carbocation Stability

To solve our problem, we must transform our three alkyl halides into their respective carbocations and judge them in a stability contest. Let's bring out our contenders.

Let's look at our second halide, propyl chloride ($C{H}_{3}C{H}_{2}C{H}_{2}Cl$). When the chloride ion packs its bags and leaves, we are left with a primary ($1^{\circ }$) carbocation: $C{H}_3-C{H}_2-C{H}_2^{+}$.

This positively charged carbon is attached to only one other carbon atom. To stabilize its positive charge, it relies on hyperconjugation—the delocalization of electrons from adjacent $C-H$ bonds. However, it only has 2 $\alpha$-hydrogens available to help out. In the harsh world of carbocations, this is barely enough to survive. Consequently, this primary carbocation is highly unstable and forms very slowly.

Next up is 2-chlorobutane ($C{H}_{3}CH(Cl)C{H}_{2}C{H}_3$). The loss of chloride here yields a secondary ($2^{\circ }$) carbocation: $C{H}_3-C{H}^{+}-C{H}_2-C{H}_3$.

Things are looking up! The positively charged carbon is now flanked by two alkyl groups. If we count the adjacent $C-H$ bonds, we find 5 $\alpha$-hydrogens (three from the methyl group and two from the methylene group). More $\alpha$-hydrogens mean more hyperconjugative structures, which translates to a much better distribution of the positive charge. This secondary carbocation is significantly more stable than our primary contender.

Finally, we examine the third halide, $p-H_{3}CO-C_6{H}_4-C{H}_{2}Cl$. Removing the chloride gives us a p-methoxybenzyl carbocation: $p-H_{3}CO-C_6{H}_4-C{H}_2^{+}$.

This is where the magic happens. The positive charge is located on a benzylic carbon, meaning it is directly adjacent to a benzene ring. This allows the positive charge to be delocalized around the ring via resonance.

But wait, there's more! At the para position sits a methoxy group ($-OC{H}_3$). The oxygen atom has lone pairs of electrons that it can generously donate into the ring through the $+M$ (mesomeric) effect. This pumps electron density directly toward the electron-deficient benzylic carbon, providing massive stabilization. Thanks to the combined forces of resonance and the $+M$ effect, this is undoubtedly the most stable carbocation of the trio.

We have our stability rankings: 1. Least Stable: Primary carbocation (from Compound II) 2. Moderately Stable: Secondary carbocation (from Compound I) 3. Most Stable: p-Methoxybenzyl carbocation (from Compound III)

Since $S_{N}1$ reactivity perfectly mirrors carbocation stability, the increasing order of reactivity is II < I < III.

Whenever you encounter an $S_{N}1$ problem, your immediate reflex should be to draw the carbocation intermediate. Look for resonance first, then hyperconjugation, and finally inductive effects.

As a thought experiment, ask yourself: What if the reagent was a strong nucleophile in a polar aprotic solvent, favoring an $S_{N}2$ mechanism? In that scenario, carbocation stability goes out the window, and steric hindrance becomes the deciding factor. The bulky benzyl and secondary halides would struggle, making the primary halide the undisputed champion. Always read the reaction conditions carefully!