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Total Exams Organic Chemistry I,II in MIT
Citation preview
TUYN TP
THI HA HU C
MIT
Massachusetts Institute of Technology 5.12, Spring 2005 Dr. Kimberly L. Berkowski Organic Chemistry
PRACTICE EXAM IIa
Books, notes, and calculators will not be allowed in the exam room.
Molecular model kits will be allowed during the exam.
You will be given a periodic table.
The exam will focus on Chapters 5-8 in McMurry as well as all topics covered in lecture and any previous material.
This practice exam is longer than the real exam.
HAVE FUN!
Tim JamisonText BoxActually, not at all since Me groupswould have to leave as methyl cations. However, does react quickly withcertain "E+" (electrophile) to form a semi-stable ionic species: [(CH3)6(Ph)E]+
Massachusetts Institute of Technology Organic Chemistry 5.13
Friday, September 29, 2006 Prof. Timothy F. Jamison
Hour Exam #1
Name _____________________________________________________
(Please both print and sign your name) Official Recitation Instructor __________________________________ Directions: Closed book exam, no books, notebooks, notes, etc. allowed. However, calculators, rulers, and molecular model sets are permitted. Please read through the entire exam before beginning, in order to make sure that you have all the pages and in order to gauge the relative difficulty of each question. Budget your time accordingly. Show all your work if you wish to receive partial credit. You should have 14 pages total: 8 exam pages including this page, 4 pages of reference information, and 2 blank pages for scratchwork. Question: Grader: 1. _________/ 10 points _________ 2. _________/ 25 points _________ 3. _________/ 25 points _________ 4. _________/ 25 points _________ 5. _________/ 15 points _________ Total: __________/ 100 points _________
1
KEY
1. (10 points total, 2 points each) For each set of compounds below, circle the one in which the indicated hydrogen is the furthest upfield in a 1H NMR spectrum.
2
H2C = CH2
O
O
HO
O
H H
O
CH3CH3CH4 (CH3)2CH2 (CH3)3CH
CH3OCH3 CH3OCH2OCH3 O
O
H3CCH3
CH3Cl CH4CHCl3 CH2Cl2
HH HH
NO2 OMeNMe26.77 6.59
Both Awarded Full Credit
A.
B.
C.
D.
E.
Figure by MIT OCW.
2. (25 points total) Answer the questions below about the structure that has the following data: EA C, 66.62; H, 11.18; N, 22.20 MW (g/mol) 126.20 13C NMR (ppm) 140.3, 48.0, 24.7 IR (cm-1) 2116 (strong more intense than the C-H stretches 1500 a1H NMR spectrum:
between 2800 and 3100); no other peaks between nd 4000.
11 10 9 8 7 6 5 4 3 2 1
ppm0
N = C = N
1H, J=6.9 HZ
6H, J=6.9 HZ
Figure by MIT OCW.
a. (3 points) To what structural fragment does the signature splitting pattern in the 1H NMR correspond? Circle your final answer.
b. (2 points) Which peak or peaks in the 13C NMR correspond(s) to the fragment you identical in a, above. List the chemical shift(s) of the peak(s), and circle your final answer(s).
CH
CH3 CH3
CH3
or = c-Pr etc.
Figure by MIT OCW.
49.0, 24.7
Figure by MIT OCW.
3
c. (5 points) Determine the molecular formula of this compound. Circle your final answer.
d. (5 points) Calculate the Index of Hydrogen Deficiency (IHD) of this unknown compound. Circle your final answer.
e. (10 points) Draw the structure of the unknown compound. Circle your final answer.
C7H14N2
2
N=C=N
N NC (Also full credit)
Figures by MIT OCW.
4
3. (25 points total) Answer the questions below about the structure that has MW = 107 and the following NMR spectra:
5
012345678910
2 H, J=7.9 HZ
1 H, J=7.9 HZ
PPM
6 H,s
200 180 160 140 120 100 80 60 40 20 0PPM
Figure by MIT OCW.
a. (10 points) Determine the molecular formula of this compound. Circle your final answer.
b. (5 points) Calculate the Index of Hydrogen Deficiency (IHD) of this compound. Circle your final answer.
c. (10 points) Draw the structure of the unknown compound. Circle your final answer.
C7H9N
.107 - ODD # OF N
.CONSIDER 1 N:
.107 -14 = 9393/13 = 7 + 2/13 => C7H9N
4
MeMe N
Figures by MIT OCW.
6
4. (25 points total) An unknown compound (X) contains only carbon and hydrogen, has MW = 112, and exhibits the spectral data below. In addition to the IR signal listed below, there are only peaks corresponding to C-H stretches (between 3300 and 2900) and several peaks in the fingerprint region. Please note that there are no overlapping peaks in either the 1H NMR or the 13C NMR spectra. In other words, what you see is all there is! IR (cm-1) 2145 13C NMR (ppm) 77.8, 70.1, 30.2 1H NMR (ppm) 2.45 (s) When compound X was treated with excess n-Buli (n-butyllithium) in tetrahydrofuran and then excess CH3I (iodomethane), a new compound (Y) with MW = 168 and 4 signals in its 13C NMR spectrum was formed. What are the structures of X (15 points) and Y (10 points)? (Show your work in the space below for partial credit consideration.) Write your final answers in the boxes provided below.
C
C
C C
CC
HH
CH
CC H
X
Y
1. 4 n-BuLi
2. 4 CH3I
CCC CH3( )4
1. n-BuLi (excess), THF2. CH3I(excess)
X (MW=112) Y (MW=168)
Figure by MIT OCW.
5. (15 points) In one of our problem sets, cubane (C8H8) was one of the possible answers to a structure elucidation problem. Based on the formula for the Index of Hydrogen Deficiency, the IDH of cubane is 5. However, as you know, a cube has six sides. In other words, it looks like cubane has 6 rings and thus that its IHD should also be 6.
Please provide an explanation (not the formula used to calculate the IHD) for this apparent discrepancy in the spaces below.
8
1
2
3
4 5
H H
HHH
H
H
H
H3C
H3C
CH2
CH3
C8H8
C8H8
H2 543
21
CHEMICALY, CAN DO 5 "HYDROGENATION" RXNS TO OBTAIN "SATURATED" (ACYCLIC, NO T BOND) BOND :
IHD=5
IHD=0REDRAW WITHDISTORTED C-C
BONDS, VOILACUBANE IS IN
FACT A PENTACYCLIC MOLECULES !
1 1
2
3
5
4
1
2
11
3
4
5
6
7
9
8
1012
ALSO, CAN DRAW AS CUBE (BONDS 1-12,ORDER 1 AT LEFT) AND SEE THAT DRAWING 5 RING GIVES CUBANE.
Figure by MIT OCW.
Massachusetts Institute of Technology Organic Chemistry 5.13
Wednesday, October 25, 2006 Prof. Timothy F. Jamison
Hour Exam #2
Name _____________________________________________ (Please both print and sign your name) Official Recitation Instructor _________________________________ Directions: Closed book exam, no books, notebooks, notes, etc. allowed. Calculators are not permitted for the exam. However, rulers, and molecular model sets are permitted. Please read through the entire exam before beginning, in order to make sure that you have all the pages and in order to gauge the relative difficulty of each question. Budget your time accordingly. Show all your work if you wish to receive partial credit. You should have 8 pages total: 6 exam pages including this page and 2 blank pages for scratchwork. Question: Grader: 1. _________/ 14 points (page 2) _________ 2. _________/ 16 points (page 3) _________ 3. _________/ 48 points _________ 4. _________/ 22 points Total: __________/ 100 points _________
1
KEY
1. (30 points total, 2 points per box) In each box below, draw the structure of the major product of the reaction. Indicate relative stereochemistry where appropriate. If no reaction occurs, put a large X in the box. (Note: D = deuterium, 2H)
H3C
CH3
H3CO
CH3
CH3
CH3
CH3
O
O
O
O
O
O
O
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me
CO2Me
O3
(CH3)2S
H
CHO
OH
OH
OH
NaBH4
(R)
HO
OE+
H H
H MgBrEtOH
H+
(S) (S)
OPh
H3C+
D
D
D
D
a.
b.
c.
d.
e.
f. g.
HH
+
BF3
O
O
Figure by MIT OCW.
2
(1., continued see previous page for directions)
3
H
H H
H H
H
H
H H
H
MeMe
Me
Me
C=N=O
O
OO
C
O
OMe
Me
Me
Me
Ph
Ph
Ph
SH 1.NaOHNaOH
H3C
CH3
CH3
CH3CH3
CH3
CH3
CH3
CH3
H3CH3C
Me
Me MeOPhCH2OH,
Hg(OAc)2
NaBH4
1.
