303 10 3rd examkey.pdf

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III. ( 08 pts). Identify the most likely product for each of the steps given by placing the appropriate compound letter in

each of the circles provided; you may use a product only once. If the choice seems ambiguous, explain your thinking.

H3CCH3

OS

CH3

O O

CH3NaI

THF

KOt -Bu

THF

NaCN

CH2Cl2

H3CCH3

CN

CH3

H3CCH3

CH3

H3C

CH3

H3CCH3

OMe

CH3

H3CCH3

I

CH3

H3CCH3

CN

CH3

H3C

CH3

H3CCH3

CH3

AB C D

E FG H

I

CH3

I

H3CCH3

OMe

CH3

(racemic mixture)

MeOH

AgNO3

H3CCH3

CH3J

OMe

CG

F I I by Sn1. But B is possible by a Ag-assisted SN2 for partial credit.

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IV. (08 pts). Deuterium serves as a useful hydrogen analog when studying reactions, as it has the same steric influence on

reactions and participates in the same types of reactions as hydrogen. There is a difference in its BDE.

(The C-H BDE=104 kcal/mol, the C-D BDE=106 kcal/mol.)

A. (03 pts)

D3C CD3

Br

D2C CD3

H2O

heat+ D-Br

compared toH3C CH3

Br

under the same conditions?

D

H

Would D react faster slower same rat e

Circle single best answer

Give the name of the mechanism operating here and explain your choice.

This is an E1 reaction with ionization of the bromide as the rate-determining step. In the second step, the C-H (C-D) bond

is broken, but that is irrelevant to the RDS.

The E2 reaction requires a strong base to be favorable.

_____________________________________________________________________________________________________

B. (03 pts).



D3C CD3

Br

D2C CD3

KOtBu+ KBr + DOtBu

D

Osolvent

KOt-Bu = K O compared toH3C CH3

Br


H


This now becomes an E2 reaction with the strong base promoting it. The mechanism has C-H (C-D) bond breaking in the

RDS, and therefore, the deuterium labeled version (shown) reacts more slowly.

_______________________________________________________________________________________________________

C. (02 pts).

D3C CD3

Br

D3C CD3

Me-S

+ Br

compared toH3C CH3

Br


D

H


SMe



DMSO

With a powerful nucleophile/weak base, the SN2 reaction is most likely. But even if it was an SN1 reaction, this is a

substitution process and there is no C-H or C-D bond breaking, so there is not difference in rate due to deuterium

substitution.

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VII. (11 pts). The following reaction yields a number of products, one of which is alcohol M.

H3C CH3CH3

BrAgNO3

H2O

M

13C NMR for M shows 4 peaks

[!]D = 00

L

A. (07 pts). What is the structure of the product M and how does it form (write the mechanism)? Clearly draw all of the

intermediates in this transformation. Explain how your structure is compatible with the observed 13C NMR data and optical rotation

data.

H3C CH3

CH3

Br

L

Ag+

H3C CH2

CH3

HH

H3C CH3

CH3

HH

hydride shift togive more stablecation

OH2

H3C CH3

CH3

OHH

H3C CH3

CH3

OH -H+

M

M has a plane of symmetry;not chiral.

4 different carbon environmentsno optical rotation

B. (02 pts). Re-draw L here and label the stereogenic carbon(s) as R or S.

H3C CH3

CH3

Br

S

S

C. (02 pts). What would you predict as the sign of the optical rotation for L (not M!)? positive negative cannot tell

(Circle single best answer)

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X. (11 pts). Note that both of the isomers X and Y can be converted to the epoxide Z by treatment with sodium hydride

but the rates are very different. Note that the planar depiction does not give a useful picture of reactivity.

OH

Me

BrO

Me

Br

Me

OH

FAST SLOW

X YZ

NaH NaH

A. (04 pts). Draw the two chair representations

for X. Use the Data on p 14 to calculate thedifference in torsional strain energy between the

two. Circle the more stable chair.

_____________________________________________________________

B. (02 pts). Draw the two chair representations for Y.

Circle the more stable one.

