The molecular formula C4H8 represents an alkene, which is a hydrocarbon with a carbon-carbon double bond. To determine the maximum number of isomers for this alkene, we need to consider the different possibilities for the arrangement of the carbon atoms and the double bond.

There are four carbon atoms in the molecule, so we can arrange them in different ways. Let's consider each possibility:

1. Straight-chain isomer: The four carbon atoms can be arranged in a straight line, with the double bond between the second and third carbon atoms. This is called but-2-ene.

CH3-CH=CH-CH3

2. Branched isomer: One of the carbon atoms can form a branch with the other three carbon atoms. The double bond can be between the second and third carbon atoms of the main chain or between the second and fourth carbon atoms of the main chain.

a) Isobutene: The branch is formed by the first carbon atom, and the double bond is between the second and third carbon atoms of the main chain.

CH2=C(CH3)2

b) 2-Methylpropene: The branch is formed by the second carbon atom, and the double bond is between the second and fourth carbon atoms of the main chain.

CH3-C(=CH2)-CH3

3. Cycloalkene isomers: The four carbon atoms can form a cycloalkene, which is a ring structure with a double bond. There are two possibilities for the arrangement of the double bond:

a) Cyclobutene: The four carbon atoms form a four-membered ring, and the double bond is between the first and second carbon atoms.

CH2=CH-CH2-CH2

b) Cyclobutene (different isomer): The four carbon atoms form a four-membered ring, and the double bond is between the second and third carbon atoms.

CH3-CH=CH-CH3

So, the maximum number of isomers for an alkene with the molecular formula C4H8 is 4. These isomers are:
- But-2-ene
- Isobutene
- 2-Methylpropene
- Cyclobutene
- Cyclobutene (different isomer)

Therefore, the correct answer is option C) 4.