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The structure of one particular isomer of 4-chlorohepta-2,5-diene is shown here. I am confused about how two of the propenyl groups attached to the central carbon have the same atoms with different configurations.

Structure of (Z,E)-4-chlorohepta-2,5-diene with isomeric propenyl groups highlighted

More specifically:

  1. Is the carbon shown in the image is a chiral carbon or pseudo-chiral carbon?

  2. Will this compound show optical isomerism, i.e. does it have an enantiomer?

  3. Are the two alkenyl groups considered to be different from each other, or the same?

  4. If the attached side centers were chiral centers (e.g $\ce{-CHClCH3}$) with different configurations, instead of geometrical isomers, then would this compound show optical isomerism?

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    $\begingroup$ 1) the carbon connected to 4 different groups is chiral, because it's connected to 4 different groups. 2) weather the whole molecule shows optical isomerism depends on one parameter. if the molecule has reflective symmetry or not. A molecule with no reflective symmetry is chiral. So I think this would be a chiral molecule. 3) the 2 circled groups are different. they are also stereoisomers E and Z or cis and trans. Why are they that? because you can't turn one group into another without breaking the double bonds. 4) I don't understand what you're asking here. An edit is needed. $\endgroup$
    – bobsburger
    Commented Nov 6, 2020 at 16:11

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Let's address your concerns one at a time:

  1. The structure 1 you have drawn is chiral with $\ce{C4}$ having four different groups attached to it. $\ce{C4}$ has the (S)-configuration with the CIP priorities Cl>(Z)-propenyl>(E)-propenyl>H.
  2. The structure has an enantiomer of the (R)-configuration. Just switch any two groups: Cl, (Z)-propenyl, (E)-propenyl or H. Each enantiomer will have the same specific rotation but opposite in sign.
  3. The two propenyl groups are different stereoisomers, just as different as the (Z)- and (E)-2-butenes.
  4. If the two propenyl groups were changed to -CHClCH3, say, (2R,4R)-2,3,4-trichloropentane 2a ($\ce{CH3CHClCHClCHClCH3}$), then the central carbon, $\ce{C3}$, is chirotopic but non-stereogenic. That is to say, if the chlorine and hydrogen at $\ce{C3}$ were switched as in 2b, the compound would still be chiral but not a different stereoisomer. Rotate 2a by 180o about a vertical axis and you will obtain 2b. They are identical.
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  • $\begingroup$ If the two propenyl groups were changed to -CHClCH3, say, (2R,4R)-2,3,4-trichloropentane 2a (CH3CHClCHClCHClCH3), then the central carbon, C3, is chirotopic but non-stereogenic. I am unable to understand your this line. I want to know if these geometrical centrical were replaced by chiral centres like in 2a and 2b then the C3 carbon will remain chiral or not? $\endgroup$
    – Arun Bhardwaj
    Commented Nov 7, 2020 at 4:38
  • $\begingroup$ I chose the 2R,4R-isomer. If I had chosen the 2R,4S-isomer, then C3 would be achirotopic and stereogenic. Two diastereomers would arise: 2R,3r,4S and 2R,3s,4S. This link should help for both your original diene and the trichloro examples: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/… $\endgroup$
    – user55119
    Commented Nov 7, 2020 at 12:17

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