Chem S-20ab
Purple Book Solutions
Week 2 - Page 1 of 27
R and S Nomenclature
1. Circle all the stereocenters in the following molecules. For each stereocenter, assign an R or S
designation according to the standard rules.
Cl
H
R
H
R
H
R
OH
H3C
S
S
H
S
O
S
R
S
HO
N
2. Draw an unambiguous structural representation of the following molecules:
(S)-4,5-dimethyl-(Z)-2-hexene
(R)-3,5,6-trimethyl-(E)-3-heptene
H
H
3. Provide the best possible systematic name for the following compound:
(S, E)-5-isopropyl-2,3,7-trimethyl-4-octene
Chem S-20ab
Purple Book Solutions
Week 2 - Page 2 of 27
Chirality
1. For each of the following species, indicate if it is chiral or achiral. For those molecules that are chiral,
circle all stereocenters, and draw the structure of the enantiomer.
H
H
F
H3C
F
Cl
Cl
CH3
chiral
H
CH3
H3C
H
Br
H
chiral
H
CH3
achiral
H
H
Br
Br
Br
H
chiral
H
Br
Br
H
achiral ("meso")
H3C
CH3
H3C
CH3
H3C
CH3
H3C
CH3
chiral
Chem S-20ab
Purple Book Solutions
Week 2 - Page 3 of 27
Enantiomers and Diastereomers
1. For each of the following pairs of stereoisomers, indicate if they are enantiomers, diastereomers, or
identical compounds.
diastereomers
identical
enantiomers
identical
CH3
CH3
Br
Br
CH3
H3C
Br
Br
H3C
CH3
diastereomers
Chem S-20ab
Purple Book Solutions
Week 2 - Page 4 of 27
Identifying Isomers
1. For each of the following pairs of isomers, indicate whether they are Identical, Enantiomers,
Diastereomers, or None of the above by circling the appropriate term for each.
HN
Me
HN
H
H
HN
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Me
HN
H
H
Me
Me
H
H
Chem S-20ab
Purple Book Solutions
Week 2 - Page 5 of 27
Stereochemical Concepts
1. Which of the following are true? Give counterexamples for those that are false.
(a) In some cases, constitutional isomers are chiral.
T
(b) In every case, a pair of enantiomers have a mirror-image relationship.
T
(c) Mirror-image molecules are in all cases enantiomers.
F
identical
CH3
(d) If a compound has an enantiomer, it must be chiral.
CH3
T
H
(e) Every chiral compound has a diastereomer.
F
(f) If a compound has a diastereomer it must be chiral.
has no diastereomer
F
H3C
Cl
F
and
are diastereomers
H
(g) Any molecule containing an asymmetric carbon must be chiral.
Br
F
achiral ("meso")
Br
(h) Any molecule with a stereocenter must have a stereoisomer.
(i) Some diastereomers have a mirror image relationship.
(j) Some chiral compounds are optically inactive.
(k) All chiral molecules have no plane of symmetry.
H
T
F; no diastereomers have a mirror image
relationship.
F; all PURE chiral compounds are optically active
T
H
(l) If a compound has a stereocenter, it has an enantiomer.
Br
F
achiral ("meso")
Br
(m) A meso compound will necessarily have at least two diastereomers.
H
T
Chem S-20ab
Purple Book Solutions
Week 2 - Page 6 of 27
Working with Stereoisomers
1. Consider the species 2,3-butanediol, whose structural formula is:
HO
OH
2,3-butanediol
H3C
CH3
a) There are three distinct, pure, configurational stereoisomers of 2,3-butanediol, all of which have the same
structural formula above. Using any unambiguous notation, draw the three different stereoisomers in the
boxes below. (The boxes have been arbitrarily labeled A, B, and C.)
A
B
C
HO
OH
HO
OH
HO
OH
H3C
CH3
H3C
CH3
H3C
CH3
b) By circling the appropriate terms below, indicate whether each of the three species you drew above is
chiral or achiral.
