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I was thinking about an organic chemistry problem a few days ago but reached a dead end. There were 2 parts to this question.

The first was thinking about the isomers of 'C3H4X2' [where X could be anything like a halogen or a deuterium etc].

I came up with the following 7 isomers:

enter image description here

(I've really tried to make it as clear as possible! Hopefully it has worked!)

...but a website I found stated that there were only 4:

A - enter image description here

B - enter image description here

C - enter image description here

D -enter image description here

;with their 'D' representing a deuterium.

So this is the first aspect of my quandary - what am I doing wrong when I draw these isomers!

The second aspect to this question was about considering the symmetry of these organic molecules. In particular, considering firstly 'plane symmetry' [which I understand] and 'rotational symmetry' [which I thought I understand but clearly I don't!]

The same website says that molecule A (from above) plane symmetry and rotational symmetry, which is fine.

But it also says that molecule C has plane symmetry but not rotational. Why is this? Surely if I rotate this molecule 60 degrees I get the same shape?

And then finally, they say that molecules B and D have rotational symmetry but not plane symmetry which I think I understand but would appreciate any further comments if anyone had them.

Ultimately, I would be very grateful if someone could point me towards something which could help me understand rotational symmetry in organic molecules and in particular why these examples given above have their particular symmetries! Thank you very much, everyone!

Best wishes,

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    $\begingroup$ Your molecules are all $\ce{C3H4X2}$ and not $\ce{C4H3X2}$ !!! What are you looking for ? Anyway, 1) and 2) are impossible. 4) and 6) are identical, $\endgroup$ – Maurice Dec 26 '20 at 12:44
  • $\begingroup$ Sorry, @Maurice - typo!! I am meant to be looking for C3H4X2 not C4H3X2...why are 1) and 2) impossible? And yes, I do see that 4 and 6 are identical now - thanks!...what about the rest? $\endgroup$ – dubstep Dec 26 '20 at 13:05
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    $\begingroup$ 1) and 2) are impossible, because the four bonds C-C or C-X must define a tetrahedron centered on the C atom. One X must be in front of the triangle plane. The second X cannot be on the same side of this plane. It must be on the other side of this plane, backwards. $\endgroup$ – Maurice Dec 26 '20 at 14:42
  • $\begingroup$ @Maurice - Oh right, I see! Now the question is about the symmetry... $\endgroup$ – dubstep Dec 26 '20 at 15:33
  • $\begingroup$ A has a plane and an axis of symmetry . B and D have an axis of symmetry defined by the bisecting line coming from the third Carbon atom. C has a vertical plane of symmetry including the bisecting line coming from the third Carbon atom. $\endgroup$ – Maurice Dec 26 '20 at 16:47
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If the website diagram didn't clarify the situation immediately, the issue is not to answer questions about 1 to 7, but to understand why the question is wrong.

My organic chemistry course had us playing around with molecular models - 3D representations of reality, one type with a metal framework representing the multivalent nucleus and plastic tubes representing bond lengths, the other type with wooden balls and springs. They are a far cry from reality - but molecular models give an added dimension of understanding.

Lacking such 3D models, and lacking an answer sheet, one possibility would be to go to 3D images on the web: they are actually in 2 dimensions, but with perspective in the drawing, so it could be sort of 2.5 D, like this:

enter image description here

Only after knowing what is represented by the dotted lines and tilted bond can you understand what you are drawing. The proposed differences between structures 1, 2 and 3 indicate an unfamiliarity with what the bond representations actually mean. One look at the picture above should clarify everything. Perhaps one more comment should be added: the bonds between a carbon and its two substituents are never coplanar with the 3-carbon ring. Interestingly, I did not find any other similar images (of dichlorocylopropane). Everything was of the straight line or dashed line/tilted bond type. This just assumes that the lines/dashes have an obvious meaning. Sometimes a good picture is worth a thousand words. (This answer is worth 1254 words, if you include 1000 words for the picture.)

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  • $\begingroup$ Thanks @James Gaidis - I get it now! 1254 words well spent! $\endgroup$ – dubstep Dec 26 '20 at 15:35
  • $\begingroup$ would you be able to help me with the second part of the question now? The one about the symmetry? $\endgroup$ – dubstep Dec 26 '20 at 15:36
  • $\begingroup$ Symmetry operations are ways to move the molecule and compare the new position with the original. Plane symmetry of C is shown by putting a vertical mirror thru the altitude of the triangle containing the CH2 carbon and the midpoint of the CDH--CDH bond. The mirror makes the molecule look the same - unshifted. C cannot be rotated less than 360 degrees without being able to notice it. A link to rotational symmetry: mathsisfun.com/geometry/symmetry-rotational.html . Note: math is fun - till they make it complicated! $\endgroup$ – James Gaidis Dec 27 '20 at 14:52
  • $\begingroup$ @JamesGaidis You should look at pubchem.ncbi.nlm.nih.gov/compound/74970#section=3D-Conformer $\endgroup$ – LDC3 Jan 3 at 0:06
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In your drawings you are correct that 5 and 7 are enantiomers. But 1 and 2 are inconsistent with tetrahedral bonding at the carbon atom (when two bonds are in the plane of the paper the others must be one above and one below), and 4 and 6 are just rotations of each other (around a twofold axis in the plane of the paper, not the perpendicular threefold one). Therefore the only actual and distinct isomers are 3, either 4 or 6, and the 5/7 enantiomeric pair.

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For the molecule to have symmetry, the atoms need to match after the symmetry operation. For A, there are 2 planes of symmetry (the plane of the 3 carbons and the plane through the 2 hetero atoms and the attached carbon) and 1 axis of rotation symmetry (the intersection of the 2 planes). With B (and D), there are no planes since the atoms cannot be reflected through a plane and have the same element. It has axis of rotation through the carbon without the hetero atom and the middle of the bond in the ring. For C, there is neither a plane of symmetry nor an axis of symmetry.

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