Animated Solution for Chemistry - Alcohols, Phenols, Ethers: The structure of the compound that gives a tribromo derivative on treatment with bromine water is

A Tale of Steric and Electronic Directing Effects

Electrophilic aromatic substitution is a cornerstone of organic chemistry, and understanding how different substituents direct incoming electrophiles is crucial. In this problem, we are tasked with identifying which compound yields a tribromo derivative upon treatment with bromine water.

Bromine water (${\text{Br}}_2/{\text{H}}_2\text{O}$) is a potent brominating agent. When phenol or its derivatives are treated with bromine water, the polar solvent facilitates the ionization of the phenol into a phenoxide ion. The phenoxide ion is an exceptionally strong activating group due to the full negative charge on the oxygen atom, which donates electron density into the benzene ring via resonance.

This immense electron density makes the ring highly susceptible to electrophilic attack, specifically directing the incoming bromine atoms to the ortho and para positions. Because the activation is so strong, polybromination occurs rapidly, substituting all available ortho and para positions simultaneously.

Let's examine m-cresol (3-methylphenol). In this molecule, the $-\text{OH}$ group is at position 1, and the $-{\text{CH}}_3$ group is at position 3.

To determine if a tribromo derivative can form, we must check the availability of the ortho and para positions relative to the strongly directing $-\text{OH}$ group: - Ortho positions: Carbon 2 and Carbon 6. - Para position: Carbon 4.

Looking at the structure of m-cresol, the $-{\text{CH}}_3$ group is at Carbon 3. This means that Carbon 2, Carbon 4, and Carbon 6 are all completely unoccupied! Since all three of these highly activated positions are free, bromine will substitute at all of them, yielding 2,4,6-tribromo-3-methylphenol.

What happens if we apply this same logic to the other isomers?

- o-Cresol (2-methylphenol): The $-{\text{CH}}_3$ group occupies Carbon 2, which is one of the ortho positions. This leaves only Carbon 4 (para) and Carbon 6 (ortho) available. Thus, it can only form a dibromo derivative. - p-Cresol (4-methylphenol): The $-{\text{CH}}_3$ group occupies Carbon 4, which is the para position. This leaves only Carbon 2 and Carbon 6 (both ortho) available. It, too, can only form a dibromo derivative. - Benzyl alcohol: The $-{\text{CH}}_2\text{OH}$ group is not a strongly activating group like the phenolic $-\text{OH}$, and it does not undergo rapid polybromination with bromine water.

Therefore, m-cresol is the only compound among the choices that possesses three free, activated positions, making it the correct answer.