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Question: simplest optically active alkene...

simplest optically active alkene

Answer

3-methyl-1-pentene

Explanation

Solution

To be optically active, a molecule must be chiral. The most common criterion for chirality in organic molecules is the presence of a chiral carbon atom, which is a carbon atom bonded to four different groups.

Let's systematically determine the simplest alkene that can be optically active:

  1. Chirality Requirement: A carbon atom involved in a double bond (sp2 hybridized) cannot be a chiral center because it is bonded to only three groups. Therefore, the chiral center must be an sp3 hybridized carbon atom that is not part of the double bond.

  2. Smallest Alkene Group: To minimize the number of carbons, we should use the smallest possible alkene group. This is the vinyl group (-CH=CH2), which contains 2 carbon atoms.

  3. Constructing the Chiral Center: Let's consider a carbon atom (C*) that will be our chiral center. It needs to be bonded to four different groups. One of these groups must be an alkene group. Let's make it the vinyl group (-CH=CH2).

    Now, we need three more groups (G1, G2, G3) attached to C* that are different from each other and from the vinyl group. To make the molecule as simple (i.e., with the fewest carbons) as possible, we should choose the smallest possible different groups:

    • G1 = -H (hydrogen atom)
    • G2 = -CH3 (methyl group, 1 carbon)
    • G3 = -CH2CH3 (ethyl group, 2 carbons)

  4. Assembling the Molecule: If we attach these four groups to the chiral carbon C*, the structure becomes:

          CH3
      |
CH2=CH-CH-CH2CH3
      |
      H

  5. Nomenclature and Carbon Count:

    • The longest carbon chain containing the double bond is 5 carbons long (pentene).
    • The double bond is at position 1.
    • The methyl group is at position 3.
    • Therefore, the IUPAC name of this molecule is 3-methyl-1-pentene.
    • Total number of carbon atoms: 2 (from vinyl) + 1 (chiral C*) + 1 (methyl) + 2 (ethyl) = 6 carbons.

  6. Verification of Chirality: The carbon at position 3 is bonded to:

    • A hydrogen atom (-H)
    • A methyl group (-CH3)
    • An ethyl group (-CH2CH3)
    • A vinyl group (-CH=CH2) Since all four groups are different, the carbon at position 3 is a chiral center, making the molecule optically active.

  7. Simplest Check:

    • Can a 5-carbon alkene be optically active? Let's list common 5-carbon alkenes: 1-pentene, 2-pentene, 2-methyl-1-butene, 3-methyl-1-butene, 2-methyl-2-butene. None of these have a chiral center.
    • Therefore, the simplest optically active alkene must have at least 6 carbon atoms. Our candidate, 3-methyl-1-pentene, has 6 carbons. This confirms it is the simplest.

The simplest optically active alkene is 3-methyl-1-pentene.

Structure:

(Where B is the chiral carbon)

Or, more explicitly:

      CH₃
      |
CH₂=CH-CH-CH₂CH₃
      |
      H

The carbon atom marked with an asterisk (*) is the chiral center.

Explanation of the solution:

To find the simplest optically active alkene, we identify the conditions for optical activity (presence of a chiral center). A chiral carbon must be sp3 hybridized and bonded to four different groups. We then systematically build the smallest possible alkene molecule satisfying these conditions by attaching the smallest possible alkene group (vinyl, -CH=CH2) and the smallest distinct alkyl groups/hydrogen to a central carbon. This leads to 3-methyl-1-pentene, which has 6 carbons and a chiral center, making it the simplest.