Question

Alkyl halides react with metallic sodium in dry ether producing?
a.) Alkanes with same number of carbon atoms
b.) Alkanes with double number of carbon atoms
c.) Alkenes with triple number of carbon atom
d.) Alkenes with the same number of carbon atoms.

Answer 1

Hint: In order to deal with this question we will first go for defining the alkyl halides and then we will see its properties and its chemical reaction with metals and others in presence of some catalyst like ether.

Complete answer:
Alkyl halides, or haloalkanes are chemical compounds often derived from alkanes which contain one or more halogens. Alkyl halides are also called halogens. Alkyl halides are also a subset in the overall halocarbon class. They are formed by replacement of hydrogen atoms with halogen atoms.
The alkyl halides can be represented as RX.
The reaction of alkyl halides with sodium in presence of dry ether is known as wurtz reaction. This reaction is used to prepare higher alkanes. In this reaction two symmetric alkyl halides are used so that the product formed contains double the number of carbon atoms.
$2RX + 2Na\xrightarrow{{dry\, - ether}}R - R + 2Na - X$
Hence, the correct option is B.

Note: The reaction will begin in the presence of dry ether as a catalyst. If dry ether is not present then sodium metal will react with oxygen to form oxides. Also, if the same alkyl halides are not used, then the product formed will contain alkanes with different numbers of carbon atoms which is very difficult to differentiate. Even with fractional distillation it is not possible to separate because alkanes have the same boiling point and there is very little difference in the boiling point when few carbon atoms increase or decrease.