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Shouldn't the N pointed to by the red arrow also be a stereocenter? I am asking this because I think it's lone pair cannot take part in nitrogen inversion.

ampicillin

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    $\begingroup$ This is a very good question. I'm not an expert at N stereochemistry but have you considered the iminol form of the amide/iminol tautomer? It would have a double-bond to the N and would not be chiral. I know the amide form predominates but if there is any contribution from the iminol it would "destroy" the chirality wouldn't it? $\endgroup$
    – Curt F.
    Commented Jul 28, 2015 at 5:43
  • $\begingroup$ Why do you think that nitrogen inversion cannot take place there? $\endgroup$ Commented Jul 28, 2015 at 6:08
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    $\begingroup$ Probably because N is part of a ring but I am not sure if any strain is involved? $\endgroup$
    – fidgetyphi
    Commented Jul 28, 2015 at 6:25
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    $\begingroup$ The two connected rings and the attached sterically demanding groups will certainly hinder inversion $\endgroup$ Commented Jul 28, 2015 at 7:10
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    $\begingroup$ Amide groups are flat! There's no point in talking about inversion. $\endgroup$
    – Mithoron
    Commented Jul 28, 2015 at 10:27

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After reading the inputs from @Curt F., @PH13, @Martin and @Mithoron, I infer that N won't be involved in nitrogen inversion due to hindrance caused by connected rings and attached sterically demanding groups. Hence, 'N' pointed to by the red arrow will be considered as a stereocenter in ampicillin.

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  • $\begingroup$ You didn't get my comment - it's flat so there can't be stereocenter! $\endgroup$
    – Mithoron
    Commented Jul 28, 2015 at 11:13
  • $\begingroup$ @Mithoron Correct me if I am wrong, you said that amides are flat and therefore nitrogen inversion won't be possible, right? Now, the absence of nitrogen inversion facilitates in N being a stereocenter given that all the substituents attached to it are different as well (which is also satisfied in this particular case). $\endgroup$
    – fidgetyphi
    Commented Jul 28, 2015 at 11:25
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    $\begingroup$ All substituents of N atom are in one plane - thats why speaking of prescence of inversion or lack of inversion doesn't make sense. $\endgroup$
    – Mithoron
    Commented Jul 28, 2015 at 13:06
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    $\begingroup$ @Mithoron I think you should write up an answer. Also everyone please avoid extended discussions in comments. Use the chatroom instead. $\endgroup$
    – bon
    Commented Jul 28, 2015 at 13:20
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    $\begingroup$ @theprogrammer see discussion starting here chat.stackexchange.com/transcript/message/23058334#23058334 $\endgroup$
    – Mithoron
    Commented Jul 28, 2015 at 14:06

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