Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 1) For the following molecule, 1) find all the stereocenters, 2) provide R or S labels for each, 3) draw
the enantiomer of the molecule (and give R and S designations), and 4) draw at least one
diastereomer.
2) Consider the molecule represented by the name (S)-4,5-dimethyl-(Z)-2-hexene.
Draw an unambiguous structural representation:
a. of this molecule.
b. of one enantiomer of this molecule, if it exists.
c. of one diastereomer of this molecule, if it exists.
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 3) Determine how many distinct stereoisomers each of the following compounds has. Circle and
assign configurations to all stereogenic centers for each.
The first molecule has 2 stereocenters, yielding 4 different chiral isomers. I included the
relationships among them below.
The second molecule also has 2 stereocenters. In theory, there are 4 possible stereoisomers, but since
one of them is meso, the (R,S) and (S,R) isomers are actually the same molecule. So there are only three
stereoisomers.
HOOC
COOH
HOOC
(R, R)
COOH
(R, S)
HOOC
COOH
(S, S)
meso
4) For each of the following molecules:
a) Find and circle any stereocenters in the molecule.
b) Indicate if the molecule is chiral or achiral by circling the appropriate term.
c) For those molecules that are chiral, draw the structure of the enantiomer.
Cl
Cl
achiral (meso)
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab Cl
chiral – flip both stereocenters for enantiomer
achiral
achiral (meso)
Note: the carbon circled in green is a “stereocenter,” because switching two groups yields a different
stereoisomer. However, it is not an asymmetric carbon, because two of the groups it is bonded to are
the same. It’s not a distinction we’ve stressed, so don’t sweat it.
5) Identify the stereochemical relationship between each of the following pairs of molecules. (i.e. label
each pair as enantiomers, diastereomers, or same molecule.)
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 6) For each of the following reactions, provide all of the expected major organic products. Where
there is a choice, include only the major regioisomeric product and show all stereoisomers that you
would expect to be formed.
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 7) For each of the following reactions, provide the missing reactant or product. Be sure to indicate
stereochemistry if it is relevant!
a.
Br
CH3
Br2
H
H
(+/-)
H3C
Br
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab d.
8) Draw the lowest- and the highest-energy chair conformations of the following cyclohexane
derivatives.
Me
Me
Me
Note: There are many other ways to draw the correct answer for these.
Answer to the supplemental problem (see review slides):
Br
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 9) Provide the major organic product(s) you would expect from each of the following reactions. In each
case, be sure to identify the mechanism by which each proceeds, and take stereochemistry into account
wherever it is relevant.
a) For both (a) and (b), note that the methyl group as a wedge on the left side is just a distractor. That’s
not a stereocenter, so we can ignore the wedge. For part (a), we have an SN2 reaction, which will invert
stereochemistry.
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab b) For this one (poor sterics + strong base), we’re going to get an E2 elimination. We should predict
that the Zaitsev product will be the major product. If this were a test question, that’s what we’d expect
you to put in the box. Note that we need to account for the anti stereochemistry of the H and the I in the
transition state. In this case, since there are two -hydrogens, we can make either the cis or the trans
product. Since the trans product is more stable, we expect it to be the major product. Recognize also
that with a bulky base like tert-butoxide, we may get a significant amount of the Hofmann (lesssubstituted) product as well, but it will be minor.
I showed mechanisms for the formation of all three products below. The little curly arrow in the first
step is just to indicate that I’m rotating around that C-C bond to get a better perspective.
H
I
H
H
t
H
H
H
BuO
Zaitsez (major)
t
KO Bu
H
Hofmann (minor)
H
+
I
H
t
BuOH
H
H
H
I
Zaitsev (minor)
t
H
BuO
H
H
H
I
H
H
t
H
BuO
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab c) Note that since we have a very poor nucleophile (acetic acid), SN2 is not an option. Since we don’t
have a strong base, E2 is not an option. So we’re left with SN1 or E1. In that case, SN1 will usually
dominate, unless it’s an alcohol with conc. H2SO4 or H3PO4, which it’s not.
CH3COOH
O
O
O
O
Br
Attack f rom either face
of carbocation
O
O
O
O
O
H
H
Sol
Sol
OH
Note: The lone pair f rom the double-bonded oxygen
is higher in energy than the one on the other oxygen
due to the contribution f rom the alternate res. structure:
O
OH
10) Put the three compounds below in order of their expected rate of reaction with the strong base NaOEt.
