In the molecule below, chlorine atoms may be attached to any of the 8 outer carbon atoms.
combinatorics/chlorinated.dioxins New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
In the molecule below, chlorine atoms may be attached to any of the 8
outer carbon atoms. Counting molecules that are the same under the
molecule's symmetries only once, how many different molecules have at
least one chlorine atom?
C C
/ \ / \
C C - O - C C
| || || |
C C - O - C C
\ / \ /
C C
[ Editor: problem statement reconstructed from Bill Taylor's reply in
the thread "Chemical combinatorics" (1998); the original question and
the second case it asked about are not in the collected material. The
molecule is dibenzo-p-dioxin, with "quadruple symmetry" assumed. ]
Show solution
Well this has been a fun little thread:- and I've particularly enjoyed
the comments on the physical chemistry of the situation. However,
there doesn't seem to have been a correct solution posted yet, so I'll
bung one into the fray, at least for the first case.
I'll leave the other as an exercise for the reader. ;-)
> C C
> / \ / \
> C C - O - C C
> | || || | (noted the appended double bonds! :) )
> C C - O - C C
> \ / \ /
> C C ...and assuming quadruple symmetry.
We're looking for the number of cases where at least one of the 8 outer
C atoms has a Cl added on.
I've probably screwed up, but I make the number of cases 75.
2 cases where 1 Cl has been added,
10 cases where 2 Cl has been added,
14 cases where 3 Cl has been added,
22 cases where 4 Cl has been added,
14 cases where 5 Cl has been added,
10 cases where 6 Cl has been added,
2 cases where 7 Cl has been added,
1 case where 8 Cl has been added.
There doesn't seem to be any reasonably quicker way than just
*counting* the damn things; though using that deBruijn-Polya thingy
would be slicker.
The symmetry between the 2 and 7 cases is obvious I hope.
The 10 2-cases include 6 where they're on opposite sides, 4 on the
same side.
The 14 3-cases include 2 with all on the same side, 12=2x6 with a 2/1
split.
The 22 4-cases include 1 with all same side, 8=2x4 with a 3/1 split,
and 13 with a 2/2 split.
Those 13 cases (the only non-obvious case) can be checked as follows:-
2 each with EE, EA, AA; and 1 each with EX, EC, AX, AC, XX, XC, CC.
E = 2 at same end; X = 2 eXtremes; C = 2 Central; A = "Alternate"
positions.
Howzat?
-------------
Hey! I just checked it by Polya - it was a lot slicker AND quicker,
though (as always) less informative in detail.
Here is the skeletal calculation...
group element I preserves 2^8 = 256 positions;
group element <-> preserves 2^4 = 16 positions;
group element ^v^ preserves 2^4 = 16 positions;
group element + preserves 2^4 = 16 positions;
-----
total # = 304 positions;
divided by the number
of group elements, 4, gives # = 76 distinct positions.
(And remove one for the forbidden chloro-free position.)
Howzatagain!
(If anyone asks for an explanation of this Polya mumbo-jumbo, could
someone else :) give it please...)
-- Bill Taylor <W.Taylor@math.canterbury.ac.nz>