A:
chiral
achiral
B:
chiral
achiral
C:
chiral
achiral
c) Two of the above species have the same melting point (19°C). The other species has a different melting
point (34°C). Circle the letter of the species which has a melting point of 34°C:
A
B
C
What characteristic of the other two stereoisomers causes them to share the same melting point (19°C)?
B and C are enantiomers, and enantiomers always have identical physical properties;
for instance, they will have the same melting point.
Chem S-20ab
Purple Book Solutions
Week 2 - Page 7 of 27
Principles of Reaction Stereochemistry
1. Each of the following statements is false. Discuss why the statement is false. Then make the statement
true by changing only the underlined word or phrase.
Two molecules which are diastereomers should be expected to have the same melting point.
enantiomers
Two molecules which are enantiomers would exhibit different rates of reactivity with an achiral reagent.
diastereomers
An achiral starting material can react with an achiral reagent to give a single chiral product.
single achiral product
or
racemic mixture of products
Two molecules which are enantiomers would exhibit identical reaction rates when reacting with a chiral
reagent (such as a biological enzyme).
different
When an achiral starting material is treated with an achiral reagent to yield a pair of diastereomers, both
products must always be produced in equal amounts.
enantiomers
Chem S-20ab
Purple Book Solutions
Week 2 - Page 8 of 27
Mechanism and Stereochemistry
1. a) In the box below, draw the product of the following reaction. Your product must match the
chemical formula provided. Indicate stereochemistry clearly, if appropriate.
O
Br2
OH
Br (+/-)
Chemical Formula: C6H11BrO
b) Provide a complete curved arrow mechanism for the transformation indicated above. Explain
the stereochemical outcome, if any.
Br
Br
Br
OH
OH
Bromonium ion can form
above or below the
plane, leading to a
racemic mixture of
bromonium ion
intermediates. This in
turn leads to a racemic
mixture of products.
O
Br
H
solv
OH attacks from the opposite
plane, leading to the
diastereoselectivity (although,
since only one sterocenter is
formed in this reaction, only one
pair of enantiomers is possible.
O
Br
c) Explain the regioselectivity of this reaction; in other words, why is the particular isomer you
drew in (a) the one that's formed.
The intermediate bromonium ion is unsymmetrical: there is much more partial
positive charge on the tertiary carbon. This is similar to the relative stabilization
conferred by tertiary vs. secondary carbocations. The nucleophile (ROH) then
reacts by breaking the weaker of the two C–Br bonds.
Chem S-20ab
Purple Book Solutions
Week 2 - Page 9 of 27
Stereochemistry of Alkene Additions
1. Fill in each box with the organic product(s) of the indicated transformation. Be sure to indicate the correct
stereochemistry. (If the product is formed as a racemic mixture, please indicate this!)
Br
Br
+
Br
Br2
Br
(racemic mixture)
Br
** Note: The chair forms are shown just
to give you extra practice converting
planar drawings to chairs.
Br
+
Br
Br
Br
Br
+
OH
Br2, H2O
OH
(racemic mixture)
OH
Br
+
Br
OH
OH
OH
+
1. BH3, THF
2. H2O2, OH–
(racemic mixture)
H
OH
+
H
(achiral)
H2 , Pt/C
OH
Chem S-20ab
Purple Book Solutions
Week 2 - Page 10 of 27
Stereochemistry of Alkene Additions
1. Fill in each box with the starting materials that would lead to the indicated product. Be sure to indicate the
correct stereochemistry.
Br
H
Br2
H
Br
HO
1) BH3, THF
H
2)H2O2, OHMe
H
(+/-)
H
D2, Pt/C
D
D
(+/-)
HO
Me
O
Br2
Me
Et
Et
Br
H
(+/-)
Chem S-20ab
Purple Book Solutions
Week 2 - Page 11 of 27
Cyclohexanes
1. Look at a chair. Notice the axial and equatorial substituents. Pay close attention to the parallel lines!
A chair has 3 sets of parallel lines, plus vertical lines for axial groups.
