I was on my high school chemistry class when I came across this problem, which, I think, belongs to group theory. The problem is that it is not possible to label the carbons as reflection should not be counted twice. (That is, 3,3-dichloro-2,2-difluoropropane is called 1,1-dichloro-2,2-difluoropropane) (The carbon chain in the case of propane is fixed.) And, how to solve the general problem of counting different isomers of $\require{mhchem}\ce{C_{n}H_{x}Cl_{y}F_{2n +2-x-y}}$ ? Thank you.

  • $\begingroup$ Can you elaborate the group structure you're talking about, for those who don't favor chemistry that much ? $\endgroup$ – Emre Jun 2 '16 at 2:04
  • $\begingroup$ @Emre : $C$ has valence $4$, $H : 1$ , $Cl : 1$, $F:1$, and the problem is that a chain of $C$ doesn't have to be linear (see isopentane or en.wikipedia.org/wiki/Cyclohexane). then we must check if the $C-C$ bounds are all simple (no double bounds as in en.wikipedia.org/wiki/Ethylene or even Benzene ), then we have to describe all the possible $C-C-\ldots$ skeletons (by taking in account the symmetries : en.wikipedia.org/wiki/Chirality_(chemistry) ) and finally place the $H,Cl,F$ $\endgroup$ – reuns Jun 2 '16 at 2:09
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    $\begingroup$ if you forget the chirality, this is purely graph theory $\endgroup$ – reuns Jun 2 '16 at 2:19
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    $\begingroup$ and in the case of propane ($n=3$) everything is easy : choose the number of $F,Cl,H$ for each $C$, and take in account the chirality of the $C-C(HClF)$ groups (two ways for arranging such groups spatially), and take in account the symmetry of the $C-C-C$ chain $\endgroup$ – reuns Jun 2 '16 at 2:24
  • $\begingroup$ So if you were to take propane $(C_3 H_8)$ and replace 2 with fluorines and 2 with chlorines, isn't it possible with casework. $\endgroup$ – Saketh Malyala Jun 3 '16 at 22:40

$\rm C_3H_4Cl_2F_2$ is an alkane.   There are no multiple bonds or rings to worry about.   Further only one skeleton is permissible here: a three carbon chain.   The end two carbons have three groups attached, the middle carbon but two.   The carbon-carbon bonds can rotate freely so we only need worry about possible carbon chirality, and any mirror symmetry.

Noting that we have the same number of chlorine and fluorine atoms simplifies things as we can just check to see if exchanging them produces an isomer; but watch out for interaction with other symmetries.

$\begin{array}{ccc:ccc} &&& \text{middle C} & \text{end C} & X\leftrightarrow Y \\ (\;)&(X)&(XY_2)& \times 2 & & \times 2\\ (\;)&(X_2)&(Y_2)&&&\times 2\\ (\;)&(XY)&(XY)&\times 2&\times 2&\div 2 & \text{the X-Y interchange cancels isometry}\\ (X)&(\;)&(XY_2)&&&\times 2\\ (X)&(X)&(Y_2)&\times 2&&\times 2\\ (X)&(Y)&(XY)&\times 2&\times 2&\times 2\\ (X)&(XY)&(Y)& \times 2\\ (X)&(Y_2)&(X)&&&\times 2 \\ (X_2)&(\;)&(Y_2)& \\ (XY)&(\;)&(XY) & \times 4&&\div 2 \end{array}$

I count 2729 isomers, but may well have missed others.

The general case gets much more complex.

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  • $\begingroup$ So after all it's just a case bash?(ie brute force enumeration) $\endgroup$ – Zhipu 'Wilson' Zhao Jun 2 '16 at 22:30
  • $\begingroup$ Are you not missing the (XY) () (XY) case? $\endgroup$ – gilleain Jun 3 '16 at 16:28
  • $\begingroup$ (it is an halogeno-alkane : a chloro,fluoro-alkane, or a chlorofluorocarbon (CFC) en.wikipedia.org/wiki/Chlorofluorocarbon used as refrigerants, and responsible of the ozone depletion en.wikipedia.org/wiki/Ozone_depletion ) $\endgroup$ – reuns Jun 3 '16 at 21:44
  • $\begingroup$ @gilleain Indeed. I'm rusty on this. $\endgroup$ – Graham Kemp Jun 3 '16 at 21:48

Odd, I get 30 33 32 (last one is a meso isomer) stereoisomers from 16 constitutional isomers. Like so (red asterisks indicate a stereocenter) :

enter image description here

Sadly, the method to get this count was 'brute force enumeration' (in fact, a variant of the canonical path augmentation method of McKay for multigraphs).

The stereocenters were deterined by eyeballing the structures and thinking about it - so they may well be wrong!

Edit : Indeed, I missed the chiral centers on FC(Cl)CC(F)Cl. As to the more general question, I'm not sure how to 'extend' the series in some natural way.

For example, here are some counts of constitutional isomers for various similar formulae:


  • C4H5Cl3F2 86
  • C4H6Cl2F2 59
  • C5H6Cl3F3 632

All found using the same method.

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  • $\begingroup$ I don't get how you get different chirality for other groups than $C-C-(HClF)$ as I wrote $\endgroup$ – reuns Jun 3 '16 at 21:43
  • $\begingroup$ @user1952009 I don't really understand what a "C-C-(HClF)" group is. In any case, the chiral centers marked with an * in the diagram all have four different groups attached. eg the bottom left image has C(H)(CHF2)(Cl)(CH2Cl) to abuse SMILES notation a bit. $\endgroup$ – gilleain Jun 3 '16 at 22:04
  • $\begingroup$ Also I notice that the last one is a meso isomer! Gah. So the count is actually more like 32 isomers. $\endgroup$ – gilleain Jun 3 '16 at 22:05
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    $\begingroup$ you are right, sorry I got it, I forgot the case $G_1-C(XY)-G_2$ which is chiral whenever $G_1 \ne G_2, X \ne Y$ (and I meant $C-C(HClF)$) $\endgroup$ – reuns Jun 3 '16 at 22:06
  • $\begingroup$ and the last one $CHClFH-CH_2-CHClF$ there are only two conformations, because of the symmetry $\endgroup$ – reuns Jun 3 '16 at 22:09

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