Exams Organic Chemistry MIT

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.


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


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


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


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)


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.



(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.



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.


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


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


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


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


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


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.


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


5.13: Organic Chemistry II December 19, 2005

Final Exam

Question 1 __________/10 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.





















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:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


1685 cm-1

Molecular weight

273.17

Data for A:


Claisen Rearr


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


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


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


MeOH


MeOH


Me

1. NaH, THF2. 1-bromo-2-butyne

chirality 2pts.


chirality 2pts.


chirality 2pts.


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)




















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


H H


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


H

H


H

H


H

H


H

H


H

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



2pts.

2pts. Add'l



2pts.

2pts. Add'l



2pts.

2pts. Add'l



2pts.

2pts. Add'l



2pts.

2pts. Add'l



2pts.

2pts. Add'l

Y and Z have no H directly attached

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