2.
[3,3]COPE
2S + 2S
h
Br
Br
2. excess EtBr
S Na
Must have "x" in boxNo Reaction
or
h. i.
j. k.
l. m.
n. o.
ON
S
(or (IS, cis)
Figure by MIT OCW.
2. (48 points total) a. Draw the orbitals (by shading the lobes appropriately) at each energy level for
1,3,5- hexatriene (2 points each). b. Write the number of nodes in the box to the left of each orbital array (1 point each). c. For the ground state of 1,3,5-hexatriene, draw the electron population for each
orbital on the line to the right of each orbital array. For each electron, clearly indicate whether it is spin up or spin down. If there are no electrons given orbital, leave it blank (1 point each).
electron # of nodes orbitals population
1 point per box 2 points per orbital array 1 point each
5
4
3
2
1
Orbitals# of Nodes
0
or
1 point per box 2 points per orbital array 1 point each
Electron population
E
Figure by MIT OCW.
4
2. (continued) d. For each reaction shown below, indicates which energy level is used to predict the
stereochemical outcome by shading the appropriate lobes of the entire orbital array. (The methyl groups are omitted for clarity; you do not have to draw them.)
e. In the box under each reaction arrow, write conrotatory or disrotatory, as appropriate. f. In the box to the right of each reaction arrow, draw the major product of the reaction,
clearly indicating the relative stereochemistry.
5
MeMe
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me(trans)
(trans)
(cis)
(cis)
heat
heat
h
h
Disrotatory
Disrotatory
Conrotatory
Conrotatory
(Shade appropriate lobes) (write "conrotatory" or disrotatory") (draw major product- show stereochemistry)
2 points each 1 points each 3 points each
Figure by MIT OCW.
(shade appropriate lobes) (write conrotatory or disrotatory) (draw major product show stereochemistry) 2 points each 1 point per box 3 points per box
3. (22 points total) Using retrosynthetic analysis, propose a synthesis of the molecule to the right (A). You may use any reagents you wish, as long as your starting materials and any other reagent that is used to install a carbon that is found in the final product (target molecule A) have no more than 6 carbon atoms. For example, 1,3-butadiene and benzene would be
OH MeH
H MeO
target molecule (A) acceptable, but benzyl bromide (PhCH2Br) would not be. Write your synthesis in the forward direction, showing all Steps and reagents necessary. (You may include solvents, but you are not required to do so.) Draw a box around or circle Your final synthesis. Hint: Use a Diels-Alder reaction.
+
+
HO
HO
HO
HO
HO
H
HH
H
H
H
H
HO
O O
O
O
O O
O
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me
Me
CH3
Product
IntramolecularOxy-mercuration
1.Hg(OAc)2
LiAlH4
Diels- Alter
Forward Synthesis:
Target
(endo)
O
O O
O
2. NaBH4
Figure by MIT OCW.
6
O
First Three Letters of Last Name: TA Name: Hour Exam #3 5.13 Fall 2006
Organic Chemistry II
November 15, 2006
Name______________________________________________________________
Signature___________________________________________________________
ID#________________________________________________________________
1. Make sure your exam has 9 numbered pages plus a periodic table.
2. Write your initials on each page.
3. Look over the entire exam before you begin to familiarize yourself with its
length. Do what you know first, then attempt the harder problems.
4. Show all of your work. Partial credit receives points!
KEY
NN
N
H3C CH3
CH3 CH3
CH3
H
N
NH3C CH3
N
NH3C CH3
H+H+
Hconjugate acidnot formed
1 2 34
O
Me
O
OMe Cl
O
Me SCH3
O
Me
The Sp3 Nitrogen is actually Sp2. The lone pair on nitrogen is in a P orbital &can delocalize into the pyridine ring. The lone pair of electrons on the Nitrogen atom in the ring are orthogonal to the ring & cannot delocalize .Those electrons are more available for bonding and more basic.
2. (4 pts) Rank the following molecules in order of electrophilicty(1= most electrophilic )
1. (4 pts) When N,N-dimethylaminopyridine reacts with one equivalent of acid,the sp2 nitrogen becomes protonated.Why don't you see protonation at the sp3 nitrogen when you know that the more p character an orbital has, the more stable it is with a positive charge?
Figure by MIT OCW.
3. (18 pts) Provide the missing products for each reaction. Indicate no reaction with N.R.
H3C H3C
O
HH
+ HNMe2NaBH3CN
H+
Me Me
+:N(Me)3
1.CH3l(excess)2.Ag2O3. D
1.CH3l(excess)2.Ag2O3. D
NH3C CH3
NCH3
H3C
ONR
NH2NaBH4
CH3
CH3
CH3H3C
H3C NH2O2
D
(d)
(c)
(b)
(a)
(e)1. HNMe2,pyridine2.LiAlH43. H2OH3C
O
Cl+ H3C N
CH3
CH3
N
2Figure by MIT OCW.
4. (21 pts) Provide the missing reagents for each reaction. Several steps may be needed for some transformations.
3
O
O
OH
OH
NH2
NH2
H3C
H3C
H3C
NH2
O
H3C
O
Pr OEt
O
H3C
O
Br H3C
N
O
O
NH2
NH2
or
1.
2. H2N
K+
1. NaN32. LiAlH43. H2O
1.SOCl22. NH33. NaOH, Br2
1. H3O+/H2O, D2. EtLi (2 eq.)
CH3 H3CCH3
or 1. POCl3 2.EtMgBr 3. H3O+/H2O
MeMgBr 2 eq. Pr Me
MeHO
H
1.H2N OH,H+
2. LiAlH43. H2O
(b)
(c)
(d)
(e)
(f)
or 1.LiAlH4 2. PCC 1. SOCl2
2.LiAl(OtBu)3H
H3C NH2
1.LiAlH42. H2O
H3C C N(a)
Figure by MIT OCW.
5. (1 2 pts) Consider the labeling experiment outlined below. What level of 18O incorporation do you expect in the recovered anhydride (high or low)? Your answer should include a mechanism of hydrolysis and a detailed explanation.
O
O O
MeMe
O
O O
O
OHO O
O
OO OHk1
k-1 k1
k-1OH
O
O
+
+O
O+
O
O O Pt
OH
O
k2
OH
O
k2
OH/H2O stop at 50% conversion of acetic anhydride
18O
18O
Incorporation is expected to be low in recovered anhydride. is a better leaving group than
OH.
O
OThere fore, k2 >> k-1, or elimination of faster than revesal to starting material. Most
will be found in the acetic acid product.
O
O
of the
*
*
* *
* **
*
4
Figure by MIT OCW.
6. (10 pts) Provide a mechanism for the following transformation.
Me
Me
MeO OHO
Me
Me
MeO OHO
H3O+/H2OMe
MeO
O
+ + OH2OH2H
Me
Me
MeO OHO H
MeOOH
O
H
Me
Me
MeO OH
O HMe
Me
O
O
H
Me
Me
OH2 O
Me
Me
O + H3O
5
Figure by MIT OCW.
PT
7. (10 pts) Under basic hydrolysis conditions, a nitrile goes through a primary amide intermediate before becoming a carboxylate. Show the mechanism for this reaction and explain why it is NOT a facile method for converting nitriles into carboxylates.
O
OR NH3R C N
-OH/H2O +
+ OHOR H
HN
C+
OHR
OHN H
O
O
R
R C N
OH
R
HN
O
R
HN
O
HO H
NH3+O
O
R HNH2+
R NH2
OOH
Base catalyzed hydydrolysis of nitriles is not a facile process because NH2is not good leaving group( -OH is better!) There, fore, the reaction is slow.
HO
R NH2
O
6
Figure by MIT OCW.
8. (9 pts) Provide a synthesis for the following compound.
OH
Cl
Cl2
Cl
NO2NO2
AlCl3
HNO3
H2SO4
H2SO4
Cl
NH2NaNO2
HCl
Cl
N N
Cl
Cl
OH
H2, Pd
7
Figure by MIT OCW.
9. (12 pts) Provide a selective synthesis for ONE of the following compounds. Circle the molecule that you want graded. All of the carbon atoms of the product should come from either ethanol or compounds that contain just one carbon atom.
