To determine if the molecule [tex]\(CH_3CH_2CH=CH_2\)[/tex] (1-butene) shows stereoisomerism, let’s analyze its structure and the conditions required for stereoisomerism.
1. Identify the double bond:
The double bond is located between the second and third carbon atoms in the molecule:
[tex]\[
CH_3-CH_2-CH=CH_2
\][/tex]
2. Recall the requirements for stereoisomerism:
For a molecule to exhibit stereoisomerism, particularly geometric (cis/trans) isomerism, the carbon atoms involved in the double bond must each have two different substituents. This is because the restricted rotation around the double bond is what allows for different spatial arrangements.
3. Examine the substituents on the double-bonded carbons:
- The second carbon (C2) double-bonded to the third carbon (C3):
[tex]\[
C2: \quad CH_2 \text{ group and } H
\][/tex]
- The third carbon (C3) double-bonded to the second carbon (C2):
[tex]\[
C3: \quad H \text{ and } H
\][/tex]
The full structure simplifies to this consideration:
[tex]\[
CH_3-CH_2-CH=CH_2
\][/tex]
4. Verify the substituents:
- On the second carbon (C2), there is a hydrogen atom and an ethyl group ([tex]\(–CH_2CH_3\)[/tex]).
- On the third carbon (C3), there are two hydrogen atoms.
5. Determine if different groups are present:
- Since the third carbon (C3) has two identical substituents (two hydrogen atoms), it does not meet the requirement for stereoisomerism.
6. Conclusion:
Given that one of the carbons in the double bond (C3) has two identical substituents, [tex]\(CH_3CH_2CH=CH_2\)[/tex] (1-butene) does not have the necessary condition of having different groups attached to each double-bonded carbon. Hence, it cannot show geometric stereoisomerism.
Thus, [tex]\(CH_3CH_2CH=CH_2\)[/tex] (1-butene) does not exhibit stereoisomerism.
