Question

In the reaction,\[2nR - X\xrightarrow[{{\text{Dry ether}}}]{{2{\text{n Na}}}}product\] the product obtained is:
A.2n alkene
B.n sodium halide
C.n alcohol
D.n alkane

Answer 1

Hint:We can see in the question, the given reactant is alkyl halide, and reaction takes place in the presence of two moles of metallic sodium in dry ether. Therefore, the reaction should be Wurtz reaction.

Complete step by step answer:Wurtz reaction is named after Charles Adolphe Wurtz. A coupling reaction where two alkyl halides are made to react with metallic sodium in the presence of dry ether solution to obtain higher alkane is Wurtz reaction.
In this reaction, alkyl halides are reacted with metallic sodium in dry ethereal solution to obtain higher alkanes.
We can use it in the preparation of higher alkanes composed of even numbers of carbon atoms. We can write the chemical reaction is,
$2R - X + 2Na\xrightarrow{{{\text{dry ether}}}}R - R + 2N{a^ + }{X^ - }$
Here $R - X$ is the alkyl halide and $R - R$ is an alkane.
Other metals like silver, indium, iron, activated copper, and a mixture of manganese and copper chloride are used to exhibit the effect of the Wurtz coupling.
So, 2 units of alkyl halides reacting with 2 units of sodium in presence of dry ether would give one unit of alkane. Thus, 2n units of alkyl halides and sodium would give n units of alkane.
Therefore, the option (D) is correct.

Note: We should not confuse Wurtz reaction and Wurtz-Fittig reaction. Aryl halides with alkyl halides are reacted with metallic sodium in the presence of dry ether to obtain substituted aromatic compounds is Wurtz-Fittig reaction. Some of the limitations of Wurtz reactions are,
-Seldom used due to side reactions
-We cannot obtain methane by Wurtz reaction
-Wurtz reaction fails for tertiary halides
-Since, Wurtz reaction goes by free radical mechanism there is a possibility to obtain alkene as a product because of side reactions.