Chemistry 303 1fall, 2001

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)

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:

1___________/21

2___________/21

3___________/34

4___________/08

5___________/16

Total: /100

There are 10 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 onTuesday. 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 itprecisely, carefully, and correctly. Show charges carefully. Distinguish between anintermediate and a transition state. A "good" mechanism is short and proceeds through lowerenergy intermediates/transition states.

NOTE: you may buy structures A, B , or C (Problem I) at a price of 4 pts each. you may also buy structure T (Problem III) at a price of 10 pts. Just ask the proctor.

PLEDGE:_________________________________________________________________

I. (21 pts) Consider the following reaction of X, and the products indicated. 2

ClCH3

Na :OMeether

A C6H10 major productB C6H10 minor productC C7H14O minor product

X

Note: you may purchase the structures of A, B, or C at a cost of 4 pts each

A. (1 pt) What is the most likely mechanism for the formation of A , the major product?

SN2 SN1 E2 E1 none of these (circle single best answer)

B. (1 pt) What is the configuration at the carbon bearing the methyl group in X ? R S cannot tell

C. (1 pt) What is the configuration at the carbon bearing the Cl substituent in X ? R S cannot tell

D. (1 pts) Draw carefully the enantiomer of X . E. (1 pts) Draw carefully one diastereoisomer of X .

ClCH3

mirror image of X ClCH3

ClCH3

or

same connectivity, non-superimposable, non-mirror images of X

cis isomers

F. (4 pts) Using the arrow formalism carefully, write the mechanism for the process giving A , showing clearly the structure of the major product.

Cl

Me

Na :OMeH

H

Me

H

A

H anti (trans) to the Cl is removed in an anti-parallel transition state for the E2

+ MeOH + Br-

G. (2 pts) Draw a careful representation for the transition state for this process, showing clearly all partial bonds and partial charges.

Cl

Me

H

H

Cl

Me

H

H

OMeδ−

δ− H

Cl

Me

OMeδ−

δ−

or

4 partial bonds, 2 being broken and 2 being formed. 2 partial (-) charges, from electrons leaving the OMe, and

arriving at the Cl 3H. (3 pts) There is an isomer of A which could be formed by the same mechanism, shown as B .

Please draw here the structure of B and indicate why it is less preferred than the major product.

Cl

Me Me

This would require a syn E2 elimination; higher energy than the anti parallel transition state.

HHO-

B

I. (3 pts) There is also another minor product, C, obtained as a single enantiomer. What is the structure of C? Write the mechanism for its formation.

ClCH3

Na :OMeOMe

CH3 cSN2

J. (4 pts) The first two products, A and B , are obviously isomers.1. (1 pt) Can you use Mass Spectrometry to differentiate them? yes noExplain, if necessary.

Since the compounds have the same molecular formula, a mass measurement is not sufficient to distinguishthem. However, you could use the fragmentation pattern differences, in principle, so a "yes" answer isallowed, with proper explanation.

2. (3 pts) 1H NMR will certainly be helpful. Each isomer contains a methyl group. Please explain how the chemical shift and pattern of peaks for the methyl group in A will differ from that for themethyl

group in B .

In A, the methyl group is adjacent to one H, and will therefore appear about 1 ppm at a doublet with J ca. 6-7 Hz.

In B , the methyl group is on a double bond, and therefore the signal is a little more downfield, perhaps about 1.8ppm. It has no H on the adjacent carbon, so it should be a singlet. However, there can be allylic coupling to theH on the double bond, with a small J value, and it could appear as a doublet. Either answer is acceptable.

II. (21 pts) Consider each of the following pairs of reactions. 4 Please decide which reaction is faster for each pair and write the product(s) for that reaction. Give the singlemost important reason why you think that one is faster. You probably have to draw the mechanism to make yourpoint. Give the name of the mechanism. If you have trouble deciding which mechanism is operating, specify areasonable mechanism, and then indicate which example would be faster by that mechanism. For each of thefaster reactions, indicate clearly which is the nucleophile and which is the electrophile.

Br HO NaHC4H8O

BrHO NaH

C3H6O

A. 1.

2.

FASTER

Hdeprotonation Br O SN2 O

+ Br-

The corresponding product here would be:O

The SN2 transition state would involve more bond angle strain compared to the TS leading to the five-membered ring, therefore this process is slower.

rate-determining step is the internal SN2

The nucleophile is the R-O and the electrophile is the carbon bearing the Br leaving group

NOTE: an E2 mechanism is also not completely crazy, to give a different product with the correct formula. If you did this, you should have gotten some partial credit. It does not normally happen with this base (NaH), but you don't know that.

