Chem140b Exam Study Info

Practice Exam Packet for Dr. Hoeger’s Organic Chemistry Attached to this cover sheet are a number of items:

1. General Study Questions (i.e. old exam questions; they give you a good idea of what I like to ask;

2. Practice Exams for an entire YEAR of Organic Chem, broken up by semester (these are ACTUAL exams that were given).

Because of changes in textbooks used, changes in coverage, school differences, etc you must keep this in mind at ALL TIMES: these are meant to be REPRESENTATIVE of past exams and, depending on the school you are at or the quarter or semester you are in you may or may not know how to do a particular problem. Always use your homework and lecture notes s a guide. IMPORTANT NOTE 1: NO ANSWERS TO THE QUESTIONS INCLUDED HEREIN WILL BE PROVIDED (so don’t ask). You want to know how to do something, come and see me or a TA. IMPORTANT NOTE 2: these exams are light on NMR and structural determination. Do NOT take this to mean that you will not see it! It just so happens that coverage of NMR was done in a different setting when these exams were administered. You can BET that if we covered it in lecture, and homework was assigned, it WILL be covered on an exam. FINALLY: No practice final exams have been included. Nor will any. They just look like long exams. Good luck, and remember: ALWAYS RELY ON FIRST PRINCIPLES FOR PROBLEMS YOU ARE UNSURE OF!

SEMESTER 1 BASIC STUDY QUESTIONS

EXAM 1 QUESTIONS 1. Consider the molecule diazomethane (CH2N2) for which the sigma (�) network is shown below. Complete the structure, drawing in all remaining bonds, lone pairs and formal charges found on the atoms (Hint: the net charge on diazomethane is zero).

C N N

H

H 2. a) Draw all isomers having the molecular formula C6H14. Name them. b) Which one of these isomers will give two (and only two) monochloro products upon free radical chlorination? Draw these monochloro products and name one of them. 3. Consider the following equlibrium:

H

H3C

H3C

HD

H

D

HKeq=17.336at 23°C

Given that AMe=1.70 kcal, calculate the A value for a deuterium (NOTE: Be careful about rounding off!). 4) Compare and contrast reactivity and selectivity in terms of energy and stabilities. Use examples! 5) Consider the molecule diisopropylcarbodiimide (I) and its reaction with the anion of acetic acid(II):

Using bond polarity, electronegativity, partial charges, etc. predict where II will react with I and attempt to draw the product. For full credit show your work. 6. Provide a mechanism for the free-radical bromination of 2,2-dimethylpropane. 7. Calculate the relative percentages of 1°, 2°, and 3° monochloro products one would get from the free radical chlorination of 1-methylcyclohexane. Show your work. 8. How does one use pKa values to predict whether a reaction will occur? 9) Consider the molecular formula C4H8O. Draw, but do not name, an isomer that is: i) a ketone ii) an aldehyde iii) an alcohol iv) an ether 10. a) What is meant by torsional energy? b) What factors account for ring strain in cyclic molecules? c) Why are the A values for methyl (1.7 kcal), ethyl (1.75 kcal) and isopropyl (2.15 kcal) similar, while that for tert-butyl is so much larger (>4 kcal)?

Carl Hoeger

Note

NOTE:Not all questions are fair game! For example, no conformational analysis will be on the Exam! Answers follow!

11) An O’Chem student from Mesa needed a quantity of 4-amino-1-butanol (A) and decided to make it as outlined below. However, the only product he obtained was compound B. Explain what must have happened and postulate a mechanism for this process.

CH2

CH2

O

CH2

CH2

Br-CH2CH2CH2CH2-OH H2N-CH2CH2CH2CH2-OHNH2

NH2

B

A

12. Calculate the formal charges and/or hybridizations for the indicated atoms:

HC CH2 CH CH2 N C

CH

BH2C

F

F

O 13. Name the indicated functional groups:

O

OO

OO

H

OH

OH

O

O

Ph

O

O

NH

O

OH

Ph

Ph

O

14. Given the molecular formula C7H14: a) Draw any four non-cyclic isomers. b) Draw and name any four non-alkene isomers.

