Animated Solution for Chemistry - Organic Chemistry: p-hydroxybenzophenone upon reaction with bromine in carbon tetrachloride gives

Electrophilic Aromatic Substitution in p-Hydroxybenzophenone

Imagine you are an electrophile, specifically a positively charged bromonium ion ($B{r}^{+}$), looking for a place to attack. You encounter a molecule of p-hydroxybenzophenone. This molecule presents an interesting dilemma: it has two distinct benzene rings connected by a central carbonyl bridge. Which ring do you choose, and exactly where do you strike? Let's break down the thought process behind this classic electrophilic aromatic substitution reaction.

Our starting material, p-hydroxybenzophenone, consists of two phenyl rings. One ring is unsubstituted, while the other bears a hydroxyl ($-OH$) group at the para position relative to the carbonyl bridge. We are reacting this molecule with bromine ($B{r}_2$) dissolved in carbon tetrachloride ($CC{l}_4$).

This setup is a textbook example of an electrophilic aromatic substitution. The key to solving this problem lies in understanding the competing electronic effects of the substituents attached to the benzene rings.

Activating vs. Deactivating Groups

To determine which ring will be attacked, we must evaluate their relative electron densities.

The left ring features the $-OH$ group. Oxygen has lone pairs of electrons that it can donate into the benzene ring through resonance. This is known as a positive resonance effect ($+R$). This effect strongly activates the ring, making it highly attractive to electrophiles.

Conversely, the carbonyl group ($-C(=O)R$) linking the two rings is an electron-withdrawing group. It pulls electron density away from the rings via a negative resonance effect ($-R$). This deactivates the rings.

When comparing the two, the strong $+R$ effect of the $-OH$ group easily overpowers the deactivating nature of the carbonyl group. Therefore, the reaction will exclusively occur on the highly activated, hydroxyl-bearing ring.

Now that we know which ring will react, we must determine where the bromine will attach. The $-OH$ group is an ortho/para director. Through resonance, it significantly increases the electron density at the positions ortho and para to itself.

However, if we look closely at our molecule, the para position relative to the $-OH$ group is already occupied by the bulky carbonyl bridge. This is a classic case of steric hindrance and structural blocking.

With the para position unavailable, the incoming electrophile has no other choice but to attack one of the available ortho positions.

The bromine atom attaches to the position ortho to the hydroxyl group. Following standard IUPAC numbering for benzophenone derivatives, this gives us our final major product: 3-bromo-4-hydroxybenzophenone.

It is also crucial to note the role of the solvent. We used $CC{l}_4$, a non-polar solvent. This restricts the reaction to monobromination. If we had used a polar solvent like water (bromine water), the phenol ring would become even more activated (forming a phenoxide-like character), which typically leads to rapid polybromination. Always pay attention to the solvent in electrophilic aromatic substitutions!