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Hint: This reaction is used to prepare higher alkanes using a pure metal to aid the process. Think about how this may be achieved and what kind of compounds do metals readily form, the answer will be apparent once this is understood.
Complete step by step solution:
Wurtz reaction is used to prepare higher, long chain alkanes from short chain alkyl halides with the aid of $Na$ (sodium metal) in the presence of dry ether. Since sodium easily forms salts after combining with the halogens in the alkyl halides, the alkyl chains join together to form a higher alkane.
The general reaction is:
\[2RX+2Na\xrightarrow{\text{dry ether}}R-R+2NaX\]
Here, we have considered that the 2 alkyl halides that are going to form the alkane are the same, but it may happen that the 2 alkyl halides involved in the reaction may have different alkyl substituents. In that case the following reaction takes place:
Let us assume that $R-X$and $R'-X$are the 2 alkyl halides. Then the reaction will be-
\[3R-X+6Na+3X-R'\xrightarrow{\text{dry ether}}R'-R+R-R+R'-R'+6NaX\]
Here, we can see that the alkyl halides may form symmetrical product alkanes by combining with a molecule that resembles them, or a cross-coupling reaction may also occur. The percentage of the products will depend on the stability of the higher alkanes.
Additional information:
The Wurtz reaction only includes aliphatic alkane formation. A reaction that is similar to the Wurtz reaction but the parent hydrocarbons are aryl halides is known as the Fittig reaction. The products formed are aromatic compounds having multiple rings. When the parent hydrocarbons include both alkyl and aryl halides, the reaction is called the Wurtz-Fittig reaction. Compounds like toluene can be formed.
Note: Just like the parent alkyl or aryl group can be different during the reaction, the halogen attached to both the parent alkyl halides can also be different. Please do not assume that the halogens have to be the same in both the reactant molecules.
Complete step by step solution:
Wurtz reaction is used to prepare higher, long chain alkanes from short chain alkyl halides with the aid of $Na$ (sodium metal) in the presence of dry ether. Since sodium easily forms salts after combining with the halogens in the alkyl halides, the alkyl chains join together to form a higher alkane.
The general reaction is:
\[2RX+2Na\xrightarrow{\text{dry ether}}R-R+2NaX\]
Here, we have considered that the 2 alkyl halides that are going to form the alkane are the same, but it may happen that the 2 alkyl halides involved in the reaction may have different alkyl substituents. In that case the following reaction takes place:
Let us assume that $R-X$and $R'-X$are the 2 alkyl halides. Then the reaction will be-
\[3R-X+6Na+3X-R'\xrightarrow{\text{dry ether}}R'-R+R-R+R'-R'+6NaX\]
Here, we can see that the alkyl halides may form symmetrical product alkanes by combining with a molecule that resembles them, or a cross-coupling reaction may also occur. The percentage of the products will depend on the stability of the higher alkanes.
Additional information:
The Wurtz reaction only includes aliphatic alkane formation. A reaction that is similar to the Wurtz reaction but the parent hydrocarbons are aryl halides is known as the Fittig reaction. The products formed are aromatic compounds having multiple rings. When the parent hydrocarbons include both alkyl and aryl halides, the reaction is called the Wurtz-Fittig reaction. Compounds like toluene can be formed.
Note: Just like the parent alkyl or aryl group can be different during the reaction, the halogen attached to both the parent alkyl halides can also be different. Please do not assume that the halogens have to be the same in both the reactant molecules.
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