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First member of optically active alkene ...

First member of optically active alkene is :

A

`C_(2)H_(5)- underset(H) underset(|)overset(CH_(3)) overset(|)(C)-CH=CH_(2)`

B

`C_(3)H_(7)- underset(H) underset(|)overset(C_(2)H_(5)) overset(|)(C)-CH=CH_(2)`

C

`C_(3)H_(7)- underset(CH_(3)) underset(|)overset(C_(2)H_(5)) overset(|)(C)-CH=CH_(2)`

D

`C_(6)H_(11)- underset(CH_(3)) underset(|)overset(C_(2)H_(5)) overset(|)(C)-CH=CH_(2)`

Text Solution

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The correct Answer is:
To determine the first member of optically active alkenes, we need to identify an alkene that has a chiral carbon atom. A chiral carbon is one that is bonded to four different groups, which results in non-superimposable mirror images, making the compound optically active. ### Step-by-Step Solution: 1. **Understanding Chiral Centers**: - A chiral carbon is a carbon atom that has four different substituents attached to it. This creates two non-superimposable mirror images, known as enantiomers. 2. **Identifying Alkenes**: - Alkenes are hydrocarbons that contain at least one carbon-carbon double bond (C=C). 3. **Finding the First Member**: - The simplest alkene is ethene (C2H4), which does not have a chiral center. The next alkene is propene (C3H6), which also does not have a chiral center. - The first optically active alkene is 2-butene (C4H8), but it has a plane of symmetry and is not chiral. 4. **Identifying the Correct Structure**: - The first member of optically active alkenes is 2-methylpropene (C4H8), which has a chiral center. The structure can be represented as follows: - CH2=C(CH3)CH2CH3 - Here, the carbon atom at the double bond is bonded to four different groups: a hydrogen atom, a methyl group (CH3), an ethyl group (C2H5), and a propyl group (C3H7). 5. **Conclusion**: - Therefore, the first member of optically active alkenes is 2-methylpropene. ### Final Answer: The first member of optically active alkene is **2-methylpropene (C4H8)**. ---
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