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How to convert $ 2-chlorobutane $ to $ 3,\text{ }4-dimethyl\text{ }hexane $ ?

Answer
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Hint :Conversion of alcohols to alkyl halides, the addition of halogens to alkenes, and hydrohalogenation of alkenes are all methods of preparation. The techniques for preparation were so dependable and efficient that they became an unavoidable part of industrial chemistry.

Complete Step By Step Answer:
 $ 2-chlorobutane $ is a chlorine, hydrogen, and carbon compound. $ Sec-butyl\text{ }chloride $ is another name for it. At room temperature, it is a colorless, volatile liquid that is not miscible with water. At room temperature, it is a colorless, volatile liquid that is not miscible with water.
 Because it is an alkyl chloride, its boiling point varies according to the type of halide attached and where it is attached. Because of chlorine's small size in comparison to other halogens and its weaker intermolecular forces, chlorides have lower boiling points than bromides or iodides.
When $ 2-chlorobutane $ is treated with solid sodium in the presence of dry ether, $ 3,\text{ }4-dimethyl\text{ }hexane $ is formed.
The reaction described above is known as the Wurtz reaction. This reaction occurs when the number of carbon atoms must be increased. Two moles of sodium halide are released as a result of this.
The Wurtz reaction, named after Charles Adolphe Wurtz, is a coupling reaction in organic chemistry, organometallic chemistry, and recently inorganic main-group polymers in which two alkyl halides react in dry ether solution with sodium metal to form a higher alkane. In this reaction, alkyl halides are treated with sodium metal in a dry ethereal (moisture-free) solution to produce higher alkanes containing an even number of carbon atoms.
Thus, in this way we can convert $ 2-chlorobutane $ to $ 3,\text{ }4-dimethyl\text{ }hexane $ .

Note :
Silver, zinc, iron, activated copper, indium, and a mixture of manganese and copper chloride have all been used to achieve the Wurtz coupling. The Wurtz–Fittig reaction is a related reaction that involves aryl halides. This is explained by the formation of a free radical intermediate and its subsequent disproportionation to yield an alkene. The Wurtz reaction occurs via a free-radical mechanism, which allows for side reactions that produce alkene products.