8
MeOH
MeOH
PBr3
PBr3
PBr3
PBr3
MeBr
MeBr
1. Mg0
2. CO23. H2O
2. H2O
O
OH
OH
OH
OH OH
O
O
SOCl2
SOCl2
Cl
Cl
O
O
O
O
O
O
pyridine
O
Clpyridine
N
N
N
N
EtOH EtBr1. NaN32. LiAlH4
1. LiAlH4
2. LiAlH4
LiAl(OtBu)3H
1. LiAlH4
3. H2O
2. H2O
1. NaN32. LiAlH43. H2O
3. H2O
2. H2O
Et NH2
NH2
H H ,H+1.
O
H H1.
H
H
H
H
OR
Me +
JonesOR
OR
MgBrOH PCC
HCNCN
O
NMe Me MeMe
OOH
OH
OH
NH2
NH2
1. CO22. H2OMgBr
Mg0
Br
SOCl2
A
A
B
B
Figure by MIT OCW.
8
H
EXTRA CREDIT
(5 pts) Synthesize methamphetamine (crystal meth) from benzene and any other reagents. All the carbon atoms in the product should come from reagents that only contain one carbon atom.
9
NH
MeMe
Br2FeBr3
1. Mg
1. Mg
2.CO2
2.CO2
3.H2O
3.H2O
OH
OOH1.LiAlH4
1.LiAlH4
2.H2O
2.H2O
1.MeLi
2.H2O
OHO
OHO
PBr3
OHN N
H
H2NMeH+
Br
Figure by MIT OCW.
Massachusetts Institute of Technology 5.13, Fall 2006 Dr. Kimberly L. Berkowski Organic Chemistry II
EXAM #3 EXTRA PROBLEMS
What to expect on Exam #3: 1. ~1 Labeling experiment 2. ~2 Mechanisms 3. ~2 Syntheses 4. ~5 transformations supply missing product 5. ~5 transformations supply missing reagents 6. ~3 General questions
KEY
1. (4 points each, 8 points total) In the boxes, please provide the reagents for the illustrated transformations. More than one step may be required.
2. (2 points each, 8 points total) Please provide the products of the following reactions. If no reaction is expected, write NR.
NAME____________________________
1
Figure by MIT OCW.
Figure by MIT OCW.
O
OHi-Pr
O
Hi-Pr
O
OHi-Pr
i-Pr Br
(a)
(b)
1. KCN
2. H3O+ or
or
1. Mg ,ether2. CO23. H3O
+/workup
1. SOCl22. LiAl(OtBu)3H3. workup
1. LiAlH4 (XS)2. H2O3. PCC
Et
O
Et Cl
O
OEt OH
O
Et OMe
O
(a)
(b)
(c)
(d)
Et NMe2
O
1. Excess Na BH4
2. Workup
2. Workup
1. Excess MeMgBr
2. Workup
1. Excess MeLi
2. Workup
1. Excess LiAlH4
OH
NRor
HO Me
MeEt
EtNMe2
Et
3. (2 points each, 16 points total) Please provide the requested products or reagents. If no reaction is expected, write NR.
Name_________________________
2
Figure by MIT OCW.
NH2 N2 Br(d)
(c)
NH2
O
O
n-BuH2O
H3O+
Me Me
H2N
cat. H+
(b)1. LiAlH4
2. workup
Me
MeMe
Me NH2NH2 H2O2,
(a)
Br2,NaOH
N-OH
POCl3
NaNO2HCl
CuBr
O
n-Bu
NH2
O
OHnBuCN
OH
4. (11 points) Please provide a detailed mechanism for the illustrated conversion of acetic acid (A) to acetyl chloride (B).
Name_______________________
3
Figure by MIT OCW.
OHMe
O
OHMeMe
O
O
OHMe
O
Me
O
ClMe
OO+ S
Cl Cl
OSCl
Cl
OS Cl
OS ClCl
+ +
+
+
SO2
SO2
HCl
BA
Cl
Cl
HCl
OH
cc_mphatoLine
5. (11 points each, 22 points total) Please provide syntheses for only two of the three indicated compounds. All the carbon atoms should be derived from the allowed starting materials. You may use any common reagents.
Name__________________________
4
Figure by MIT OCW.
Allowed Starting Materials:
Me
MeMe Me
Me
Me CO2OH OH
Me
Me
CN
CN
Me
Me
HO
MeMeNH
A
A
B
B C
Pick Two:
Synthesis # 1:1.O32.MeS
3.Na2Cr2O7H2SO4
OH
O2
KMnO4D
Cl
O
O
NH2NH3
SOCl2
POCl3
1.BH3, THF2. H2O2,OH
orH3O+/H2O
OH 1.PBr32.Mg
MgBr1.CO22. H
O
OH
O
Cl
SOCl2
1.MeMgBr2.H2O
(excess)OH
+
5. (Continued)
Name________________________________
5
Figure by MIT OCW.
Allowed Starting Materials: Me
Me Me
Me Me
Me
Me Me
CO2 OH
Me
Me-OH
NH2BrOHPBr3or
or
orTsCl, Pyridine
Me
Me
CN
HO
NHPick Two:
Synthesis #2:
O
O
N K
2. H2NNH21. NaN32. LiAlH43. H3O
+
1.
NH2N N
H
H
H
OMeOH
PCCCat H
1. LiAlH42. H2O
XS NH3
A B C
C
6. (11 points) Provide a synthesis that will selectively convert A to B. Show all the key intermediates and furnish all the important reagents. This is not a one-step process.
Me
O
Me
O
Me
O
A
HO
Me
O
B
H2N
HNO3H2SO4
Me
O
Me
O
H2, Pd
NaNO2, HCl
O2N
N2
Me
O
H2SO4, H2O
HO
Name________________________________
6
Figure by MIT OCW.
7. Methyl acetimidate (A) is hydrolyzed in aqueous sodium hydroxide to give mainly acetamide and methanol (eq 1). In aqueous acid, A hydrolyzes to give primarily methyl acetate and ammonium ion (eq 2).
a) Provide a detailed mechanism for the illustrated process. Please show all arrow
pushing.
b) Provide a detailed mechanism for the illustrated process. Please show all arrow pushing.
c) Briefly explain why the two reactions provide different products.
7
Figure by MIT OCW.
Figure by MIT OCW.
Figure by MIT OCW.
OMe NH2
OMe
Me
Me
AH2O
NH
Me
NH2
O
O
NH2Me
O
+
OMeMeN-H
OH
PT
(1) MeOH
MeOH + OH+ MeO H
HO-
OH
Basic conditions: NH2 worse L.G. than OMe . Elimination favors amide
Acidic conditions:
L.G. Also, NH4 is not nucleophilic + formation of ester is reversible
Acid/base equilibrium favors protonation of ntrogen making it a good
OMe OMe
OMe
Me
Me
A
H2O
NH
OMeMe
N
Me
NH2
OH2
O
OMeMe
NH4+
OMeMeNH3
OMeMe
O
NH4+
OH
PT O
(2)
H H
H2O
NH3
excess H+
+
H
Massachusetts Institute of Technology
5.13: Organic Chemistry II 8. Synthesize the indicated compounds from the allowed starting materials shown below. All of
the carbons of the target compounds should be derived from the allowed starting materials.
8
Figure by MIT OCW.
Me OH
Me OH
Me OH
Me
Me OHKMnO4, H
+O
H
H
O
Me H
O
Me
O
SOCl2 Cl
Me
O
Cl
Me
O
PBr3Br
Me Br Me Me
MeMe
Me
PCC
C N NH2
O
N
Me
Me NH
H
HLiAlH4
LiAlH4
CN
T
3
1
2
3
24
1
T
N
MeOH MeOTs MeCN1) LAH
2) H2O
NaCNNH28.
EtOH EtBr EtMgBr
EtNH2, H+PCC
OHOMgHBr
TsCl
H O1)2)
O O
c)
b)
O
,3
+Na2Cr2O7
N
Massachusetts Institute of Technology
5.13: Organic Chemistry II
9
Figure by MIT OCW.
Cl Me
Me
AlCl3
O
Me
O
H-CN Me
CNOH
OH
LiAlH4
NH2
NH2
Br Br
BrNH2NO2
Br Br
BrN N
Br Br
BrC N
H2SO4HNO3
H2/Ni cat Br2(excess)FeBr3
NaNO2HCl
CaCNH2O, H
Br Br
Br O
OH
(d)
(e)
1 from part c
Br Br
Br O
Cl + H2N Me
Br Br
Br O
N MeH
4 from part c
SOCl2
Massachusetts Institute of Technology
5.13: Organic Chemistry II 9. Provide the best stepwise mechanism for the illustrated process. Please show all arrow
pushing.
