2
$\begingroup$

Please consider 2-acetyl-1-pyrroline: http://en.wikipedia.org/wiki/2-Acetyl-1-pyrroline

1-(3,4-dihydro-2H-pyrrol-5-yl)ethan-1-one

What is the molecular geometry of the nitrogen lone pair? Is the lone pair oriented in-plane with the ring? How is this distinct from the orientation carbon-acetyl bond, which should presumably point out-of-plane with the ring?

$\endgroup$
3
$\begingroup$

The nitrogen lone pair is in plane with the ring because it is in one of the three $sp^2$ hybrid orbitals of N. The other two hybrid orbitals form $\sigma$ bonds to the adjacent carbons, and all three are in plane, with an angle of 120° between them. The $p_z$ orbital of N which is perpendicular to the ring plain forms the $\pi$ bond of the C=N double bond. The acetyl substituent should also lie in plane with the ring because this would allow energetically favorable conjugation between the two double bonds (C=N and C=O).

$\endgroup$
  • $\begingroup$ Thank you for your answer! Would the carbon substituent group still lie in-plane with the ring if we replaced the acetyl with something like a methane? $\endgroup$ – user5463 May 11 '14 at 15:06
  • $\begingroup$ There would be no conjugation, but the methyl group would still lie in plane because it is smaller, and it does not contain more substituents (only 3 H) which can rotate out of plane and thus significantly affect conformation of the molecule (like in the case of C=O, where the orientation is significant due to the possible conjugation). $\endgroup$ – Jannis Andreska May 11 '14 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.