Once that is drawn in, every subsequent line must be parallel to a line that
has already ben drawn! Try drawing t-butyl groups in the equatorial positions.
2. Draw each of the following dimethyl cyclohexanes (on the planar ring), then identify whether the
substituents would be either: "Axial and Equatorial" or "Both Equatorial"
1,2-cis
Ax. and Eq.
1,2-trans
Both Eq.
1,3-cis
Both Eq.
1,3-trans
Ax. and Eq.
1,4-cis
Ax. and Eq.
1,4-trans
Both Eq.
3. Draw each of the following in the most stable chair conformation:
Et
Chem S-20ab
Purple Book Solutions
Week 2 - Page 12 of 27
Chair Conformations
1. Consider the following molecule:
CH3
Cl
H3C
CH3
a) This molecule is expected to have two relatively stable "chair" conformations. Draw clear representations
of the two different conformations in the boxes below. (The boxes have been arbitrarily labeled A and B.)
In addition, please be sure to draw the isopropyl substituent in its best conformation in each case.
A
B
H3C
H3C
Cl
H
H
CH3
H
H3C
Cl
H3C
H
CH3
b) Which of the two chair conformations is more stable? (circle one of the statements below)
1) A is more stable than B.
2) Both conformations are equally stable.
3) B is more stable than A.
Chem S-20ab
Purple Book Solutions
Week 2 - Page 13 of 27
Bicyclic Compounds and Bredt's Rule
1. Draw each of the following bicyclic compounds in a good "perspective" drawing.
CH3
CH3
trans-decalin: two chairs
CH3
CH3
CH3
norbornane skeleton:
H3C
CH3
CH3
2. The molecule trans-cyclooctene is known to exist. (It is chiral, by the way). Why is the analagous molecule
trans-cyclohexene unstable?
Far too strained to have a trans-alkene in a six-membered ring.
H
H
H
H
=
H
H
trans-cyclooctene
(planar representation)
trans-cyclooctene
(perspective representation)
trans-cyclohexene????
3. Draw each of the following bicyclic alkenes in a good "perspective" representation. Only one of these
three compounds actually exists. Which one, and why?
BAD!
In these two,
the alkenes
can't be planar.
Highly strained.
(Try to model!)
Or, notice the
trans-alkenes
in small rings.
BAD!
Ok!
** Bredt's Rule:
Can't have sp2 carbon at
a bridgehead. (No alkenes,
no carbocations!)
Chem S-20ab
Purple Book Solutions
Week 2 - Page 14 of 27
More Practice With Reaction Stereochemistry
1. Consider compounds A and B shown on the right:
a) Is A chiral? (circle)
Yes
No
b) Would a solution of B in a
polarimetry cell at room temperature
rotate plane-polarized light? (circle)
Yes
No
c) What is the stereochemical relationship of A and B?
A
B
Diastereomers
Draw the products of the following transformations. Be sure to include all unique stereoisomers that will
be produced in each reaction.
Br
Br
Br2
d)
A
Br
Br
CH2Cl2
achiral
diastereomers
achiral
OH
H2O
e)
B
H3O+
diastereomers
chiral
H2
f)
A
Pd/C
achiral
achiral
diastereomers
H2
g)
B
Pd/C
same compound
achiral
h) Label each of the products above as chiral or achiral as appropriate. (You do not need to worry about
labeling anything meso.)
i) For all products of the same formula in d-g above, identify pairwise stereochemical relationships.
Chem S-20ab
Purple Book Solutions
Week 2 - Page 15 of 27
Introduction to SN2, E2, SN1, and E1 Mechanisms
1. Provide complete curved-arrow mechanisms for the following transformations.
NaOH
Br
OH
:OH
Br
H
KOtBu
:OtBu
H2O
Br
OH
:OH2
H
:OH2
O
H
conc. H2SO4
OH
H
:sol
OH2
H
Chem S-20ab
Purple Book Solutions
Week 2 - Page 16 of 27
Mechanisms and Stereochemistry
1. Predict the product.
H
CH3S–
CH3
H
CH3
I
SCH3
2. Provide a curved-arrow mechanism, and explain the observed stereochemistry.
Br
S
:OH
:OH–
S
O
HO
SH
H
O
O
With SN2, "retention" is ALWAYS "double inversion"
H+
:O
O
:OH
Br
O
O
:O
O
O
SH
S:
SH
O
3. Explain the difference in reactivity.
Br
Br
but
t
KOtBu
KO Bu
no reaction!
