Question

Wurtz’s reaction is suitable for preparing unsymmetrical alkanes. If true enter $ 1 $ , else enter $ 0 $ .

Answer 1

Hint :Wurtz reaction is a coupling reaction that is used in the preparation of symmetrical alkanes. Reaction of alkyl halides with sodium metal in the presence of dry ether produces symmetrical alkanes. Wurtz-Fittig reaction is a modification of Wurtz reaction wherein both aryl and alkyl halides are used to react with sodium metal.

Complete Step By Step Answer:
Wurtz reaction is the reaction of alkyl halides with sodium metal in the presence of dry ether to form a symmetrical alkane with double the number of carbon atoms in alkyl halide and sodium halide as by-product. The exchange of halogen and metal takes place in the reaction with the involvement of R radicals to form carbon-carbon bonds.
The equation is:
 $ 2R - X\,\,\, + \,\,2Na\,\,\,\,\xrightarrow{{dry\,\,\,ether}}\,\,\,\,R - R\,\, + \,\,2NaX $
Alkyl halide alkane
If we use two different alkyl halides, we get a combination of symmetrical alkanes and unsymmetrical alkanes. Therefore, the yield for unsymmetrical alkanes would be quite low.
 $ C{H_3}Br\, + \,{C_2}{H_5}Br\, + \,2Na\,\,\,\xrightarrow{{dry\,\,ether}}\,\,\,{C_3}{H_8}\, + \,2NaBr $
Here, along with propane we get a mixture of ethane and butane as products. The alkanes thus produced will have similar boiling point and physical properties and it is seemingly difficult to separate the unsymmetrical ones from the mixture. Thus, the Wurtz reaction is not suitable for the preparation of unsymmetrical alkanes.

Note :
Sodium metal is highly reactive. In order to reduce its reactivity that may disturb the reaction, we use dry ether as a solvent. And in case the ether is not dry, sodium will react with water to form oxides and hydroxides and we will not get the desired products.
The Wurtz reaction cannot be used to prepare methane $ (C{H_4}) $ and the lowest alkane that can be made through the Wurtz reaction is ethane $ ({C_2}{H_6}) $ .