Question

The compound which is not formed when a mixture of n-butyl bromide and ethyl bromide treated with sodium metal in presence of dry ether is:
A. butane
B. octane
C. hexane
D. ethane

Answer 1

Hint: Higher alkanes are produced by heating alkyl halides with sodium metal in dry ether. This reaction is called the Wurtz reaction. Two molecules of alkyl halides lose their halogen atoms as sodium halide. From this reaction, only symmetrical alkanes are obtained.

Complete step by step answer:
Alkyl halides are the derivatives of alkanes in which the hydrogen atom is replaced by a halogen atom. They are represented by \[{\text{R}} - {\text{X}}\], where ${\text{R}}$ is the alkyl group and ${\text{X}}$ is the halogen group. There are different types of reactions with alkyl halides like nucleophilic substitution, reduction, elimination etc.
When alkyl halides are treated with sodium in dry ether it produces higher alkanes. This net result is the joining of two alkyl groups to yield symmetrical alkane having an even number of carbon atoms. The general chemical equation is given below:
$2{\text{R}} - {\text{X + 2Na}} \to {\text{R}} - {\text{R + 2NaX}}$
When n-butyl bromide reacts with sodium in dry ether, it produces octane.
${\text{2C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{Br + 2Na}}\xrightarrow{{{\text{ether}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3} + {\text{NaBr}}$
n-butyl bromide octane sodium bromide
When ethyl bromide reacts with sodium in ether, it produces butane.
$2{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{Br + 2Na}}\xrightarrow{{{\text{ether}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3} + {\text{NaBr}}$
Ethyl bromide butane
When a mixture of n-butyl bromide and ethyl bromide reacts with sodium in ether, it produces hexane.
${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{Br + C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{Br + 2Na}}\xrightarrow{{{\text{ether}}}}{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3} + 2{\text{NaBr}}$
n-butyl bromide ethyl bromide hexane
Hence we can tell that by reacting ethyl bromide or n-butyl bromide with sodium in ether, only higher alkanes can be obtained. Ethane cannot be obtained.
Hence, Option D is the correct option.

Additional information:
The limitations of this reaction are:
We cannot obtain methane from this reaction.
Tertiary halides do not undergo this reaction.

Note:
The mechanism involves three steps:
The electron from the metal transfers to the halogen which produces alkyl radical and metal halide.
${\text{R}} - {\text{X}} + {\text{Na}} \to {{\text{R}}^ \bullet } + {\text{N}}{{\text{a}}^ + }{{\text{R}}^ - }$
This alkyl radical is reacted with another sodium which produces alkyl anion.
${{\text{R}}^ \bullet } + {\text{Na}} \to {{\text{R}}^ - }{\text{N}}{{\text{a}}^ + }$
The carbon in the alkyl group is nucleophilic in nature, thereby displaces the halogen in alkyl halide. Thus the carbon forms a covalent bond with the alkyl group.
${{\text{R}}^ - }{\text{N}}{{\text{a}}^ + } + {\text{R}} - {\text{X}} \to {\text{R}} - {\text{R}} + {\text{N}}{{\text{a}}^ + }{{\text{X}}^ - }$