10
Figure by MIT OCW.
O O O OH H H H
HH
NH2 NH2
H2O
H2O
+
H
H
+H
OH
H HH
HN
OH
HH NHO
O
O
H
HH H
N
OH
H
N
H
OH
HH
NH
HH
H
H NOH
NNaBH3CN
HH N " "
OH
HHN
HNaBH3CN
" "
NTarget
(9)
Massachusetts Institute of Technology
5.13: Organic Chemistry II 10. (a) Provide the best mechanism. Please show all arrow pushing.
O O
OH
OHOH
OH
N C
C C
C
O H
H
HH
CH3
N
O
CH3
CH3
NN
N Me
etc
OH
benzylic cation
Me
Target
or workup
(a)
11
Figure by MIT OCW.
Massachusetts Institute of Technology
5.13: Organic Chemistry II
(b) Provide the best mechanism. Please show all arrow pushing.
N
O
Me N
O
Me
H2O
H
H
H
OH2
N
OOMe
H
HH
N
OO
O
Me
Me
NH2O
H H
H
HO
Me
NH3O
H
OMe
NH3O
HNH3
OO
O OH2
OH2
OH
H
HH
NH3
OO
H
H
H
H
Me
Me
Me
O HNH3
OH
NH2
O
O
H
H
workup
+
12
Figure by MIT OCW.
Massachusetts Institute of Technology
5.13: Organic Chemistry II
11. Consider the labeling experiments outlined below:
Name_______________
13
Figure by MIT OCW.
HO
NHMe k1
k1
k2
NHMeHO
NHMe
H2O
NHMe
PT H
OMe
OMe OMeOMe OMe
O
OH
NH2Me
H
OH
HO
O
HOPT PT
k2HOPT PT
H
OH
OH
H
H
O
H
Proton transfer are very fast.
Acid/base equilibria favor the protonatedN (not O). Therefore, once the tetrahedralintermediate forms, loss of NH2Me (k2) isfaster than loss of OH (k1 .Very little is incorporated into the unreacted starting material.
All of the oxygens in the tetrahedralintermediate are roughly equally basic.Therefore, each protonated form 13present in the same concentration ,andk2 =
~ k1. as a result, you would expectmore incorporation into the ester startingmaterial than you would into the correspondingamide.
11.Start with the mechanisms:
H2O H2OO
O
O
O
HO
O
HOk1
k1HO H2OPT HH2O H2O
O
O
O
O
HO
HO
HO
HO
HO
HO
)
First Three Letters of Last Name: TA Name: Hour Exam #4 5.13 Fall 2006 Organic Chemistry II December 6, 2006 Name____________________________________________________________________ Signature_________________________________________________________________ ID#______________________________________________________________________
1. Make sure your exam has 7 numbered pages plus a periodic table.
2. Write your initials on each page.
3. Look over the entire exam before you begin to familiarize yourself with its length. Do
what you know first, then attempt the harder problems.
4. Read the instructions carefully and budget your time.
5. Show all of your work. Partial credit receives points!
1. (3pts) For each molecule, write the correct pKa value for the most acidic proton.
OO OOO
9 20 25
OMe
2. (16 pts) Fill in the correct reactants for the following transformations. Be specific about quantities, where relevant.
Figure by MIT OCW.
Figure by MIT OCW.
O
H
OO
O
OO
O
O
PhPh
Ph Ph
CH3
CH3 CH3
Br
a)
b)
c)
d)
4.NaNO2,HCl3.H2O2.LiAlH4 or CH2N2
1.KCN, HCN
Br2
1. LDA (1 eq)2. Br (1 eq)
MgBr2
AcOH
3. (24 pts) Provide the missing products for each reaction. Indicate no reaction with
N.R.
2
Figure by MIT OCW.
Ph CH3
O
O O
O
O
O
O
Ph
O
OExcess I2
Excess NaOH(a)
(b)
(c)
(d)
(e)
(f)
+ HCI3
OMe
OH3O
+/H2OD
CH3CH3
CH3
CH3
NaOEtEtOH
O
OEt
O
OH3C mCPBA
CH3
CH3
H3CN
N
HO
HO
O
O
Me
H3O+/H2O
H3O+/H2O
OH
or
Figure by MIT OCW.
4. (15 pts) Provide a mechanism for the following transformation.
Figure by MIT OCW.
OO O
OEt
OEtOEt
O
Me1 equiv.Me
O O
OEtO O
OEtOEt
OEt
Me
OH2C
Me
O
+
HOEt+++
O
O
H
OO
OEt
OEt+O H
OO
OEt
OEt+O H
+
OEt
OEt
O
O +H
OEt
O
O
OH
HOEt
OEt
O
O
OHH
EtO
OEt
OEt
O O
O
HO
O
O
OEt
+ HOEt
3
Figure by MIT OCW.
5. (14 pts) Provide a mechanism (steps 1 and 3 only)
4
Figure by MIT OCW.
OH OH
OH
OH
HOHO
HO
H2O
H2O
OH2H OH2
OH2
H O
H3O +
+
H2O H
O
NaBH4
1. H3O+/H2O
2.NaBH4(no mech)3.H3O
+/H2O
+ H2O
+
HH2O
H3O +
Figure by MIT OCW.
6. (14 pts) Synthesize the target molecule from methyl acetate. Partial credit will be given for a retrosynthetic analysis.
OO
Me
MeMe
Me
target
OMe
O
Me OMe
O
OMe
OMe
O O
OMe
O O
OMe
O O
OMe
methyl acetate
2(claisen)
1 eq 1.
2. H3O
1.LiAlH42. H2O OH
PBr3Br
Br
O O
OMeNaOMe
Br
NaOMe
H3O+/H2OD
O O
OH
OCO2
Figure by MIT OCW.
5
7. (14 pts) Synthesize the target molecule from methyl acetate and 2-butanol. Partial credit will be given for a retrosynthetic analysis.
OH OH
OHMeMe
MeMeMe
OMeOMe
O
O
target
OO
OO
2-butanol methyl acetate
OO
Cl
OOH
aOHAcOH
Cl2
PCC
MeO
OO
MeOMeO
O
MeO
NaOMeMeOH
NaOMeMeOH
2
O O
O
MeO
O
HO
O OH
MeO
O
+
H2O
NaBH4
6
Figure by MIT OCW.
EXTRA CREDIT (5 pts) Propose a reasonable mechanism for the following transformation.
MeMe
Me
Me
Me
Ph Ph
O
Me Ph
O O
O+ +H H
O
H H
O
H
HH
H
H
N NH+, cat
OH
NMe2
H H
OH2
N
N
Me
Me Me
Me
O
H H
H H
NMe Me
H
H
H
H
PhPh
OH+OH2+
H
PT
+ +H HNMe2
Ph
OMe
MeN
Ph
OMe
MeN
H2OH
7
OH OH
OHMeMe
MeMeMe
OMeOMe
O
O
target
OO
OO
2-butanol methyl acetate
OO
Cl
OOH
aOHAcOH
Cl2
PCC
MeO
OO
MeOMeO
O
MeO
NaOMeMeOH
NaOMeMeOH
2
O O
O
MeO
O
HO
O OH
MeO
O
+
H2O
NaBH4
Figure by MIT OCW.
Massachusetts Institute of Technology 5.13, Fall 2006 Dr. Kimberly L. Berkowski Organic Chemistry II
EXAM #4 MORE PROBLEMS
DO THESE PROBLEMS BEFORE THE OTHER SET OF EXTRA PROBLEMS!
(they are more relevant to the exam material) What to expect on Exam #4:
1. pKas of ketones, diketones, esters, etc. 2. ~3 Transformations supply missing reagents 3. ~10 Transformations supply missing product 4. ~2 Mechanisms 5. ~2 Synthesis
What NOT to expect on Exam #4: 1. Determine mechanism by crossover and stereochemical experiments
(end of Fridays lecture) 2. Neighboring Group Participation Do not work through problems
#8, 24 & 25 on the Extra Problem Set.
1. Please provide a detailed mechanism for the following transformation. Show all arrow pushing.
O O
H
OH
+ MeO
O O
OMe O O
OMe cat. MeO
OMe
H
O
O
O
O OMe
OO OMe
H
O
O O
H
O
OMe
OMe
H
O
O O
O
OMe
O
OH
OMe
O
O
O
OH O
O
H
O
CO2Me
O
O HHO H
OMe
OMe
Figure by MIT OCW.
