My question is why isn't the resonance structure drawn in pencil on the right a valid one? My review book states that after initial protonation on the indicated nitrogen no resonance is possible.
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$\begingroup$ What's the valency of N? $\endgroup$– Black Jack 21Feb 21, 2017 at 7:22
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$\begingroup$ Well N is 5 valence electrons $\endgroup$– l33tcodesFeb 21, 2017 at 7:22
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$\begingroup$ The top C=C is less stable than N so it can definitly rotate its pi electrons to meet N's valence deficiency .. right? $\endgroup$– l33tcodesFeb 21, 2017 at 7:23
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$\begingroup$ It has 5e, so it makes 3 bonds, getting 1 election from each to complete octet. $\endgroup$– Black Jack 21Feb 21, 2017 at 7:23
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$\begingroup$ The amine is partially positive while the imine has electron pair. In order to satisfy the partial positive, the the double bonded alkene can donate its pair to the N and have its positive deficiency satisfied by the electron pair of the Nitrogen on the bottom? $\endgroup$– l33tcodesFeb 21, 2017 at 7:39
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3 Answers
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You need to understand the basic reason for the aromatic nature of materials. A molecule is said to be aromatic only if:
- A ring in it is planar,
- Continuous delocalisation is possible throughout the concerned cyclic ring, and
- This delocalisation stabilises the ring.
Now, for continuous delocalisation to be possible, the $\mathrm{\pi}$ orbitals of all the elements in the ring need to be parallel to each other ( They need to overlap sideways for formation of $\mathrm{\pi}$ bonds).
In your example, consider the carbons in the ring. All of their $\mathrm{\pi}$ orbitals are parallel to each other and PERPENDICULAR to the plane of paper. On the other hand, the imine nitrogen is $\mathrm{sp^2}$ hybrid. The lone pair of electrons is in an $\mathrm{sp^2}$ hybrid orbital, i.e, IN the plane of the paper. And mind you, it's fixed (can't rotate without distorting the geometrical shape of the ring, that is, which would be too much to expect from the poor guy.)
So the imine Nitrogen cannot give the lone pair anywhere. it can only play around with the $\mathrm{\pi}$ bond already on it.
Moreover, as soon as the amine Nitrogen takes up a proton, it loses its ability to use the lone pair for delocalisation. The condition for planarity itself is not fulfilled ($\ce{(R)2-N+-(H)2}$ has tetrahedral geometry), so there's no use of discussing aromaticity. Before the proton uptake, the ring was aromatic because the amine Nitrogen was using the lone pair for continuous delocalisation, effectively attaining planar geometry. But now... It can no longer be done.
Conclusion: Resonance is still possible(between all atoms of the ring except the amine Nitrogen), but Continuous Delocalisation is not. So, the system loses its aromaticity.
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The key to understanding why your proposed resonance structure is not possible is geometry. The five-membered ring is planar, or otherwise there wouldn’t be any aromaticity in the first place. The aromatic system is made up of a set of p orbitals perpendicular to the ring plane — so if you decided that your ring is lying in the $x,y$-plane then the aromatic system consists of $\mathrm{p}_z$ orbitals. For the atoms to have available $\mathrm{p}_z$ orbitals, they need to have $\mathrm{sp^2}$ or lower hybridisation; in this case, all atoms of the aromatic ring are $\mathrm{sp^2}$ hybridised, giving them three $\mathrm{sp^2}$ orbitals lying fully within the $x,y$ plane or ring plane.
One of these orbitals contains the lone pair of the non-protonated nitrogen. Thus, this lone pair is perpendicular to the aromatic system and cannot take part in resonance. The situation is similar to furane, in which oxygen has two lone pairs but only one of the two is in a $\mathrm{p}_z$ orbital and therefore able to participate in resonance.
Now that we have taken care of that, I need to point out a textbook-erratum — at least I would have to point it out from what I see, I don’t know what else the book writes on the page.
They drew side-chain protonated histidine as a nonaromatic system with one nitrogen having two protons. That is not what side-chain protonated histidine looks like. The lone pair of the other nitrogen (the one you wanted to use for resonance) is
- pointing away from the ring
- not participating in resonance
- and therefore much more basic
So before the $\ce{R2N-H}$ nitrogen’s lone pair — which, as you should remember, is delocalised — is ever protonated, protonation will happen at the other nitrogen giving $\ce{{Ring}-\overset{+}{N}H={Ring}}$. This protonated histidine still features an aromatic imidazole ring.
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2$\begingroup$ I'm guessing that the question was something like, "Why is this nitrogen much less basic than the other one?" We are seeing the answer key that explains it. I don't think this is textbook erratum. $\endgroup$ Feb 21, 2017 at 16:18
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1$\begingroup$ @jerepierre I think you're right, especially given what it says at the top of the picture. The diagram above will be the one with the correct nitrogen protonated. $\endgroup$– bonFeb 21, 2017 at 17:54
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$\begingroup$ @jerepierre I don’t really like guessing. I added it to guard OP from any misconceptions that may occur. If further textbook context is given that confirms your guesses, I’ll happily edit. $\endgroup$– JanFeb 21, 2017 at 22:38
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Very simple: The resonance structure in pencil on the right is not possible because you have five bonds to nitrogen, breaking the octet rule. This is the one and only reason why the structure drawn in pencil is not correct. Realizing this, the point that the review book is trying to make (about why the imine nitrogen is protonated) should become even more clear.
The other answers are not actually addressing the question you asked.
