Animated Solution for Chemistry - Organic Chemistry: Major product $P$ of above reaction is

The journey of a molecule through a chemical reaction is often a tale of selectivity. When a molecule possesses multiple reactive sites, the reagents act as the directors, choosing exactly which part of the molecule to transform and which to leave untouched. In this beautiful problem, we are presented with a dual-threat molecule: 1-bromo-4-allylbenzene.

A Tale of Two Halogens Our starting material is fascinating. On one end, we have a rigid, unyielding aryl bromide—a bromine atom attached directly to a benzene ring. On the other end, we have an allyl group, featuring a reactive, electron-rich carbon-carbon double bond.

When we introduce our first set of reagents, HBr and benzoyl peroxide, we are setting the stage for a classic showdown.

The Kharasch Effect Normally, the addition of hydrogen halides to an asymmetric alkene follows Markovnikov's rule, where the halogen attaches to the more substituted carbon. However, the presence of a peroxide completely flips the script. This is the legendary Kharasch Effect, or anti-Markovnikov addition.

The peroxide undergoes homolytic cleavage to generate free radicals. These radicals abstract a hydrogen atom from HBr, creating a highly reactive bromine radical (${\text{Br}}^{\bullet }$).

When this bromine radical approaches our allyl group, it faces a choice. It attacks the terminal carbon of the double bond. Why? Because doing so leaves the unpaired electron on the secondary carbon, creating a secondary radical which is significantly more stable than a primary radical.

After abstracting a hydrogen atom, the double bond is fully saturated, and we are left with our intermediate: 1-bromo-4-(3-bromopropyl)benzene.

The Art of Halogen Exchange Now, we enter the second phase of our synthesis. We introduce Cobalt(II) fluoride (${\text{CoF}}_2$).

Direct fluorination of organic compounds is notoriously violent and uncontrollable. To safely introduce fluorine, chemists rely on halogen exchange methods like the Swarts reaction. Heavy transition metal fluorides, such as ${\text{CoF}}_2$, $\text{AgF}$, or ${\text{Hg}}_2{\text{F}}_2$, are perfect for this job. They smoothly swap their fluorine atoms with the heavier halogens (chlorine or bromine) of alkyl halides.

Chemoselectivity in Action Here is where the problem tests your deep conceptual understanding. Our intermediate now has two bromine atoms. One is on the alkyl chain, and the other is still attached to the benzene ring. Which one will the cobalt fluoride attack?

This is a question of chemoselectivity. The bromine attached to the benzene ring is not just sitting there; its lone pairs of electrons are actively participating in resonance with the pi-electron cloud of the aromatic ring. This delocalization gives the $\text{C}-\text{Br}$ bond a partial double bond character. It becomes shorter, stronger, and incredibly resistant to nucleophilic substitution.

Furthermore, the $s{p}^2$ hybridized carbon of the benzene ring is more electronegative than an $s{p}^3$ carbon, holding the electrons even tighter.

Because of this immense stability, the aryl bromide acts as a silent spectator. The Swarts reaction exclusively targets the $s{p}^3$ hybridized alkyl bromide.

The aryl bromide remains completely untouched.

Our final major product is 1-bromo-4-(3-fluoropropyl)benzene.

This elegant sequence of reactions highlights the importance of understanding not just what a reagent does, but where it does it. By mastering regioselectivity (anti-Markovnikov) and chemoselectivity (alkyl vs. aryl halides), you can predict the outcome of even the most complex multi-step syntheses.