
Wurtz reaction may be used to unite:
A.Two alkyl halides
B.Two aryl halides
C..Alkyl and aryl halides
D.Two benzene units.
Answer
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Hint: We must know that the Wurtz reaction used to prepare alkane in a simple method. Therefore, two halides are brought together to give an alkane chain in the presence of sodium and dry ether.
Complete step by step answer:
We know that the Wurtz reaction is a coupling reaction. It unites two alkyl halides with metallic sodium in the presence of dry ether solution to form hydrocarbons.
The general equation for Wurtz reaction is,
$2R - X + 2Na\xrightarrow{{dry\,\,ether}}R - R + 2N{a^ + } + {X^ - }$
Therefore, we can understand from the reaction that the Wurtz reaction uses two alkyl halides for coupling. Hence, option (A) is correct.
Additional Information:
Wurtz reaction is a coupling reaction between alkyl halides with metallic sodium in the presence of dry ether solution.
Other than sodium, a metal such as silver, indium, activated copper could also be used in this reaction to form alkanes.
If an aryl halide is used in the place of alkyl halides, then the reaction is called a Wurtz-Fittig reaction.
The mechanism involves free radical species that are a part of an exchange between halogen and metal.
Free radicals are involved in the reaction, permitting the possibility of side reactions, which leads to the formation of the alkene as the product.
Compounds containing at least two carbon atoms must be used for the coupling reaction to take place. Therefore, Methane cannot be prepared by this method.
This method does not give satisfactory products when tertiary alkyl halides are used as one of the reagents.
Only symmetric alkanes could be produced by this method.
Note:
One should not confuse between Wurtz reaction and Wurtz Fittig reaction. In the Wurtz reaction, alkyl halides are used whereas in the Wurtz Fittig reaction aryl halide is used. Alkyl halide consists of alkane with a halogen group. Aryl halide is a halogen group attached to the aromatic ring.
Complete step by step answer:
We know that the Wurtz reaction is a coupling reaction. It unites two alkyl halides with metallic sodium in the presence of dry ether solution to form hydrocarbons.
The general equation for Wurtz reaction is,
$2R - X + 2Na\xrightarrow{{dry\,\,ether}}R - R + 2N{a^ + } + {X^ - }$
Therefore, we can understand from the reaction that the Wurtz reaction uses two alkyl halides for coupling. Hence, option (A) is correct.
Additional Information:
Wurtz reaction is a coupling reaction between alkyl halides with metallic sodium in the presence of dry ether solution.
Other than sodium, a metal such as silver, indium, activated copper could also be used in this reaction to form alkanes.
If an aryl halide is used in the place of alkyl halides, then the reaction is called a Wurtz-Fittig reaction.
The mechanism involves free radical species that are a part of an exchange between halogen and metal.
Free radicals are involved in the reaction, permitting the possibility of side reactions, which leads to the formation of the alkene as the product.
Compounds containing at least two carbon atoms must be used for the coupling reaction to take place. Therefore, Methane cannot be prepared by this method.
This method does not give satisfactory products when tertiary alkyl halides are used as one of the reagents.
Only symmetric alkanes could be produced by this method.
Note:
One should not confuse between Wurtz reaction and Wurtz Fittig reaction. In the Wurtz reaction, alkyl halides are used whereas in the Wurtz Fittig reaction aryl halide is used. Alkyl halide consists of alkane with a halogen group. Aryl halide is a halogen group attached to the aromatic ring.
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