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Question: n – Propyl chloride reacts with sodium metal in dry ether to give: A.\(\text{C}{{\text{H}}_{\text{...

n – Propyl chloride reacts with sodium metal in dry ether to give:
A.CH3 − CH2 − CH2 − CH2 − CH2 − CH3\text{C}{{\text{H}}_{\text{3}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{3}}}
B.CH3 − CH2 − CH3\text{C}{{\text{H}}_{\text{3}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{3}}}
C.CH3 − CH2 − CH2 − CH3\text{C}{{\text{H}}_{\text{3}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{3}}}
D.CH3 − CH2 − CH2 − CH2 − CH2 − CH2 − CH3\text{C}{{\text{H}}_{\text{3}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{3}}}

Explanation

Solution

Hint: In order to answer this question, we need to first write down the reaction. Once we know the structure of the reactant, we can predict the product from the reaction. Moreover, knowing the structure of the reactant also gives us an idea about the structure of the product, in the presence of any other compound.

Step by step answer:
Let us first write down the reaction and then we can move on for further explanation. So, here is the equation:
n ( CH3CH2CH2Cl ) →Dry EtherNa CH3 − ( CH2)4 − CH3\text{n ( C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl ) }\xrightarrow[\text{Dry Ether}]{\text{Na}}\text{ C}{{\text{H}}_{\text{3}}}\text{ }-\text{ ( C}{{\text{H}}_{\text{2}}}{{\text{)}}_{\text{4}}}\text{ }-\text{ C}{{\text{H}}_{\text{3}}}
The above reaction can also be written as:
n ( CH3CH2CH2Cl ) →Dry EtherNa CH3 − CH2 − CH2 − CH2 − CH2 − CH3\text{n ( C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl ) }\xrightarrow[\text{Dry Ether}]{\text{Na}}\text{ C}{{\text{H}}_{\text{3}}}\text{ }-\text{ C}{{\text{H}}_{_{\text{2}}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{2}}}\text{ }-\text{ C}{{\text{H}}_{\text{3}}}
In the above mentioned reaction, when propyl chloride reacts with sodium metal, in the presence of dry ether and results in the coupling of two molecules of propyl chloride to produce n – hexane. This reaction is also known as the Wurtz reaction.
Wurtz reaction involves the reduction of an alkyl halide with Sodium in ether. Wurtz reaction is a convenient method of preparation of symmetrical alkanes (R – R). This means that the alkane will contain an even number of C atoms.
Therefore, the correct answer to the question is Option A.
The best way of defining the Wurtz reaction in an organic compound is that it is a coupling reaction where there is reaction between sodium metal and two alkyl halides in an environment which has the presence of a solution of dry ether. The main of the reaction is to form a higher alkane along with a compound, which contains sodium and the halogen.

Note: Wurtz reaction can't be used for the preparation of unsymmetrical alkanes. This is because, if two dissimilar alkyl halides are taken as the reactants, a mixture of alkanes is obtained as the products. As the reaction involves free radical species, a side reaction also takes place producing an alkene.