15. In conformational analysis of cycloalkanes, the tert-butyl group is used quite frequently as a 'conformational lock'. Consider both cis 1-methyl-4-t-butylcyclohexane and trans 1-methyl-4-t-butylcyclohexane.

a) Draw the conformational equlibrium for both. b) Calculate the equlibrium constant (Keq) at 23˚C for both equlibria shown in a). SHOW ALL

WORK FOR FULL CREDIT! Note: AMe=1.7 and AtBu=4.0. c) For the cis compound, calculate the relative percentages of each conformer at equlibrium. 16. cis 3-Amino-1-cyclohexanol has a high propensity to exist in a diaxial orientation in aprotic

solvents and in a diequitorial orientation in protic solvents. Explain. 17. Draw all the mono-bromo products formed when 2-methylbutane reacts with Br2 in the presence

of light. Name any three. 18. Draw the following structures: a) 5-chloro-7-ethyl-5-iodo-3-isopropyl-6-methyldecane b) trans 1-methyl-3-secbutylcyclohexane c) 1-cyclobutyl-2-cyclopropyl-3-methylcyclopentane d) 2-isopropyl-3-methyl-1-cyclohexanol (draw in chair conformation with methyl and isopropyl

axial and hydroxyl equitorial) e) What is wrong with the name 3-iodo-5-isopropylhexane? 19. a) Describe in detail (using examples) how pKa is used to predict whether a reaction will go or

not. Use appropriate labels, distinctions, etc. b) Explain why CH3CH2- is more basic than CH3CH2O- ("because of pKa" is not an appropriate

answer) c) CH3CH2O- is much more basic than CH3CO-, although simple principles would not predict this.

Explain.

20. Complete and then predict the outcomes of the following reactions using any principles you want. NR=No Reaction

CH3CH2S CH3CH2Br+

H3C

C

CH2

C

CH3

O O

CH3Li+

CH3CH2OCH2 + C CHBrCH2

CH3

C

CH3

O

CH3MgBr+

a)

b)

c)

d)

21. Starting from any organic starting materials you want of three carbons or less, prepare 2,5-dimethylhexane.

ANSWERS TO EXAM 1 STUDY QUESTIONS 1. A: There are two possibilities:

C N N

H

H

C N N

H

H

or

First-count total # of valence e- there should be (16; 4 for C, 5 for each N, 1 for each H). Second, count how many e- are accounted for with the � structure (8; for bonds at 2 e- each). Finally, fill out structure with remaining e-’s. 2. A: C6H14-no double bonds or rings a)

hexane

2-methylpentane

3-methylpentane

2,3-dimethylbutane

2,2-dimethylbutane

b) 2,3-Dimethylbutane is the only one that will give only two products because it is the only one with ONLY 2 different types of hydrogens (all others have 3 or more different types of hydrogens). The products one would get are:

ClCl

and

3.

H

H3C

H3C

HD

H

D

HKeq=17.336at 23°C

I II A: As the equlibrium is defined, methyl is going from axial to equitorial, and deuterium is going from eq. to ax. Therefore AT = AMe + [-AD] = RTlnKeq ; substituting in numbers, we get: 1.70 - AD = (1.986 x 10-3)(300)(ln 17.336). Simplifying we get -AD = 1.6996895 - 1.70 or AD=3.104 x 10-4. As one might predict, the A value for D is very small, as expected. 4) A. Reactivity and selectivity are inversely proportional; as a species has higher E, it is more reactive (remember everything is trying to get to lower E). As reactivity increases, selectivity decreases (no time to be ‘picky’). Best example is Cl• vs. Br•. See lecture discussion for more details.

