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प्रश्न
Predict the possible product of the following reaction.
bromination of phenol
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उत्तर
Phenolic –OH group is ortho and para directing group. So, the bromination of phenol gives p-bromophenol and o-bromophenol.


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संबंधित प्रश्न
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\[\begin{array}{cc}
\phantom{..............................}\ce{CH3}\\
\phantom{...........................}|\\
\ce{CH3 - C = CH2 + HBr->[Benzoyl][peroxide] H3C - CH - CH2Br}\\
|\phantom{....................................}\\
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\end{array}\]
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\[\begin{array}{cc}
\phantom{..............................}\ce{CH3}\\
\phantom{............................}|\\
\ce{CH3 - C = CH2 + HBr ->[benzoyl][peroxide] CH3 - CH - CH2Br}\\
|\phantom{....................................}\\
\ce{CH_3}\phantom{.................................}
\end{array}\]
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- State the rule that governs the formation of this product.
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\[\begin{array}{cc}
\phantom{..............................}\ce{CH3}\\
\phantom{...........................}|\\
\ce{CH3 - C = CH2 + HBr ->[benzoyl][peroxide] CH3 - CH - CH2Br}\\
|\phantom{....................................}\\
\ce{CH3}\phantom{.................................}
\end{array}\]
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\[\begin{array}{cc}
\phantom{..............................}\ce{CH3}\\
\phantom{............................}|\\
\ce{CH3 - C = CH2 + HBr ->[benzoyl][peroxide]CH3 - CH - CH2Br}\\
|\phantom{....................................}\\
\ce{CH3}\phantom{.................................}
\end{array}\]
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\[\begin{array}{cc}
\phantom{..........................}\ce{CH3}\\\phantom{........................}|\\\ce{CH3 - C = CH2 + HBr ->[benzoyl][peroxide] CH3 - CH - CH2Br}\\
|\phantom{......................................}\\
\ce{CH3}\phantom{....................................}\end{array}\]
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- State the rule that governs the formation of this product.
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\[\begin{array}{cc}
\phantom{..............................}\ce{CH3}\\
\phantom{............................}|\\
\ce{CH3 - C = CH2 + HBr ->[benzoyl][peroxide] CH3 - CH - CH2Br}\\
|\phantom{....................................}\\
\ce{CH3}\phantom{..................................}
\end{array}\]
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