For each one, predict the major product.
In order to have any H’s lined up anti to the leaving group, the LG must be in the axial position. So I’ve
drawn each molecule in the chair with Cl axial.
A
B
C
CH3
CH3
CH3
Cl
Cl
Cl
CH3
CH3
H
H
CH3
H
CH3
CH3
Cl
Will react slowly: Only one
anti H, and it must have Me
group in axial position. Since
the molecule spends relatively
little time in this conf ormation,
reaction will be slow.
Cl
Will not react: can't do E2
because no anti H's
Cl
Will react most quickly: Two
options f or anti H's, and this
conf ormation is about equally
stable with the other chair, so
the molecule spends lots of
time in this conf ormation.
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 11) Put the four compounds below in order of increasing rate of reaction with water in an SN1 substitution
reaction.
slowest = B < C < A < D
For SN1, the rate-limiting step is the formation of the carbocation. So we need to compare the stability of
the carbocations that would be formed in each case.
Review 2 Weekly Review Lecture ‐ Answers 12) Propose a mechanism for the following transformation.
Chem S‐20ab Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 13) Predict the stereochemistry of the following product. Is it a single diastereomer, enantiomerically pure,
racemic, or something else? Draw a complete mechanism.
Note: The stereochem is S on the lef t, R on the right to begin with.
OTs
H2S
OTs
H3C
achiral (meso)
KOH
CH3
H3C
S
CH3
redrawing
SH
H
OTs
OH
S
S
The stereochem on the lef t is now R.
H3C
CH3
The stereochem on the right is now S.
OTs
S
OTs
H3C
CH3
redrawing
redrawing
S
CH3
OTs
S
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 14) The following reactions are very similar, yet proceed with completely different stereochemical results.
Provide curved-arrow mechanisms that account for these results.
S
Cl
Cl
O
Ph
OH
Ph
Ph
Ph
OtBu
H and Cl must be
anti. This leads to
the (E) product.
pyr
H
H
Ph
O
Cl
Ph
S
Ph
Cl
Ph
O
rotate
O
Cl
Ph
Stereochem is inverted.
S
Cl
Ph
Ph
O
Ph
Cl
O
O
S
R
Cl
OH
Ph
Ph
Ph
pyr
H
Ph
OtBu
H and OTs must be
anti. This leads to
the (Z) product.
H
H
O
R
Ph
S
Ph
Ph
O
OTs
O
Ph
rotate
OTs
Ph
Ph
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab 15) Provide a complete curved-arrow mechanism that accounts for the stereochemical outcome of the
following reaction.
This is essentially an SN1 mechanism with a resonance stabilized carbocation intermediate. The Cl next
to O gets replaced first because of the stabilization.
16) Propose a synthetic route for each of the following transformations.
a.
Review 2 b.
Weekly Review Lecture ‐ Answers Chem S‐20ab c. I’m looking at this problem now and thinking, “My bad.” You could actually carry out this synthesis in a
pretty straightforward way as shown below. However, at this point, I don’t think we’d expect you to
isomerize the alkene just by adding H2SO4. So this problem would be unfair. Maybe if I modified it as
shown below, it would be better.
Answer to problem as given in packet:
Modified problem:
Answer to modified problem:
Notice that in this case, we have a clear difference between the two carbons of the double bond (one is
secondary, one tertiary). This makes it more obvious what the results of each step will be.
Review 2 Weekly Review Lecture ‐ Answers d. Two equally good pathways:
Chem S‐20ab e.
3
aN
N
f.
MeO
CN
Br2, MeOH
NaCN
MeO
OMe
rotate
Only included to show perspective,
not a required step.
Br
(+/-)
g.
Br
Review 2 Weekly Review Lecture ‐ Answers Chem S‐20ab EtO
as a mixture of
diastereomers
Br2, EtOH
CH2I2, Zn-Cu
EtO
EtO
KOtBu
Br
(+/-)
h.
i.
(+/-)
Review 2 Weekly Review Lecture ‐ Answers j.
H
(+/-)
H
SMe
1. BH3, THF
2. H2O2, NaOH
NaSMe
H
H
TsCl, pyr
HO
(+/-)
H
TsO
(+/-)
H
Chem S‐20ab