Lecture #35
Review for Exam IV
  CURMUDGEON GENERAL'S WARNING. These "slides" represent highlights from lecture and are neither complete nor meant to replace lecture. It is advised not to use these as a reliable means to replace missed lecture material. Do so at risk to healthy academic performance in 09-105.
The list from which questions will be drawn for Exam IV on Weds, Dec 1.
...topics, topics,
A review problem for stimulating thought and discussion. It will take several steps to work through. It is much too long and complex to worry about as a possible exam question.
From the position of Mn on the periodic table and understanding of oxidation numbers, the possibilities are +7 through 0, ruling out all but +3.
Two geometries and the extreme weak/strong field arrangements in combination with the given magnetic property allow the illustrated conclusion (highlighted).
Another practice problem
And another
This is superimposable on the original geometry. Not a chiral isomer!
..on and on we go....
If this were the correct geometry for the ligands, three geometric isomers would have been found. Since "two" is the given result, this cannot be the correct arrangement.
If each asymmetric carbon gives you two isomers (enantiomers/chiral isomers), then two such carbons in the same molecule gives you four possible isomers. How many isomers of taxol are possible?

There are 11 asymmetric centers giving 2048 total isomers.

Rank the following six chlorobenzenes in order of expected increasing boiling points.
(This was not included from an earlier lecture.) Here's another attempt to show how crystal field splitting works, but with coordination number 2, also known as linear geometry.
This orbital's electron, pointing directly at the ligands (with their electrons) is raised in energy.
On the other hand, this orbital does not point towards the ligands and is unaffected by their presence. Likewise for the dxy.
Putting it all together, the crystal field idea rearranges the five d-orbitals into the illustrated diagram.