________________________________________________________________________________________

Br2, H2O

CH2Cl2 solvent

CF3

1.

2.

B.

CH2Cl2 solvent

Br2, H2O

Br--BrBr Br Br

etc

This cation intermediate is stabilizedby resonance interaction with the phenyl group

In this reaction, theintermediate cation wouldbe destabilized by the inductivewithdrawing effect of the CF3 group

H2O

Br OH2

Br OHNucleophiles: Ph

and H2OElectrophiles: Br2 or HOBr and

Ph

Br_________________________________________________________________________________________

Br

Br (CH3)2NH

C.CH3OH

NH

Primary carbon undergoing substitution: SN2

N is a better electron donor than O (better base)

The rate-determining step involves the addition of the nucleophile to the electrophile; better nucleophile, higher rate.

The MeOH is smaller than Me2NH and for that reason might react faster, but the electronic effect is very strong. Partial credit for a well-reasoned argument that MeOH would be faster.

Nucleophile: Me2NH. Electrophile: -CH2Br

III. (34 pts) An Orgo student planned a synthesis of W from U , and figured that one of our 5standard substitution mechanisms would be perfect. He simply heated U in acetic acid and expected W wouldform in a fast, favorable reaction. The reaction was fast but, unfortunately, W was only a minor product. Themain product was an isomer, T. Another minor product, S , was also observed. Dr. Gingrich appeared andasked cryptically "Aren't you forgetting that the rate-determining step is not necessarily the product-determiningstep?" and then went off to have a cigarette. Note: you may buy the structure of T at any time for 10 ptspenalty.

I

CH3CO2HO

O

WU

+ T + S(major)

(minor)

(minor)

Spectral data for T: IR: 1743(s), 1676(m)1H NMR(CDCl3): δ 1.71(s,3H); 1.77(s,3H), 2.05(s,3H), 4.57(d, J=6Hz, 2H), 5.34(t, J=6Hz, 1H) ppm13C NMR(CDCl3): δ18.0, 21.0, 25.8, 61.4, 118.6, 139.0, 171.0 ppm

heat

A. (6 pts) Draw the mechanism which leads to W . Show any intermediates. What is the general name of thismechanism? Why was this reaction expected to have a particularly favorable rate-determining step?

I

SN1 CH2

cation intermediate especially stabilized by resonance. Also: tertiary and great leaving group

OOH

oxoniumion intermediate

OOH

OO

-H+

W

Acetic acid can be a nucleophile using the lone pair on the carbonyl oxygen (best, but not shown here) or using the a lone pair from the -OH group (shown)

B. (5 pts) Draw the structure for T and correlate it with the 1H NMR data. That is, make clear which H (or set of equivalent H) give rise to which of the 5 peaks. Make clear which H are coupled to which other H. Indicate any ambiguous assignments.

H3C

H3C O

O

CH3

H

HH

a

b cd

e a or b at δ 1.71 and 1.77 (ambiguous) no coupling, singletsc at δ 2.05, s, no couplingd at δ 4.57, coupled to e, J= 6 Hz, de at δ 5.34, coupling to d, J= 6 Hz, , t

continued....

III, cont. C. (5 pts). Write the best mechanism for the formation of T, showing all intermediates. 6Explain why this is the major product, in preference to W .

ICH2

O

O

H

O

O

intermedidate

intermediate

O

OT

The cation intermediate has (+) charge atboth ends of the allyl system. The nucleophilecan add to either end. The preferential formationof T is based on minimizing steric repulsion as the nucleophile approaches the cation, and the formation of the more stable alkene (same factors stabilize thetransition state a bit).

SN1resonance stabilized

H-H

+ H+

D. (6 pts) The only spectral data collected for S were the mass spectum [m/z 69(5.6%), 68(100%)] and the UV spectrum [λmax 223, ε 21,500]

1. Write the structure for S .

S2. Explain how this structure is consistent with the UV spectrum.

Two conjugated double bonds give an observable pi-pi* absorption at ca 220 nm, witha high ε.

3. Write a mechanism for the formation of S . What is the name of this mechanism?

I S

H H

SN1

-I- + H+

E. (4 pts). Compared to the rate of formation of W, is the rate of formation of T: higher lower same?

circle single best answer and explain your choice briefly.SAME

Prob III, cont. 7F. (8 pts). Now draw carefully two reaction coordinate diagrams, one for the conversion of U to W ,

and a second one for the conversion of U to T. On the diagrams, take care to make the energy levels qualitativelycorrect between the two so that the energy of transition states and intermediates can be compared between the two;the reactants start at the same energy.

