Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

Imagine you are looking at a molecule that is a bit of a chemical chimera. On one side of the benzene ring, we have a methoxy group ($-OC{H}_3$), making it an ether. On the other side, we have a vinyl group ($-CH=C{H}_2$), making it an alkene. This molecule is $m$-methoxystyrene.

We are throwing this molecule into a flask with an excess of concentrated hydrobromic acid ($HBr$) and turning up the heat. When you have multiple functional groups and an excess of a highly reactive reagent, you must evaluate how each group will respond to the conditions.

Ethers are notoriously unreactive. They are the couch potatoes of organic chemistry, often used as inert solvents. However, when you introduce a strong acid like $HBr$ and apply heat, they are forced to react. Alkenes, on the other hand, are electron-rich and eagerly undergo electrophilic addition reactions with hydrogen halides. Let's break down the reaction of each functional group step by step.

The $S_{N}2$ Pathway

Let's focus on the ether group first. The oxygen atom in the methoxy group has two lone pairs of electrons. In the presence of a strong acid like $HBr$, the first thing that happens is an acid-base reaction. The oxygen atom grabs a proton ($H^{+}$) from $HBr$, becoming protonated. This transforms the methoxy group into an oxonium ion ($-O^{+}(H)C{H}_3$), which is a much better leaving group.

Now, the bromide ion ($B{r}^{-}$) is floating around, looking for a positively charged or partially positively charged carbon to attack. It has two choices: the carbon of the benzene ring or the carbon of the methyl group.

Here is where the structure of the molecule dictates the chemistry. The bond between the benzene ring and the oxygen atom has partial double bond character. Why? Because the lone pairs on the oxygen can delocalize into the pi system of the benzene ring through resonance. This makes the aryl-oxygen bond incredibly strong and very difficult to break. Furthermore, an $S_{N}2$ attack on an $s{p}^2$ hybridized carbon of a benzene ring is sterically and electronically forbidden.

Therefore, the bromide ion takes the path of least resistance. It attacks the less sterically hindered $s{p}^3$ hybridized methyl carbon via an $S_{N}2$ mechanism. The carbon-oxygen bond breaks, releasing methyl bromide ($C{H}_{3}Br$) as a byproduct and leaving behind a phenol group ($-OH$) attached to the benzene ring.

Markovnikov's Rule

Now that the ether has been cleaved, let's turn our attention to the alkene group. Because we are using an excess of $HBr$, the carbon-carbon double bond will also react. This is a classic electrophilic addition reaction.

The pi electrons of the alkene are nucleophilic. They reach out and attack the electrophilic proton ($H^{+}$) from another molecule of $HBr$. But which carbon gets the proton? According to Markovnikov's rule, the proton adds to the carbon that has more hydrogens, which in this case is the terminal carbon ($=C{H}_2$).

This regioselectivity is driven by the stability of the intermediate carbocation. By adding the proton to the terminal carbon, a secondary carbocation is formed at the benzylic position ($-C^{+}H-C{H}_3$). This benzylic carbocation is exceptionally stable because the empty p-orbital can overlap with the pi system of the adjacent benzene ring, allowing the positive charge to be delocalized over the ortho and para positions. If the proton had added to the internal carbon, a primary carbocation would have formed, which is highly unstable and not resonance-stabilized.

With the highly stable benzylic carbocation formed, the final step is a rapid nucleophilic attack. The bromide ion ($B{r}^{-}$) attacks the positively charged carbon, forming a new carbon-bromine bond. This completes the hydrobromination of the alkene, resulting in an alkyl bromide side chain ($-CH(Br)C{H}_3$).

Bringing it all together, both functional groups have successfully reacted. The methoxy group was cleaved to form a phenol, and the vinyl group underwent electrophilic addition to form a 1-bromoethyl group. The final major product is 1-(3-hydroxyphenyl)ethyl bromide.

It is crucial to note the importance of the reaction conditions. If we had used only one equivalent of $HBr$ at a cold temperature, the ether would have remained intact, and only the alkene would have reacted. Ether cleavage requires the brute force of excess acid and high heat. Always pay close attention to the reagents and conditions provided in the problem, as they are the directors of the chemical symphony!