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I learnt in my chemistry class that if a molecule has an element of symmetry (either a plane of symmetry or a centre of symmetry), it will not be chiral and hence will not show optical activity.

My professor told me that in alanine anhydride the carbon atoms with methyl groups would be non-planar, hence it will be non-symmetrical:

3,6-dimethylpiperazine-2,5-dione

But if I consider one of the above-mentioned carbon atom above the plane of the ring, and the other below the plane of ring, a centre of symmetry will still exist. Where am I going wrong?

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    $\begingroup$ Well You are assuming that the ring is planer which is not the real case $\endgroup$ – Pj30 Mar 31 at 16:25
  • $\begingroup$ @Pj30 The trans isomer has a centre of symmetry even though the ring is non planar(the answer posted by someone has its image)..How then is the compound optically active? $\endgroup$ – Nathalia Apr 1 at 4:41
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There is no evidence (e.g. aromaticity) to support your claim that the compound is planar. In fact, crystalographic studies [1] revealed existence of both cis- and trans-3,6-dimethylpiperazine-2,5-dione. Corresponding CCDC numbers are LCDMPP01 and TRDMPP01, respectively — feel free to open the links and play around with the 3D structures in JSmol to understand the difference.

According to Sletten, in cis (ʟʟ) modification the ring is puckered resembling a skewed boat conformation, whereas the ring in trans (ᴅʟ) molecule is nearly planar:

*cis*-3,6-dimethylpiperazine-2,5-dione
Figure 1a. cis-3,6-dimethylpiperazine-2,5-dione (drawn from LCDMPP01)

*trans*-3,6-dimethylpiperazine-2,5-dione
Figure 1b. trans-3,6-dimethylpiperazine-2,5-dione (drawn from TRDMPP01)

References

  1. Sletten, E. Conformation of Cyclic Dipeptides. The Crystal and Molecular Structures of Cyclo-D-Alanyl-L-Alanyl and Cyclo-L-Alanyl-L-Alanyl (3,6-Dimethylpiperazine-2,5-Dione). J. Am. Chem. Soc. 1970, 92 (1), 172–177. DOI: 10.1021/ja00704a028. (Open Access)
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  • $\begingroup$ This is still rather nitpicking then full answer. Thing is if trans isomer is centrosymmetric or not. I wager it is and therefore OP's right. (Perhaps some additional effects would come with interactions in solid state or temperature, though.) $\endgroup$ – Mithoron Apr 1 at 0:18
  • $\begingroup$ Yes the trans isomer has a centre of symmetry.How is the compound still optically active then?(I got this question wrong in a test where they had asked whether the compound shows optical activity or not) $\endgroup$ – Nathalia Apr 1 at 4:38
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    $\begingroup$ If you flip the directions of the methyl groups (make the sticky-down one on the left stick up, and vice versa), can you superimpose that on the original? If not, then you have two enantiomers. If you can, then you don't. $\endgroup$ – Will Crawford Apr 1 at 9:29
  • $\begingroup$ The sufficient condition for a compound to be optically inactive is that it should have a centre of symmetry..and i think the trans isomer has one.Can you explain to me why is the trans isomer optically active. I agree that the cis isomer is optically inactive due absence of any element of symmetry $\endgroup$ – Nathalia Apr 1 at 12:08
  • $\begingroup$ The transfer isomer does not have C2 symmetry or a centre of inversion. If you try to superimpose a mirror image, the carbonyl oxygen would be where the NH is and vice versa. Plus, the methyls will be in opposite configuration. Has the possibility of amide tautomerism been considered too, just to muddy the water ? $\endgroup$ – Beerhunter May 23 at 13:30

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