Examkin

Question

Difficulty: HardStructural Isomerism and Stereoisomerism

Which of the following open-chain isomeric alkenes with the molecular formula C5H10C_5H_{10} exhibits geometric (cis-trans) isomerism?

  1. Pent-2-eneAnswer
  2. B
    Pent-1-ene
  3. C
    2-Methylbut-2-ene
  4. D
    3-Methylbut-1-ene

Answer

Pent-2-ene is the correct answer because neither of its double-bonded carbon atoms carries two identical attached groups.
Geometric (cis-trans) isomerism requires restricted rotation around a carbon-carbon double bond (C=CC=C) along with two different substituents attached to each of the double-bonded carbon atoms. In pent-2-ene (CH3−CH=CH−CH2−CH3CH_3-CH=CH-CH_2-CH_3), C2 is bonded to −H-H and −CH3-CH_3, while C3 is bonded to −H-H and −CH2CH3-CH_2CH_3. Because neither carbon carries identical groups, pent-2-ene can form distinct cis and trans stereoisomers.

Step-by-Step Solution

1
Identify the structural condition required for geometric (cis-trans) isomerism in alkenes.
For an alkene R1R2C=CR3R4R_1R_2C=CR_3R_4 to exhibit geometric isomerism, R1≠R2R_1 \neq R_2 and R3≠R4R_3 \neq R_4 must hold for both carbon atoms involved in the double bond.
If either carbon atom in the double bond is bonded to two identical atoms or groups, rotating structural representation does not create distinct non-superimposable stereoisomers.
2
Analyze the structural formula of Pent-1-ene.
Pent-1-ene is CH2=CH−CH2−CH2−CH3CH_2=CH-CH_2-CH_2-CH_3. C1 is attached to two hydrogen atoms (−H-H and −H-H).
Since C1 has identical hydrogen atoms, it cannot exhibit cis-trans isomerism.
3
Analyze the structural formula of Pent-2-ene.
Pent-2-ene is CH3−CH=CH−CH2−CH3CH_3-CH=CH-CH_2-CH_3. C2 is bonded to −H-H and −CH3-CH_3. C3 is bonded to −H-H and −CH2CH3-CH_2CH_3.
Both double-bonded carbons have two distinct attached groups, permitting the existence of cis-pent-2-ene and trans-pent-2-ene.
4
Analyze the structural formulas of 2-Methylbut-2-ene and 3-Methylbut-1-ene.
2-Methylbut-2-ene has two methyl groups on C2 ((CH3)2C=CH−CH3(CH_3)_2C=CH-CH_3), and 3-Methylbut-1-ene has two hydrogen atoms on C1 (CH2=CH−CH(CH3)2CH_2=CH-CH(CH_3)_2).
Both compounds violate the condition of having distinct groups on each doubly bonded carbon atom.

Key Concept

Geometric (cis-trans) Isomerism in Alkenes
Estimated Time:1m 30s
Rate this question

Topics

Structural Isomerism and Stereoisomerism

More questions from this topic