Phenol (C_6H_6O) xrightarrow{Br_2 text{ in } | vedclass

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(D) $1$. When phenol reacts with $Br_2$ in a non-polar solvent like $CS_2$ or $CHCl_3$ at low temperature,the reaction is monobromination. The major p

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(D) $1$. When phenol reacts with $Br_2$ in a non-polar solvent like $CS_2$ or $CHCl_3$ at low temperature,the reaction is monobromination. The major product is $p$-bromophenol due to less steric hindrance compared to $o$-bromophenol. Thus,$X$ is $p$-bromophenol.$2$. When phenol reacts with $Br_2$ in water $(H_2O)$,the phenol ionizes to form the phenoxide ion,which is highly activating. This leads to rapid electrophilic substitution at all ortho and para positions,resulting in the formation of $2,4,6$-tribromophenol as the major product. Thus,$Y$ is $2,4,6$-tribromophenol.

Phenol $(C_6H_6O)$ $\xrightarrow{Br_2 \text{ in } CS_2}$ $X$ (Major product)
Phenol $(C_6H_6O)$ $\xrightarrow{Br_2 \text{ in } H_2O}$ $Y$ (Major product)
What are $X$ and $Y$ in the above reaction respectively?

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Phenol $(C_6H_6O)$ $\xrightarrow{Br_2 \text{ in } CS_2}$ $X$ (Major product)Phenol $(C_6H_6O)$ $\xrightarrow{Br_2 \text{ in } H_2O}$ $Y$ (Major product)What are $X$ and $Y$ in the above reaction respectively?