1
O OMe Me O
1. MeMgBr Me Me2. H3O+
Me Me
Me Me Me MgBr
H2O H
OMe
O Me Me
Me Me
Me
H2O H
OMe
HO Me
Me
Me
Me Me
PT
Me Me
Me MeO OH
H
Me
Me
Me O H
Me OH
Figure by MIT OCW.
Me
Me
Me
Me O H2O
Me
Me
Me
Me
Me O
OR:
Me
Me
Me
OMe H2O Me
Me
Me
Me
OMe Me
H2O
Me
Me
Me
Me MeO OH2
2. (10 points) Please provide a detailed mechanism for the following transformation. Show all arrow pushing.
2
3. (10 points) Please provide a detailed mechanism for the following transformation. Show all arrow pushing. Hint: This mechanism is from problem set 6.
O
O Me
OMe
1. NaOMe, MeOH 2. H+ workup
O O
Me
O
Me O
OMe
OMe
O
O
Me
O
Me
O OMe
OO
Me
H
OO
Me
H
OMe
Figure by MIT OCW.
3
4. (10 points) Diastereomers A and B provide different products upon diazotization. Please explain why only one product is formed selectively in each reaction. Your explanation should include a 3-dimensional mechanism for the formation of each product from the corresponding diazonium salt.
OH OH O
NaNO2
HCl
t-Bu NH2 t-Bu N2 t-Bu A
O
OH OH
NaNO2
t-Bu
H
HCl
t-Bu NH2 t-Bu N2 B
In the concerted Tiffeneau-Demjanov rearrangement, the migrating bond must be antiperiplanar to the leaving group.
(A) H O H
:B H OH
-N2 product
N2
O H
(B) OH
:B
H -N2 product N2 H
H H
(Bolded bonds are antiperiplanar)
Figure by MIT OCW.
4
5. Please provide a detailed mechanism for the following transformation. Show all arrow pushing.
O
Me
cat. -OH H2O
Me
O
Me
O
O O
O
retro-aldol (aldol)
Think about common disconnection...
O
OH
O
OH
H OH
O
OH
OH
O
O
O
O H2O
O
O H
OH
O
O
O
O
H-OH
HO
H
O OH O
HO Me
Me
O Target
Figure by MIT OCW.
5
O Cl
O
O OH
O
O O
Ar O
O HH O O HO
O O O Ar
O O
O Ar
6. Please provide a detailed mechanism for the following transformation. Show all arrow pushing.
Figure by MIT OCW.
6
Target O
O
Me Me Me OMe
methyl acetate Me
O O O O O O
H++ 1 equiv
MeO-
OMe OMe MeO MeO
2.1. 1 equiv Br NaOMe (below)
O O O O O 1. 1 equivH3O, NaOMe 2.MeO BrMeO
(below)
O O H3O + MeOH
PBr3OMe OH
1. LiAlH4 2. workup MeBr
OH PBr3
Br
7. Please provide a synthesis of the indicated compound. All of the carbon atoms should be derived from methyl acetate.
Figure by MIT OCW.
7
Target
OH
OH
Me Me
isopropanol
OH Br OPCC O LDA OH
(1 equiv.) (below)
OH Br 1. LiAlH4PBr3 2. workup
Target
OR:
OH O OPCC H2 O Pd/C
1. cat. -OH 2.
1. LiAlH4 2. workup
Target
8. Please provide a synthesis of the indicated compound. All of the carbon atoms should be derived from isopropanol.
Figure by MIT OCW.
8
Me
O
O Me
O
OMe
Target
methyl acetate
O O 1. 1 equiv. O O
+ NaOMe 2. H+ wkup
Me OMe Me OMe OMe 1. 1 equiv. NaOMe
2. EtBr
O O O
O O 1. 1 equiv.H3O+ NaOMe
OMe2. EtBr OMe
mCPBA
Target
9. (12 points) Please provide a synthesis of the indicated compound. All of the carbon atoms should be derived from methyl acetate. You will receive partial credit for a complete retrosynthesis
Figure by MIT OCW.
9
Target O OO O
HO R
MeO OMeMeO O alcohols containingdimethyl malonate three or fewer carbons
O OH O 1. cat. H+, excess O formaldehyde PCCPCC
MeOH 2. H+,
H H
O O O O
O O 1. 1 equiv. NaOMe MeO OMe 1. NaBH4O MeO OMeOH2. O 2. workup
MeO OMe
cat. H+
O O
MeO
Figure by MIT OCW.
10. (12 points) Please provide a synthesis of the indicated compound. All of the carbon atoms should be derived from dimethyl malonate and alcohols containing three or fewer carbons. You will receive partial credit for a complete retrosynthesis.
10
Massachusetts Institute of Technology Organic Chemistry 5.13
Friday, September 30, 2005 Prof. Timothy F. Jamison
Hour Exam #1
Name ______________________________________________________
(please both print and sign your name)
Official Recitation Instructor ____________________________________ Directions: Closed book exam, no books, notebooks, notes, etc. allowed.
However, calculators, rulers, and molecular model sets are permitted.
Please read through the entire exam before beginning, in order to make sure that
you have all the pages and in order to gauge the relative difficulty of each
question. Budget your time accordingly.
Show all of your work if you wish to receive partial credit.
You should have 11 pages total: 6 exam pages including this page, 3 pages of reference information, and 2 blank pages for scratchwork.
Question: Grader: 1. ________/ 40 points _______ 2. ________/ 30 points _______ 3. ________/ 30 points _______
Total: _________/ 100 points _______
1
SOLUTIONS
1. (40 points total 5 points each) The molecular formulas and 1H NMR spectra of 8 common organic solvents are provided below and on the following 2 pages. For each, neatly draw the entire structure (i.e., not the acronym) in the box provided. In some cases, relative integration values (circled numbers) and/or other information have been provided.
Note: Do not represent functional groups with partial molecular formulas or other abbreviations. For example, do not use Ph or C6H5 for a phenyl group. Draw the entire group (including hydrogen atoms).
2
012345678910
CH3
35
a. C7H8
Draw structure here
ppm
Figure by MIT OCW.
OH
CH3 CH3
Draw structure here
11 10 9 8 7 6
6
5 4 3 2 1
1
1
ppm
,d
Septet
0
b. C3H8O
Figure by MIT OCW.
310 9 8 7 6 5 4 3 2 1 0ppm
c. C3H6O
Draw structure here
H3C CH3
O
Figure by MIT OCW.
10 9 8 7 6 5 4 3 2 1 0
ppm
CH3C N
d. C2H3N
Draw structure here
Figure by MIT OCW.
O
N
CH3
CH3H
e. C3H7NO
Draw structure here
01
1
23
3
3
45
s
s
s
67891011
,,
,
Figure by MIT OCW.
4
10 9 8 7 6 5 4 3
3
2
2
1 0ppm
,qt,
O
CH3CH2 CH2CH3
Draw structure here
g. C4H10O
Figure by MIT OCW.
012
2
3
3
3
45678910ppm
f. C4H8O2
O
H3C OCH2CH3
Draw structure here
,t
q,
Figure by MIT OCW.
h. C4H8O
Draw structure here
1 1
ppm11 10 9 8 7 6 5 4 3 2 1 0
O
Figure by MIT OCW.
5
2. (30 points total) Answer the questions below about the structure that has the following data:
EA C, 81.61; H, 11.06; N, 7.32 MS 191, 176. 13C NMR 162.7, 136.5, 118.9, 35.1, 31.9 1H NMR 7.59 (t, J = 7.8, 1H), 7.14 (d, J = 7.8, 2H), 1.34 (s, 18H)
a. (10 points) Determine the molecular formula. Circle your final answer.
b. (5 points) Calculate the Index of Hydrogen Deficiency (IHD). Circle your final answer.
13 -21/2+1/2+1= 4
c. (2 points) How many types of carbon (chemically non-equivalent) does this compound have? Circle your final answer.
C13H21N
Figure by MIT OCW.
N
Figure by MIT OCW.
5
d. (3 points) How many types of hydrogen (chemically non-equivalent) does this compound have? Circle your final answer.
3
e. (10 points) In the space below, draw the structure of the molecule that is consistent
with all of the data provided. Circle your final answer.
