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(B) The reaction shows the monobromination of phenol to form $p$-bromophenol as the major product.$(1)$ Bromine water $(Br_2/H_2O)$ is a polar solvent

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$(2)$,$(3)$ and $(4)$ only

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(B) The reaction shows the monobromination of phenol to form $p$-bromophenol as the major product.$(1)$ Bromine water $(Br_2/H_2O)$ is a polar solvent that causes the ionization of phenol into phenoxide ion,which is highly activating,leading to the formation of $2,4,6$-tribromophenol.$(2)$ $Br_2$ in $CS_2$ at $273 \ K$ is a non-polar solvent condition that favors monobromination,yielding $p$-bromophenol as the major product.$(3)$ $Br_2/FeBr_3$ is a standard electrophilic aromatic substitution condition,but for phenol,it is generally not used as it can lead to polybromination or oxidation; however,in the context of competitive exams,it is often considered to favor monobromination similar to other non-polar conditions.$(4)$ $Br_2$ in $CHCl_3$ at $273 \ K$ is a non-polar solvent condition that favors monobromination,yielding $p$-bromophenol as the major product.Therefore,conditions $(2)$,$(3)$,and $(4)$ favor the formation of the monobromo product.

The given reaction can occur in the presence of:
$(1)$ Bromine water
$(2)$ $Br_2$ in $CS_2, 273 \ K$
$(3)$ $Br_2 / FeBr_3$
$(4)$ $Br_2$ in $CHCl_3, 273 \ K$
Choose the correct answer from the options given below:
[Image: Phenol reacting to form p-bromophenol as the major product]

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The given reaction can occur in the presence of:$(1)$ Bromine water$(2)$ $Br_2$ in $CS_2, 273 \ K$$(3)$ $Br_2 / FeBr_3$$(4)$ $Br_2$ in $CHCl_3, 273 \ K$Choose the correct answer from the options given below:[Image: Phenol reacting to form p-bromophenol as the major product]