Question

How many minimum carbons required for Chain isomerism and Position isomerism in alkanes?

Answer 1

$\underset{–}{\text{One carbon}}:-$ One carbon does not really form chains, So, there is no chain or position isomerism,

$C{H}_4$ (methane) has one carbon.

$\underset{–}{\text{Two carbons}}:-$ Two carbon form a chain, but there is only one way to form this chain, so no isomerism as well.

$C{H}_3-C{H}_3$ (ethane) has two carbons.

$\underset{–}{\text{Three carbons}}:-$ One carbon has to be in the middle and the other two at the ends, so only one possible chain, no isomerism.

$C{H}_3-C{H}_2-C{H}_3$ (propane) has three carbons.

$\underset{–}{\text{Four carbons}}:-$ One carbon has to be in the middle, two at the ends, but the fourth carbon could be in the middle (forming isobutane) or at the end (forming $n-$ butane), so there is a chain as well as position isomerism.

See the figure.

So the minimum numbers of carbon are $4.$