Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Alkenes of the same formula can differ in the position of the double bond, in the branching of the chain and in the arrangement about the double bond.
Geometrical isomers need two different groups on each carbon of the double bond. Alcoholic KOH removes HX from an alkyl halide, and the more substituted alkene is the main product (Saytzeff). The more alkyl groups on the double bond, the more stable the alkene and the less heat it gives out on hydrogenation.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. C4H8. But-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene. That is .
Step 2. C5H10, straight chain. Pent-1-ene, cis-pent-2-ene and trans-pent-2-ene: .
Step 3. C5H10, branched. 2-Methylbut-1-ene, 3-methylbut-1-ene and 2-methylbut-2-ene: . The total is .
Step 4. Cis-trans. Only pent-2-ene has two different groups on each carbon of its double bond. In 2-methylbut-2-ene one carbon holds two methyl groups. So of the six shows geometrical isomerism.
Answer: , and .
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