Animated Solution for Chemistry - Organic Chemistry: ${\text{CH}}_3\text{Br}+{\text{Nu}}^{-}⟶{\text{CH}}_3-\text{Nu}+{\text{Br}}^{-}$ The decreasing order of the rate of the above reaction with nucleophile (${\text{Nu}}^{-}$) A and D is $[{\text{Nu}}^{-}=(\text{A}){\text{ PhO}}^{-},(\text{B}){\text{AcO}}^{-},(\text{C}){\text{HO}}^{-},(\text{D}){\text{CH}}_3{\text{O}}^{-}]$

Unmasking the $S_{N}2$ Reaction Rate

Welcome to a fascinating exploration of nucleophilic substitution! In this problem, we are tasked with determining the rate of an $S_{N}2$ reaction where various nucleophiles attack methyl bromide. The core principle here is simple: the rate of an $S_{N}2$ reaction is directly proportional to the strength of the attacking nucleophile.

But what exactly makes a nucleophile "strong"? A nucleophile is an electron-rich species seeking a positively charged center to donate its electrons. If the negative charge on the nucleophile is highly stable and comfortable where it is, it will be reluctant to donate those electrons. Conversely, a less stable, highly concentrated negative charge makes for an aggressive, strong nucleophile.

Delocalized Charges

Let's divide our four candidates into two distinct groups based on how their negative charge is distributed.

On one side, we have methoxide (${\text{CH}}_3{\text{O}}^{-}$) and hydroxide (${\text{HO}}^{-}$). In these ions, the negative charge is strictly localized on a single oxygen atom.

On the other side, we have phenoxide (${\text{PhO}}^{-}$) and acetate (${\text{AcO}}^{-}$). These ions enjoy the stabilizing effect of resonance, meaning their negative charge is delocalized over multiple atoms. Because delocalization significantly increases stability, the localized anions (C and D) are inherently much stronger nucleophiles than the delocalized ones (A and B).

Methoxide vs. Hydroxide

Now, let's compare the two localized champions: methoxide (D) and hydroxide (C).

Look closely at the methyl group in methoxide. Alkyl groups are known for their positive inductive effect ($+I$). The methyl group pushes electron density towards the already negatively charged oxygen atom. This intensifies the electron density on the oxygen, making it even more unstable and reactive compared to the oxygen in hydroxide, which only has a neutral hydrogen atom attached.

Therefore, methoxide is a stronger nucleophile than hydroxide ($D>C$).

Phenoxide vs. Acetate

Finally, let's evaluate the resonance-stabilized contenders: phenoxide (A) and acetate (B).

Both are stabilized by resonance, but the quality of that resonance differs drastically. In phenoxide, the negative charge is delocalized over one electronegative oxygen atom and several less electronegative carbon atoms in the benzene ring.

In acetate, however, the negative charge is shared equally between two highly electronegative oxygen atoms. This creates equivalent resonance structures, which are exceptionally stabilizing. Because the acetate ion is so incredibly stable and "happy" with its charge distribution, it is a very weak nucleophile.

Therefore, phenoxide is a stronger nucleophile than acetate ($A>B$).

Combining all our logical deductions, we find that the localized anions beat the delocalized ones, methoxide beats hydroxide, and phenoxide beats acetate.

The final decreasing order of nucleophilicity, and thus the rate of the $S_{N}2$ reaction, is D > C > A > B. This perfectly matches option (a).