Chapter 5 – Stereochemistry

Chirality
o For a carbon to be chiral, it must have four different groups bonded to it
 Must be tetrahedral, but that’s necessary for the above to be true
anyway.
C
C
CH3
CH3
chiral


achiral
Thus, CH3 or CH2 will never be chiral
The groups do not have to be different atoms
C


This means that if a carbon has a methyl, ethyl, propyl, and butyl
group coming off it, then it is chiral, even though it has four
carbons attached to it.
R/S – absolute configuration
o Assign priorities to all the atoms coming off the chiral carbon one atom at a time
1
4
3
2
o

Based on Mass, not electronegativity
 The heaviest atom is given priority #1
o If there’s a tie, go out one more atom
o Then, make sure your lowest priority atom is facing away from you
o Make a circle from 1 to 2 to 3
1
4
2
3

o If the circle was clockwise, then assign R
o If counterclockwise, then assign S
What do I do if the #4 group is coming forward?
o Still assign priorities to the different groups
2
3
1
4
o Find R/S as though #4 was going back.
2
3
1
4

o Switch it.
 This one looks like S, so we have an R.
What do I do if the #4 group is in the plane?
4
3
2
1
o You have several options.
o 1) Look at the molecule from another direction so that the #4 is facing away
from you.

This method seems to be easiest for those of you who like this chapter/
find this stuff easy.
4
3
look from over here
2
1
o 2) Rotate around a single bond to get the #4 group in the back.
2
4
4
3
2
3
1
1
o 3) Switch two groups, find R/S for the new molecule, and then switch it back.
 This one seems to be easiest for those of you who hate this chapter.
1
4
3
3
2
2
1
4
looks like an S, so it's an R

Enantiomers vs. Diastereomers
o An enantiomer is a non-superimposible mirror image


All the R/S designations will be changed in enantiomers
If there is a double bond, then its stereochemistry (cis/trans) will remain
the same.
o A diastereomer is a non-superimposible non-mirror image.



Some but not all of the R/S designations will change.
Cis/trans double bonds make your molecules diastereomers as well.
How do enantiomers differ from one another?
o Physical properties
 Most are identical
 Boiling point
 Melting point
 Density
 Optical activity is opposite
 Chiral compounds rotate plane-polarized light
o If a compound rotates light clockwise, then it is called the
+ enantiomer.
o Likewise, if it rotates light counterclockwise then it is
called the – enantiomer.
o Sometimes +/- is called d/l
 Older chemists use this, but you’ll see +/- in this
course.

Enantiomers rotate light the same amount, but in opposite
directions.
o So if one stereoisomer rotates light 4° to the right, then its
enantiomer will rotate light 4° to the left.
o It is important to understand that R/S configuration does
not tell you whether you have the + or – enantiomer
 R/S configuration is a manmade convention,
whereas +/- is a physical property.
o It’s important to distinguish that R/S configuration has no
relationship to +/-.
 50/50 mixtures of enantiomers are called racemic mixtures or
racemates.
o These do not rotate plane-polarized light because the two
enantiomers cancel each other out.
 Calculation of enantiomeric excess
o We will not do this in this class.
 You are responsible for this in your lab lecture.
o If you see a problem on the test which looks like it requires
any sort of calculation with optical rotation, then you need
to look at the question again.
 For example, sometimes you’ll see a question like
“which of the following would rotate light 1°
clockwise?” followed by a series of solutions.
 All but one of the answers will be achiral, so
the answer is the only chiral answer.
 You might have an achiral molecule, an
equal mixture of enantiomers, an equal
mixture of diastereomers, and another
achiral molecule.
 The correct answer would be the mixture of
diastereomers because it is the only chiral
solution given.
o Chemical properties
 Chiral materials react with other chiral reactants selectively
 Think of the glove and hand model here
 If you have a bunch of right-handed gloves only right hands will
fit.
o Separation of enantiomers






Very often tested!
There are two ways to separate enantiomers
1) Convert to diastereomers, Separate, convert back to enantiomers
2) Chiral chromatography
Mr. Baker wants to see all of the above if he asks this.
 If you skip “convert back” in number 1, you don’t get credit
 If you skip “chiral” in number 2, you don’t get credit
Meso compounds
o Compounds containing chiral carbons but which are not overall chiral
 This means a meso compound has not enantiomer.
 It will also not rotate plane-polarized light.
o There must be a plane of symmetry present
meso
chiral


Be sure you’re looking for a plane of symmetry and not an axis of
rotation; the compound on the left has the plane of symmetry
and the compound on the right has an axis of rotation.
If no stereochemistry is shown, you cannot say whether it is meso or
chiral.
o Be careful not to be tricked by cyclic compounds with no chiral centers.
These are achiral because there are no chiral carbons present.
o Meso compounds do not have to be cyclic
 What is the relationship between these two compounds?