5) A:

N C N O CH3

O

N C N

O O

CH3

II will react at central carbon of I; product should look like:

d dd

6. A. 2,2-Dimethylpropane [(CH3)4C] has only one type of hydrogen (1°). The mechanism is as follows: Br2 — 2Br• (Initiation; done with light {h�}) (CH3)3C-CH3 + Br• — (CH3)3C-CH2• + HBr (First propagation step) (CH3)3C-CH2• + Br2 — (CH3)3C-CH2Br + Br• (Second propagation step) Chain termination steps need to be shown as well (see notes and book) 7. A: 1-Methylcyclohexane has three 1°, ten 2° and one 3° hydrogen(s). Using a Cl• selectivity ratio of 1 : 3.6 : 5 and multiplying the number of a given type of hydrogen by its’ selectivity, one obtains the following sets of numbers:

1° pdt:2° pdt:3° pdt is (3 x 1):(10 x 3.6):(1 x 5) = 3:36:5. Therefore the relative percentages are 6.8% 1° product, 81.8% 2° product and 11.4% 3° product. 8. A: Compare the pKa value of the acid with the conjugate acid that is derived from the base used; if going from a strong acid to a weaker acid the reaction will go (i.e. equlibrium will lie on the right). If you find you are going from a weak acid to a stronger acid the reaction will NOT go (i.e. equlibrium will lie on the left).

9) A:

H

OH

O

H

O O

or

OH OH

OH

O

O

O

i) Ketone

ii) Aldehyde

iii) Alcohol (there are MANY possibilities)

iv) Ether (there are MANY possibilities)

etc.

etc.

10) A: a) Torsional energy is the E required to move one group past another while rotating around a carbon-carbon single bond; in a general sense it is the energy necessary for two groups to be in close proximal space to one another (later we will refer to this as steric strain or steric hinderence). Torsional energy is derived from van der Waals repulsions, etc. b) Ring strain = Torsional strain + angle strain. The torsional energy in rings comes from eclipsing interactions while angle strain comes from trying to put a bond angle of 109.5° into a limited, smaller space. c) t-Butyl is the only one that can’t ‘hide’ the branching methyl group:

CH

H

H

H

H

CH

H

H

CH3

CH3

CH

H

H

H

CH3

CH

H

CH3

CH3

CH3

vs.

vs.

vs.

11) A: This is a pKa problem, first and simply!

Br-CH2CH2CH2CH2-OH Br-CH2CH2CH2CH2-ONH2+ NH3+

pKa ! 16 pKa ! 33

Strong to weak; therefore this reaction will go! (Equlibrium will lie FAR on the right). After this reaction occurs, an internal substitution reaction occurs that leads to the final product:

Br-CH2CH2CH2CH2-O

O 12 A: Remembering that formal charge = group # of atom - [0.5 x # of shared e-] - [# of unshared e-], we get the following results:

HC CH2 CH CH2 N C

CH

BH2C

F

F

O 13) A: Here are all the functional groups:

O

OO

OO

H

OH

OH

O

O

Ph

O

O

NH

O

OH

Ph

Ph

O

14) A: This problem is similar to #2 and #9 above. The answers in a) can not have rings, while the answers in b) can not have double bonds. 15) A: a) The conformational equlibrium for both is as follows:

CH3

C

H3C

H3C

H3C H

H

CH3

H

C

H

H3C

CH3

CH3

Cis compound

H

C

H3C

H3C

H3C CH3

H

H

CH3

C

H

H3C

CH3

CH3

Trans compound

I II

III IV b) As the equlibria above are defined, for the Cis compound methyl is going from axial to equitorial and t-butyl is going from eq. to ax; for the Trans compound both methyl and t-butyl are going from eq. to ax. Therefore, for the Cis compound, ATcis = AMe + [-AtBu] = -2.3; for the Trans compound, ATtrans = [-AMe] + [-AtBu] = -5.7. Using 23°C (296°K) and substituting numbers into AT = RTlnKeq, we get: Keq(cis) = 2.0 x 10-2 and Keq(trans) = 6.15 x 10-5. c) Using Keq(cis) = [II]/[I] = 2.0 x 10-2 we get [II] = 0.02[I] or [I] = 50[II]. Based on these ratios we get relative percentages of 98% of I and 2% of II at equlibrium. 16) A: In aprotic solvents cis 3-Amino-1-cyclohexanol will undergo internal hydrogen bonding; in protic solvents, it will hydrogen bond with the solvent making the group much bulkier and thus less likely to be axial:

OH

H

OH

H

HN

H

H

vs. NH2

H

HOR

HOR

HOR

ROH

ROH

ROH

17) A:

Br

Br

Br

Br

1-bromo-2-methylbutane 1-bromo-3-methylbutane

2-bromo-2-methylbutane2-bromo-3-methylbutane or 3-bromo-2-methylbutane

18) A: a) thru d)--Come on, you can do these! e) If you draw the structure and count the longest chain in what you drew, you will find it is 7 not 6! 19) A: a) See Q # 8 above. b) Oxygen is more electronegative than carbon and as such can support (and therefore stabilize) a negative charge better. c) The second compound can stabilize the negative charge through resonance while the first one can not (dispersal of charge):

H3C

O

OH3C

O

O

(Yes, I did have a misprint in the set of questions I gave you...sorry!) 20) A: Here are the answers; the principles used to solve them are as follows: a) bond polarity/partial charge; b) pKa; c) pKa; and d) bond polarity/partial charge:

CH3CH2S CH3CH2Br+

H3C

C

CH2

C

CH3

O O

CH3Li+

CH3CH2OCH2+ C CHBrCH2

CH3

C

CH3

O

CH3MgBr+

a)

b)

c)

d)

CH3CH2-S-CH2CH3 + Br

H3C

C

CH

C

CH3

O O

+ CH4

Li

CH3CH2OCH3+ C CBrCH2

CH3

C

CH3

O

CH3

MgBr

21) A: We will do this in class...

SEMESTER 1 EXAM 1 QUESTIONS

1.(15 pts) a. Calculate the formal charge for the indicated atoms; put answer next to arrow. Show work for the two atoms in II in space below (no others). NOTE: All lone pairs are shown if present, except forFl which has an octet!

CH3

CH2

C

N

O

F3BC

C

O

OH

H

CH OCH3

O O

I) II)III)

b. Fill in the blanks:

The compound in I) has _____ sp3 carbons and _____ sp2 carbons; The compound in II) has _____ sp3 carbons, _____

sp2 carbons, and _____ sp carbons.

Compound I has how many H’s that are NOT shown?_______

The compound with the LEAST primary hydrogens is _______

2) (8 pts) Provide names for the following:

I

Cl

F

3) (10 pts) Draw all resonance structures possible for the anion shown below:

NO N

CH3

4) (5 pts) Cipro® (Ciprofloxacin) is a synthetic broad spectrum antibacterial agent. It was most recently in the news as the

antibiotic of choice for the treatment of anthrax. The structure of Cipro is shown below. Circle and identify as many functional groups in the Cipro representation below as you can.

5). (20 pts) Consider the molecule 1-iodo-2-chloropropane with respect to rotations about C1-C2. Given the torsional

energy values below, answer the following questions, putting your answers on the next page. a) Draw the line-angle structure of this molecule: b) Draw the THREE MOST stable conformations expected for this molecule using Newman projections looking

down the C1-C2 bond (for all htree conformations keep the CH3 on carbon 3 ‘up’); Rank then from MOST stable (I) to LEAST (III).

c) Draw the LEAST stable possible conformation expected for this molcule using a Newman projection (HINT: NOT one of the ones you drew in (b)).

d) Calculate the equlibrium constant at 27°C for the following interconversion:

MOST Stable Rotomer (I) !"!

#!! LEAST Stable Rotomer (III)

e) Calculate the relative percentages of the two conformers at equlibrium Note that the values given are estimates of the per interaction energy cost; if a value is not given, assume it is ≈0.

