KEY
Massachusetts Institute of Technology
5.13: Organic Chemistry II
December 19, 2005
Final Exam
Question 1
__________/10 points
Question 2
__________/15 points
Question 3
__________/30 points
Question 4
__________/10 points
Question 5
__________/10 points
Question 6
__________/15 points
Question 7
__________/10 points
Question 8
__________/12 points
Question 9
__________/10 points
Question 10
__________/12 points
Question 11
__________/12 points
Question 12
__________/12 points
Question 13
__________/12 points
Question 14
__________/14 points
Question 15
__________/16 points
TOTAL
_________/200 points
Name (printed) ________________________________
Name (signed) __________________________________
T.A
_____________________
There are 18 pages (2-19) of questions in this exam.
1
1.
(10 points total) Write an arrow-pushing mechanism for the reaction below.
Note: Aste risk(*)=13C.
*
O
N
H
Me
+
H
H
O
heat
Me
O
OH
N
H
O
O
R2N+H2
+
*
O
H
O
+
Me
O
H * H
OH H
N
H
CO2
IMINIUM FORMATION +3
*
OH2
H
N
Me
CO2
*
Aza-cope +4
O
N
Me
O
Me
O H
*
H * H
H O
N
Me
*
CO2
CO2
N
Me
H2O
HYDROLYSIS +3
N
O
O
H
H2O
*
N
Me
CO2
13
Solution must account for C in formaldyde, otherwise no more than 5 pts should be awarded.
2
Figure by MIT OCW.
2. (15 points total) Compound A is prepared from B and C and has the spectroscopic data listed
below. Draw the structure of A in the box provided, and write an arrow-pushing mechanism for
its formation from B and C in the space below.
2. points 2. points
Ph
OH
O
+
N
N
+
O Catalytic H
Heat
CH3
B
O
O
N
N
A
C
N
N
1. point
OH
Ph
O
H
N
O
Ph
N
O
H
O
Partial credit for another
mechanism leading to the
right molecule that doesn't
include [3.3] sigmatropic
REARR: [0-3 points].
∆
O
N
N
7
O
2. points
Data for A:
1H NMR (ppm)
IR
7.05-7.15, m, 5H
1685 cm-1
5.80, t, J = 6.3, 1H
3.67, t, J = 6.5, 4H
Molecular weight
3.47. t , J = 6.5, 4H
3.22, d, J = 6.3, 2H
2.34, t, J = 7.4, 2H
273.17
2.12, t, J = 7.4, 2H
1.71, s, 3H
O
O
O
Ph
N
O
Ph
O
MECHANISM 8
Ph
Claisen Rearr
O
N
O
Ph
N
Ph
O
O
3
Figure by MIT OCW.
3. (30 points total, 1 point per box) For the following 15 structures, write the number of
chemically non-equivalent (number of “different types”) of hydrogens and carbons in the
appropriate boxes below. (Be careful to put the numbers in the correct boxes – we can’t read your
mind, i.e. wrong numbers will receive no credit – no exceptions.)
# non-equivalent H
# non-equivalent C
CH3
a.
2
All
or
nothing
3
CH3
Cl
Br
b.
c.
CH3
CH2
CH3
d.
CH3
CH2
CH2
CH3
Me
6
2
2
2
2
2
3
1
Me
2
e.
4
Me
3
Me
Me Me
Figure by MIT OCW.
4
# non-equivalent H
f.
Me
Me
Me
3
3
Me
HO
2
Me
1
6 Me
H
7
3
3
6
7
3
4
5
7
5
7
O
Me
2
i.
2
1
Me
g.
h.
Me 1
# non-equivalent C
4
5
3
4
5
6
H
3 2
4
Me
7
H
Me
1
2
1
j.
6 5
7
3
4
Figure by MIT OCW.
5
# non-equivalent H
O
k.
2
O
N
Me
H
3
Me
Me
N
3
(DIASTEREOTOPICITY)
O
m.
2
restricted rotation
O
i.
(3 was
accepted)
# non-equivalent C
O
9
8
Me
Me
(Me's are diastereotopic)
O
n.
Me
6
4
1
1
Me
o.
(10e-,
aromatic)
Figure by MIT OCW.
6
4. (10 points) An alcohol (R-OH) was treated with sodium hydride and 1-bromo-2-butyne to give
compound D (molecular weight = 166.10). Using the 1H NMR data listed below, determine the
structure of the product and the starting alcohol. Draw the structures in the boxes provided.
1. NaH, THF
2. 1-bromo-2-butyne
1
chirality 2pts.
R-OH
OH
5.68, ddd, J= 17.0, 10.5, 8.5, 1H
5.27, dd, J=10.5, 1.5, 1H
5.19, dd, J=17.0, 1.5, 1H
4.14, d, J=15.0, 1H
3.94, d, J=15.0, 1H
3.42, d, J=8.5, 1H
1.86, s, 3H
0.91, s, 9H
Me
O
draw D here
draw R-OH here
H NMR data for D (ppm)
= From ROH
= From ALKYNE
Correct ether synthesis +2 pts.