Br
equatorial Br
can't get antiperiplanar
with any H, so can't
do E2 elimination.
axial Br
perfect for
antiperiplanar
E2 elimination
Br
H
:B
4. Explain the difference in reactivity.
conc.
conc.
but
H2SO4
HO
H2SO4
OH
E1 elim. OK, with
secondary carbocation
which can be planar
(as required)
+
This tertiary carbocation
cannot be planar,
and the product violates
Bredt's Rule
+
no reaction!
Chem S-20ab
Purple Book Solutions
Week 2 - Page 17 of 27
Choosing a Reaction Pathway 1
Consider the following conditions:
for SN2 reactions:
need good steric interaction (Me, 1° is good, 2° OK, 3° bad)
for E2 reactions:
need strong attacking base (at least as strong as OH–)
for SN1/E1 reactions:
need stable carbocation (3° is good, 2° is OK, 1° is bad)
All things being equal, choose the reaction higher up on the above list.
For each of the following, predict the major product, and indicate the type of mechanism.
1.
CH3I + OH–
2.
CH3I +
3.
4.
5.
CH3OH
–
O
excellent SN2
horrible f or SN 2, but no other reactive pathway
exists
OCH3
Br
+ OH–
OH
good SN 2
Br
+ CN–
CN
excellent SN2
Br
+
nucelophile much too bulky f or SN 2, so
E2 prevails
–
O
6.
Br
+ OH–
7.
Br
+ CN–
when nucleophile is also a strong base, secondary
substrates will pref er E2 to SN2
tertiary substrate cannot undergo SN 2 reaction, so
E2 is the only pathway available
+ OH–
8.
excellent nucleophile that is not basic, so SN 2
will dominate
CN
Br
9.
Br
OH
major
10.
Br
+ CH3OH
SN 1 and E1 always compete, but SN 1 will
always dominate in such situations
+
+ H2O
minor
SN 1 and E1 always compete, but SN 1 will
always dominate in such situations
+
OCH3
major
minor
alcohols dehydrate in mineral acids only via E1
11.
OH
+ H2SO4
Chem S-20ab
Purple Book Solutions
Week 2 - Page 18 of 27
Choosing a Reaction Pathway 2
1. Which product (or) products would you expect to obtain from each of the following reactions? In each case
identify the mechanism by which each product is formed (SN1, SN2, E1, or E2) and predict the relative amount
of each (i.e. would the product be the only product, major product, minor product, etc?).
a)
NaOMe
Br
OCH3
+
SN2 major
E2 minor
KOtBu
b)
Br
E2 only product
t
BuOH
c)
NaOEt
(CH3)3CCl
d)
E2 only product
EtOH
+
E1 minor
( trans or E f avored
over cis or Z)
***Not always true in E2
elimination, where the
anti-periplanar conf ormation
is required!!!***
OEt
SN 1 major
Br
KOtBu
e)
+
+
t
Br
BuOH
E2 major,
Zaitsev product
E2 minor,
E2 minor,
Hof mann product
Zaitsev product
( KOt Bu is
bulky enough
*both Zaitsev products are
to give some
possible here since there are
two protons available f or
Hof mann product)
deprotonation, each giving
rise to a dif f erent alkene
geometry
Chem S-20ab
Purple Book Solutions
Week 2 - Page 19 of 27
Substitution and Elimination: Synthesis
1. Each of the following products can be synthesized in one step from an alkyl halide. Show the starting
material that could serve as an immediate precursor to the indicated target, and fill in the reactions conditions
required to obtain the desired product. Be sure to consider stereochemistry where relevant!
