Animated Solution for Chemistry - Organic Chemistry: Identify the major products A and B respectively in the following reactions of phenol.

Phenol consists of a hydroxyl group ($-\text{OH}$) attached directly to a benzene ring. This $-\text{OH}$ group is a strongly activating group due to its $+R$ (resonance) effect, which increases the electron density on the benzene ring, particularly at the ortho and para positions. This makes phenol highly reactive towards electrophilic aromatic substitution.

Bromination in a Non-Polar Solvent In the first reaction, phenol is treated with bromine (${\text{Br}}_2$) in the presence of carbon disulfide (${\text{CS}}_2$) at a low temperature of $273\text{K}$.

Why is the solvent important here? ${\text{CS}}_2$ is a non-polar solvent. In a non-polar medium, the ionization of phenol into the highly reactive phenoxide ion is suppressed. Consequently, the electrophilic attack is less vigorous, leading to monobromination rather than polybromination.

The incoming bromine electrophile (${\text{Br}}^{+}$) can attach to either the ortho or the para position. However, the bulky $-\text{OH}$ group creates steric hindrance at the ortho position. Therefore, the attack at the para position is more favorable, making p-bromophenol the major product. Thus, we identify A as p-bromophenol.

The Reimer-Tiemann Reaction In the second reaction, phenol reacts with chloroform (${\text{CHCl}}_3$) and sodium hydroxide ($\text{NaOH}$), followed by acidic hydrolysis (${\text{H}}_3{\text{O}}^{+}$). This specific combination of reagents is the hallmark of the famous Reimer-Tiemann reaction.

The strong base ($\text{NaOH}$) deprotonates chloroform to generate a highly reactive intermediate called dichlorocarbene ($:{\text{CCl}}_2$), which acts as the electrophile. Simultaneously, the base converts phenol into the phenoxide ion, which is even more reactive than phenol itself.

The dichlorocarbene attacks the phenoxide ring. Interestingly, in this reaction, the ortho product is the major one. This ortho-selectivity is driven by the stabilization of the transition state through chelation with the sodium ion, and later, the product is stabilized by intramolecular hydrogen bonding. After hydrolysis, the final product is o-hydroxybenzaldehyde, commonly known as salicylaldehyde. Thus, we identify B as o-hydroxybenzaldehyde.

By carefully analyzing both pathways, we have determined that: A is p-bromophenol. B is o-hydroxybenzaldehyde.

Looking at the given options, this perfectly matches option (b).

Always remember to pay attention to the reaction conditions! If the bromination had been carried out in a polar solvent like water (${\text{Br}}_2/{\text{H}}_2\text{O}$), the phenol would have ionized completely, leading to the formation of a white precipitate of 2,4,6-tribromophenol.