CH3
Br
+
Me
RO
RO
H H
RO
Has : 9 H S 2,(tBu) 2pts.
2pts. 3 H IN Alkene region, All Coupled to each other
2H, S OR 2 1H IF R IS CHIRAL
WHICH IT IS IN THIS CASE
H
H
H
2pts. Add'l 1H is coupled to 1 Alkene H
H
H
H
H
Z
Y
Y and Z have no H directly attached.
3.42ppm
O
CH3
ONLY tBr and O Remain
Figure by MIT OCW.
7
5. (10 points) At room temperature, compound E is converted to compound F in high yield. Using
the data provided, determine the structure of F (and draw the structure in the box provided), and
write an arrow-pushing mechanism for its formation from E.
H2N
N
O
S
NO2
Me
O
23 0C
Me
O
Ph
or
E
H
N
H
Ph
SO2
N
Data for F :
+3
Me
Ph
Ph
coupled to *
+ 4 if not concerted
but correct arrow
pushing over all
O
Me
IR
1H NMR (ppm)
1955 CM
7.15-7.30, m, 5H
5.99, d, J = 2.0, 1H
5.25, dq, J =7.0, 2.0, 1H
1.59, d, J =7.0, 3H
Molecular
weight
+ 2 for
allene
-1
allene
130.19
CH3
C10H10 = MW
130
+ HN=N-SO2Br
Me
Ph
CORRECT
ARROW PUSHING AND STRUCTURE =10 pts
1 OF ABOVE CORRECT: 5pts
PARTIAL CREDIT FOR PARTIALLY CORRECT MECH AND/OR STRUCTURE BOTH OK.
Figure by MIT OCW.
8
6. (15 points) Propose a synthesis of G from H, maleic anhydride, and benzyl bromide (BnBr =
PhCH2Br). (All of the substituents on the five-membered ring in G are cis to one another, and your
synthesis must establish this relative configuration.) Your synthesis must use H, maleic anhydride,
and BnBr. You may use any other reagents in addition to these. Write your synthesis neatly in the
forward direction, and for each transformation, write the reagents necessary over the arrow.
BnO
BnO
OBn
OBn
BnO
BnO
OO
HH
BnO
BnO
OBn
OBn
OO
OO
OO
+3
+3pts
pts
HH
HH
HH
OBn
OBn
OHC
OHC
OO
+2
+2pts
pts
OO
OO
+2
+2pts
pts
HH
OO
maleic
maleicanhydride
anhydride
+6
+6pts
pts
HH
endo
endo
HH
2
HH
away
awayfrom
fromCH
CH2OBn
2OBn
OH
OH
(1)
(1)OO3 3
(2)
(2)Me
Me2SS
OO
OBn
OBn
OBn
OBn
G
G
(all
(allcis)
cis)
OO
BnO
BnO
LiAlH
LiAlH4 4
CHO
CHO
OO
All
CIS
AuH's are
H' s
CiS
OO
OBn
OBn
HO
HO
OH
OH
OH
OH
OH
OH
MANY POSSIBLE CORRECT SEQUENCES AFTER
MANY POSSIBLE CORRECT SEQUENCES AFTER DIELS
DIELS - ALDER
+2
+2pts
pts
NaH
NaH
BnBr
BnBr
GG
(x.s. of each)
9
Figure by MIT OCW.
(7) (2 points for each box; 10 points total) Please provide the indicated information. If you use a
base or an acid, please specify whether a “catalytic amount”, “1 equivalent”, etc. is required.
cat. HO
(a)
Ph
Ph
CH3
Ph
CH3
O
(c)
N OH
(d)
H
O
Ph
cat. HO
or RO
+
or H
H
O
Me
OH
O
Ph
O
CH2N2
O
H2SO4
NH
H 2O
(e)
O
Ph
O
O
(b)
or RO
O
O
O
MCPBA
O
Figure by MIT OCW.
10
(8) (12 points) Please provide an efficient synthesis of the indicated target compound. All of the
carbon of the target compound must come from methyl acetate.
O
O
O
Me OMe
methyl acetate
Me Me
Target compound
Various routes were possible and partial credit was given depending on efficiency.
Below are the most common:
O
1 eq MeO
OMe MeOH
H
O O
+
OMe
O-
LDA
(B)
Route 2:
O
Route3:
(A)
OMe
D
A
O-
cat. MeO
OMe
OMe
OMe
H
O (C)
∆
(-H2O)
O- (A)
OMe
O (C)
O
(A)
H
H
+
O
OMe
OMe
1 eq MeO
2. H+, H2O
∆
O
O
+
O
OMe
O
(D)
O-
C
O O
O O
O
∆
(-H2O)
H2O
∆
B
OMe
Route 1:
H
cat. H+
or OH-
O
+
O
+
(C)
O
1 eq MeO
2. H
+
cat MeO
MeO
O
O O
1 eq MeO,
+
2. H , H2O, ∆
(B) OMe
Figure by MIT OCW.
