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प्रश्न
Why can aryl halides not be prepared by reaction of phenol with \[\ce{HCl}\] in the presence of \[\ce{ZnCl2}\]?
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
The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ______.
Which of the following compounds can be classified as aryl halides?
(i) \[\ce{p-ClC6H4CH2CH(CH3)2}\]
(ii) \[\ce{p-CH3CHCl(C6H4)CH2CH3}\]
(iii) \[\ce{o-BrH2C - C6H4CH(CH3)CH2CH3}\]
(iv) \[\ce{C6H5 - Cl}\]
Classify the following compound as a primary, secondary and tertiary halide.
4-Bromopent-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 halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:
\[\ce{CH3C(C2H5)2CH2Br}\]
Name the following halide according to the IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:
\[\ce{CH_3CH_2C(CH_3)_2CH_2I}\]
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
CH3C(Cl)(C2H5 )CH2CH3
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
\[\ce{CH3C(Cl)(C2H5)CH2CH3}\]
Give the IUPAC name of the compound.
\[\ce{CICH2 C ≡ CCH2 Br}\]
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
\[\ce{CH3C(C2H5)2CH2Br}\]
