Problem: R or S?
S
R
S
1) Draw the structure of (S)-1-bromo-1-chlorobutane.
2
Cl
H
Br
3
S
1
2) Draw the structure of (2R,3R)-2,3-dichloropentane.
Cl
1
R
2
3
Cl
3
2
R
1
4
3) Draw the S isomer of phenylalanine (the compound shown).
CO2H
CO2H
Ph
Ph
NH2
NH2
S
3
2
1
4) Label all of the stereogenic atoms in the following compound with map number 1.
Cl
1
1
Cl
5
5) Label all of the stereogenic atoms in the following compound with map number 1.
Br
OH
1
1
6) Label all of the stereogenic atoms in the following compound with map number 1.
OH
1
H3C
H
6
7) Draw the enantiomer of the following structure.
CH3
CH3
Br
H
Cl
CH3
H
Br
H
3
Cl
S
1
Br
Cl
CH3
2
Cl
3
H
2
Br
1
R
7
8) Draw all stereoisomers of 2,3,4-tribromopentane. You should find two meso structures
and one pair of enantiomers.
Br
Br
Br
Br
Br
Br
Both meso due to plane of symmetry through 3rd carbon
Br
Br
Br
Br
Br
Br
8
9) Submit as your answer the structures of all the chiral molecules among the ones
shown.
Cl
Br
Cl
Cl
chiral
Cl
Cl
chiral
Cl
Cl
9
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
chiral
Cl
Cl
Cl
Cl
Cl
Cl
Cl
10
10) Submit as your answer the structures of all the molecules among the ones shown
that are chiral in their lowest-energy conformations.
NH2
NH2
NH2
NH2
chiral
H
H
1
NH2
H
2
NH2
NH2
H2N
NH2
2
NH2
H2N
H
1
NH2
Enantiomers
11
NH2
NH2
NH2
chiral
NH2
H2N
NH2
Achiral
chiral
NH2
H2N
H
NH2
H
NH2
NH2
H2N
Axis to cut
H
H
12
NH2
NH2
H2N
H2N
NH2
H2N
Achiral
chiral
Achiral
NH2
H2N
NH2
H2N
13
11) Submit as your answer the structures of all of the compounds among the ones
shown that may be optically active.
Br
F
Br
Cl
B
Cl
H
Cl
Cl
H
H
H3C
Br
Br
Trigonal
planar
Cl
Cl
Cl
H
H3C
H
14
Finding stereocenters
O
O
O
CH3
*
*
CH3
plane
( indicates no stereoisomers )
CH3
Finding stereocenters
O
O
O
CH3
*
*
CH3
CH3
*
*
CH3
CH3
*
*
CH3
The cis isomer (two methyl
groups) - achiral!
Thetrans isomers has two
stereocenters!
Draw all of the stereoisomers!
CH3
CH
CH
CH
Cl
OH
Cl
CH3
CH 3
Cl
OH
Cl
CH 3
CH 3
Cl
HO
Cl
CH 3
meso
meso
Same (upside down)
CH 3
CH 3
Cl
OH
Cl
HO
Cl
CH 3
Cl
CH 3
CH 3
Cl
NO!
HO
Cl
CH 3
Same (upside down)
CH 3
Cl
OH
Cl
CH 3
CH2Cl
H3C
H
CH2
CH3
Cl2, hv
(S)-(-)-1-chloro-2-methylbutane
C5H10Cl2
+ HCl
six fractions:
four optically active
two optically inactive
CH2Cl
H3C
H
CH2
CH3
Cl2, hv
CHCl2
H3C
H
CH2
CH3
A
No bonds to the chiral center
are broken, configuration is
retained. Product is optically
active
CH2Cl
H3C
H
CH2
CH3
Cl2, hv
CH2Cl
H3C
Cl
CH2
CH3
+
CH2Cl
Cl
CH3
CH2
CH3
B
A bond is broken to the chiral center.
Stereochemistry depends on the
mechanism. Here, the intermediate
free radical is flat and a racemic
modification is formed. This fraction
is optically inactive.
CH2Cl
H3C
H
CH2
CH3
Cl2, hv
CH2Cl
ClH2C C H
CH2
CH3
C
The product no longer has
a chiral center. It is achiral
and optically inactive.
CH2Cl
H3C
H
CH2
CH3
Cl2, hv
CH2Cl
H3C
H
H
Cl
CH3
D
CH2Cl
H3C
H
+
Cl
H
CH3
E
No bond is broken to the chiral center
and a new chiral center is formed. The
products are diastereomers and each
fraction is optically active.
CH2Cl
H3C
H
CH2
CH3
Cl2, hv
CH2Cl
H3C
H
CH2
CH2Cl
F
No bonds to the chiral center
are broken, configuration is
retained. Product is optically
active
CH2Cl
H3C
H
CH2
CH3
Cl2, hv
CHCl2
H3C
H
CH2
CH3
CH2Cl
H3C
Cl
CH2
CH3
CH2Cl
Cl
CH3
CH2
CH3
(S)-(-)-1-chloro-2-methylbutane
A
CH2Cl
ClH2C C H
CH2
CH3
C
CH2Cl
H3C
H
CH2
CH2Cl
F
B
CH2Cl
H3C
H
H
Cl
CH3
D
CH2Cl
H3C
H
Cl
H
CH3
E
optically active
optically inactive
Example: 0.5g (-)-epinephrine-HCl in 10mL H2O measured in 20 cm cell (25o/D) obs = 5.0o, []D =?
Information given
0.5g (-)-epinephrine-HCl in 10mL H2O .
Amount per mL = 0.5/10 = 0.05 g /mL.
Length =20 cm cell
Length in dm = 20/10 = 2 dm.
25o
[] D =
 ( obs)
l(dm) [ g/ mL]
25o
[] D =
 obs)
( 2dm) [ 0 .0 5g/ mL]
25o
[] D =
-50o
[] = deg (cm2g-1 )