Question:
Which of the following compounds will exhibit geometrical isomerism?
Which of the following compounds will exhibit geometrical isomerism?
Show Hint
Check the double-bonded carbons. If either carbon is bonded to two identical groups (like $=\text{CH}_{2}$), it cannot form geometrical isomers.
Updated On: Jun 3, 2026
- 1-Butene
- 2-Butene
- Propene
- 2-Methylpropene
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Geometrical isomerism ($cis/trans$) occurs in alkenes when rotation around the carbon-carbon double bond is restricted, and each carbon atom in the double bond is attached to two different groups.
Step 2: Meaning
For a molecule structured as $xy\text{C}=\text{C}ab$ to show geometrical isomers, $x \neq y$ and $a \neq b$.
Step 3: Analysis
Let us evaluate the options: * *1-Butene:* $\text{CH}_{2}=\text{CH}-\text{CH}_{2}-\text{CH}_{3}$. The terminal carbon carries two identical hydrogen atoms, preventing cis/trans structures. * *Propene:* $\text{CH}_{2}=\text{CH}-\text{CH}_{3}$. It also features a terminal $=\text{CH}_{2}$ group with identical hydrogens. * *2-Methylpropene:* $(\text{CH}_{3})_{2}\text{C}=\text{CH}_{2}$. One double-bonded carbon holds two identical methyl groups, while the other holds two identical hydrogens. * *2-Butene:* $\text{CH}_{3}-\text{CH}=\text{CH}-\text{CH}_{3}$. Each double-bonded carbon is bonded to one distinct $-\text{H}$ atom and one distinct $-\text{CH}_{3}$ group. This satisfies the criteria, allowing for *cis*-2-butene and *trans*-2-butene.
Step 4: Conclusion
2-Butene is the molecule that exhibits geometrical isomerism.
Final Answer: (B)
Geometrical isomerism ($cis/trans$) occurs in alkenes when rotation around the carbon-carbon double bond is restricted, and each carbon atom in the double bond is attached to two different groups.
Step 2: Meaning
For a molecule structured as $xy\text{C}=\text{C}ab$ to show geometrical isomers, $x \neq y$ and $a \neq b$.
Step 3: Analysis
Let us evaluate the options: * *1-Butene:* $\text{CH}_{2}=\text{CH}-\text{CH}_{2}-\text{CH}_{3}$. The terminal carbon carries two identical hydrogen atoms, preventing cis/trans structures. * *Propene:* $\text{CH}_{2}=\text{CH}-\text{CH}_{3}$. It also features a terminal $=\text{CH}_{2}$ group with identical hydrogens. * *2-Methylpropene:* $(\text{CH}_{3})_{2}\text{C}=\text{CH}_{2}$. One double-bonded carbon holds two identical methyl groups, while the other holds two identical hydrogens. * *2-Butene:* $\text{CH}_{3}-\text{CH}=\text{CH}-\text{CH}_{3}$. Each double-bonded carbon is bonded to one distinct $-\text{H}$ atom and one distinct $-\text{CH}_{3}$ group. This satisfies the criteria, allowing for *cis*-2-butene and *trans*-2-butene.
Step 4: Conclusion
2-Butene is the molecule that exhibits geometrical isomerism.
Final Answer: (B)
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