6
3. (30 points total) Answer the questions below about the structure that has the following data: EA C, 75.69; H, 8.80 M+ 206 IR 3430 (broad), 1705 (strong) 13C NMR 181.4, 140.9, 137.0, 129.5, 127.4, 45.9, 44.1, 30.3, 22.5, 18.2 1H NMR 11.9 (broad s, 1H), 7.21 (d, J = 7.7, 2H), 7.09 (d, J = 7.7, 2H), 3.70
(q, J = 7.0, 1H), 2.44 (d, J = 6.8, 2H), 1.84 (nonet (9 lines), J = 6.8, 1H), 1.49 (d, J = 7.0, 3H), 0.89 (d, J = 6.8, 6H)
a. (7 points) Determine the molecular formula. Circle your final answer.
b. (5 points) Calculate the Index of Hydrogen Deficiency (IHD). Circle your final answer. c. (8 points) Which protons are coupled to which? Complete the tables below using the
NMR data above. Write H1, H2, etc. or none, as appropriate, in the box provided, and list all protons to which a given proton is coupled.
Proton(s) (ppm) Coupled to Proton(s) (ppm) Coupled to
H1 11.9 H5 2.44
H2 7.21 H6 1.84
H3 7.09 H7 1.49
H4 3.70 H8 0.89
d. (10 points) Draw all of the possible enantiomers and diastereomers of the unknown compound that are consistent with all the data given. Circle your final answers.
e. (Extra credit 5 points total) What is the common name of this over-the-counter pharmaceutical (3 points), and for which symptoms is it indicated (2 points)?
13 -18/2+1= 5
C13
H18
O2
CH3
CH3 H
OH3C
OH CH3
CH3H
OH3C
OH
Figure by MIT OCW.
IBUPROFEN; PAIN
none
H3
H2
H7
H6
H4
H6
H5, H8
1
Massachusetts Institute of Technology Organic Chemistry 5.13
Wednesday, October 26, 2005 Prof. Timothy F. Jamison
Hour Exam #2
Name _________________________________________________ (please both print and sign your name)
Official Recitation Instructor ____________________________________ Directions: Closed book exam, no books, notebooks, notes, etc. allowed.
Calculators are not permitted for this exam. However, rulers and molecular model
sets are permitted.
Please read through the entire exam before beginning, in order to make sure that
you have all the pages and in order to gauge the relative difficulty of each
question. Budget your time accordingly.
Show all of your work if you wish to receive partial credit. You should have 7 pages total: 5 exam pages including this page and 2 blank pages for scratchwork.
Question: Grader: 1. ________/ 42 points (page 2) _______ 1. ________/ 30 points (page 3) _______ 2. ________/ 28 points _______
Total: _________/ 100 points _______
SOLUTIONS
1. (72 points total, 3points per box) in each box below, draw the structureof the reagent or major product of the reaction, where appropriate. If no reaction occurs,put a large X in the box. Clearly indicate the double bond geometry and relativestereochemistry of the major product, where appropriate.
(a.)
(b.)
(c.)
(d-f.)
(g-h.)
(i-n.)
CH3OH
1. NaH
1. NaH
1. NaHCO3
1. NaOH
2. PhCH2Br CH3O
OCH3OH2. CH3l
2. CH3l
Ph
2. CH2=CHCH2CIt-BuSH t-BuSH
Me
Me
Me
Me MeOMe OMe
Me
Me Me
MeO
O OO
O
O3
O OO
O tBu SO O O
DMDO (excess)
Hg(OAc)2MeOH NaBH4
NaBH4
+D
Hg(OAc)
(draw the structure)
HH
HH
H H
OHOH
OH H
OHOO
O H
HCl
m-CPBA(draw the structure)
Me O
O
O
H
HH
H(CH3)2S
Me
(o.)
Figure by MIT OCW.
OH
3
D
D
D
+O
OMe
Me
O
HMe OMe
O
O
O
O
O
OH
H
Cy
O
O
OO H
HMe
Me
EtPh
Me
+
Me Me
CO
+
Me
EtPh
hv
Dis(r).
(q).
(p).
(o).
Cy = Cydohexyl =
EtPh
Me
hv
hv
Dis
con Dis
Me
PhEt
+
+
C N O
CO2Me
CO2Me CO2Me
CO2MeC10H12O4
Me
Me
Me
Me
CO2MeMe
NO
Ph
Me CO2Me
NO
Ph
MeO2C
Me
Me
C14H18O4
CO2Me
CO2Me
orD
D
D
D
(s)
(t-u)
(v-w)
(x)
O
Me
Figure by MIT OCW.
2. (28 points total) In a Nazarov Cyclization (below), treatment of a dienone with a strong Lewis acid effects a thermal 4 electrocyclic ring closure, giving intermediate A, and an aqueous workup affords the final product (B), the thermodynamically most stable cyclopentenone.
a. In the diagram below, draw the atomic orbitals (by shading the lobes appropriately) that represent the system of C (the precursor to A) in the reaction above (2 points each).
b. Write the number of nodes in the box to the left of each orbital array (1 point each). c. For the ground state of C, draw the electron population for each orbital on the line to the
right of each orbital array. Clearly indicate whether each electron is spin up or spin down. If there are no electrons in a given orbital, leave the line blank (1 point each).
4
Orbitals# of Nodes
1 point per box 2 points per orbital array 1 point each
Electron population
E
Figure by MIT OCW.
O O
C
OO
B
H2OTiCl4TiCl4
TiCl4
A
Figure by MIT OCW.
4. (continued) d. (4 points each) For the example of the Nazarov cyclization below, in the indicated
boxes draw the direct product of the electrocyclic ring closure and the cyclopentenone final product after the aqueous workup. In both cases, clearly indicate stereochemistry and double bond geometry, as appropriate.
5
Me
Me
Me
O
Me
Me
Me
OH
OMe
MeMe
TiCl4
Me
Me
Me
O TiCl4
direct cyclization product
H
Me
Me
Me
O
cyclopentenone
H2O
LESS STABLE CYCLOPENTENONES (2 Pts PARTIAL CREDIT):
Figure by MIT OCW.
KEY
Massachusetts Institute of Technology 5.13, Fall 2006
Dr. Kimberly L. Berkowski Organic Chemistry II
PRACTICE EXAM #3
Hour exam #3 will be held on Wednesday, November 15, from 12:0512:55.
Books, notes, and calculators will not be allowed during the exam.
Molecular model kits will be allowed during the exam. You will be given a periodic table and blank pages.
Material Covered on Exam #3: Everything presented in lecture related to Amines, Carboxylic
Acids, and Carboxylic Acid Derivatives Reaction and Drill Problems Problem Sets 5 and 6 McMurry Chapters 20, 21, 24 All 5.12 material.
The answer key will be posted on Monday
1. Rank the following acyl derivatives based on their reactivity as electrophiles toward hydroxide ion (1 = most reactive, 5 = least reactive).
O O O O O O
Me NMe2 Me O Me Me Cl Me O Me OMe
4 2 1 5 3
2. In the boxes, please provide the reagents for the illustrated transformations. More than one step may be required
O
Me Me NH2
(a)
Br
1. NaN3 2. LiAlH4 3. H2O
N K 1.
2. H2NNH2
oror Excess NH3
Without Over-Alkylation
Me Me Me Me
O HO NH2
(b) 1. HCN, Cat. KCN 2. LiAlH4 3. H2O
Figure by MIT OCW.
O
3. Please provide the requested products. If no reaction is expected, write NR.
O O
HO
O
O
O
O
O
Me
n-Bun-Bu
n-Bu n-Bu
n-Bu
H
Et
Et
NR
1. Li(t-BuO)3AlH
2. workup
2. workup
2. workup
2. workup
1. excess EtMgBr
1. excess NaBH4
1. excess MeLi
EtO -, EtOH
Na2Cr2O7
H2SO4
Cl
OMe
n-Bu OMe
n-Bu OH
n-Bu OH
n-Bu OH OH
O
O
O
n-Bu
n-Bu
NR or
Figure by MIT OCW.
4. Please provide the requested reagents.
O
O
O
(a)
(b)
(c)
(d)
(e)
n-Bu
n-Bu
NH2
NH2
CH2
n-Bu NH2
n-Bu CN
CN
n-Bu
n-Bu
CN
Me Me NMe2
N2 Cl
Br2, NaOH
H2O
1. Excess MeI 2. Ag2O,
H2O2, or
CuCN
POCl3 or
P2O5
H+/H2O
-OH/H2O or
OH
Figure by MIT OCW.
5. (12 points) Consider the labeling experiment outlined below:
O
OH
H2O Stop the reaction at 50% conversion and examine the recovered acyl chloride for incorporation of
Me Cl O
O = isotopically labeled oxygen (18O)
(a) Please provide the mechanism for the hydrolysis reaction shown above, including the pathway for incorporation of O into the acyl chloride.
