Animated Solution for Chemistry - Organic Chemistry: The major product in the following conversion is

Imagine you are presented with a molecule that has multiple reactive sites. It's like a puzzle waiting to be solved. In this problem, we are given an aryl alkyl ether that also features an alkene side chain. The reagent is excess $HBr$ accompanied by heat. The word "excess" is a massive clue—it tells us that the reaction won't stop after just one functional group reacts. We must evaluate every part of the molecule that can interact with a strong acid.

The Ether Cleavage (Zeisel's Method)

Let's start with the ether group on the left side of the molecule: $C{H}_3-O-Ar$. Ethers are generally quite stable and unreactive, but when subjected to strong acids like $HBr$ and high temperatures, they undergo cleavage. This classic reaction is known as Zeisel's method.

The first step is the protonation of the ether oxygen. The lone pair on the oxygen atom acts as a base and grabs a proton ($H^{+}$) from $HBr$, forming a highly reactive oxonium ion: $C{H}_3-\stackrel{+}{O}(H)-Ar$.

Now, the bromide ion ($B{r}^{-}$) must attack to cleave the ether. But which carbon-oxygen bond will it break? The bond between the oxygen and the benzene ring is exceptionally strong because the oxygen's lone pairs delocalize into the aromatic ring, giving the $O-Ar$ bond partial double-bond character. Therefore, the $B{r}^{-}$ nucleophile takes the path of least resistance. It attacks the less sterically hindered methyl carbon via an $S_{N}2$ mechanism, kicking off the phenol group as the leaving group.

The result of this first phase is the formation of methyl bromide ($C{H}_{3}Br$) and a phenol derivative: $HO-C_6{H}_4-CH=CH-C{H}_3$.

The Alkene Addition

Because we used excess $HBr$, the reaction continues. We now turn our attention to the alkene side chain on the right. Alkenes readily undergo electrophilic addition with hydrogen halides.

The $\pi$ electrons of the double bond attack another proton from the excess $HBr$. According to Markovnikov's rule, the proton adds to the carbon atom that already has more hydrogen atoms, ensuring the formation of the most stable carbocation intermediate.

In our molecule, protonation of the terminal carbon yields a secondary carbocation at the benzylic position: $HO-C_6{H}_4-\stackrel{+}{C}H-C{H}_2-C{H}_3$.

Resonance Stabilization

Why is this specific benzylic carbocation so incredibly favored? This is where the magic of organic chemistry happens. Look back at the phenol group we formed in Phase 1. The oxygen atom of the $-OH$ group has lone pairs that can delocalize into the benzene ring.

Through the $+R$ effect (positive resonance effect), electron density flows from the oxygen, through the conjugated $\pi$ system of the benzene ring, all the way to the positively charged benzylic carbon at the para position. This extensive delocalization provides massive stabilization to the carbocation, making its formation highly favorable.

With our super-stable benzylic carbocation formed, the final step is a simple nucleophilic attack. Another bromide ion ($B{r}^{-}$) swoops in and bonds with the positively charged carbon.

The final major product is $HO-C_6{H}_4-CH(Br)-C{H}_2-C{H}_3$.

This problem is a beautiful demonstration of how different functional groups in a molecule can react sequentially, and how an intermediate formed in one step (the phenol) can dramatically influence the stability and outcome of the next step (the benzylic carbocation). Always keep an eye out for resonance—it is the ultimate stabilizing force in organic chemistry!