11
(9) (10 points) The Strecker reaction, followed by a hydrolysis reaction, is an excellent method for
synthesizing amino acids, which are the building blocks of proteins. Provide the best mechanism
for this process. Please show all arrow pushing. Note: You do NOT have to draw the mechanism
for the hydrolysis reaction.
O
R
HCN
H CN
O
R
H
NH3
Strecker
H2N CN
reaction
R
H
HO
O H
R
H
hydrolysis H2N CO2H
R
NH3
R
H
H2N
CN
R
H
H
H+
transfer
H
H2O NH2
R
H
NH3
2pts for each step !
NH2
R
H
CN
Figure by MIT OCW.
12
(10) (12 points) Provide the structure of A and the best mechanism for both of the illustrated
transformations. Please show all arrow pushing.
HO OH
H
1) LiAlH4
2) H
H
A
C8H12O
+
HO OH
HO OH2
O
OH2
H
O
O
O
AlH3Li
H
workup
OH
H
H2O
H
OH
H
4
POINTS
+
2
POINTS
A
HO
H
+
H
OH2
OH2
H
+
+
4
POINTS
H2O
* FULL CREDIT FOR ALL STEPS AND CLEAN, THROUGH MECHANISM
Figure by MIT OCW.
13
Bu3SnI
I
Figure by MIT OCW.
14
(12) (12 points) Provide the best mechanism for the illustrated process. Please show all arrow pushing. Your
mechanism should rationalize why the reaction proceeds with complete retention of stereochemistry.
O
O
HO
HO
Me PT
NH2
N
O
HO
HO
HO
(2) H
Me
N
Me
Cl
N O
H2O N O
O
Me (1) PT
NH
N
O
O
H+
HO N O
O
HO
NaNO2 HCl
NH2
H
Na -O N O
O
Me
O
HO
Me
+
N2
N OH2
4
inversion
inversion
Me
+4
Cl
Overall retention
H O
O
Me
+
4
H
Cl
Neighboring Group Participation
O
Me
+ 4 pts formation of HO
O
+ 10 pts if via HO
N2
Me
NH3
+8 pts if double inversion mechanism involving NO2 as nucleophile
Figure by MIT OCW.
15
(13) (10 points) Please provide a detailed mechanism for the illustrated transformation. Show all arrow
pushing. (Bn = CH2Ph)
Hint #1: Number your carbons! Hint #2: PhSH is catalystic!
OBn
O
+
O
cat. AIBN, ∆
OMe
OBn
MeO
cat. PhS-H
Initiation:
NC
N
∆
N
CN
CN
+ N2
2
CN
+
CN
H
PhS
H
+ PhS
Propagation:
OBn
PhS
OBn
PhS
OBn
PhS
O
+
MeO
OBn
PhS
OBn
PhS
O
MeO
O
OMe
OBn
O
MeO
+ PhS
Figure by MIT OCW.
16
(14) (10 points) Compound A is converted to B, C, and D upon heating. The reaction is accelerated by
irradiation. Provide the structures of B, C, and D, and provide the mechanisms by which they are formed
(please show all arrow pushing).
Me
O
Me Me
A
Cl
∆
C4H9Cl
B
C3H6O
C
+
+
C7H15ClO
D
O
+ Cl
O
+
C
O
Cl
Cl
+
O
B
H
OH
O
O
Cl
Cl
OH
+
O
D
Figure by MIT OCW.
17
(15) (16 points total) In an amazing process, Nature transforms squalene oxide into steroids (as a single
stereoisomer!). For each of the process illustrated below, provide the best mechanism. Please show all arrow
pushing.
Me
squalene oxide
Me
Me
Me
Me
O
Me
Me
Me
H2O H
Me
Me
Me
Me Me
Intermediate l
HO
Me
Me Me
Me
Me
Me
Me
Me
Ianosterol, a steroid
HO
Me
Me Me
(a). (10 points) Squalene oxide into intermediate l:
Me
Me
Me
Me
Me
O
Me
Me
Me
H
Me
Me
Me
Me
Me
HO
Me
Me
Me
Me
Me
Me Me
Me
Me
HO
* or Show Stepwise C
HO
Me
Me
Me
Me
Me
Me
Me Me
Me
Charges
Figure by MIT OCW.
18
+2 for each step: Protonation opening of epoxide each cation π − Cyclization
* if no errors
-2 if show wrong connectivity after cation π − Cyclic.
(b) (6 points) Intermediate 1 into lanosterol:
Me
Me
+2 for deprotonation
H
H
Me
+1 for every 1,2-shift
Me Me
HO
Me
Me Me
H2O
H
Me
HO
Me H
Me
Me
Me
Me
Me
Me
HO
Me
Me
Me
Me
Me Me
Me Me
* or show stepwise C
charges !
-1 if show deprotonation in same step at alkyl shift.
-1 if show formation of double bond s deprotonation, but show arrow
-2 for every 1,3 shift
-1 for every 2 missing mech. arrows
Figure by MIT OCW.
19