O
OH
Me Cl Cl
O
Me
HO
Cl Me ClMe
Me
O
Me
k1k1
k2k2
PT
(b) What level of O incorporation ("high" or "low") you would expect to observe in the recovered acyl chloride? Explain briefly.
Very low incorporation of labeled Oxygen into acid chloride Cl is a much better leaving group than OH. Hydrolysis will take place much faster than label incorporation.
k2 >> k1
(c) Based on your answer to part b, do you think the results of this labeling study definitively prove the mechanism of this reaction? Explain briefly.
No. It is impossible to definitely prove a mechanism incorporation of the label is consistent with both SN2 and addition elimination mechanisms.
Figure by MIT OCW.
Name_______________
O
O O
OH
OH
OH
Name_______________
(a) Me H+, H2O
Me ON
C
MeMe NH2 A B
H Me Me N Me NH2H
PT
Me Me OH2 Me
C N
OH H2O
Me NH2 Me NH2 H3O + H OH2
Me O Me O
(b) Me O Me O H+, H2O
Me NH2 Me OH B C
OHH Me OH Me Me O
NH2 PT NH3
Me OH2 Me OH Me NH2 H2O
H NH3 Me O Me O
NH4 +
Me OH Me OH
6. (12 points) The hydrolysis of a nitrile (A) to a carboxylic acid (C) involves initial formation of a primary amide (B). Provide a detailed mechanism for each the following transformations.by MIT OCW.
Name_______________ Figure by MIT OCW.
O NaOH n-Bu NH2+ Br2 H2O
n-Bu NH2
OO O H H H
OH n-Bu N
Br Br
n-Bu N n-Bu N
H
O
n-Bu N
O O Br Br H
OH+n-Bu Nn-Bu N Br
O n-Bu
n-Bu O n-Bu H O NO C N C N
OH OH HOH
H OH
H2O + CO2 + HNn-BuH2N n-Bu +
H OH OH
7. Provide a mechanism for the Hofmann elimination. Please show all arrow pushing.
Figure by MIT OCW.
HNO3 MeO H2SO4
OMe
H2, Pd
MeO OMe MeO NO2
Br
MeO
OR
OMe HNO3 O2N H2SO4 fuming
MeO OMe MeO
Br
MeO
OMe Br
OMe MeO A OMe
Br Br2
OMe MeO NH2
OMe Br Br
H3 PO2
OMe MeO
OMe NO2 H2NH2, Pd
OMe MeO
OMe ClBr
CuBr N N
OMe MeO
OMe Br
OMe B OMe
Br
OMe NH2
NaNO2, 2HCl
OMe Br
OMe N Cl
N
OMe NH2
OMe NaNO2 HCl
OMe Cl N N
OMe
8. Provide a synthesis that will selectively convert A to B. Show all the key intermediates, and
furnish all of the important reagents.
Figure by MIT OCW.
9. Provide synthesis for the following compounds. All of the carbons in the target molecules should be derived from the allowed starting materials. You may use any common reagents.
O
OO
O
MeOH
Me
Me
OH
OH
OH
Me
OH
Me
Me NH2
MeMe Me
Me
Me
CN
Cl
Cl
CO2 H H
O
H H
H H
H+
H H
Allowed Starting Materials:
EtOH
MeOH
Me
MgBr Me MgBr
Me CN
NH
Me
Me
Me N Me
Me Me
Me
Me N
O Me
H
+
+
MeMgBr MeOH
OH Me
Me H+ workup
O
H
1. PBr3
1. PBr3
1. PBr3 2. Mg, Et2O
1. PBr3 2. Mg, Et2O
2. Mg, Et2O 1. CO2, Et2O
SOCl2
1.
2.
H+2.
H+2.
PCC
PCC
1. LiAlH4
H+ H+
Cat. 2.
1. LiAlH4
2. KCN
A
B
A
B
(a)
(b)
from(a)
Figure by MIT OCW.
Massachusetts Institute of Technology 5.13, Fall 2006 Dr. Kimberly L. Berkowski Organic Chemistry II
PRACTICE EXAM #4
Hour exam #4 will be held on Wednesday, December 6, from 12:05-12:55. Books, notes, and calculators will not be allowed during the exam. Molecular model kits will be allowed during the exam. You will be given a periodic table and blank pages. Material Covered on Exam #4:
Everything presented in lecture related to Enols and Enolates and Carbocations
Recitation and Drill Problems Problem Sets 7 & 8 McMurry Chapters 22 & 23 All 5.12 materials.
The answer key will be posted on Monday
1
KEY
2
OO
Me MeH
OO
OMeOMe MeOH
OO
MeH
9 11 13
4-6 9-11 16-23 23-27
MeCO2H Me3NH Me
OH
O
EtOH
(1) (1 point each, 7 points total) Please provide the pKa value for the indicated H.
(2) (2 points for each box; 20 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.
(a) (b) (c)
O
Ph
O
Ph CH3
O
Ph CH3
O
Ph CH3
H2O
O
Ph
OH
O
Ph
OH
(a)
(b)
(c)
(d)
CHI3
also acceptable:l eq. LDA; H workup
excess I24 eq. NaOH
1eq.LDA
or H
or H
O
H H
O
OPh
O
Ph
Figure by MIT OCW.
Figure by MIT OCW.
cat. OH
cat. OH
3
(f)
CH3 CH3
O CH3 CH3
O
CO2CH3
O
CH3
O
CH3MeS
O
CH3Ph
O
Ph
O O
CO2CH3
OO
OH3CO
O
CH3MeS
Ph PhO O
MeMeS
also accepted:
also accepted:
O O
MeMeS
1 equiv. of base
cat.OMe
1 eq. LDA; H workup
(g)
(h)
(i)
O O
H3CO CH3
1 eq. OMe H
+O
CH2 H3CO(e)
1 eq. LDAalso accepted
cat.OMe
cat.OMe
Figure by MIT OCW.
O
(3) (12 points) Please provide an efficient synthesis of the indicated target compound. All of the carbons of the target compound must come from ethyl acetate and 1, 5-dibromopentane.
4
Br
Br Br
BrO
MeEtO
OEtO
O
Me
O
EtO
O-
Me
O
MeEtO EtO
O
Me
2
target compound
1 equvEtO
Br
O OO
1 equiv. LDA
H2O
cat H+ or HO-D
Figure by MIT OCW.
(4) (12 points) Please provide an efficient synthesis of the indicated target compound. All of the carbons of the target compound must come from the three illustrated alcohols.
5
Me MeMe
OMe
Me O
target compound
OHOH
OH
OOH PCC
SYNTHESIS:
(1)
OH OPCC
PCCOH
O O
O
O
OLi
O
H1.
O
cat. NaOMe
1. LDA(1.1 eq.)
}(2)(3) O
O
OH, 2.
Figure by MIT OCW.
(5) (12 points) Please provide an efficient synthesis of the indicated target compound. All of the carbons of the target compound must come from acetone and diethyl malonate.
6
Me Me
OO
O
OO
O
Me
O
O O
O O
Me
XS
EtO
EtO
OEt
O
O
O
EtO OEt
O O
EtO OEt
O O
OEtO
EtO2C
OEt
(-H2O)
,-H2O
H2O
H2O
EtO
1 eq. EtO
EtO
1 eq.
H+,
H+
OO O
EtO OEt
O
O
O-
O-
EtOOEt
O
O
O
EtOOEt
OEt
O
O
EtO2C HH
O
O
EtO2COO
Mechanism not
necessary
target compound
,
cat H+
Figure by MIT OCW.
O
(6) (12 points) Provide a mechanism for the conversion of A to B and B to C. Please show all arrow pushing.
7
OMe
O
O
OH
OH
OOH
OH
O
H
OOH
O
O
H
O OOH
O H
CH2 HH 2O
OH
OOH
OH
OHHO
HO
H
H OH
H OHO
OHOH
O
OHHH
O
OH
OH
OH
OH
O
O
O
O
O
O
cat. OH cat.
A
B
A B C
C
Figure by MIT OCW.
(7) (12 points) Provide the best mechanism for the illustrated transformation. Please show all arrow pushing.
8
catalystOH
O
HOH
O
O
O
O
O O
O
H OH
OH
OH
OHOH
HO
H O
O
O
+
O
Figure by MIT OCW.
(8) (13 points) Provide the best mechanism for the illustrated reaction. Please show all arrow pushing. Hint: RS can serve as a nucleophile and add to the carbon of Michael acceptors.
