I.
(30 points) Name _______________________
Page 1
A. Reaction analysis: provide the requested information in each of the following chemical
transformations. Show all missing starting materials and products unless otherwise
indicated, and if any reaction results in a stereoisomeric mixture, draw one and indicate "+
enantiomer," "+ diastereomer," or "+ more than one stereoisomer" in the box.
(a)
(i) C12H16 (ii) The product shown forms with (circle
the best answer):
HO no other stereoisomers
KMnO4
its enantiomer
its diastereomer
H
OH with > 2 other stereoisomers
3 2
(b) (i) (ii) Major regioisomer only
H H 1) BH3
HO HO HCl Cl Cl
O
O O
2) H2O2/NaOH O O
+ enantiomer + enantiomer
either one may be drawn
either one may be drawn 4
4
(c) (i)
(ii) Draw the organic products with the info given:
2 equiv 1 equiv 1 equiv
1H-NMR: 1 peak 1H-NMR: 2 peaks 1H-NMR: 1 peak
NH3 1) O3
O O O O O
Na 2) (CH3)2S C C C C
H C H H H
H3C H
6
3
(d)
(ii)
(i)
right C does not have to be 3D O OR
O
O H H2
O O Pt
OR
draw one and "+ enantiomer" draw one and "+ enantiomer 4
4
II. (26 points) Name _______________________
Page 2
In the following reactions, an optically pure starting material is transformed using slightly
different reagents and/or conditions. Predict the product(s) for each transformation, using
the information given. Then consider how the changes might affect the rate of that same
reaction path.
(i) an optically inactive mixture containing two
(a) chiral compounds: C7H14OS;
draw each
O Cl CH3S Na
O (R) O (S) SCH3
CH3CH2OH
1 SCH3
4
assuming that the same reaction mechanism occurs (with identical or analogous product(s) forming)...
(ii) if 1 was changed to (iv) if
the reaction rate would
the reaction rate would Na (circle the best answer):
Cl (circle the best answer): CH3S
increase
increase decrease
was doubled in
decrease
concentration remain the same
remain the same
3 3
(iii) if 1 was changed to (v) if
the reaction rate would
O Br the reaction rate would (circle the best answer):
CH3CH2OH
(circle the best answer): increase
increase was changed to decrease
decrease remain the same
CH2Cl2
remain the same
3 3
(i) a single chiral product: C7H11NO
(b)
Cl O O
NaCN CN
H CH3CH2OH H
2
4
(ii) if (iii) if the reaction rate would
the reaction rate would
(circle the best answer): (circle the best answer):
CH3CH2OH NaCN
increase increase
was changed to was doubled in decrease
decrease concentration
remain the same remain the same
CH2Cl2
3 3
III. (30 points) Name _______________________
Page 3
A. DBN is a strong, bulky base that is often used to encourage elimination reactions over
substitution reactions. In both of the following cases, the resulting bimolecular elimination
reaction yields a single major product which can be explained by conformational favorability
in the starting material. Draw the product for each and draw the conformation that favors its
formation. (i) Using a chair, draw the conformation (ii) Selected product
that is favored and leads to the product
(a)
Cl Cl
N CH 3
N
Cl H
Cl
DBN
CH 3 1 equivalent CH 3
Cl H
Draw the conformation appropriately:
draw both axial and equatorial positions
for each ring atom that has a substituent. 5 3
(b) CH 3 (i) Using a Newman projection, draw the (ii) Selected product
O conformation that is favored and leads to
S the product
O OSO2CH 3
O N
N H
DBN H
H
View the C-C bond as indicated 5 3
B. Complete the following regioselective reactions by drawing the intermediate of highest
energy that forms during the reaction (best resonance contributor if applicable), and draw the
final product(s) as well. If either the intermediate or the regioselected product(s) forms as a
stereoisomeric mixture, draw one and write "+enantiomer" or "+diastereomer" in the box.
(a) (i) highest energy intermediate (ii) Final product(s)
predicted to form
Cl
HOCl
Cl
OH
no stereochem
no partial -1 if "+ enantiomer" 3 -1 if "+ enantiomer" 4
(i) highest energy intermediate (ii) Final product(s)
(b) predicted to form
O
H 2O O OH
O H 2SO 4 (cat.)
no partial -1 if "+ enantiomer" 3 + diastereomer 4
IV. (28 points) Name _______________________
Page 4
A. Consider the reaction scheme below, and provide the missing information. Note that
stereochemistry is not considered here.
(a) C8H16O
13C-NMR: 6 signals
HCl CH3OH
OH
heat OCH3
H2SO4 (cat.)
dehydration
Major product C9H18O
3 Electrophilic addition with rearrangement
(b) It has been proposed that the addition product forms as a result of several mechanistic
steps, including two carbocation shifts. Draw this (the most straightforward) curved arrow
mechanism for this transformation. You should choose the appropriate acid/conjugate base
pair for any proton transfers, and be sure to regenerate your acid.
H3C
1 2
O H
H
3
O H
4 H
H3C
O
CH3
O H 5
H3C
2 points/intermediate OCH3
2 points/mechanism (per step)
Grading stops when a chemically illogical step is shown
18
(c) The final product shown will form as (circle the best answer):
a single unique compound a racemic mixture a diastereomeric mixture a mixture of 3 or more
stereoisomers 2
B. The same starting material can be transformed to give a structural isomer of the compound
formed above with no evidence of rearrangement. Draw only the connectivity of the organic
product (C9H18O) that forms, and answer the question about how it forms.
C9H18O The final product here will form as
1) BH3 OCH3 (circle the best answer):
2) H2O2/NaOH a single unique compound
a racemic mixture
3) NaH
a diastereomeric mixture
4) CH3I
a mixture of 3 or more stereoisomers
3 2
V. (26 points) Name _______________________
Page 5
Predict the product(s) in each of the following reactions, using the information given.
(a) If either product forms as a stereoisomeric mixture, draw one and write "+ enantiomer" or "+ diasteromer".
HO OH
I I
I2 OsO4
-1pt
if "+ enantiomer"
"+ enantiomer" 3 3
(b) + other byproducts a single optically active
O substitution product
Cl
S 1) NaBr
H3C
O
2) NaOCH3
OH O
O
N S OCH3
3 O 4
CH3
both products have molecular formula C10H12
13C-NMR: 5 peaks 13C-NMR: 10 peaks
(c)
I NaOCH2CH3
+
3 3
(d) Provide all the missing information for the following transformation. Each reaction should be balanced.
O
O
O O S CH3 O
O O S CH3
O O
O O
CH3CH2CH2CH2Li O
C C
C C C Li
CH Li
3
+ 4
CH3CH2CH2CH3
