1Chemistry 303

fall, 2002

THIRD EXAMINATION7:30 pm December 3rd

Duration: 2.5 hr

Name______________________KEY________________________________

Lab TA__________________________________________________________(if you do not know his/her name, give day of lab section. NOT Hooley nor Kim)

This is an "open book" examination; you may use anything which is not alive.

Note: if you do not know the complete or specific answer, give a partial or general answer--

WRITE SOMETHINGWrite only in the space provided for each question.

Score:

I _____/31

II _____/12

III _____/28

IV _____/11

V _____/05

VI _____/05

VII _____/08

Total: /100

There are 13 pages in this exam. Please check now to be sure you have a complete set.

If you are using a resonance argument in your answer, draw the relevant resonance structures.

If you are asked to analyze a structure and you have no idea what it is, do a general analysis of the data andpropose partial structures.

Please be aware that a small number of students will be taking the exam at different times up until late afternoon onWednesday. It would be well not to discuss the exam until after that time.

NOTE: You will be asked to write mechanisms and use the arrow formalism. Please use it precisely,carefully, and correctly. Show charges carefully. Distinguish between an intermediate and a transitionstate. A "good" mechanism is short and proceeds through lower energy intermediates/transition states.

PLEDGE:_________________________________________________________________

ClX

I. (31 pts) Consider the molecule X.

Ph

or

ClX

Note: Ph is larger than Cl2

A. (04 pts). Draw the most stable chair form of X here and label the Ph and Cl groups as being axial or equatorial. State any assumptions. Take care that you do not draw the enantiomer of X.Then draw the least stable chair form of X. Indicate clearly which is the more stable. Take care that youdo not draw the enantiomer of X.

more stable less stable

Ph

HCl

H

H

Cl

Ph

H

The smaller substituent is axial The larger substituent is axial

equatorial axial

equatorialaxial

B. (03 pts). In the planar representation of X (above),what is the absolute configuration of the carbon bearing the Cl: R S cannot tell

In the most stable chair form of X (above),what is the absolute configuration of the carbon bearing the Cl: R S cannot tell

In the leas stable chair form of X (above),what is the absolute configuration of the carbon bearing the Cl: R S cannot tell

The configuration have to be the same, since the chair flip does not change the configuation, only the conformation.

C. (02 pts). Draw the enantiomer of X, in either the planar representation or a chair representation.Your drawing must make clear the exact configurations of the stereogenic carbons, as in the structure at thetop of this page. Sloppiness will cost you points.

Ph

HCl

HH

Cl

Ph

H

Cl

Ph

Just the mirror image, obviously, Easy to draw if you just copy one through an imaginary mirror.

D. (02 pts). Draw one of the diastereoisomers of X here, in its most stable chair form. State any assumptions.You may use the template or make your own.

Ph

HH

Cl

Cl

H

Ph

Hmore stable

any cis arrangement is OKcont inue

E. (20 pts) Imagine the four main reaction types applied to X, SN2, SN1, E1, and E2. 3Each of the reactions below follows a different one of the primary mechanisms. In three cases, two isomers areformed and in the other case, only one. For each of the examples below,

a. name the mechanism being followed.b. draw the mechanism, using the usual arrow formalism. You need not show transition states.c. draw the product(s) carefully. Specify if the products are stereoisomers and, if so, whether they are

enantiomers or diastereoisomers.d. explain briefly why you think these conditions favor the mechanism you chose.

ClX

Ph

CH3-S Nadiethyl ether solvent

1.

SMe

Ph

SN2 mechanism.

A single product is formed, due to a one-step substitution with inversion.

The sulfur anion is very polarizable and therefore a good nucleophile. The rate of SN2 depends on the reactivityand concentration of the nucleophile. The non-polar solvent and a secondary chloride do not favor ionization. Thesulfur anion is only a moderate base, and therefore E2 is not favored.

______________________________________________________________________

ClX

Ph

Ag NO3

CH3NH2 solvent

2. Ph

Hplanar cationintermeidate

Me-NH2

Me-NH2

N

Ph

N

Ph

NHMe

Ph

NHMe

Ph

MeH H

MeHH

SN1

-H

-H

Two stereoisomers are formed. Either as the ammonium ions or free amines after loss of a proton.They are diastereoisomers and not mirror images; one stereocenter is the same in each, the other inverts. Theyneed not form in equal amounts.

The silver ion strongly accelerates ionization, and with only a moderately reactive nucleophile, the usual SN2cannot compete with the ionization process. But the nucleophilic solvent is sufficiently reactive to trap the cationintermediate before significant E1 elimination can occur.

cont inued . . .