9
O
Ph
O OOH
H
O
H Ph
Me
O
Me
O
Me
Ph Me
OO
Ph Me
RS RSSR
RS
RSRS
OOH
Ph Me
SR
SR
OOH
Ph Me
O O-
Ph Me
SR
OO
Ph Me
OOH
Ph Me
Tautomerization mech. (not necessary)
H
RSH
Keto Enol Protonated enol
catalytic
A
A
Figure by MIT OCW.
(9) BONUS question (10 points) The process shown below is an example of a Mannich reaction. Nature uses this reaction to synthesize alkaloids (natural product that contain a basic nitrogen). Suggest the best mechanism for this process. Please show all arrow-pushing.
10
Figure by MIT OCW.
O
Ph Ph H
HB
B
H+O
Ph
OH
Ph
Ph Ph
HO OH2
OH
Ph
O
Ph Ph
NMe2O
PhPh Ph Ph
NOH
MeMe2NH
H+O
O
Ph
Ph
Ph
A
A B
(H2O or Me2NH)
Me2NH
NMe2
NMe2
NMe2
H
NN
Ph Ph
O
H
+
+ H
H Cat.
OH
1
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.
KEY
2
1. (10 points total) Write an arrow-pushing mechanism for the reaction below.
Figure by MIT OCW.
OH
ONH
OH
ONH
OH
ONH
OH
ONH
Me ++
+
Me ++
+
Me ++
+
Me ++
+
H H
O
*
heatH H
O
*
heatH H
O
*
heatH H
O
*
heatH H
O
*
heat
R2N+H2 HR2N+H2 HR2N+H2 HR2N+H2 HR2N+H2 HR2N+H2 HR2N+H2 H
O
O
O
O
O
HH
HH
H
NMe
NMe
O
O
O
*
O
O
O
*
O
O
O
*
O
O
O
*
Me HN
Me HN
Me HN
OH HOH H
CO2CO2*
MeNMeN
OH2 HOH2 HOH2 H
CO2CO2*
Me
CO2CO2*
NMeN
*N
*CO2CO2
H2OH2OH2O
IMINIUM FORMATION +3IMINIUM FORMATION +3IMINIUM FORMATION +3
MeNMeN
O
O
O
OH H
HO
*
*H H
HO
*
*H H
HO
*
*H H
HO
*
*H H
HO
*
*H H
HO
*
**
HO
ON
O
ON
O
ON
OH OH
Me
NMe
H2O CO2Me
H2O CO2Me
H2O CO2Me
H2O CO2Me
H2O CO2Me
H2O CO2
HYDROLYSIS +3
Aza-cope +4
HYDROLYSIS +3
Aza-cope +4
HYDROLYSIS +3
Aza-cope +4
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
Solution must account for 13C in formaldyde, otherwise no more than 5 pts should be awarded.
Note: Aste risk(*)=13C.
3
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.
OOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
NOOHPh
Ph
Ph
PhPh
CH3
O
C
A
B
N
ON
N
ON
N
ON
NO
O
NO
O
NO
O
N
O ON N
O ON N
O ON N
O ON N
O ONOH
NO O
NOH
NO O
NOH
NO O
NOH
NO O
NOH
N
Ph
OO
O
O
N
O
Ph
OO
O
O
N
O
Ph
OO
O
O
N
O
Ph
OO
O
O
N
O
Ph
OO
O
O
N
O
Ph
OO
O
O
N
O
Ph
OO
O
O
N
O
PhO
N
H
H
O
N
H
H
O
N
H
H
O
N
H
H
ON
ON
ON
ON
O
O
NO
O
NO
O
N
Ph
1. point
2. points 2. points
2. points1. point
2. points 2. points
2. points1. point
2. points 2. points
2. points1. point
2. points 2. points
2. points
Catalytic H+
Heat
Catalytic H+
Heat
Catalytic H+
Heat
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:
Partial credit for anothermechanism leading to theright molecule that doesn'tinclude [3.3] sigmatropicREARR: [0-3 points].
Claisen Rearr
7.05-7.15, m, 5H5.80, t, J = 6.3, 1H3.67, t, J = 6.5, 4H3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H2.34, t, J = 7.4, 2H 2.12, t, J = 7.4, 2H1.71, s, 3H
1685 cm-1
Molecular weight
273.17
Data for A:1H NMR (ppm) IR
MECHANISM
7
1H NMR (ppm) IR
MECHANISM
7
1H NMR (ppm) IR
MECHANISM
7
1H NMR (ppm) IR
MECHANISM
7
1H NMR (ppm) IR
MECHANISM
7
8
+
+
Figure by MIT OCW.
4
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 cant read your mind, i.e. wrong numbers will receive no credit no exceptions.)
Allor
nothing
CH3
CH3
Cl
Br
a.
b.
c.
d.
Me
Me
Me Me
Me
Me
3
2
1
e.
2
22
2
2
2
4
3
3
6
# non-equivalent H # non-equivalent C
CH3 CH2 CH2 CH3
CH3 CH2 CH3
Figure by MIT OCW.
5
4
33
3
6
# non-equivalent H # non-equivalent C
MeMe
MeMe 1
23
f.
Me
Me Me
O1
2 34
56
7
g.
h. MeMe
12 3 4
HO H
i.
12
3
456
7j.
5
5
7
7
7
Me
Me1
234
HH
56
7
Figure by MIT OCW.
6
k.
i.
m.
n. MeMe
O
# non-equivalent H # non-equivalent C
O
O
H NMe
Me
O restricted rotation
MeMe
Me N O
O(DIASTEREOTOPICITY)
(Me's are diastereotopic)
(10e-,aromatic)
o.
1 1
6 4
9 8
3 3
2 2(3 was accepted)
Figure by MIT OCW.
7
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.
Figure by MIT OCW.
OOH
draw D heredraw R-OH here
MeOH
draw D heredraw R-OH here
MeOH
draw D heredraw R-OH here
MeOH
draw D heredraw R-OH here
Me
1. NaH, THF2. 1-bromo-2-butyne
chirality 2pts.
1. NaH, THF2. 1-bromo-2-butyne
chirality 2pts.
1. NaH, THF2. 1-bromo-2-butyne
chirality 2pts.
1. NaH, THF2. 1-bromo-2-butyne
chirality 2pts.R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
R-OH1H NMR data for D (ppm)
5.68, ddd, J= 17.0, 10.5, 8.5, 1H5.27, dd, J=10.5, 1.5, 1H5.19, dd, J=17.0, 1.5, 1H4.14, d, J=15.0, 1H3.94, d, J=15.0, 1H3.42, d, J=8.5, 1H1.86, s, 3H0.91, s, 9H
From ROHFrom ALKYNE
== From ALKYNE== From ALKYNE== From ALKYNE== From ALKYNE== From ALKYNE==
CH3BrCH3BrCH3Br + RO+ RO
MeRO
MeRO
H H
Correct ether synthesis +2 pts.
H H
Correct ether synthesis +2 pts.
H H
Correct ether synthesis +2 pts.
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
ROHas : 9 H S 2,(tBu) 2pts.
3 H IN Alkene region, All Coupled to each other
H
H
H1H is coupled to 1 Alkene H
H
H
H1H is coupled to 1 Alkene H
H
H
H1H is coupled to 1 Alkene H
H
H
H1H is coupled to 1 Alkene H
H
H
H1H is coupled to 1 Alkene H
H
H
H1H is coupled to 1 Alkene H
H
H
H1H is coupled to 1 Alkene H
OCH3CH3
H
H
H
H
ZY
H
H
H
H
ZY
H
H
H
H
ZY
H
H
H
H
ZY
H
H
H
H
ZY
H
H
H
H
ZY
3.42ppm
2pts. Add'l
Y and Z have no H directly attached.
ONLY tBr and O Remain
2pts.
2pts. Add'l
Y and Z have no H directly attached.
ONLY tBr and O Remain
2pts.
2pts. Add'l
Y and Z have no H directly attached.
ONLY tBr and O Remain
2pts.
2pts. Add'l
Y and Z have no H directly attached.
ONLY tBr and O Remain
2pts.
2pts. Add'l
Y and Z have no H directly attached.
ONLY tBr and O Remain
2pts.
2pts. Add'l
Y and Z have no H directly attached.
ONLY tBr and O Remain
2pts.
2pts. Add'l
Y and Z have no H directly attached.
ONLY tBr and O Remain
2pts.
2pts. Add'l
Y and Z have no H directly attached