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I'm learning organic chemistry from Clayden's book (2e). I've noticed they draw the cyclic hexenone below like this:

enter image description here enter image description here

However, since the carbon on the carbonyl group is sp2 hybridized I am wondering if this is correct. Shouldn't this just be planar and look like this:

enter image description here

Ie, more of an envelope conformation? Or am I not grasping something correctly, like is it maybe not convenient to draw it like an envelope? The below structure was made in ChemSketch and using the 3D optimization function.

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    $\begingroup$ ChemSketch is not a thing for making 3D structures, at least realistic ones. $\endgroup$
    – Mithoron
    May 3 '21 at 17:00
  • $\begingroup$ So, yeah, Clayden's right. $\endgroup$
    – Mithoron
    May 3 '21 at 17:05
  • $\begingroup$ @Mithoron Actually, I can see an envelope now if I Iook at it from a different angle. But because it is not mentioned before it confused me, since you were looking at cyclohexenes from the "front" in all depictions of a regular cyclohexene before in the book like shown here: i.imgur.com/vMNnAIK.png . So I think it is a bit misleading in the depictions above, which give the impression that the carbonyl carbon is at an angle up (right conformation in the figure) vs down (left conformation in the figure).with the next carbon. $\endgroup$
    – user21398
    May 3 '21 at 17:14
  • $\begingroup$ Simply, three carbons in the ring are $\mathrm{sp^2}$ while other three are $\mathrm{sp^3}$. Thus, it can't be planner without high strain ($109^\circ$ going on $120^\circ$). $\endgroup$ May 3 '21 at 18:17
  • $\begingroup$ Anyway in the red draws the carbonyl C doesn't really depart from planarity. $\endgroup$
    – Alchimista
    May 4 '21 at 9:57

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