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प्रश्न
Explain why Lewis acid is not required in bromination of phenol?
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
Which of the following compounds is NOT prepared by the action of alcoholic NI3 on alkyl halide?
(a) CH3NH2
(b) CH3- CH2- NH2
(c) CH3 - CH2 - CH2 - NH2
(d) (CH3)3 C- NH2
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(1) formation of carbonation
(2) formation of an ester
(3) protonation of the alcohol molecule
(4) elimination of water
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Write IUPAC names of the following
n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent?
Give IUPAC name of the compound given below.
\[\begin{array}{cc}
\phantom{}\ce{CH3 - CH - CH2 - CH2 - CH - CH3}\phantom{.}\\
\phantom{.........}|\phantom{...................}|\phantom{...........}\\
\phantom{..}\ce{Cl}\phantom{.................}\ce{OH}\phantom{..}
\end{array}\]
Which of the following compounds will react with sodium hydroxide solution in water?
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) | \[\begin{array}{cc} \ce{CH3}\phantom{....}\\ |\phantom{.......}\\ \ce{CH3-C-OH + CH3I}\\ |\phantom{.......}\\ \ce{CH3}\phantom{....} \end{array}\] |
Give the structures of Thiosulphuric acid and Peroxy monosulphuric acid.
Write the IUPAC name.
\[\begin{array}{cc}
\phantom{................}\ce{CH3}\\
\phantom{.............}|\\
\ce{CH3 - CH - CH - C -CH3}\\
\phantom{.}|\phantom{......}|\phantom{......}|\\
\phantom{....}\ce{CH3\phantom{...}\ce{OH}\phantom{...}\ce{CH3}}\
\end{array}\]