1.
NaCN
X
( where X = Cl, Br, I)
X
( where X = Cl, Br, I)
H3C
H
CN
*SN 2, so remember to account
f or inversion!
KOH, NaOH, NaOR,
t
1. or KO Bu etc.
*Since no reaction pathway
is possible except f or E2,
so any strong base is okay.
X
1.
( where X = Cl, Br, I)
H 2O
*Stereochem is unimportant
here since this is a tertiary
center, making only an SN 1
mechanism operative.
OH
Chem S-20ab
Purple Book Solutions
Week 2 - Page 20 of 27
Carbenes and Cyclopropanes
1. Chloroform can react with strong bases to yield dichlorocarbene, :CCl2. Provide a curved-arrow
mechanism for this transformation:
Cl
H
Cl
t
KO Bu
C
C
Cl
Cl
Cl
:OtBu
Cl
C
Cl
Cl
This is an alpha-elimination.
2. Dichlorocarbene can react with alkenes to yield cyclopropanes. For instance:
H
Cl
+
Cl
C
Cl
Cl
H
This reaction involves two significant donor-acceptor orbital interactions between the two reactants. Identify
the donor and acceptor orbitals in each interaction.
• Include both a verbal description and a picture of each orbital.
Cl
Donor:
Interacts W ith
Acceptor:
Cl
Alkene
Carbene
nC
*
C-C
Cl
Donor:
Alkene
C-C
Interacts W ith
Acceptor:
Carbene
2pC
Cl
Chem S-20ab
Purple Book Solutions
Week 2 - Page 21 of 27
Putting It Together: Products
1. Fill in each box with the major organic product of the indicated transformation. Be sure to indicate
stereochemistry when relevant.
Br
NaOH
CH3
CH3
The Zaitsez product will dominate.
H+
HO
O
OH
Mechanism? Intramolecular SN1
1. Zn
CH2I2
2.
Form carbenoid, then use it to make
cyclopropane
1. Mg, Et2O
Br
2. H2O
Form Grignard, then protonate it-get an alkane!
Chem S-20ab
Purple Book Solutions
Week 2 - Page 22 of 27
More Putting It Together: Products
1. Fill in each box with the single major organic product of the indicated transformation. (Any chiral starting
materials are provided as single, pure enantiomers.) Be sure to give the stereochemistry of the product if
it is relevant!
H3C
H3C
a)
OH
Br
HBr
Br
CH3
Racemic mixture (from SN1 reaction)
H3C
CH3
b)
I
H3C
NaN3
I
CH3
N3
I
(those are
iodine substituents)
CH3
CH3
KOtBu
c)
Cl
Note: The Zaitsev product is not possible here,
since there is no H anti to the leaving group
on the more substituted carbon.
H
1. KOtBu
d)
H3C
H
(+/-)
2. IZnCH2I
Cl
Note: While you will get some of the Hofmann
elimination product, Zaitsev will still be the most
prominent.
Chem S-20ab
Purple Book Solutions
Week 2 - Page 23 of 27
Making Alcohols into Leaving Groups
1. The OH from an alcohol is not a good leaving group. Why?
Good leaving groups are weak bases, but OH - is a strong base!
One way of making it a good leaving group is to protonate it. This often creates problems with carbocations,
rearrangements, etc. It can be used in limited cases:
2. Provide curved-arrow mechanisms for the following transformations:
HCl
OH
HCl
OH
Cl
Cl
H
H
Cl
Cl
OH2
:Cl
OH2
:Cl
The best ways of turning alcohols into leaving groups (at least for SN2 reactions) are to use either SOCl2 or to
form a sulfonate ester (most often a "tosylate"):
3. Provide the intermediates and products in the following synthetic sequences. Watch stereochemistry!
OH
OH
SOCl2, pyridine
Cl
NaCN
CN
OTs
TsCl, pyridine
CN
NaCN
Chem S-20ab
Purple Book Solutions
Week 2 - Page 24 of 27
Making Alcohols into Leaving Groups: Mechanisms
1. Provide a complete curved-arrow mechanism for each of the following transformations. Be sure to pay
attention to stereochemistry!
