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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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संबंधित प्रश्न
Write the IUPAC name of the given compound:
Write the IUPAC name of the following compound:
\[\begin{array}{cc}
\ce{H3C - CH - CH2 - CH - CH - CH2 - CH3}\\
|\phantom{....................}|\phantom{.........}|\phantom{.............}\\
\ce{OH}\phantom{................}\ce{OH}\phantom{...}\ce{C{_2}H5}\phantom{.........}\\
\end{array}\]
Write the IUPAC name of the following compound:
Write the IUPAC name of the following compound:
Write the IUPAC name of the following compound:
\[\begin{array}{cc}
\ce{CH3 - O - CH2 - CH - CH3}\\
\phantom{................}|\\
\phantom{....................}\ce{CH3}
\end{array}\]
Natalite is a mixture of
(a) diethyl ether and methanol
(b) diethyl ether and ethanol
(c) dimethyl ether and methanol
(d) dimethyl ether and ethanol
How is phenol converted into the following?
benzoquinone
In the dehydration of alcohols to alkenes by heating with concentrated sulphuric acid, the initiation step is:
(1) formation of carbonation
(2) formation of an ester
(3) protonation of the alcohol molecule
(4) elimination of water
Resorcinol on distillation with zinc dust gives _________.
Write IUPAC name of the following
\[\begin{array}{cc}\ce{CH3-CH-CH-CH2-OH}\\|\phantom{.....}|\phantom{.......}\\\ce{OH}\phantom{..}\ce{CH3}\phantom{.....}\end{array}\]
Give IUPAC names of the following compound:
The compound HOCH2 – CH2OH is __________.
IUPAC name of m-cresol is ____________.
Among the following sets of reactants which one produces anisole?
IUPAC name of the compound is:
\[\begin{array}{cc}
\ce{CH3-CH-OCH3}\\
|\phantom{....}\\
\ce{CH3}\phantom{..}
\end{array}\]
Write the IUPAC name of the compound given below.
\[\begin{array}{cc}
\phantom{}\ce{CH3 - CH2 - C = C - OH}\\
\phantom{........}|\phantom{....}|\phantom{}\\
\phantom{..............}\ce{CH3 CH2OH}
\end{array}\]
Match the starting materials given in Column I with the products formed by these (Column II) in the reaction with HI.
| Column I | Column II | ||
| (i) | CH3—O—CH3 | (a) |  |
| (ii) | \[\begin{array}{cc} \ce{CH3}\phantom{..................}\\ \backslash\phantom{.............}\\ \ce{CH-O-CH3}\\ /\phantom{..............}\\ \ce{CH3}\phantom{..................} \end{array}\] |
(b) | \[\begin{array}{cc} \ce{CH3}\phantom{....}\\ |\phantom{.......}\\ \ce{CH3-C-I + CH3OH}\\ |\phantom{.......}\\ \ce{CH3}\phantom{....} \end{array}\] |
| (iii) | \[\begin{array}{cc} \ce{CH3}\phantom{.}\\ |\phantom{....}\\ \ce{H3C-C-O-CH3}\\ |\phantom{....}\\ \ce{CH3}\phantom{..} \end{array}\] |
(c) |  |
| (iv) |  |
(d) | CH3—OH + CH3—I |
| (e) | \[\begin{array}{cc} \ce{CH3}\phantom{.....................}\\ \backslash\phantom{.................}\\ \ce{CH-OH + CH3I}\\ /\phantom{.................}\\ \ce{CH3}\phantom{.....................} \end{array}\] |
||
| (f) | \[\begin{array}{cc} \ce{CH3}\phantom{.....................}\\ \backslash\phantom{.................}\\ \ce{CH-I + CH3OH}\\ /\phantom{.................}\\ \ce{CH3}\phantom{.....................} \end{array}\] |
||
| (g) | \[\begin{array}{cc} \ce{CH3}\phantom{....}\\ |\phantom{.......}\\ \ce{CH3-C-OH + CH3I}\\ |\phantom{.......}\\ \ce{CH3}\phantom{....} \end{array}\] |
Assertion: IUPAC name of the compound
\[\begin{array}{cc}
\ce{CH3 - CH - O - CH2 - CH2 - CH3}\\
|\phantom{....................}\\
\ce{CH3}\phantom{.................}
\end{array}\] is 2-Ethoxy-2-methylethane.
Reason: In IUPAC nomenclature, ether is regarded as hydrocarbon derivative in which a hydrogen atom is replaced by —OR or —OAr group [where R = alkyl group and Ar = aryl group]
Write a chemical reaction for the following conversion:
Acetic acid into ethyl alcohol.
