Question

Minimum no. of carbon atoms required for an alkane to exhibit chain isomerism is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

One carbon doesn't form any chain. So there is no chain isomerism in case of alkane having one carbon atom.
$C{H}_4$ methane has one carbon atom

Two carbons : Two carbon form a chain, but there is only one way to form this chain, so no isomerism is there here as well.
$C{H}_3-C{H}_3$ ethane has two carbon atoms

Three carbon atoms :
One carbon has to be in the middle and the other two at the ends . So, only one possible chain is there which contains all the three carbon atoms. So, chain isomerism is not there.

Four carbon atoms :
One carbon has to be in the middle, two at the ends, but the fourth carbon could be in the middle (forming 2-methylpropane) or at the end (forming butane).


So , minimum four carbon atoms show chain isomerism.