R
O2S
OH
Cl
OTs
TsCl, pyridine
O
TsCl = Cl
S
CH3
= ClSO2R
O
pyr:
O
Pyridine is:
H
O
O
S
O
O
S
R
O
R
N
O
OH
S
Cl
Cl
SOCl2, pyridine
Cl
+ SO2 (g)
+ Cl–
pyr:
O
H
O
S
O
S
Cl
O
Cl
SN2 displacement
inverts stereochemistry
Cl:–
Br
P
OH
Br
Br
PBr3
Br
Br
H
Br:–
+ HPOBr2
P
O
Br
SN2 displacement
inverts stereochemistry
Chem S-20ab
Purple Book Solutions
Week 2 - Page 25 of 27
Putting It Together: Reagents
1. Each of the following transformations can be carried out in one or two steps. Fill in the reagents required
for each step. If a second step is not needed, please put an "X" through the second box.
N
OH
SOCl 2 OR T sCl, pyridine
N
N
1.
2.
OH
1.
2.
OH
1.
NaN 3
conc. H 2SO4
H 2, Pd/ C
PBr 3 OR conc. HBr
Br
2.
OH
1.
conc. H 2SO4
OH
Br
2.
OH
1.
2.
Br 2, H 2O
conc. H 2SO4
H 2, Pd/ C
(plus enantiomer)
Chem S-20ab
Purple Book Solutions
Week 2 - Page 26 of 27
Ether Syntheses 1
1. Devise efficient syntheses for each of the following types of ether.
Primary—Primary
example:
O
Choose either SN2 route (typical Williamson synthesis):
+
O:–
Primary—Secondary
CH3I
example:
O
Use the secondary alkoxide as the nucleophile, and a primary halide in an SN2 process:
+
O:–
Br
Secondary—Secondary: tough! Either way gives a strong probability of elimination! Show SN2 and SN1
example:
O
+
+
OH
with acid,
likely to give some E1
+
–
O:
Br
HO
HO
with acid, use large excess
of the alcohol.
(Can you show
mechanism?)
likely to give E2
Tertiary—anything
Use a tertiary alcohol or an alkene, a trace of acid, and lots of the other alcohol, in an SN1-type route:
example:
O
+
+
OH
HO
with acid,
use large excess of
the primary alcohol
HO
with acid, use large excess
of the alcohol.
(Can you show
mechanism?)
Chem S-20ab
Purple Book Solutions
Week 2 - Page 27 of 27
Ether Syntheses 2
1. a) The following ethers can be synthesized by an SN2 reaction from two different combinations of an
alkoxide and an alkyl halide. For each molecule, show the combinations of alkyl halide and alkoxide that could
combine to form bond 1 and the combination that could combine to form bond 2 and circle the pair that will
react to provide a higher yield of the ether.
SN2 Starting Materials:
X
SN2 Starting Materials:
2
1
1
O
Me
O
2
+
+
O
X
SN2 Starting Materials:
X
CH3
SN2 Starting Materials:
1
+
H3C
O
Me
2
Me
O
CH3
O
CH3
+
Me
1
CH3
Me
H3C
2
X
CH3
CH3
b) Consider using an SN1 reaction to form the same ethers. For each molecule, show the combination of
alkyl halide and alcohol that could combine to form bond 1 and the combination that will combine to form bond
2 and circle the pair that will react to provide a higher yield of the ether.
SN1 Starting Materials:
X
SN1 Starting Materials:
2
1
1
O
Me
OH
2
+
+
OH
X
SN1 Starting Materials:
X
CH3
SN1 Starting Materials:
1
+
H3C
HO
Me
2
Me
O
CH3
CH3
Me
1
Me
2
HO
CH3
+
H3C
X
CH3
CH3