They’re the same because if you rotate around the single bonds you can
get them into symmetrical forms.
 They’re two different drawings of the same meso compound.
Maximum number of stereoisomers possible for a compound
o If n number of chiral carbons, then 2n possible stereoisomers
 Be sure to subtract any meso compounds from this total.
 How many possible stereoisomers are there of this compound?

There are 2 chiral centers, so 2n=4.
o But wait!
o Here are the four versions given by this formula
o The top two of these structures are meso, so we have to
subtract one from the four.
 There are three total stereoisomers of 1,2dibromocyclohexane.
o Common question for this material

How many monochorinated products, including stereoisomers, will you
get when you chlorinate a particular compound.
Ex.


First, find all the positions that give you distinct structural isomers.
Then, go through and see if adding a chlorine at each spot makes any spot
chiral.



If it doesn’t, then you only get one version of that chlorination.
If it does, you get 2n products where n=the number of chiral carbons.
Back to our example, here’s how I would do it.
 If I put a chlorine here, nobody becomes chiral. That said, you can
still get the chlorine placed either cis or trans to the alkyl group
that’s on the ring, so you get two versions.
o I almost always miss this cis/trans possibility so make sure
you don’t make my mistake!

If I put a chlorine here, two positions are now chiral so there are four
versions of this chlorination.
*
*

The same thing happens here.
*
*

No new chirality, so just one version here.

Here, you get one chiral center so there are two versions
*

Here you have no chirality.

Here you get four products again.
*
*

And at the last position you get one chiral center so 2 products.
*

If you kept track of this as you did this, then you would have something
that looks like this:
1,2
3,4,5,6
7,8,9,10
11
12, 13
14
15,16,17,18
19,20
o If you see, I just kept a tally of how many isomers we had gotten
to at each spot.
o So here, there are 20 total monochlorinated products possible.

Allenes
o Chiral compounds with no chiral carbons
o Two double bonds directly next to each other
A
C
B
D
o For the molecule to be chiral, A≠B AND C≠D

Fischer Projections
CO2H
H2 N
H
CH3
o Every intersection is a chiral carbon.
o Everything on the horizontal is coming forward and everything on the vertical is
going back
CO2H
H2N
H
CH3

If you are asked to convert from Fischer to line-angle you cannot just
draw the dashes and wedges like I’ve done here. Two of the pieces
coming off any atom must be drawn flat (in the plane).
o Still assign priorities the same way as before
2
CO2H
H2 N
1
H
4
3
CH3
o Draw your circle from 1 to 2 to 3
2
CO2H
H2 N
1
H
4
3
CH3
o If H (or any lowest priority group) is on the Horizontal, it’s Horribly Wrong!
 It looked like an R, but H was on the Horizontal so it’s an S.

Comparing two molecules drawn as line-angle and Fischer
What is the relationship between these two molecules?
CH3
H
Br
HO
H
HO
CH2CH2CH3
o Method #1 – Assign R/S
 This is the easiest way for most people.
CH3
S
H
Br
S
HO
R
H
HO
R
CH2CH2CH3
 Both configurations are reversed, so these are enantiomers.
o Method #2 – Advanced method which is quicker but harder for most
 Remember that with a Fischer you are looking at the molecule so that the
vertical pieces are going back and the horizontal pieces are coming
forward.
 Stay with me here.
 Because of the implied stereochemistry of a Fischer, you are basically
looking at the “peaks” of the chain.
eye
HO
eye

Using this, you can easily convert to a Fischer from the line-angle and
then compare the two Fischers.

When you look down at the carbon with the bromine, you see
that you have a CH3 on top, an H on the right, a Br on the left, and
the rest of the molecule going down.
CH3
Br
H
rest

o View of this carbon from above.
When you look up at the carbon with the OH, you have the OH on
the right and an H on the left.
CHBrCH3
OH
H
rest
o View of this carbon from below.
Special stereochemistry designations for biological molecules
 Sugars and amino acids are chiral
 When you draw each in Fischer form, they have special stereochemical
designations.
 Amino acids I
 If the amino group is on the left, it is an L-amino acid.
CO2 H
H2 N
H
CH3
L-alanine

o This is the natural form.
If the amino group is on the right, it is a D-amino acid.
CO2 H
H
NH2
CH3
D-alanine

Sugars
 If the last chiral carbon has the OH on the left, then it is an Lsugar.
CHO
HO
H
H
OH
HO
H
HO
H
CH2 OH
L-glucose

If the last chiral carbon has the OH on the right, then it is a Dsugar.
CHO
H
HO
OH
H
H
OH
H
OH
CH2 OH
D-glucose

To switch between the L and D form, you switch all the chiral
carbons, not just the bottom one.
CHO
H
HO
H
HO
OH
H
OH
H
CH2 OH
not glucose (but definitely L)