Draw structures of intermediates but not transition states. Use these diagrams to explain what Henry meant by "the rate-determining step is not necessarily the

product-determining step."Identify the activation energies in the product-determining step for formation of W and for T. Your

diagram should make clear their relative magnitude.

U

G

extent of reaction

W

O

OH

ΔG# for the rate-determining step

ΔG# for the product-determining step

U

G

extent of reaction

T

O

OH

ΔG# for the rate-determining step

ΔG# for the product-determining step

In this diagram, the product-determining step is of lower ΔG# compared to that for W, while the ΔG# for the rate-determining steps are the same. Therefore, the formation of T is dominant.

IV. (8 pts). Consider the following conversion of M to N . 8

Na :CNO

H

CN

M N

Br

OH

H

(no stereochemistry implied)

A. (1 pts) Identify clearly the stereogenic centers in M with a circle.

B. (7 pts) The conversion of M to N is a multi-step process and leads to a single enantiomer of N . Draw the best mechanism here, showing all intermediates (but not transition states).Label each step as SN2, SN1, E2, or E1 where relevant.Draw clearly the exact structure (stereochemistry) of the product N .

Br

O

HH

:CN

M

+

SN2 at the less hindered carbon, with the better leaving group

Br

O

HH

O

HH

inversion here

SN2

SN2

N

cis configuration of the epoxide due to the inversionin the SN2 process

V. (16 pts) The following substitution reaction was studied in an effort to prepare one 9enantiomer of E. The desired pure enantiomer of E has a molecular rotation [α]D = +68o.

HI

CH3

CN

[α]D = +45o

CH3-NH2

[α]D = see conditionsD

in a solvent

HN

H3C

NCH

H

CH3E

A. (4 pts) When the reaction was run in hexane (100 mL) as a solvent the rate was very low, and the product had[α]D = +68o. Draw carefully the primary mechanism which must be operating in hexane solution. Show thestructure of the main enantiomer of E that is formed. Name the mechanism.

HI

CH3

CN

[α]D = +45o

CH3-NH2

Din a solvent H

CH3

NCN

H3CH

H

E

Since the optical rotation is exactly that of the desired enantiomer, the reaction must have proceeded with perfect inversion (or retention). That implies a pure SN2 process (we have no retention mechanism).

SN2

B. (5 pts) When the reaction was run in DMSO as solvent (100 mL) with everything else the same, the rate wasmuch higher but the product had [α]D = +18o. Use a mechanistic analysis with words and pictures to explainwhy the change in solvent to DMSO resulted in a higher rate and a lower molecular rotation for the product.The spectral data in all cases were completely consistent with the general structure E.

HI

CH3

CN

[α]D = +45o

CH3-NH2

D

H

CH3

NCN

H3CH

H

E

H

CH3

CNSN1

E + enantiomer (50:50)

SN2The reaction is faster because the more polar solvent favors the SN2 transition state, but it also (even more strongly) favors the ionization pathway, SN1. Both mechanisms operate, and the rotation is less than the maximum because of partial racemization.

C. (4 pts) In an effort to increase the [α] value, the reaction was tried again in DMSO, but this time 10 with ten times less solvent (e.g., with 10 mL instead of 100 mL). Everything else was the same.

Rewardingly, the [α]Dwas now higher, at +51o. Please explain in terms of your mechanism analysis in part B.

The SN2 reaction is bimolecular, and very sensitive to concentration effects--moreconcentrated, faster due to more frequent collisions between the reactants. Rate isproportional to the product of the concentrations.

The SN1 is unimolecular, related only to the conc of one reactant

Therefore at higher concentrations, the rate of SN2 is relatively faster than the SN1,and the rotation is closer to a pure SN2 result.

D. (3 pts) When the methylamine was replaced as nucleophile by aniline (Ph-NH2), reaction with D in DMSOgave a product (F) for which [α]D = 0o. Explain why this nucleophile gives a product with [α]D = 0o.

HNNCH

H

FAniline is a weaker nucleophile (base) because the non-bonding electron pair is heldmore tightly compared to methylamine. This is a resonance effect:

NH

HN

H

H

etc (two more strucutres delocalizing the electron pair into the benzene ring.

A weaker nucleophile will slow down the SN2 process, but will not affect the rate ofthe SN1. The rate of the SN1 depends only on the ionization of the halide. In thiscase, the SN2 is slowed sufficiently so that only the SN1 operates, and completeracemization is observed.

_________________________________________________________________________________________End Exam 3

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