Like, the structure of lines of the rings of crown ether looks very different from regular cycloalkane rings. For example, the oxygen in crown ether is pulled more inside the ring. The line between carbon atoms consecutively bonded look almost horizontal. Why?enter image description here

  • 4
    $\begingroup$ What do you mean? 6-crown-2, a.k.a. 1,4-dioxane is drawn exactly like cyclohexane, and the drawings we put on paper only have a very limited correlation to the actual molecular geometry … $\endgroup$ – Jan Feb 4 '17 at 16:06
  • $\begingroup$ Oh okay. Yeah, it does look very similar. I guess what I am asking is as crown ethers become larger, why does ring structure start looking so different. I am trying to understand why they look the way they so, so I can find some rationality while drawing them on paper $\endgroup$ – TLo Feb 4 '17 at 16:09
  • 5
    $\begingroup$ Cyclooctadecane, the oxygen-free form (18-crown-0?) of 18-crown-6. Notice any difference except for the missing oxygens? Note again that these are all representations on paper, only loosely connected to actual geometry. $\endgroup$ – Jan Feb 4 '17 at 16:12
  • $\begingroup$ Oh wow. I am new to organic chemistry so I always through that actual structure of ring would be quite closer to drawn structure. But wouldn't drawing them in such different way distort like bond angles and stuff. And I know there are conformation and stuff but but wouldn't drawing them in such vastly different bond angles distort what sturucutre looks like in reality. $\endgroup$ – TLo Feb 4 '17 at 16:16
  • 2
    $\begingroup$ Not more and not less than the distortions already given to linear molecules in the skeletal formula formalism. Angles are typically drawn as $120^\circ$ while angles on non-double bonded carbons are actually closer to $109^\circ$ and bond lengths are standardised rather than having some shorter than others. But I reiterate: these are representations and only meant to loosely represent the actual structure. $\endgroup$ – Jan Feb 4 '17 at 16:19

Crown ethers don't look like that except in structure drawings

You confuse two completely different things: the drawings we use to describe chemical structures and what molecules actually look like in real chemical environments. We draw crown ethers like this because the drawing accurately describes the connectivity in the structure. In reality crown ethers are flexible and floppy and certainly not flat. we don't add that extra complexity to 2D structure drawing as they get very messy very quickly and obscure the essential features that describe the connectivity in a simple way.

When they are in solution with appropriate ions (for 18-crown-6, potassium ions would be a good example) they form complexes driven by the attraction between the oxygens and the charged ions. These complexes have 3-dimensional structures (with K+ the complex is approximately octahedral as potassium fits neatly into the cavity created by the oxygens when the molecule folds up). It looks a little like this as a 3D structure:

wikipedia picture of 18-crown-6 potassium complex

Wikipedia has some good descriptions of this and further pictures.

Crown ethers are not rigid molecules and the actual shapes they take up in 3D space demand a lot on what else is in solution with them. Other, similar, complexing agents are far more rigid. Calixarenes, for example, have a 3D shape like a chalice which is even more misrepresented by a 2D drawing.

| improve this answer | |
  • 3
    $\begingroup$ Is there a reason why you’re not linking to Wikipedia proper? $\endgroup$ – Jan Feb 4 '17 at 16:30
  • 1
    $\begingroup$ @jan No reason other than the fact that I use Wikiwand as it gives neater formatting of wikipedia content (it doesn't alter the content otherwise). And, therefore, including a native wikipedia link would take more effort on my part. And I'm lazy when there is no meaningful advantage in putting extra effort in. $\endgroup$ – matt_black Feb 4 '17 at 16:33
  • $\begingroup$ I always thought that crown ethers were drawn like that to emphasize the special coordinating effects of the oxygens on a central ion. $\endgroup$ – ron Feb 4 '17 at 17:24

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