Examkin

Question

Difficulty: MediumStructural Isomerism and Stereoisomerism

Which of the following 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 only isomer listed that exhibits geometric (cis-trans) isomerism because each carbon of the double bond is attached to two distinct groups.
Pent-2-ene has a double bond between carbon-2 and carbon-3. Carbon-2 is attached to a hydrogen atom and a methyl group (CH3CH_3), while carbon-3 is attached to a hydrogen atom and an ethyl group (CH2CH3CH_2CH_3). Because neither carbon atom of the double bond holds two identical groups, spatial restriction gives rise to distinct cis and trans stereoisomers.

Step-by-Step Solution

1
Recall the necessary structural condition for geometric (cis-trans) isomerism in alkenes.
For a molecule to show cis-trans isomerism around a double bond C=CC=C, each of the two carbon atoms in the double bond must be attached to two different atoms or groups.
If either carbon in the double bond has two identical groups attached, rotating spatial arrangements results in identical molecules.
2
Examine the connectivity of each option at the C=CC=C double bond.
Pent-2-ene has C2C_2 bonded to −H-H and −CH3-CH_3, and C3C_3 bonded to −H-H and −CH2CH3-CH_2CH_3. Both carbons have two different groups.
This satisfies the criteria for both cis and trans geometric arrangements.
3
Check the remaining options for duplicate attached groups on double-bonded carbons.
Pent-1-ene and 3-methylbut-1-ene both have a terminal =CH2=CH_2 (two H atoms on C1C_1). 2-Methylbut-2-ene has two methyl groups on C2C_2.
None of these three options satisfy the non-identical substituent requirement on both double-bonded carbons.

Key Concept

Geometric Isomerism Requirements in Alkenes
Estimated Time:1m 0s
Rate this question

Topics

Structural Isomerism and Stereoisomerism

More questions from this topic