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प्रश्न
Explain why Lewis acid is not required in bromination of phenol?
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उत्तर
Usual halogenation is carried out in the presence of Lewis acid, \[\ce{FeBr3}\] which polarises the halogen molecule. In case of phenol, the polarisation of bromine occurs even in the absence of Lewis acid. This is because of highly activating effect of –OH group on the benzene ring. The reaction follows:
Note: In aqueous solution, phenol ionizes to give phenoxide ion. Due to the presence of the negative charge, the oxygen atom of the phenoxide ion donates electrons to the benzene ring to a large extent. As a result, the ring gets highly activated leading to the formation of trisubstituted product. On the other hand, in the non-polar solvents, the ionization of phenol does not occurs to a large extent. As a result, the -OH group donates electrons to the benzene ring only to a small extent. Consequently, the ring is activated slightly and, therefore, only monosubstitution occurs.
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\[\begin{array}{cc}
\phantom{.....................................}\ce{CH2OH}\\
\phantom{............................}|\\
\ce{CH3 - CH - CH2 - CH - CH - CH3}\\
|\phantom{....................}|\phantom{............}\\
\ce{CH3}\phantom{..............}\ce{OH}\phantom{.........}\\
\end{array}\]
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\ce{CH3 - CH - CH - CH3}\\
|\phantom{.........}|\phantom{...}\\
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\end{array}\]
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\[\begin{array}{cc}
\phantom{................}\ce{CH3}\\
\phantom{.............}|\\
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\phantom{................}\ce{CH3}\\
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|\phantom{......}|\phantom{......}|\\
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