MeOH

Br

Me OH

Br

axial Me (-1.7) +axial Br(-0.5) +axial OH (-1.0) +1,3-diaxial Me-Br(approx -1.9) = approx -5.1torsional strain

One gauchediequatorialapprox. -0.9torsional strain

Y

C. (05 pts). Write the mechanism for the formation of Z from Y. Show all intermediates. Explain carefully with reference to yourmechanism why X reacts much faster than Y.

MeOH

Br

Me O

Br

Na H:

Me O

Br

1,2-diaxial arrangementis crucial for goodtransition state for

the SN2

MeO

Me

O

H

The transition state for the SN2 requires a 180

o relationship for the nucleophile (alkoxide, in this case) and the leaving

group (Br, in this case). This, in turn, requires the trans diaxial orientation, and not trans diequatorial nor cis-

axial/equatorial.

Structure X has the requisite trans-diaxial relationship in the more stable isomer. However, Y must flip to a very strained

chair conformation in order to have the trans diaxial relationship. This extra energy to get to the correct conformation

increases the overall activation energy for the RDS compared to the process from X.

Me

Br

OH Me

Br

OH

axial Br (-0.5) + axial OH (-1.0) = -1.5 torsional strain

axial Me (-1.7) +gauche diequatorial (approx -0.9) =-2.6 torsional strain

X

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XII. (14 pts). Lab-related question.

Treatment of !-chloroallylbenzene (1) with silver acetate (the silver salt of acetic acid, CH3CO2 – Ag+) affords two isomeric products, 2

and 3 (C11H12O2).

CHCH

Cl

CH2

AgOAc2 + 3

1 The 1H NMR spectrum of the major isomer 2 is summarized here:

1H NMR (CDCl3 300 MHz)*: " 2.08 (s, 3H); 4.71 (d, J = 6.0 Hz, 2H); 6.27 (dt, J = 15.9 Hz, J = 6.0 Hz, 1H); 6.63 (d, J = 15.9 Hz,

1H); 7.2-7.4 (m, 5H)*This spectrum was simplified by omitting coupling (long-range or otherwise) of less than 2 Hz.

A. (09 pts). Deduce the structure of isomer 2. Draw the structure of isomer 2 and label each unique hydrogen on your proposed

structure. Under the structure, indicate your chemical shift assignments for all of the 1H NMR resonances. Also, carefully explain the

splitting patterns and make coupling constant assignments.

A)

The HB protons at " 4.71 ppm are split into a doublet by the HC proton with J BC = 6.0 Hz.

The HC proton at " 6.27 ppm is first split into a doublet by the HD proton with J CD = 15.9 Hz, then

into a doublet of triplets by the two HB protons with J BC = 6.0 Hz. The magnitude of J CD (= 15.9 Hz) isdiagnostic for trans vicinal alkene coupling.

The HD proton at " 6.63 ppm is split into a doublet by the HC proton with J CD = 15.9 Hz.

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B. (05 pts). Explain how isomer 2 was formed (i.e., write a detailed arrow formalism for the formation of 2 from 1

and silver acetate. Please show all intermediates and every step in exquisite detail. Also, please indicate the role played by the silver

ion.) Why is the formation of isomer 2 favored over isomer 3 in this reaction? (Hint: You should propose a structure for isomer 3,

based on mechanistic considerations.)

B) The presence of the silver ion should surely suggest an SN1 ionization, assisted by the silver ion.

The resulting doubly resonance-stabilized benzylic/allylic carbocation can be captured by the acetateanion at either of the carbons sharing the positive charge to give isomers 2 and 3. Isomer 2 is themajor product because the C=C bond in 2 is more substituted than the C=C bond in 3, but, more

importantly, the C=C bond in 2 is now conjugated with the benzene ring. (Or, if you thought thatisomer 3 was the cis (Z ) isomer of 2, then the trans (E ) isomer is favored over the cis (Z ) isomer forsteric reasons.)

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Typical gauche interactions: 0.9 kcal/mol

Typical 1,3-diaxial interactions: approx 2.0 kcal/mol

Glossary:

Me = methyl Et = ethyl Ph = phenyl tBu = tert-butyl

O THF

303 10 3rd examkey.pdf

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