ClX

PhAg NO3

CH2Cl2 solvent(non-nucleophilic)

3.Ph

H

H

H

Ph

HHPh

H

HThe products are not stereoisomers--they have differentconnectivity.

4

Ionization is favored by the presence of silver ion. In the absence of any serious nucleophiles, the E1 processwill predominate over the SN1.

______________________________________________________________________

ClX

Ph Na

diethyl ether solvent

CH3C

H3CH3C

O4.

H

Cl

HPh

CH3C

H3CH3C

O

HPh

PhHPh

Ph

The products are not stereoisomers; theyhave different connectivity. They need not be formed in equal amounts, but it is difficult to predict which would be preferred and by how much.

E2

With the Cl axial, there are two adjacent axial and either one of them can be removed by base to initiate elimination.

In addition, the bulky base molecule, tert-butoxide, tends to disfavor the SN2, and not significantly disfavor the E2.

II. (12 pts) Consider each of the following pairs of reactions. Please decide which reaction 5is faster for each pair and write the product(s) for that reaction. Give the single most important reason why youthink that one is faster. You probably have to draw the mechanism to make your point. Give the name of themechanism. If you have trouble deciding which mechanism is operating, specify a reasonable mechanism, andthen indicate which example would be faster by that mechanism. For each of the faster reactions, indicate clearlythe nucleophile and the electrophile.

Br

OH

AgNO 3

AgNO 3

MeOH, heat C11H14O

MeOH, heatC11H14O

HO

Br

OH

SN1

O H

O

-H+

The silver-promoted reaction is much slower with a primary bromide compared to a tertiary bromide, because a primary cation is less stable than a tertiary cation.

Another possibility is the E1 elimination, giving HO also: C 11H14O

However, the intramolecular addition of OH to give a 5-membered ring and is a very favorable process. This answer was given -1 for the wrong product, but full credit for the right mechanism.

continued

H H

Br NaCN

DMSO solventC11H19N

H H

Br NaCN

DMSO solventC11H19N

Br

Br

CN

Br

H

H

CN

δ-

δ-

transition state

CN

CN

NC H

Br

δ-

δ-

transition state

This conformer with the axial bromide is higher energy than the configurational isomer below with the equatorial bromide. The transition states are essentially identical in energy, so the activation energy from the axial isomer is less compared to that for the equatorial isomer. Therefore the axial isomer goes faster in an SN2.

NOTE: Cyanide is a so-called "ambident" anion, as is obvious from the resonance structure here:C N

Therefore it is not unreasonable to imagine SN2 reaction via the nitrogen end to give:

NEither product would be acceptable, the analysis is the same.

axial

equatorial

C

This is an unusual but real functional group, and "isocyanide"

6

III. (28 pts). HBr adds to allylbenzene (Y) in dichloromethane solvent to give three isomers in 7varying amounts. The major isomer accounts for about 96% of the product, while two minor isomers appearto the extent of 1-2% each. The isomers were identified by 1H NMR.

Y

BrBrCH2Cl2

H-BrBr

1 2 3+ +

(96%; racemic) (1-2%) (1-2% racemic)

A. (03 pts) Draw structure 2, showing all H. Note from the NMR data below that, in addition to thephenyl group at 7.2-7.5, there are 3 types of H in the molecule. Label those H as a,b, and c on your structure.Correlate the NMR spectral data with your structure: explain why each proton type has the pattern listed. 1H NMR (CDCl3): 2.20 (2H, quintet, J=7 Hz); 2.75 (2H, triplet, J=7 Hz); 3.35 (2H, triplet, J=7 Hz );

δ 7.2-7.5 (multiplet, 5H)

Br

Ha Ha

Hb Hb

Hc Hc Ha at 2.75, coupled to two equivalent H (Hb), a triplet with J=7Hb at 2.20, coupled to 2 pairs of equivalent H (Ha and Hc), all with J=7, therefore appearing as a quintetHc at 3.35, coupled to two equivalent H (Hc), a triplet with J=7

Ha and Hc assigments could be reversed for full credit.

B. (05 pts). Draw the major product, 1, here, showing all H. Note from the NMR data that, in additionto the phenyl group at 7.2-7.5, there are 4 other types of H in the molecule. Label those H as a,b,c, and d.Correlate the NMR spectral data with your structure. explain why each proton type has the pattern listed. 1H NMR (CDCl3): 1.60 (3H, doublet, J=6 Hz); 3.00 (1H, doublet of doublets, J=14 Hz and 6 Hz);

3.20 (1H, doublet of doublets, J=14 Hz and 6 Hz), 4.22 (sextet, 1H, J=6 Hz), δ 7.2-7.5 (multiplet, 5H).

