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Question:
The reaction, $ {{C}_{6}}{{H}_{5}}{{N}_{2}}Cl\xrightarrow{C{{u}_{2}}C{{l}_{2}}/HCl}{{C}_{2}}{{H}_{5}}Cl+{{N}_{2}}\uparrow $ is called:
The reaction, $ {{C}_{6}}{{H}_{5}}{{N}_{2}}Cl\xrightarrow{C{{u}_{2}}C{{l}_{2}}/HCl}{{C}_{2}}{{H}_{5}}Cl+{{N}_{2}}\uparrow $ is called:
Updated On: Aug 23, 2023
Sandmeyer reaction.
- Gattermann reaction
- Gattermann-Koch reaction
- Lederrer Mannase reaction
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The Correct Option is A
Solution and Explanation
The correct option is(A): Sandmeyer reaction.
The given reaction, involving the conversion of an aryl diazonium salt (C6H5N2+Cl−) into a halogenated aromatic compound using copper(I) chloride (Cu2Cl2) as a catalyst and hydrochloric acid (HCl) as a reagent, is indeed the Sandmeyer reaction.
So, the correct option is: Sandmeyer reaction.
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Concepts Used:
Preparation of Haloalkanes
- Preparation by Alcohols: By using alcohol, we can very easily prepare Haloalkane as when R-OH is reacted with a suitable reagent it will form R-X. The following reagents will help in the formation of the reactions:
- Halogen Acids (HX)
- Thionyl chloride (SOCl2)
- Phosphorous halides (PX5 or PX3)
- Preparation by Free Radical Halogenation: The formation of alkyl bromides and alkyl chloride is very much possible with the help of free radical halogenation, but as these radicals are highly reactive and non-specific in nature, they can result in the formation of a mixture of products. Though, it's not a preferred method for the preparation of haloalkanes. For example, free radical chlorination can result in the formation of a number of haloalkanes, which in turn makes it difficult to set apart a single product.
- Preparation by Alkenes: An electrophilic addition reaction can be used to transform an alkene into a haloalkane as alkene will react with HX to form R-X.
Read More: Haloalkanes and Haloarenes
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