Animated Solution for Chemistry - Organic Chemistry: The acidic hydrolysis of ether (X) shown below is fastest when

Accelerating Ether Hydrolysis

Imagine you are a molecular architect tasked with speeding up a chemical reaction. In this problem, we are looking at the acidic hydrolysis of a very bulky ether, specifically a trityl ether, which breaks down into a trityl alcohol and another alcohol molecule. The core question is: which structural modification will make this reaction proceed at the fastest possible rate?

To answer this, we cannot just guess; we must dive deep into the reaction mechanism. The overall reaction is given by:

The Protonation

The first step in the acidic hydrolysis of any ether is the protonation of the oxygen atom. In an acidic medium, the ether oxygen uses its lone pairs of electrons to attack a proton (${\text{H}}^{+}$). This step is generally fast and reversible.

Why does this happen? An unprotonated alkoxide group ($-\text{OR}$) is a terrible leaving group because it is a strong base. By protonating the oxygen, we convert it into a neutral alcohol molecule ($\text{ROH}$), which is an excellent leaving group.

The Cleavage (The Bottleneck)

Now comes the crucial part of the mechanism. The protonated ether undergoes heterolytic cleavage. The carbon-oxygen bond breaks, and the alcohol molecule departs, leaving behind a positively charged carbon atom—a carbocation.

Because this step involves breaking a stable covalent bond to form a highly reactive, high-energy intermediate, it requires a significant amount of activation energy. Consequently, this is the slow step, or the rate-determining step (RDS) of the entire reaction.

Carbocation Stability

Since the formation of the carbocation is the bottleneck, the rate of the overall reaction depends entirely on how stable this intermediate is. The golden rule of organic mechanisms applies here: The more stable the intermediate, the faster it is formed.

In our unmodified molecule, the intermediate is a trityl carbocation (${\text{Ph}}_3{\text{C}}^{+}$). This is already a very stable carbocation because the positive charge is delocalized over three phenyl rings via resonance. But we want to make it even faster.

To increase the stability of a carbocation, we need to attach electron-donating groups (EDGs) that can pump electron density toward the electron-deficient carbon.

Let's evaluate the options:

Option A: Replacing a phenyl group with a methyl group. A methyl group (${\text{-CH}}_3$) is an electron-donating group. It stabilizes carbocations through the $+\text{I}$ (inductive) effect and hyperconjugation. However, these effects are relatively weak compared to resonance. In fact, replacing a highly resonance-stabilizing phenyl group with a methyl group would likely decrease the overall stability of the trityl system.

Option B & C: The para-methoxyphenyl group. A para-methoxy group (${\text{-OCH}}_3$) attached to a phenyl ring is a game-changer. The oxygen atom in the methoxy group has lone pairs of electrons. Through resonance, these lone pairs can be delocalized into the benzene ring and all the way to the positively charged central carbon. This is known as a strong $+\text{M}$ (mesomeric) effect.

The $+\text{M}$ effect is incredibly powerful at stabilizing positive charges. Therefore, replacing a phenyl group with a para-methoxyphenyl group will drastically lower the activation energy and speed up the reaction.

Naturally, if one para-methoxyphenyl group is good, two are even better! Replacing two phenyl groups with two para-methoxyphenyl groups provides double the mesomeric stabilization, making the carbocation exceptionally stable.

By maximizing the stability of the carbocation intermediate formed in the rate-determining step, we maximize the rate of the reaction. The strong $+\text{M}$ effect of two para-methoxyphenyl groups provides the greatest stabilization among the choices. Therefore, the acidic hydrolysis is fastest when two phenyl groups are replaced by two para-methoxyphenyl groups.