CC

C

BrHa at 3.00, coupled to Hb with a large geminal coupling constant, 14 Hz; and coupled to to Hc with a typical vicinal coupling constant of 6 Hz. Overall appearing as a doublet of doubletsHb at 3.20, coupled to Ha with a large geminal coupling constant, 14 Hz; and coupled to to Hc with a typical vicinal coupling constant of 6 Hz. Overall appearing as a doublet of doubletsHc at 4.22, coupled to two non-equivalent H (Hb, Ha), both with J=6; coupled to three equivalent H (Hd) with J=6. Therefore appearing as a sextet.Hd at 1.60, coupled to Hc with J = 6, appearing as a doublet.

Ha Hb

Hc

HdHdHd

Ha and Hb are adjacent to a stereogenic center, and are therefore diastereotopic and chemically different. Can't say which has which chemical shift position, either could be 3.00 or 3.20

C. (06 pts) Write the best mechanism for the formation of 1, using the arrow formalism. 8Then show the formation of 2 under the same conditions. Explain carefully why 2 is a minor product and 1 is themajor product. Explain why 1 is racemic; include in your answer a definition of the term "racemic".

H BrH Br

HBr H

HH Br

+

racemic mixture

H

H

H BrH

H

HBr

1

2

The major product is formed from the more stable secondary cation, formed by protonation at the CH2 group in the alkene. The transition state leading to this cation is lower energy than the one leading to the primary cation, and therefore more of product 1 is formed, relative to 2.

Racemic means that a chiral object exists as a 50:50 mixture of enantiomers. Cpn 1 is racemic because the intermediate cation is planar and there is no asymmetric influence in the reaction. The Br- MUST attack equally from either side of the planar cation.

D. (04 pts) Draw carefully the Energy/Reaction Progress diagram for the addition of HBr toallylbenzene to give 1. Indicate clearly the location of all transition states, the reactants, and the product, with theright relative positions on the chart.

cationE

rxn progress

TS-1

TS-2

reactants

products

E. (02 pts) Compared to the transition state for formation of 1, the transition state for the formation 9of 2 is: (circle single best answer) higher lower the same cannot tell

F. (04 pts). Consider the other minor product, 3. Write the best mechanism for its formation whenallylbenzene reacts with HBr.

Br

3H-Br

H

H

HH

Br

The H shift is favored by the formation of a cation adjacent to the benzene ring, with its associated resonancestructures

G. (04 pts). When the reaction of allylbenzene with HBr is carried out in methyl alcohol solvent instead ofdichloromethane, the major product does not contain Br. Instead, the 1H NMR spectrum shows a new singlet at δ3.65 (3H) and otherwise approximately the same pattern as for 1. This reaction requires only a catalytic amount ofHBr. Write the mechanism for this reaction in methyl alcohol, showing all intermediates. Explain why 1 doesnot form in significant amounts, and why only a catalytic amount of HBr is needed..

H-Br

H MeOH HO

MeH

-H+

HO

Me

The proton is transferredto start things off, and thenis regenerated from the MeOH; overall the H+ is not consumedand is therefore a catalyst.

Cpn 1 requires the addition of Br- , which cannot compete with the addition of MeOH, a moderate

nucleophile in huge excess as the solvent.

IV. (11 pts) Treatment of S-2-bromobutane, D, with EtONa in EtOH solution leads to both 10E- and Z-2-butene. The E isomer is by far the major product in this reaction.

HH

CH3H3C

HBr EtO Na

EtOH E-2-butene (major) + Z-2-butene

D

A. (03 pts) Using the Newman projection, draw here the staggered conformations of D. Label the most stableand the least stable, and explain your choices qualitatively (you need not quote energy numbers). Hint: Br islarger than CH3.

Me

H BrH

HMeMe

H BrMe

HHMe

H BrH

MeH

most stable least stable

B. (04 pts) Draw the mechanism for the formation of E-2-butene and Z-2-butene, making clear the exactly howeach is formed. It will be useful to use the "sawhorse" projection to show the mechanism.

Me

HBr

Me

H H

OEtanti-H to Br

E2 Me

H

Me

H

Me

HBr

H

Me H

OEtanti-H to Br

E2 Me

H

H

Me

more stable product

less stable product

Many answers involved the eclipsed conformation, as drawn above for D. Obviously, not only is that a highenergy, unlikely conformation, but it also does not allow the anti relationship that is ideal for breaking the C-H andthe C-Br bonds at the same time.

