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Will this compound exhibit geometrical isomerism?

1-(chloromethylidene)-3,5-dimethylcyclohexane


This question was asked in JEE (Mains) 2020, which is an entrance exam held in India annually. The official answer given for this question was "No", but some books say otherwise.

According to me, this would not show GI about the double bond, because even though the groups on one side are different, the cyclohexane ring is symmetrical and has the methyl groups on both sides. The answer might lie in the groups coming out of the plane, but am not quite sure about that.

Question: Will the above compound show geometrical isomerism?

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  • $\begingroup$ According to the NCERT (which the National Testing Agency follows religiously and accepts no other source), $\ce{XYC=CXY, XYC=CYZ, XYC=CZW}$ kinds of compounds show geometrical isomerism. Going by that alone it is seen that the cyclo ring is symmetric with respect to the doubly bonded carbon, and thus it does not exhibit geometrical isomerism. // However, a point is that you should include the sources which you have seen and quote it (...some books say otherwise... - include which books and quote them) so that the community can help you better, and thus I am downvoting the question now. $\endgroup$ Mar 24 at 11:06
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    $\begingroup$ @HarikrishnanM By "some books" I meant my institute's book, and the answer just says "Will show GI", and no further explanation is provided, so I don't think there is much to quote. $\endgroup$
    – Haider
    Mar 24 at 11:20
  • $\begingroup$ just be sure by checking the NCERT as that is THE KING in this case. From my personal experience, institute books are good only for questions. $\endgroup$ Mar 24 at 11:22
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    $\begingroup$ @HarikrishnanM According to NCERT this compound will NOT show GI, so I think I would just follow that. Thanks for the help :) $\endgroup$
    – Haider
    Mar 24 at 11:56
  • $\begingroup$ chemistry.stackexchange.com/questions/58630/… $\endgroup$
    – Mithoron
    Mar 24 at 16:40

2 Answers 2

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There are four stereoisomers of the compound (two pairs of enantiomers), shown in the first row.

enter image description here

If you swap the chlorine with the hydrogen (second row, swap from structure right above), you don't get a new isomer. In two cases, you get the other enantiomer, and in two cases you get the same compound. I wouldn't call any of these cases geometrical isomerism. Any pair of isomers is related by switching methyl groups from R to S or vice versa.

Note that without the chlorine, there would only be three isomers:

enter image description here

You could analyze the two hydrogen atoms directly attached to the double bond. Using terms from NMR spectroscopy, the two hydrogen atoms in the left structure are enantiotopic, while the ones in the other two structures are homotopic. Thus, substituting a chlorine on the leftmost structure yields one additional stereoisomer, while substituting a chlorine for the other two structures does not.

Will this compound exhibit geometrical isomerism?

No, not in the sense that we would have to break the double bond to turn one stereoisomer into another.

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  • $\begingroup$ Thanks @Karsten! Unfortunately I am not familiar with the NMR spectroscopic terms, and would prefer if you could give any source from where I can read (12th grader from India). I got the real logic behind this one. $\endgroup$ Mar 25 at 1:06
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    $\begingroup$ @HarikrishnanM e.g. here $\endgroup$
    – Karsten
    Mar 25 at 2:33
  • $\begingroup$ I get that there is no geometrical isomerism about the double bond, but wouldn't switching position of methyl groups also be a type of geometrical isomerism? $\endgroup$ Mar 31 at 8:55
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    $\begingroup$ After all, isomers of 1,3-Dimethylcyclohexane are labelled cis and trans, i.e. they show geometrical/cis-trans isomerism: stereoelectronics.org/webSC/SC108.html $\endgroup$ Mar 31 at 8:58
  • $\begingroup$ @Thecoder001 1,4-Dichlorocyclohexane uses cis and trans in its IUPAC name, but 1,3-Dichlorocyclohexane uses R,S nomenclature. For the 1,4 substitution, you can't use R,S because of the symmetry of the molecule. This is not the case for the molecule the OP asked about. $\endgroup$
    – Karsten
    Mar 31 at 13:26
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According to NCERT, which is the textbook followed by the National Testing Agency (which conducts JEE), this compound will NOT show geometrical isomerism, because both the methyl groups in the ring are identical/symmetrical, so I would just follow this as an answer.

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