Gauche Interaction

E per interaction (kJ/mol)

Eclipsed Interaction

E per interaction (kJ/mol)

Cl/H 0.84 Cl /H 4.18 H/H 0 H/H 4.18

CH3/I 5.85 CH3/I 9.20 I/H 4.18 I/H 1.67 Cl /I 5.10 Cl /I 5.10

CH3/H 1.67 CH3/H 7.53

6). (25 pts) a) Using pKa data, calculate equilibrium constants for the following reaction (note: you may have to estimate

pKa values from your table!) i)

O

CH3CH2O-

O

+CH3CH2OH+

ii)

OH + OH-

O-

+ H2O

iii) for ii, calculate the relative percentages of OH– and (CH3)3CO–.

iv) Anion A is much more stable than anion B. Explain.

O OO O

A B v) Explain the leveling effect; using your explanation, tell why NaNH2 can not be used in ethanol (CH3CH2OH).

6). (20 pts) a) Consider the following IR information for the reaction sequence provided. Use this to identify each material in the reaction: Compound A is treated with LiAlH4 to give compound B, which upon heating in the presence of acid is converted

to compound C. Compound C reacts with HBr to give compound D, which is in turn reacted with Mg metal, followed by CO2 to give, after a standard work-up, Compound E.

IR Data (only major absorbances given): Compound A: 2920 cm–1 (s); 1715 cm–1 (s); 725 cm–1 (w); Compound B: 3450 cm–1 (s, br); 2980 cm–1 (s) 1150 cm–1 (s); Compound C: 3050 cm–1 (m, sharp); 2950 cm–1 (s) 1620 cm–1 (m, sharp); Compound D: 2950 cm–1 (s) 620 cm–1 (s); Compound E: 3200 cm–1 (s, very br); 1705 cm–1 (s); 1190 cm–1 (s);

The “suspects” are below; put the appropriate leter below which compound you think is which:

OH OC

Br

O

OH

b) Interpret the following IR

WHAT LIKELY IS PRESENT WHAT CAN NOT BE PRESENT


Name: _____________________________ Page 2 of 4

1. For the reaction of (S)-2-iodobutane with CN

– in acetone, provide the following: a) product(s) expected; b) rate expression; c) mechanism; d) transition state(s) expected; and e) reaction coordinate diagram for this reaction if it progresses as an

S

N2 reaction. Assume the reaction

is exothermic. Be sure to completely label your reaction coordinate diagram. 2. a) Using transition state drawings discuss how the SN2 and E2 reactions compete and why the

stereochemical outcomes of the products are as observed when (R) 2-iodo-3-methylbutane is reacted with

CH

3O

– b) Briefly, what factors influence SN1 over E1 and vice versa? 3. a) Provide the mechanism for the addition of HI to 1-butene. b) When HI is added to 3-methyl-1-butene instead of 1-butene, a different product than expected is obtained. What is this unexpected product and provide a mechanism for its formation. 4. a) State Markovnikov’s rule and give the modern interpretation of it. Illustrate what you describe

using pictures!

b) What is Zaitsev’s rule? Provide an example. c) What is Bredt’s rule? Provide an example. 5. Provide the stereochemistry for each chiral centerin the molecules below:

CH3

H Cl

Br H

CH3

H Br

OH

NH2

CO2H

Br

Br

Cl

(Include the double bond)

Name: _____________________________ Page 3 of 4

6. Provide the expected products for the following reactions. Show Stereochemistry where needed:

HCl

Br2/CCl4

(1) B2H6

(2) H2O2, OH-

(1) O3

(2) Reductive WU

HBr/HOOH

CD3

mCPBA

OsO4

H2O2

Name: _____________________________ Page 4 of 4


1) During a study aimed at exploring the effect of steric hindrance on the reactivity of para-substituted arenes, a researcher was in need of 4-neopentyl-1-isopropylbenzene (A). This question deals with aspects of the synthetic protocols necessary to prepare it from isopropylbenzene.

a) The initial synthetic scheme required 1-bromo-2,2-dimethylpropane, the

most efficacious synthesis of this being a free-radical bromination of 2,2-dimethylpropane (neopentane).

i) Provide a complete mechanism for the monobromination of neopentane. ii) What differences would you expect to see in the proton NMR for 1-bromo-2,2-

dimethylpropane versus neopentane? What about the CMR?

