Post by: Big-Daddy on March 30, 2013, 11:47:47 AM
formula [Pt(py)(NH3)BrCl] (where py = pyridine, C5H5N).
How do I go about working this out? I know there are only 3 structures but I'm not sure how to judge the stereochemistry here. This is probably just a typical case of complex isomerism but I still don't know what to look for.
Post by: Dan on March 30, 2013, 01:10:38 PM
Start drawing, you should be able to come up with a couple of structures at least.
Post by: Big-Daddy on March 31, 2013, 10:58:43 AM
http://en.wikipedia.org/wiki/Cis%E2%80%93trans_isomerism#Coordination_complexes
Start drawing, you should be able to come up with a couple of structures at least.
Is there a way to draw a complex ion on here?
So let me check before I start: the type of isomerism is dependent on the positions of the ligands with heavier first atoms as identified by the Cahn-Ingold Prelog rules (i.e. we will have "cis-platin" if the heavier Cl atoms are on the same side, and "trans-platin" if they are not). In the case of [Pt(py)(NH3)BrCl], our heaviest priority of all is Br, followed by Cl, then C5H5N, then NH3.
The shape is square planar so there will be two bonds 'below' the Pt and two bonds 'above' the Pt on the page. (I'm not sure what the technical language for this is.)
Ok so far?
Post by: Big-Daddy on April 01, 2013, 04:15:17 PM
It might help if you would classify in what way the bonds cannot rotate in this compound. For instance, if they are all fixed in position (so if I had Cl in a certain place and C5H5N in another, there are no cases in which their positions are interchangeable), we should have 6 isomers?
Post by: Dan on April 01, 2013, 04:25:37 PM
Post by: Big-Daddy on April 01, 2013, 06:32:42 PM
Post your 6 isomers. Some of these are going to be repeats drawn in different orientations - we can help you if you show your work.
How do I draw complex ions on this forum?
Post by: Dan on April 02, 2013, 03:22:06 AM
You can use SMILES
![[NH3][Pt]([NH3])(Cl)Cl](https://www.chemicalforums.com/SMILES/b306af08b73888486003.png)
![[NH3][Pt](Cl)(Cl)[NH3]](https://www.chemicalforums.com/SMILES/74ef9feb08995051147d.png)
Or just attach an image to your post (which is probably easier) by clicking the ""Additional options..." immediately below the reply text box.
Post by: Big-Daddy on April 02, 2013, 06:58:46 AM
Post formatting guide is here (http://www.chemicalforums.com/index.php?topic=59314.msg230001#msg230001).
You can use SMILES
Or just attach an image to your post (which is probably easier) by clicking the ""Additional options..." immediately below the reply text box.
OK I just attached my attempt.
I just held Br in place (to avoid obvious repeats) and then did every configuration of the other 3 ligands in the other 3 places (3!=6 'isomers'). The question is, how could I know which ones were always repetitions and thus avoid drawing them? There must be some systematic way to come up with all the isomers which actually work.
I have left out the 3D nature of the bonds because I can't see the significance of that to cis-trans isomerism in a square planar shape. If you want me to include them then please explain to me exactly how the rotation is restricted in this molecule and then I should understand (Wikipedia was unclear on this for complex ions).
Post by: Dan on April 02, 2013, 08:10:08 AM
Example:
Post by: Big-Daddy on April 02, 2013, 01:09:57 PM
That's fine. Now rotate them to work out which are the same and which are different. You don't have to rotate any bonds, just rotate the entire molecule in 3D space. Make models if this is difficult.
Example:
OK, I see what you mean by the rotation. But I'm not sure how that applies to my molecule, because I can't see any rotation of 1 configuration that would reach any other (because if you rotate any configuration 90 degrees, the place of Br will shift, 1 place clockwise, whereas in my isomers Br is the same place for each one).
Post by: Dan on April 02, 2013, 01:39:22 PM
Post by: Sophia7X on April 02, 2013, 05:06:40 PM
Let's pick Cl.
Now look at the first structure and write the atoms by going clockwise.
This would be Cl, py, NH3, and Br.
Now look at the second structure (the one to the right of the first structure). Start with Cl again and try to duplicate the above sequence by either going clockwise or counterclockwise. You will find that it is not possible to duplicate the exact sequence. That means the 1st and 2nd structures are not equivalent.
Next, look at the third structure. Start with Cl, and you will see that you can get Cl, py, NH3, Br if you go counterclockwise. That means the first and third structures are the same.
Post by: Big-Daddy on April 03, 2013, 06:54:47 AM
The idea is that, if you can pick any two of the ligands which are 2 apart (i.e. opposite) to each other, and then rotate them through 180 degrees (i.e. switch their places on the paper), you've got the same isomer as the molecule is the same - because this type of rotation is allowed in the structure.
By that logic half of my structures should be the same, leaving me with these:
Are they correct?
On a side-note, is there any way to name these each ... I've seen the cis-trans naming on cis/trans-platin but there are 3 isomers here?
Post by: Dan on April 03, 2013, 08:14:51 AM
Are they correct?
Yes
Quote
On a side-note, is there any way to name these each ... I've seen the cis-trans naming on cis/trans-platin but there are 3 isomers here?
cis/trans doesn't work here, as you have realised.
There is nomenclature system for cases like these, see section IR-9.3.3.3, p180 in this document (http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf).
Post by: Big-Daddy on April 03, 2013, 08:36:14 AM
Are they correct?
YesQuoteOn a side-note, is there any way to name these each ... I've seen the cis-trans naming on cis/trans-platin but there are 3 isomers here?
cis/trans doesn't work here, as you have realised.
There is nomenclature system for cases like these, see section IR-9.3.3.3, p180 in this document (http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf).
Ah ok, my PDF reader won't render p180 correctly but it does render p181 and from the examples given I think I can leave this naming system aside for the time being.
Post by: Big-Daddy on April 03, 2013, 09:00:48 AM
Is it simply the same, but if rotating any opposite bonded pairs on Configuration 2 can take me back to Configuration 1, then Configuration 2 is the same molecule as Configuration 1? i.e. I follow the same process as for square planar, except I can also rotate the 2 new bonds that are present (and opposite each other) in the same way I rotate the opposite bonds for square planar molecules ...
Post by: Dan on April 03, 2013, 10:03:08 AM
Post by: Big-Daddy on April 21, 2013, 11:21:19 AM
Imagine you have the square planar complex written in 2D with ligands at position a (top left), b (top right), c (bottom right) and d (bottom left). If you have a ligand at a and a ligand at b, you cannot switch them and be left with the same isomer; however, ligands at a are freely rotatable with those at c, and ligands at b are freely rotatable with those at d. So, if by performing these free rotations you can go from one proposed isomer to another, then they are really the same isomer and you should only write down one of them (you could write down either).
Before we move onto octahedrals (where there are also problems) can you explain why this method won't work with square planar complexes that don't specifically have 4 different ligands?
An example is: (PPh3)2PtCl2, draw all isomers. (Pt is central of course.)
The mark scheme suggests we've got a=PPh3, b=Cl, c=Cl, d=PPh3 (ligands ordered clockwise from top left) as 1 isomer, but then the other is a=PPh3, b=PPh3, c=Cl, d=Cl (ligands ordered clockwise from top left). But wouldn't a simple free rotation of b and d in either one of the isomers bring us back identically to the other?
Post by: Big-Daddy on April 28, 2013, 04:51:32 PM