Animated Solution for Chemistry - Organic Chemistry: HBr reacts with ${\text{CH}}_2=\text{CH}-{\text{OCH}}_3$ under anhydrous conditions at room temperature to give

Electrophilic Addition

Imagine you are looking at a molecule of ${\text{CH}}_2=\text{CH}-{\text{OCH}}_3$. This molecule features a carbon-carbon double bond, which is essentially a dense cloud of $\pi$-electrons. Because it is so electron-rich, it acts as a nucleophile, eagerly waiting to share its electrons with an electron-deficient species, known as an electrophile.

Enter $\text{HBr}$. Under anhydrous conditions, $\text{HBr}$ acts as a strong acid, providing the highly electrophilic ${\text{H}}^{+}$ ion. The moment the ${\text{H}}^{+}$ ion approaches the alkene, the $\pi$-electrons reach out to grab it. This is the initiation of a classic electrophilic addition reaction.

Two Possible Intermediates

When the ${\text{H}}^{+}$ ion attacks the double bond, it faces a choice. The double bond connects two different carbon atoms: the terminal carbon (${\text{CH}}_2$) and the inner carbon ($\text{CH}$). Depending on which carbon the ${\text{H}}^{+}$ attaches to, we get two entirely different carbocation intermediates.

If the ${\text{H}}^{+}$ attaches to the inner carbon, the positive charge is left on the terminal carbon, forming a primary carbocation: ${\text{CH}}_2^{+}-{\text{CH}}_2-{\text{OCH}}_3$ (Let's call this Intermediate I).

Conversely, if the ${\text{H}}^{+}$ attaches to the terminal carbon, the positive charge is left on the inner carbon, forming a secondary carbocation: ${\text{CH}}_3-{\text{CH}}^{+}-{\text{OCH}}_3$ (Let's call this Intermediate II).

In organic chemistry, the path a reaction takes is almost always dictated by the stability of its intermediates. The more stable the intermediate, the lower the activation energy required to form it, and the faster that pathway proceeds. So, which of these two carbocations is more stable?

Inductive vs. Mesomeric

To determine the stability of these carbocations, we must look at the neighboring groups. In both cases, we have a methoxy group ($-{\text{OCH}}_3$) nearby. Oxygen is a highly electronegative atom, meaning it loves to pull electron density towards itself through the $\sigma$-bonds. This is known as the $-I$ (negative inductive) effect.

In Intermediate I (${\text{CH}}_2^{+}-{\text{CH}}_2-{\text{OCH}}_3$), the oxygen atom pulls electron density away from the already electron-deficient carbocation. This makes the carbocation even more positive and, consequently, highly unstable.

Now, let's look at Intermediate II (${\text{CH}}_3-{\text{CH}}^{+}-{\text{OCH}}_3$). Here, the positive charge is directly adjacent to the oxygen atom. You might think, "Oh no, the oxygen will pull electrons away and make it incredibly unstable!" But there is a catch—a beautiful, elegant catch.

Oxygen doesn't just have electronegativity; it also has lone pairs of electrons in its p-orbitals. Because the oxygen is directly adjacent to the positively charged carbon (which has an empty p-orbital), the oxygen can share one of its lone pairs with the carbon. This sharing of electrons through $\pi$-overlap is called the $+M$ (positive mesomeric or resonance) effect.

The resonance structure looks like this: ${\text{CH}}_3-\text{CH}={\text{O}}^{+}-{\text{CH}}_3$. In this resonance contributor, every single atom (except hydrogen) has a complete octet! This is a state of profound thermodynamic bliss. The $+M$ effect is so incredibly powerful that it completely overwhelms the $-I$ effect. Therefore, Intermediate II is exceptionally stable.

Nucleophilic Attack

Because Intermediate II is vastly more stable than Intermediate I, the reaction almost exclusively proceeds through the formation of Intermediate II.

Now that we have our stable carbocation (${\text{CH}}_3-{\text{CH}}^{+}-{\text{OCH}}_3$), the final step is simple. The bromide ion (${\text{Br}}^{-}$), which was left behind when the ${\text{H}}^{+}$ attacked, is a nucleophile. It sees the positive charge on the carbon and attacks it, forming a new carbon-bromine bond.

The final major product is ${\text{CH}}_3-\text{CH}(\text{Br})-{\text{OCH}}_3$.

This reaction perfectly illustrates why memorizing rules like Markovnikov's rule isn't enough. While the product does follow Markovnikov's rule (the hydrogen went to the carbon with more hydrogens), the true driving force is the stability of the carbocation intermediate. Whenever you see an atom with lone pairs adjacent to a positive charge, always remember the immense stabilizing power of resonance!