Question

When 2-chlorobutane is heated with sodium metal in THF, what would be formed (including by-products)
(This question has multiple correct options.)
A. $3,4-$ dimethylhexane
B. Butane
C. $1-$ Butene
D. $2-$ Butene

Answer 1

Hint: When alkyl halide is treated with sodium metal in presence of dry ether solution it results in the formation of higher alkane. This reaction was given by Charles Adolphe wurtz in organic chemistry.

Complete step by step answer:
The reaction consists of halogen metal exchange involving radical species ${{R}^{+}}$ with carbon-carbon bond formation that occurs in a nucleophilic substitution reaction.
The Wurtz reaction also contains some limitations. When two different alkyl halides are taken as reactants, it gives alkanes as a product that becomes difficult to separate by fractional distillation. With this reaction, the methane product is not formed.
In this given question, when $2-$ chlorobutane is heated with sodium metal in presence of THF $3,4-$ dimethylhexane, butane, $1-$ butane, $2-$ butane is formed as a product.

So, the correct answer is Option A,B,C,D.

Additional information:
In wurtz reaction, when one electron from the metal is transferred to the halogen, it results in the formation of metal halide and an alkyl radical.
$RX+M\to {{R}^{\bullet }}+MX$
After that, the alkyl halide accepts the electron from the other metal atom that results in the formation of alkyl anion.
${{R}^{\bullet }}+M\to {{R}^{-}}{{M}^{+}}$
The alkyl anion contains a nucleophilic carbon that displaces the halide in $S{{N}^{2}}$ reaction, that forms a new carbon carbon covalent bond.
${{R}^{-}}{{M}^{+}}+RX\to RR+{{M}^{+}}{{X}^{-}}$

Note: When aryl halides is treated with alkyl halides and sodium metal in presence of dry ether it results in the formation of substituted aromatic compounds. This reaction is known as Wurtz fittig reaction. It can be explained by the formation of free radical intermediate and the disproportionate of the substituent that gives alkanes.