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प्रश्न
Why can aryl halides not be prepared by reaction of phenol with \[\ce{HCl}\] in the presence of \[\ce{ZnCl2}\]?
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उत्तर
Aryl halides can't be prepared by reaction of phenol with \[\ce{HCl}\] in presence of \[\ce{ZnCl2}\] because the carbon-oxygen bond in phenols has partial double bond character and it, being stronger than a single bond, is difficult to break.
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संबंधित प्रश्न
Draw the structures of major monohalo products in each of the following reaction:
Draw the structure of the major monohalo product in the following reaction:
Name the following halide according to the IUPAC system and classify it as an alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:
(CH3)3CCH2CH(Br)C6H5
Name the following halide according to the IUPAC system and classify it as an alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:
p-ClC6H4CH2CH(CH3)2
Name the following halide according to the IUPAC system and classify it as an alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:
m-ClCH2C6H4CH2C(CH3)3
In which of the following molecules carbon atom marked with asterisk (*) is asymmetric?
(a) \[\begin{array}{cc}
\ce{H}\\
|\\
\phantom{.}\ce{C^*}\phantom{}\\
\phantom{}/\phantom{..}|\phantom{..}\backslash\phantom{}\\
\phantom{.}\ce{I}\phantom{...}\ce{Br}\phantom{..}\ce{Cl}
\end{array}\]
(b) \[\begin{array}{cc}
\ce{D}\\
|\\
\phantom{.}\ce{C^*}\phantom{}\\
\phantom{}/\phantom{..}|\phantom{..}\backslash\phantom{}\\
\phantom{.}\ce{I}\phantom{...}\ce{Br}\phantom{..}\ce{Cl}
\end{array}\]
(c) \[\begin{array}{cc}
\ce{H}\\
|\\
\phantom{.}\ce{C^*}\phantom{}\\
\phantom{}/\phantom{..}|\phantom{..}\backslash\phantom{}\\
\phantom{..}\ce{OH}\phantom{..}\ce{C2H5}\phantom{}\ce{CH3}
\end{array}\]
(d) \[\begin{array}{cc}
\ce{H}\\
|\\
\phantom{.}\ce{C^*}\phantom{}\\
\phantom{}/\phantom{..}|\phantom{..}\backslash\phantom{}\\
\phantom{..}\ce{H}\phantom{...}\ce{C2H5}\phantom{}\ce{CH3}
\end{array}\]
Which is the correct IUPAC name for \[\begin{array}{cc}
\ce{CH3 - CH - CH2 - Br}\\
\phantom{}|\phantom{.......}\\
\phantom{}\ce{C2H5}\phantom{....}
\end{array}\]?
Which of the products will be major product in the reaction given below? Explain.
\[\ce{CH3CH = CH2 + HI -> \underset{(A)}{CH3CH2CH2I} + \underset{(B)}{CH3CHICH3}}\]
Classify the following compound as a primary, secondary and tertiary halide.
1-Bromobut-2-ene
Two isomers (A) and (B) with molar mass 184 g/mol and elemental composition C 52.2%; H 4.9% and Br 42.9% gave benzoic acid and p-bromobenzoic acid, respectively on oxidation with KMnO4. Isomer ‘A’ is optically active and gives a pale yellow precipitate when warmed with alcoholic AgNO3. Isomer ‘A’ and ‘B’ are, respectively:
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
\[\ce{CH3CH2C(CH3)2CH2I}\]
Name the following halide according to the IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl, or aryl halide:
\[\ce{CH3CH2C(CH3)2CH2I}\]
Write the structure of the following organic halogen compound.
1,4-Dibromobut-2-ene
Name the following halide according to the IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:
CH3CH2C(CH3)2CH2I
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
\[\ce{CH3C(Cl)(C2H5)CH2CH3}\]
Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
\[\ce{CH3C(Cl)(C2H5)CH2CH3}\]
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
\[\ce{CH3C(C2H5)2CH2Br}\]
Name the following halide according to IUPAC system and classify as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:
\[\ce{CH3C(Cl)(C2H5)CH2CH3}\]
Which is the correct IUPAC name for 
is______.
The IUPAC name of the compound shown below is:
