Question

Write the equations for the following reactions:
(a) Wurtz reaction
(b) Wurtz-Fittig reaction

Answer 1

Wurtz reaction is the coupling of two alkyl halides to generate an alkane compound. The reaction between an alkyl halide and an aryl halide was termed as Wurtz-Fittig reaction.

Complete step by step answer:
Both the Wurtz reaction and Wurtz-Fittig reaction are alkane formation reactions means the final product is an alkane. Let us discuss the reaction, reaction conditions and key points of each reaction one by one.
(a) Wurtz reaction.
The Wurtz reaction was developed by French chemist Charles Adolphe Wurtz. The starting material of Wurtz reaction is alkyl halides and sodium metal in presence of dry ether as solvent. It is a useful reaction in organic synthesis for the preparation of alkanes. The mechanism of the reaction proceeds through radical mechanisms.
The reaction is written as:
$2R - X + 2Na \to R - R + 2Na - X$
The reaction is effectively used for the synthesis of higher alkanes from smaller alkyl halides. The steps involved in the reaction are as follows:
Step $1$: The reaction of one alkyl halide with a sodium proceeds by an electron transfer from sodium metal to alkyl halide.
$R - X + Na \to {R^ \cdot } + Na - X$
Step $2$: The alkyl radial accepts an electron from another sodium metal which leads to formation of alkyl anion as:
${R^ \cdot } + Na \to {R^ - }N{a^ + }$
Step $3$: The reaction of the alkyl anion with another molecule with alkyl halide via ${S_N}2$ mechanism leads to formation of alkane.
${R^ - }N{a^ + } + R - X \to R - R + Na - X$
The reaction is also found to work using other metals like iron, copper, manganese or zinc etc.
(b) Wurtz-Fittig reaction
Wurtz-Fittig reaction is related to Wurtz reaction. The reaction was developed by Charles Adolphe Wurtz and Wilhelm Rudolph Fittig. This reaction is an extension of Wurtz reaction. It is the coupling reaction between an alkyl halide with an aryl halide. The halides were treated with sodium in presence of dry ether to obtain the aryl-alkyl coupling product with carbon-carbon bond formation. The reaction is
$Ar - X + R - X\xrightarrow[{dry{\text{ }}ether}]{{Na}}Ar - R + 2Na - X$
The mechanism of the reaction proceeds either by organo-anion formation or by radical formation as seen above. The steps involved in the reaction are as follows:
Step $1$: The aryl halide accepts electrons from the sodium metal forming an aryl sodium complex.
$Ar - X + 2Na \to Ar - Na + Na - X$
Step $2$: An ${S_N}2$ attack of the phenyl anion to the alkyl halide leads to the formation of aryl-alkyl coupled product as:
$Ar - Na + R - X \to Ar - R + Na - X$

Note:
Both reactions are called organic coupling reactions. A side product alkene also formed during the formation of alkane, by elimination of a $H - X$ molecule.