classroom
activity
16
Chirality
Show You Know!
1. Identify any lines of internal symmetry within the molecules below.
NH2
HO
CH3
H3C
H
HO
O
O
N
H
NH2
H
NH2
Based on what you have just learned about chirality, could any of these molecules be chiral? (Remember
chiral is just Greek for having qualities similar to the human hand.)
No, they have internal planes of symmetry, and could be superimposed.
3. For each of the following compounds: identify any stereogenic centers, assign absolution configuration to those stereogenic centers, identify any plane of symmetry, and indicate whether the molecule
is chiral.
a.
OH
HO
HO
Chiral molecule, no plane of
symmetry. Chiral carbons shown with
stereochemistry.
c.
HO
R
R
OH
(1R,3R)-cyclopentane-1,3-diol
H
CH3
OH
Achiral molecule, no
chiral carbons plane of
symmetry as shown
1-methylcyclopentanol
OH
CH3
e.
CH3
OH
CH3
OH
CH3
OH
Achiral molecule, no
chiral carbons plane of
symmetry as shown
g.
CH3
HO
1-methylcyclohexanol
Chiral molecule, no
plane of symmetry.
Chiral carbons
shown with absolute
configuration.
H3C
HO
S
R
(1R,2S)-2-methylcyclopentanol
5. For each of the following compounds:
● Indicate with an asterisk any chiral centers in the molecules.
● Rank each group on the chiral centers from high to low priority.
● Determine whether the chiral center is (R) or (S).
a.
c.
F
CH2
H3C
*R
HO
H
(2R)-but-3-en-2-ol
H
g.
O
H
*R N
H
HN
Ecstacy (MDMA)
*R
O
CH3
N
*R
CH3
(1S)-1-chloro-1-fluoroethane
LSD
e.
*S
Cl
CH3
CH3
O
H3C
N
CH3
H
16 Chirality
6. Provide a proper IUPAC name for each of the following molecules including any stereochemical
designations (R,S or E,Z)
a
Br
c
Cl
b
a.
(2R,5R)-5-bromo-2-chloroheptane
c.
(4R,5R)-4,5-dimethylcyclohexene
d
CH3
F
CH3
Challenge!
Warning! Not for the faint of heart; this one is more challenging than would ever be given to you on a
test. It is only provided as a challenge.
For the following compound:
Estrogen analog
● Indicate with an asterisk any chiral center.
CH3
OH
CH
● Rank each group on the chiral centers from high
to low priority.
H
● Determine whether the chiral center is (R) or (S).
H
HO
H
CH3
OH
CH
H
We will first look at what is known as
the 17 carbon in estrogen nomenclature...
H
H
HO
Highest priority (1)
connected to three carbons
is it 2 or 3 priority?
CH3
H
connected to three carbons
is it 2 or 3 priority?
OH
C
C
CH
lowest
priority (4)
H
H
Blue carbon is bonded to 1
carbon with a "real" H
substitutent and 2 carbons
with no substituents
H3C
HO
H3C
To determine which is 2 and
3, we need to look at what
each carbon connected to 3
carbons is connected to:
1
C
4
C
C
C
C
C
Red carbon is bonded to 3
carbons that each have "real"
substitutents; it has priority.
H
C
CH3
H3C
H
We have the groups prioritized we can look at what the stereochemistry
is. One problem we have is that the lowest group is in the plane of the
page, neither back or front. We need to rotate the molecule.
Look from
this direction
1
H3C
H
H
HO
2
H
3
4
Look from
this direction
CH
1
Simplified structure
2
3
4
Rotate
3
1
2
Pseudo Newman
Clockwise so R configuration
16 Chirality
CH3
OH
CH
So, now it is easy to see that the nomenclature
for these molecules usually refers to anything
above the ring as α and anything below the ring
is β rather than R or S. Now, let’s look at the
carbon next to it: the 13 carbon. It appears to
be connected to 4 other carbons.
H
H
H
HO
Lowest priority
CH3
H
H
13
Carbon with a C and 2H
attached, 3rd priority
OH
Carbon with an O, a C and
an H attached, 1st priority
OH
4
CH
CH
Simplify
H
H
H
H
HO
Carbon with 2 C and an H, 2nd priority
HO
Simplify
3
4
Rotate
1
3
Psuedo Newman
Counterclockwise so S
2
2
Let’s look at Carbon 14 now.
This one is relatively easy.
1
CH3
OH
1
CH
H
2
H
HO
H
3
Pseudo Newman
Counter clockwise=S
OH
CH3
CH
H
Let’s look at carbon 8. This is another
bear, but priority 4 is easy!
H
H
HO
C bonded to 3 C’s, all with “real“ substituents
CH3
C with 2 C’s
and an H
C with 2H and a C
OH
C bonded to 3 C’s but not
all have “real” substituents
CH
CH
4
4
H
HO
H
OH
H3C
H
C with 2 Cs and an H
HO
C with 1 C and 2 H, priority 3!
H
C with 2H and a C
Simplify
2
Rotate
2
1
3
4
1
3
3
Psuedo Newman
Clockwise R configuration
C with 1 C and 2H, priority 3
CH3
C with 3C attached, priority 1
OH
CH3
CH
3
H
Simplify
1
H
4
HO
Finally carbon 9,
priority 4 is easy again
H
HO
2
H
Simplify
3
1
H
H
OH
CH
H
CH
HO
C with 2C and
an H, priority 2
CH3
4
OH
Pseudo Newman
Counterclockwise,
2 S configuration