
Which of the following cannot be obtained as the only product in Wurtz’s coupling reaction of halides with sodium metal in ether solution?
A.
B.
C.
D. (A) and (C)
Answer
514.2k+ views
Hint : Wurtz reaction is a type of coupling reaction in which two moles of alkyl halide reacts with two moles of sodium metal in the presence of dry ether to yield higher alkanes of even number i.e., to form alkanes having even number of carbon atoms. The mechanism involves the formation of an alkyl radical followed by nucleophilic substitution reaction.
Complete Step By Step Answer:
When two same alkyl halides participate in the Wurtz reaction, then the alkane with even number of carbon atoms is formed and it is the only product of the reaction but if in case two different alkyl halides participate in the reaction, then there is possibility of formation of more than one product. Hence, in the Wurtz reaction an alkane with an odd number of carbon atoms will never be formed as the only product after the reaction. So, among given options the structures of alkane are given as follows:
Structure-A:
Number of carbon atoms in the chain $ = 5 $
As the alkane consists of an odd number of carbon atoms, it will not be the only product in Wurtz's reaction.
Structure-B:
Number of carbon atoms in the chain $ = 6 $
As the alkane consists of an even number of carbon atoms, it will be the only product in the Wurtz’s reaction.
Structure-C:
Number of carbon atoms in the chain $ = 7 $
As the alkane consists of an odd number of carbon atoms, it will not be the only product in Wurtz's reaction.
Hence, structures A and C will not be formed as the only product in Wurtz’s reaction.
So, option (D) is the correct answer.
Note :
It is important to note that other than sodium, other metals like silver, zinc, iron and copper can also be used for this reaction. If aryl halides are considered as reactants instead of alkyl halides, then the reaction is named as Wurtz-Fittig reaction.
Complete Step By Step Answer:
When two same alkyl halides participate in the Wurtz reaction, then the alkane with even number of carbon atoms is formed and it is the only product of the reaction but if in case two different alkyl halides participate in the reaction, then there is possibility of formation of more than one product. Hence, in the Wurtz reaction an alkane with an odd number of carbon atoms will never be formed as the only product after the reaction. So, among given options the structures of alkane are given as follows:
Structure-A:
Number of carbon atoms in the chain $ = 5 $
As the alkane consists of an odd number of carbon atoms, it will not be the only product in Wurtz's reaction.
Structure-B:
Number of carbon atoms in the chain $ = 6 $
As the alkane consists of an even number of carbon atoms, it will be the only product in the Wurtz’s reaction.
Structure-C:
Number of carbon atoms in the chain $ = 7 $
As the alkane consists of an odd number of carbon atoms, it will not be the only product in Wurtz's reaction.
Hence, structures A and C will not be formed as the only product in Wurtz’s reaction.
So, option (D) is the correct answer.
Note :
It is important to note that other than sodium, other metals like silver, zinc, iron and copper can also be used for this reaction. If aryl halides are considered as reactants instead of alkyl halides, then the reaction is named as Wurtz-Fittig reaction.
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