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1-chloro-4-(chloromethylidene)cyclohexane

Would this compound show geometrical isomerism, or GI for short?

![enter image description here

There certainly wouldn't be any restricted rotation for any atoms in the ring, as for the chiral carbon having chlorine and hydrogen (I assume it is chiral, if I'm wrong, please correct me), the adjacent carbons have 2 same hydrogen atoms.

But would there be GI along the double bond? There should be GI along the double bond as for Carbon A, both groups attached to A are different, as one has chlorine and the other has hydrogen.

Is this assessment correct? Would there be GI in this compound?

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  • $\begingroup$ Try drawing the molecule in a 3D form of the cyclohexane ring. $\endgroup$
    – porphyrin
    Commented Apr 18 at 16:56
  • $\begingroup$ Yeah, I just did. Quite obvious now the compound won't show GI! Even in 2D, if I drew the chlorine horizontally and omitted the H, it's obvious the cyclohexyl group is symmetrical for carbon A $\endgroup$
    – Mr Gubbo
    Commented Apr 18 at 18:00
  • $\begingroup$ Previously addressed in a different example. $\endgroup$
    – user55119
    Commented Apr 27 at 18:10

2 Answers 2

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The starred carbon, numbered 1 in the compound name, is in fact a chiral center. With it comes the molecule not being superposed on its mirror image, even allowing for umbrella-like inversion of the cyclohexane chair (which is assumed herein).

Start from C-1 and proceed clockwise around the ring. When you get to the second, third and fourth positions from the start you first pass the hydrogen atom of the chloromethylene group and then pass the chlorine atom, both of these being coplanar with the umbrella-averaged ring. If you proceed counterclocwise from C-1 instead, you pass the chlorine first. This difference in ordering is enough to break mirror-image symmetry and define a chiral center.

So, indeed, the molecule is chiral and comes as a pair of enantiomers. If we assume that the hydrogen atom attached to C-1 is directed away from the observer in the structure shown, we can identify the orientation about that carbon as $S$ because the counterclockwise loop, passing the chlorine sooner, ranks above the clockwise loop.

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  • $\begingroup$ This wouldn’t happen if there wasn’t that double bond, correct? Because the double bond forces that planar trait whereas it could otherwise rotate on that single bond? Just asking for some clarification. This one was a bit tricky to see the chirality at first because I was expecting the chirality of “seeing” the Cl first on counterclockwise to come from the Cl’s rotation itself (and not the lack of rotation of that group from the double bond.) $\endgroup$ Commented Apr 26 at 19:25
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    $\begingroup$ That's right, if there were free rotation pendant to C-4 we would have no lasting distinction between the direction around the ring and thus no chirality. $\endgroup$ Commented Apr 26 at 19:36
  • $\begingroup$ This answer is correct; there is a chiral centre at C-1, and OP already assumed so. The question, however, was whether there is geometric isomerism (cis-trans isomerism or E-Z isomerism) around the double bond. $\endgroup$
    – Loong
    Commented Apr 27 at 8:10
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First of all, the carbon atom in the ring (with chlorine and hydrogen, marked as star in your image) isn't chiral. The reason being that if you go from either way from this carbon though the ring, you'll always end up getting the same group from up or down. So the star marked carbon isn't chiral and not a stereo-centre. The same is true for the 'A' marked carbon atom.

Another thing you need to keep in mind whenever you see a double bond directly next to a ring, or even number of consecutive double bonds/rings, they won't show geometrical isomerism either, even if both the extreme carbon atoms contain to different groups. The reason is that in such compounds, where there is consecutive even double bonds/rings, the compound goes out of the plane and becomes non-planar, thus it doesn't show geometrical isomerism either.

Now, more importantly and the answer of your question, the given compound is optically active even though it does not have any chiral centre since due to the consecutive ring and double bond, the molecule will bend out of the plane and thus will not have any element of symmetry (neither Plane of Symmetry not Centre of Symmetry).

Edit: Corrected the mistake.

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    $\begingroup$ As you said, the compound is non-planar. The molecule is different from its mirror image, so it is optically active. $\endgroup$
    – Karsten
    Commented Apr 18 at 17:29

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