A

iii) During the purification of the reaction, a small amount (< 1%) of a side product was isolated (MW < 150). NMR analysis gave a spectrum with singlet resonances at δ 1.21 and 1.06 ppm in a ratio of 1 : 4.5. What is it and how is it formed?

b) Once prepared, 1-bromo-2,2-dimethylpropane was reacted with isopropylbenzene and AlBr3;

however, after work-up of the reaction mixture, NO A was isolated. Instead, a single product (with a molecular weight identical to that expected for A) was obtained in good yield. NMR analysis showed (in addition to other expected resonances) the unmistakable presence of an ethyl group. What is this product and provide a mechanism for its formation.

c) Starting with isopropylbenzene and any other materials you wish, synthesize A (more than one step will be required).

2. Acyclic dienes of the type shown below are notoriously unreactive in Diels-

Alder reactions, especially with R1 and R2 being C2H5 or larger. a) Draw and name the diene with R1 and R2 = C2H5 using correct

stereochemical nomenclature. b) Suggest a reason for the unreactivity of dienes of this particular stereochemistry. BE CLEAR IN

YOUR ANSWER.

R1

R2

3. Conjugated trienes will undergo electrophilic addition reactions to give only 1,2 and 1,6 addition products when reacted with one equivalent of electrophile. Consider the reaction of 3,4-dimethyl-1,3,5-hexatriene with one equivalent of HI.

a) Draw and name the two addition products expected to be formed in this reaction (ignore stereochemistry).

b) Provide a mechanism for the formation of the 1,6 addition product.

c) Which product (1,2 or 1,6) is expected to be the thermodynamic product and which is expected to be the kinetic product? Briefly explain the concepts of thermodynamic and kinetic control of a reaction.

d) No 1,4 addition product is seen. Explain why.

4. Briefly, what are the four key pieces of information obtained from an NMR spectrum, and what does each tell us about the molecule?

5. Provide a possible explanation for the following trend in IR stretching vibrations for C-X bonds: Fluorides:1400-1000 cm-1; Chlorides: 800-600 cm-1; Bromides: 625-400 cm-1; Iodides: 625-400 cm-1

6. Provide products for each step in the multi-step syntheses shown below:

CH3a)

CH3

1. AlCl3/ HO2CCH2CH2COCl2. Zn(Hg)/HCl; reflux3. PCl54. AlCl3 (dilute conditions)5. NH2NH2/KOH/heat

b)

1. AlCl3 / excess CH3CH2I

2. KMnO4 / KOH/heat

3. Na metal / NH3 / EtOH

c)1. H2SO4 / heat

2. NBS / light

3. EtO-Na+ / EtOH / heat

4. NBS / light

OH

7. Devise a synthesis of ANY TWO of the following molecules, using a Diels-Alder reaction in at least

one of the steps. You will probably need more than one step for at least one of them. You may use any starting materials you wish; pay attention to final stereochemistry if indicated.

CH3

O

CO2Et

CH3

HO

CO2Et

O

CN

H

H

CN

a) b) c)

8. Attached are NMR spectra for two compounds, each with the molecular formula C4H8O2.

Interpret each spectra and come up with a possible structure for each. Remember, you will be give most of your points for your logic and interpretation NOT for the final structure! Integral heights are provided in mm.

Carl Hoeger

Note

No spectra present. Sorry


1. a) Provide the mechanism for the Wittig reaction between benzyl triphenylphosphonium bromide (I)and acetaldehyde (II) in the presence of t-BuLi. Provide names of the intermediate species whereappropriate.

CH2 PPh3 CH3C

O

HBr

I II

b) Show how one might make benzyl triphenylphosphonium bromide (I).

2. The Grignard reaction is a convenient way to make a variety of different products.

a. Consider the reaction of phenyl magnesium bromide withdiethyl carbonate (III). What product would you expect toobtain in the presence of an EXCESS of the Grignardreagent? Provide a complete mechanism for its formation.