11cont inued

C. (04 pts) Explain why the E-isomer is favored.It will be helpful to use reaction coordinate diagrams

The E isomer is more stable because the Z isomer has some steric repulsion with the cis Me groups. The transitionstate leading to the Z isomer also has gauche interactions of the Me groups which makes it higher in energycompared to the transition state leading to the E isomer, where the Me groups are anti. This is independent of anyarguments regarding the energy of the conformations of the reactants. All activation energies to the transitionstates must start from the more stable conformer.

three conformers

TS leading to Z

TS leading to E

Z alkene

E alkene

__________________________________________________________

HN

N

O

OH

O O

R-thalidomide

V. (05 pts)

HO

HN

N

O

O

O O

planar anion, stabilized byresonance interaction withthe carbonyl group

HN

N

O

O

O O

H+

HN

N

O

OH

O O

+

HN

N

O

OH

O O

racemic, equal probability of protonation from either face of the planar anion.

The stereogenic center bears aproton which is activated by the adjacent carbonyl group, and is therefore relatively easily abstractedby basic medium

S R

12

VI. (05 pts) Methyl alcohol and acetonitrile have similar polarity, but yet the rate of the reaction ofsodium cyanide with methyl iodide increases by a factor of 4 x 104 on changing the solvent from methyl alcohol toacetonitrile. Write the mechanism for this reaction, and explain carefully why the reaction is much slower inmethyl alcohol compared to acetonitrile. Use reaction coordinate diagrams to illustrate your point.

acetonitrile is CH3CN

NC + H3C-I NC-CH3 + I C INC

H H

H

δ δ transition state

This reaction starts from an anion (cyanide anion) and proceeds to a transition state where the (-) charge is spreadout over two atoms. There is a decrease in the concentration of charge as one proceeds to the transition state.

MeOH strongly stabilizes the reactant anion by H-bonding, and stabilizes the transition state less effectively.Acetonitrile has no H-bond donor capability (it is a very weak H-bond acceptor) and does not stabilize the anionreactant particularly well.. Therefore, the activation energy in MeOH is larger than the activation energy inacetonitrile, because the reactant anion is relatively more stabilized by the MeOH.

E

rxn progress

product (in either solvent)reactant anion in MeOH

reactant anionin acetonitrile

ΔG in acetonitrileΔG in MeOH

(larger)

13

VII. (08 pts) An Orgo student at Harvard was interested in producing the dibromide U,([α] = +30o), and he had available the pure enantiomer, T ([α] = +25o). Having memorized everything aboutsubstitution chemistry, he tried adding HBr, to replace the -OH with -Br and make the desired enantiomer U, but hewas concerned that there might be a little racemization. He was totally shocked when he isolated his product,verified that it had exactly the right NMR and other spectral data for U, but showed [α] = -30o.

C CH2BrHO

H

MeT

U

HBrCH2Cl2

A. (04 pts). Draw the mechanism by which the student expected to get U. Show all intermediates, you need not show transition states. Why was he worried about racemization? How might this happen?

C CH2BrO

H

MeH

HBr

C CH2BrH

Me

Brtypical SN2 with inversion

C CH2BrHMe

C CH2BrH

Me

Br

C CH2BrBr

Me

H+ racemic

SN1 is a possible mechanism, although this is only a secondary cation possibility. If it occurs, it will lead to racemization and decrease the rotation value of the product. In this relatively low polarity solvent, the SN1 is not likely to be favored

B. (04 pts). Most importantly, what went wrong? Draw the exact structure of the product he did get, and explainwith a detailed mechanism how this product formed exclusively in this reaction.

C CH2BrHO

H

MeC CH2Br

HBr

MeT U

HBrCH2Cl2

C

O

HMe

H H

Br

HH C

HMe

Br

HH

Br

C CH2BrBr

H

Meretention

The primary bromide participated in the leaving of water to give a bromonium ion; then the Br- added to thebromonium ion at the more substituted site, the site of higher positive charge. This is exactly the mechanism of

bromine addition to propene, which was discussed explicitly in class. The overall result is retention, via doubleinversion.

14

C. (BONUS: 02 pts) A brilliant Princeton student (you?) developed a process that does successfully convert T to U. What would that be?

This is not so obvious. Any idea that employed a different mechanism of substitution is a candidate to solve theproblem. For example, one might prepare the sulfonate ester of the alcohol, and then do an SN2 reaction withNaBr, avoiding any temptation for formation of a cation and active participation by the primary bromide. Ofcourse, the primary bromide might participate in any substitution, so this idea may not work. It depends on adelicate balance of rates of the competing processes. A solvent which did not particularly stabilize the Br- would bean advantage--noon-polar, aprotic solvent.

C CH2BrHO

H

Me

SO

OCl

C CH2BrO

H

Me

SO

O

a sulfonate ester

Br

C CH2BrHBr

MeOS

O

O+