Question

The maximum number of isomers possible for an alkene with molecular formula C₄H₈ is:

• A. 5
• B. 4
• C. 2
• D. 3

Hint:

List the straight chain, branched chain, and cis-trans forms of C4H8 separately.

Answer 1

The Correct Option is B

Work this out systematically by fixing the carbon skeleton first, then checking for geometrical isomerism on each skeleton.

Skeleton 1, a straight chain of 4 carbons: the double bond can sit at position 1,2 (giving 1-butene) or at position 2,3 (giving 2-butene). Placing it at 3,4 just repeats 1-butene counted from the other end, so it is not a new structure.

Skeleton 2, a branched 3 carbon chain with a methyl side group: here the double bond sits at the branch carbon, giving 2-methylpropene, and only one such isomer is possible because the branch carbon carries two identical CH3 groups, so no cis-trans pair can form.

Now test each structure for cis-trans isomerism. 1-butene fails the test because one of its double bond carbons carries two hydrogen atoms, both identical, so no cis-trans pair exists. 2-butene passes the test because each double bond carbon carries one H and one CH3, two different groups, so cis and trans forms both exist. 2-methylpropene fails the test for the same reason as before.

Counting every distinct structure: 1-butene (1) + cis-2-butene (1) + trans-2-butene (1) + 2-methylpropene (1) = 4 isomers total, confirming option B.