Draw Enantiomers For Each Of The Following Compounds Using

 3. Draw Enantiomers For Each Of The Following Compounds Using: A. Perspective Formulas B. Fischer Projections CH3 Br CH3 C. A. Он Draw The Structural Formula Of At Least One Alkene Of Molecular Formula CsH&Br2 That Shows: 4. A. Neither Cis-Trans Isomerism Nor Optical Isomerism. B. Cis-Trans Isomerism But Not Optical Isomerism. C. Optical Isomerism But Not

The Correct Answer and Explanation is:


3. Draw Enantiomers Using:

Compound: CH₃–CHBr–CH₃ (2-bromopropane)

This compound does not have a chiral center because the central carbon (C2) is bonded to:

  • a CH₃ group (on the left),
  • a CH₃ group (on the right),
  • a Br atom,
  • and an H atom.

Since there are two identical CH₃ groups, this compound does not have enantiomers. So, let’s consider a corrected example that does have enantiomers:

Corrected Example: CH₃–CHBr–OH (2-bromo-1-propanol)

This molecule has a chiral center at the middle carbon (C2) with the groups:

  • CH₃,
  • Br,
  • OH,
  • and H.

This gives rise to two enantiomers (non-superimposable mirror images).


A. Perspective Formulas:

        OH                    Br
         |                    |
H — C — Br       ⇄       H — C — OH
     /     \                /     \
  CH3     H             CH3      H
(S-enantiomer)       (R-enantiomer)

B. Fischer Projections:

Enantiomer 1:              Enantiomer 2:
     OH                         Br
      |                          |
CH3–C–Br         ⇄         CH3–C–OH
      |                          |
      H                          H

4. Draw Alkenes (C₅H₈Br₂)

The general formula for an alkene with two Br atoms and five carbon atoms is C₅H₈Br₂. Let’s go over the options:


A. Neither Cis-Trans Nor Optical Isomerism

Compound: 1,1-dibromopent-1-ene

Structure:

CH2=C(Br)2–CH2–CH2–CH3
  • No cis-trans isomerism: because both Br atoms are on the same carbon (C1), no geometric isomerism is possible.
  • No optical isomerism: no chiral centers.

B. Cis-Trans Isomerism But Not Optical Isomerism

Compound: 2,3-dibromopent-2-ene

Structure:

CH3–CH=CH–CHBr–CH3
  • Cis-trans isomerism possible: due to the restricted rotation around the double bond and different groups attached to each carbon of the double bond.
  • No optical isomerism: symmetric molecule with no chiral centers.

C. Optical Isomerism But Not Cis-Trans Isomerism

Compound: 3,4-dibromopent-1-ene

Structure:

CH2=CH–CH(Br)–CH(Br)–CH3
  • No cis-trans isomerism: double bond is terminal.
  • Optical isomerism: C3 and C4 are both chiral centers → possible enantiomers.

Explanation

Enantiomers are stereoisomers that are non-superimposable mirror images of each other. For a molecule to have enantiomers, it must have at least one chiral center—a carbon bonded to four different substituents. Using the compound CH₃–CHBr–OH, we identified a chiral center at the second carbon atom. Drawing both the perspective and Fischer projections clearly distinguishes the two mirror-image configurations (R and S forms), which differ in their spatial arrangement and thus in their interaction with plane-polarized light.

For alkenes with the molecular formula C₅H₈Br₂, the ability to exhibit isomerism depends on the position of the bromine atoms and the double bond. In 1,1-dibromopent-1-ene, both Br atoms are on the same carbon, eliminating any possibility for geometric (cis-trans) or optical isomerism. In contrast, 2,3-dibromopent-2-ene exhibits cis-trans isomerism due to different groups around the double bond but lacks chiral centers. Lastly, 3,4-dibromopent-1-ene lacks cis-trans isomerism due to a terminal double bond, but it has two chiral centers, allowing for enantiomers (optical isomers).

Understanding these differences is essential in organic chemistry as isomerism significantly affects chemical behavior, biological activity, and physical properties.

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