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Can anyone suggest a probable reason for this conversion? This seems like a simple conversion from hemiaminal to imine. For secondary amine this kind of conversion might be possible by enter image description here

but in the case of tertiary amine how can something like this form? Is there a chance of removal of RCH2OH? The metal here is copper(II). Is it possible that the metal has some role to play?

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    $\begingroup$ I think you have to consider the influence of the metal. This cannot be compared to a simple hemiaminal to imine because of the electrons donated to the metal-N complex $\endgroup$ – Waylander Nov 26 '19 at 9:30
  • $\begingroup$ The metal used here is copper. $\endgroup$ – Suchi Ghosh Nov 26 '19 at 9:50
  • $\begingroup$ I think the electron donation to copper would weaken the N-H bond and make a proton shift from N to O easier. Once that occurs the loss of water is highly favoured because of the driving force of conjugation of the C=N with the aromatic ring $\endgroup$ – Waylander Nov 26 '19 at 9:58
  • $\begingroup$ what about the first case? the nitrogen is tertiary there? this might be the case for secondary amine where there is a presence of N-H bond. $\endgroup$ – Suchi Ghosh Nov 26 '19 at 10:11
  • $\begingroup$ I think a similar weakening of the C-N bond is reasonable $\endgroup$ – Waylander Nov 26 '19 at 10:32

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