Animated Solution for Chemistry - Organic Chemistry: Increasing order of reactivity of the following compounds for $S_{N}1$ substitution is

Mastering $S_{N}1$ Reactivity

The $S_{N}1$ reaction is a classic tale of patience and stability. Unlike the $S_{N}2$ reaction, which is a single-step concerted process, the $S_{N}1$ mechanism takes its time. It proceeds in two distinct steps. The first, and most crucial, step is the departure of the leaving group to form a carbocation intermediate. This step is the bottleneck—the rate-determining step. Therefore, the golden rule of $S_{N}1$ reactivity is simple: The more stable the carbocation intermediate, the faster the $S_{N}1$ reaction.

Let's evaluate our four contenders by examining the carbocations they form.

Ethyl Chloride (B)

When ethyl chloride (Molecule B) loses its chloride ion, it forms an ethyl carbocation ($C{H}_3-C{H}_2^{+}$). This is a primary ($1^{\circ }$) carbocation. It relies solely on the hyperconjugation from three adjacent alpha-hydrogens to stabilize its positive charge. In the grand scheme of carbocations, this is very weak stabilization. Consequently, the ethyl carbocation is highly unstable, making Molecule B the least reactive in our lineup.

At first glance, 1-chloropropane (Molecule A) seems just as doomed as ethyl chloride. Upon losing its leaving group, it forms a primary propyl carbocation ($C{H}_3-C{H}_2-C{H}_2^{+}$). However, molecules are smart! This primary carbocation can undergo a rapid 1,2-hydride shift. A hydrogen atom from the adjacent carbon migrates over, shifting the positive charge to the central carbon. This creates a secondary ($2^{\circ }$) carbocation ($C{H}_3-C{H}^{+}-C{H}_3$), which is significantly more stable due to six alpha-hydrogens providing hyperconjugation. Because it can rearrange to a more stable form, Molecule A is more reactive than Molecule B.

Benzyl Chloride (D)

Now we enter the realm of resonance. When benzyl chloride (Molecule D) sheds its chloride ion, it forms a benzyl carbocation ($C_6{H}_5-C{H}_2^{+}$). The positive charge is located on a carbon atom directly attached to a benzene ring. This allows the $\pi$-electrons of the ring to delocalize and share the burden of the positive charge. This resonance stabilization is incredibly powerful, making the benzyl carbocation much more stable than any simple secondary aliphatic carbocation. Thus, Molecule D outpaces Molecule A.

Finally, we have $p$-methoxybenzyl chloride (Molecule C). Like Molecule D, it forms a benzyl-type carbocation. But it has a secret weapon: a methoxy ($-OC{H}_3$) group at the para position. The oxygen atom in the methoxy group possesses lone pairs of electrons. Through the $+R$ (resonance) effect, it donates this electron density directly into the benzene ring, which then flows to the positively charged carbon. This massive influx of electron density heavily stabilizes the carbocation. While the methoxy group also has an electron-withdrawing $-I$ effect, the $+R$ effect completely dominates at the para position. This makes the $p$-methoxybenzyl carbocation the most stable of all, crowning Molecule C as the most reactive.

By comparing the stabilities of the intermediate carbocations, we arrive at the final reactivity order:

This perfectly aligns with option (c).