CH3CH2O C

O

OCH2CH3

III

b. When methyl magnesium bromide reacts with propionitrile (CH3CH2CH2C≡N) one obtains2-pentanone (DRAW!) after work-up with H+/H2O. Provide the mechanism for this addition;be sure to point out why the reaction stops where it does.

O

H3C CH3

H CH2O-tBu

IV(optically active)

c. Now consider the reaction of ethyl magnesium bromidewith oxirane IV, followed by treatment with dilute acidin water. Draw and name (with correct stereochemicaldesignations!) the product you expect to obtain.

3. Each of the following compounds leads to either an unsuccessful preparation of a Grignard reagentor an unsuccessful reaction with Grignard reagent. Briefly explain why for each.

OHBr

CN

CHO

O-tBuI

OTs

N(Me)3

I

O

OCH3

Br

O O

4. The following reaction scheme was employed in an attempt to make hexane-1,2,6-triol (TMS =trimethylsilyl):

OH

H2COH

HOCHHC1) NaNH2

2)

OTMSBr

A1) H2/Ni2B

2) Bu4N+F-

B1) MMPP/EtOH

2) H+/H2O/50°C

a. Draw the structures of A and B .

b. What is the product obtained after the MMPP step?

O

CH2

V

c. Upon completion of the reaction none of the triol was obtained; instead, a highyield of V was obtained. Provide a possible mechanism for its formation.

d. 1-trimethylsiloxy-4-bromobutane can be obtained in two steps from tetrahydrofuran. How?

5. Complete the following reaction sequences; provide stereochemical formulas where needed:

a) Cyclopentanol C

D

H+/Cr2O72-

PBr3 Mg/etherE F G

1) D

2) H+/H2O

PCC

b) Cyclohexanol H

K

H2SO4

heat

MCPBAO

1) N

2) H+/H2O

Na (s)I

H+ (dry)CH3OH

HBr (conc)

heatL + M

N

Mg/ether

P

Add M

trans-1-cyclohexyl-2-methoxycyclohexane

6. Devise a synthesis of each of the following. Unless indicated otherwise, you may use what everreagents and/or starting materials you wish:

a) O CH2Ph

b)

OTs

c)

using ONLY cyclopentene as your sole carbon source and employing a Wittig reaction

d)

using ONLY cyclopentene as your sole carbon source and employing a Grignard reaction

e)

OH

f)

OCH2CH3


Chem 233 Exam 3 Part 1: Transformation Mechanisms. There are two primary reasons for learning all of the reactions in the detail that we have for this course. The first reason is synthesis (of course). The second reason we learn so many reactions is as important, if not more, than synthesis: the “why” certain transformations occur. For the following reactions, provide a mechanism for their occurance. You will need to rely on all reactions and principles you have learned in organic chemistry to date to solve these problems. 1.

HNOH

Br

COOEt

Et3N

N

O

COOEt

+

2.

O

NO2

OH

NaH O

O

NO2 3.

SO2

O

CN

CH3CH

2ONa

SO2

O

NH

4.

CH3O

OCH3

CH2

O

O

CH3NH2 N O

OCH3

O

CH3

5.

O O

COOH

H+

O

O

OH

+ Acetone

6.

O

Br

1) NaOH

2) H+

COOH

7.

O

Cl-CH2-COOEt

NaOEt

O

COOEt

Part 2: Synthesis What else can I say? 8. Prepare, using a Robinson Annulation as one of the steps employed in your synthesis:

O

O

9. Prepare

O

PhPh

starting from cyclohexanone and anything else you wish. 10. Prepare

O

CN

starting from cyclohexanone and anything else you wish. 11. Design a method to accomplish the following transformation:

O

CN

O

12. Starting with α-bromo ethyl acetate prepare the following lactam.

N

N

O

CH3

CH3 In this synthesis you must employ both a Reformatsky reaction and a Michael addition (in separate steps).

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Chem140b Exam Study Info