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This question already has an answer here:

How to identify an asymmetric carbon in the molecule below? Is the carbon identified by the asterisk asymmetric or not? I would say no as it has no 4 different groups attached.

2-(4-(1-bromoethylidene)cyclohexyl)acetaldehyde

Is this correct? I will greatly appreciate if anyone can confirm or otherwise explain why it can be asymmetric.

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marked as duplicate by Mithoron, Todd Minehardt, airhuff, Tyberius, pentavalentcarbon Oct 17 '17 at 4:00

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  • $\begingroup$ A simple test. Create the mirror image, using wedge and hash bonds to indicate configuration if required.. See if it superimposes on the original. If it doesn't, it's a stereoisomer (the enantiomer if there only one chiral centre). The "four different group test" can fall down for compounds with internal mirror planes e.g. meso-tartaric acid $\endgroup$ – Beerhunter Oct 16 '17 at 6:37
  • $\begingroup$ @Beerhunter How does meso-tartaric acid fail the four different substituents rule? It has two stereo-centres, i.e. asymmetric carbons, it just isn't chiral due to the mirror plane. $\endgroup$ – Martin - マーチン Oct 16 '17 at 10:10
  • $\begingroup$ related: chemistry.stackexchange.com/q/59124/4945 $\endgroup$ – Martin - マーチン Oct 16 '17 at 10:16
  • $\begingroup$ How do CIP rules deal with (E)- / (Z)- relative to a group? i.e., is this stereocenter (R) or (S)? $\endgroup$ – Zhe Oct 16 '17 at 13:38
  • $\begingroup$ Exactly, the mirror plane criteria is better $\endgroup$ – Beerhunter Oct 20 '17 at 19:08
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There is probably little reason in discussing the hydrogen and the $\ce{CH2CHO}$ group which are definitely different residues. The question remains whether the sides of the ring are identical or not.

If you go along the top side of the ring, you will encounter:

  1. $\ce{CH2}$

  2. $\ce{CH2}$

  3. $\ce{CR=CMeBr}$

Looking into it a little deeper, we see that we are coming from the (Z) side of the double bond.

Going along the bottom side you will encounter:

  1. $\ce{CH2}$

  2. $\ce{CH2}$

  3. $\ce{CR=CBrMe}$

Here, however, we are arriving from the (E) side of the double bond.

Since (E) is not equal to (Z) the two sides of the ring are not equal, therefore the carbon has four different residues, therefore it is chiral.

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Yes it does have 4 different groups attached. The sides of the ring are not equivalent. One of them is in trans position to bromine, another is in cis, and that can't be changed easily.

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