Question

The number of alkenes that exhibit cis/trans isomerism with the molecular formula C\(_5\)H\(_{10}\) is:

• A. 2
• B. 1
• C. 4
• D. 3

Hint:

Cis/trans isomerism in alkenes is possible only when each double bonded carbon has different substituents.

Answer 1

The Correct Option is B

Isomerism in alkenes requires: - A double bond between two carbons - Each carbon of the double bond must have two different substituents Only pent-2-ene (CH\(_3\)-CH=CH-CH\(_2\)-CH\(_3\)) satisfies the condition for cis-trans isomerism. Thus, only 1 alkene (pent-2-ene) shows geometrical (cis-trans) isomerism.

Answer 2

The number of alkenes that exhibit cis/trans isomerism with the molecular formula C₅H₁₀ is:

Step 1: Determine the degree of unsaturation:
C₅H₁₀ corresponds to the general formula C_nH_2n, which indicates that it is an alkene (one degree of unsaturation).
So, the compound is an acyclic alkene or a cycloalkane. Since we are considering alkenes, we will focus on open-chain compounds only.

Step 2: List possible structural isomers of alkenes (C₅H₁₀):
1. Pent-1-ene
2. Pent-2-ene
  • cis-pent-2-ene
  • trans-pent-2-ene
3. 2-methylbut-1-ene
4. 2-methylbut-2-ene
5. 3-methylbut-1-ene

Step 3: Identify which among these can show cis/trans (geometrical) isomerism:
To show cis/trans isomerism, each carbon of the double bond must have two different groups attached.
- Pent-2-ene satisfies this condition, and both cis and trans forms exist → exhibits cis/trans isomerism.
- Pent-1-ene and the methyl-substituted alkenes do not satisfy the condition for geometrical isomerism as one of the double-bonded carbons has two identical hydrogen atoms.

Step 4: Conclusion:
Only **pent-2-ene** satisfies the criteria and exhibits cis/trans isomerism.

Final Answer:
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