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Question
The correct options for the products \(A~and~B\) of the following reactions are :
\( A +\frac{ Br \text { (entess) }}{ H _{2} O } \)
\( \frac{ Br _{2}}{ CS _{2}, 5^{\circ} C }= B \)
The correct options for the products \(A~and~B\) of the following reactions are :
\( A +\frac{ Br \text { (entess) }}{ H _{2} O } \)
\( \frac{ Br _{2}}{ CS _{2}, 5^{\circ} C }= B \)
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Solution
The correct option is B
Product A:
In the water solvent when phenol treated with Br2 gives a polybromo derivative in which all hydrogen atoms at ortho, meta, and para positions with respect to the −OH group are replaced by bromine atoms. It is so because in aqueous medium phenol ionizes to form peroxide ion. Due to the presence of negative ions the ring gets highly activated and tri substitution occurs and the formation of 2,4,6 – tribromophenol takes place.
Product B:
Phenol undergoes bromination in CS2 to give para-bromophenol. Though OH is a ortho/para directing group, para product is major due to the less steric hindrance.
Hence, option (c) is correct.
Product A:
In the water solvent when phenol treated with Br2 gives a polybromo derivative in which all hydrogen atoms at ortho, meta, and para positions with respect to the −OH group are replaced by bromine atoms. It is so because in aqueous medium phenol ionizes to form peroxide ion. Due to the presence of negative ions the ring gets highly activated and tri substitution occurs and the formation of 2,4,6 – tribromophenol takes place.
Product B:
Phenol undergoes bromination in CS2 to give para-bromophenol. Though OH is a ortho/para directing group, para product is major due to the less steric hindrance.
Hence, option (c) is correct.
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