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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 IUPAC name of the following compound:
\[\begin{array}{cc}
\ce{HO - CH2 - CH - CH2 - OH}\\
|\phantom{..}\\
\ce{OH}
\end{array}\]
Write IUPAC name of the following compound:
C6H5 – O – C2H5
Give IUPAC name of the following ether:
\[\begin{array}{cc}
\ce{C2H5OCH2 - CH - CH3}\\
\phantom{.....}|\\
\phantom{.......}\ce{CH3}
\end{array}\]
Give IUPAC name of the following ether:
How is phenol converted into the following?
picric acid
Write the IUPAC name of the following compound:
Write IUPAC name of the following
The compound HOCH2 – CH2OH is __________.
Butane-2-ol is ____________.
Ethyl alcohol is industrially prepared from ethylene by:
Ethylene reacts with Baeyer’s reagent to give ______.
\[\ce{HC ≡ CH ->[HgSO4][H2SO4] ->[CH3MgBr][H2O] ->[PBr3]}\]
The product of acid catalysed hydration of 2-phenylpropene is:
The correct acidic strength order of the following is:
(I)
(II)
(III)
Which of the following compounds will react with sodium hydroxide solution in water?
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: Addition reaction of water to but-1-ene in acidic medium yields butan-1-ol.
Reason: Addition of water in acidic medium proceeds through the formation of primary carbocation.
Assertion: Phenols give o- and p-nitrophenol on nitration with conc. \[\ce{HNO3}\] and \[\ce{H2SO4}\] mixture.
Reason: –OH group in phenol is o–, p– directing.
Convert the following:
Ethyl alcohol into ethyl acetate.
