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I think that carbons with highest electron density will be both $\ce C1$ and $\ce C2$ because of the loosely held $\pi$-bond between them, but according to the answer given the carbons with most electron density should be both $\ce C3$ and $\ce C6$.

How is this possible?


2 Answers 2


:-( The book gives an answer, not necessarily the right one.

Looking at the $\ce{^13C}$ NMR of cyclohexene the carbons at positions 4 and 5, in your diagram, are shifted the least and therefore have the most shielding. Thus those carbon atoms also have the greatest electron density around them.

  • $\begingroup$ So, $C1$, $C2$, $C4$, $C5$ all have highest electron density, while $C3$ and $C6$ have the lowest? $\endgroup$ Commented Jan 25, 2017 at 16:37
  • $\begingroup$ Looking at the $\ce{^13C}$ NMR of cyclohexene the carbons at positions 4 and 5, in your diagram, are shifted the least and therefore have the most shielding. c4, c5 > c1, c2 > c3, c6 $\endgroup$
    – MaxW
    Commented Jan 25, 2017 at 16:42
  • $\begingroup$ Actually, I'm a high school student so it is tough for me to understand that 13C NMR of cyclohexene $\endgroup$ Commented Jan 25, 2017 at 16:43
  • $\begingroup$ Ugh. I should have written that (C4 or C5) > (C3 or C6) > (C1 or C2). $\endgroup$
    – MaxW
    Commented Jan 25, 2017 at 22:01

Intuitively, I would expect carbons 1/2 to have the highest electron density. MaxW's NMR spectrum does indeed suggest that 1/2 have the least density, then 3/6, with carbons 4/5 being the most electron-dense.

I suppose you could rationalize this conclusion by saying the the CH2 carbons (3/6, 4/5) will suck a bit of density from the attached hydrogens (because C is slightly more electronegative than H), whereas carbons 1/2 only have one H each to pull electron density from, therefore 1/2 have the least electron density. But my gut feels uncomfortable about this.

I attempted to calculate an electrostatic potential map of cyclohexene, but unfortunately I'm not proficient with the software. I think that's what we need here though.

For what's worth, here is a computer-generated model showing that there is more current density around the carbons you labeled 1/2:

enter image description here (http://pubs.rsc.org/en/content/articlelanding/2012/cs/c2cs35037h)

  • $\begingroup$ I appreciate your efforts, but according to MaxW, carbon atoms at 1/2 positions have least electron density, but the model suggests that carbon at 1/2 have highest electron density. What should be the ultimate correct answer? $\endgroup$ Commented Jan 26, 2017 at 6:27
  • $\begingroup$ I think the ultimate answer should come from an electrostatic potential map. I'm not sure if NMR is the most direct way to assess electron density, there may be other factors at play that could be misleading. But I'm not a physical chemist so I'm not sure. I guess I'd take the NMR data over intuition and say carbons 4/5 are most electron-dense. But that's just settling in lieu of the potential map. $\endgroup$ Commented Jan 26, 2017 at 19:36
  • 1
    $\begingroup$ This has been bugging me. I think the gist is sloppy wording of the problem. There is a difference in "electrostatic potential" and "electron density." As chemists we tend to ignore the core electrons which contribute to "electron density" but are not reactive chemically. "Electrostatic potential" is much indicative of chemical reactivity. $\endgroup$
    – MaxW
    Commented Jan 27, 2017 at 0:52

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