Question

Reaction of Ethyl chloride with sodium leads to (in presence of dry ether):
A.Ethane
B.Propane
C.N-butane
D.N-pentane

Answer 1

Hint: Recall the Wurtz reaction in which an alkyl halide combines with the same alkyl halide and Sodium metal in the presence of dry ether to form a higher alkane. Wurtz reactions can be simple or mixed. In simple two the same alkyl halides combine whereas in a mixed Wurtz reaction both the alkyl halides are different.

Complete answer:
In the given question, the alkyl halide is ethyl chloride and it will combine with another ethyl chloride. So, a Simple Wurtz reaction will take place.
The general reaction for a Simple Wurtz reaction is given by,
$2R-X\text{ + 2Na }\xrightarrow{dry\text{ ether}}\text{ R-R + 2NaX}$
The general reaction for a Mixed Wurtz reaction is given by,
$R-X\text{ + }{{R}^{'}}-X\text{ + 2Na }\xrightarrow{dry\text{ ether}}\text{ R-}{{\text{R}}^{'}}\text{ + 2NaX}$
Here,
R and R’ are Alkyl groups
X is any halide
The mechanism of this reaction is very simple. The two alkyl halides combine to form a long chain hydrocarbon by eliminating the halide group from the parent chain. The alkyl halide
The reaction between Ethyl chloride and Sodium in the presence of dry ether results in the formation of n- butane which is given below:
$2{{C}_{2}}{{H}_{5}}Cl\text{ + 2Na }\xrightarrow{dry\text{ ether}}\text{ }{{\text{C}}_{4}}{{H}_{10}}\text{ + 2NaCl }$
The correct answer is Option C: n-Butane.

Additional information: Wurtz reaction has some limitations too.
Higher alkanes cannot be formed using this reaction.
Symmetrical alkanes give an excellent yield however; with unsymmetrical alkanes the yield is low as other by-products are also formed.
Only simple chain alkyl halides form desired products. Branched chain alkyl halides fail to give the preferred alkane.

Note:
 In the Simple Wurtz reaction, to find out the resultant alkane doubles the number of carbon and hydrogen atoms in the alkyl group of reactants. Similarly, in the Mixed Wurtz reaction, add the number of carbon and hydrogen atoms of the alkyl group of reactants to form the alkane on the product side.