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Chemistry 303

fall, 2009

THIRD EXAMINATION

7:30 PM, DECEMBER 16TH, 2009

Duration: 2.5 hr

Name___________________________KEY________________________________

This is an “open book” examination; you may use anything that is not alive or connected to the Web.

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

We love to give partial credit.

If there seems to be more than one good answer, explain your thinking.

If you invoke resonance delocalization as part of your answer, draw the relevant resonance structures.

If you draw a chair cyclohexane, be sure to orient the bonds carefully.

If you do not know a structure and need to write a mechanism, write a general mechanism for partial credit.

You need not draw transition states as part of a mechanism unless expressly instructed to do so.

USE THE ARROW FORMALISM CAREFULLY FOR ALL MECHANISMS. SHOW ALL INTERMEDIATES.

BE SURE TO INCLUDE ALL FORMAL CHARGES.

Write only in the space provided for each question.

Score:

p 2___________/12 p3___________/11 p4___________/08

p5___________/14 p6___________/10 p7___________/15 Lab question _________/14

p8___________/10 p9___________/ 12 p10__________/08

Lecture Total: /100

There are 12 pages in this exam; please check now to be sure you have a complete set.

Pledge:_________________________________________________________________________________ 

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II. (11 pts). Consider the simple SN1 reaction (no rearrangements) of the

secondary p-toluenesulfonate esters represented by X, and the relative rate data forthe cases where both R = Me, and where both R = tert-butyl.

A. (03 pts).  Draw the mechanism for the reaction involving R = Me.

Me

Me

OTs

X

HMe

Me

H

OH   H

Me

Me

OH

H

H

Me

Me

OH

H

 

B. (03 pts).  Explain in terms of the mechanism the single most important reason why the case with R = t-butyl

is much faster.

OTsHH

109o bond

angles; crowded

with t-Bu in place

of methyl

120o bond

angles; reduced

crowding. A bigger factor for the R = tBu case compared to the R = Me case.

Relatively larger relief of steric repulsion during ionization; the reactant is destabilized by steric repulsion more thanthe transition state is destabilized.

C. (05 pts). Now consider the SN1 reaction of substrate Y under the sameconditions with a very different outcome in relative rates.

 Explain in terms of mechanism why Y  with R = t-butyl

is much slower than than Y  with R = Me.

RBr

Y

RH

H

H

H

R

H

H

H

resonance stabilized intermediate cationrequires planar conformation

possible steric repulsion

With R = tBu, the steric repulsion inhibits the planar conformation and prevents effectiveresonance delocalization of the cation, raising the energy of the cation and the transitionstate leading to it, relative to the reactant.  

R

ROTs

H2O

50o C

  rates

R = Me = 1

R = t-Bu = 10,000

X

 

RBr

H2O

50o C

  rates

R = Me = 600

R = t-butyl = 1

Y

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III. (08 pts).  Draw the structure of an alkyl bromide having molecular formula C 5 H 11 Br that

 fits the criterion for each part  (different structure for each answer; in each case, explain in one sentence why your structure is a particularly good choice). There may be more than one correct answer; give only one.

A. (02 pts). Undergoes E1 elimination at the fastest rate (also draw the alkene).

H

Br

most stable cation

 

B. (02 pts). Is incapable of reacting by the E2 mechanism (not just very slow).

Br

no beta H to eliminate

 

C. (02 pts). Yields only a single alkene on E2 elimination (also draw the alkene).

Br

H

Br

Br

Br

+

would giveE/Z isomers

D. (02 pts). Yields the most stable alkene upon E2 elimination (also draw the alkene).

H

Br

Br

H

most substituted alkene

Could also use:

if not used in (A). 

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IV. (14 pts). A. (07 pts).  Write the best mechanism to account for the following conversion

of G to H  and J. 

H++

G

H  J

HH   HHb

Ha

HH

  H+

H   Ji iimore stable alkene

(-Ha) (-Hb)

B. (07 pts).  Draw one reaction coordinate diagram to represent this process, leading to both products. Be

sure to indicate the relative energy of intermediates and products carefully.

Intermediate ii is more stable than i due to relief of bond angle strain ( breaking a 4-membered ring),

although both are tertiary. H is the more stable alkene and this is reflected in the TS leading to it.

E

reaction progress

iii

J

G

H

 

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V. (10 pts).  Provide the starting material, reagent(s), or major product for each reaction below.

1. O3

2. Zn

O

MeH

O

A. (02 pts).

or:etc....

 

OEt

O

O

OO

HCl

B. (03 pts).

pay attention to exact stereochemistry

one mole

one moleOEt

O

O

cis-racemic

less reactive pi bond

due to electron withdrawingeffect of the C=O  

tBu

Me

a. OsO4 

b. reduction

C. (03 pts).

most stable chair form

HO

OH

Me

tButyl must be equatorial; HO (a) must be trans to tButyl due toleast hindered approach of reagent and therefore equatorial. HO(b)must be cis to HO(a) by the mechanism of OsO4 reactions and thereforeaxial.

(a) (b)

 

O

H

H

H

HH

O

H

H

H

HH

BrOEt

D. (02 pts).

O O

Br2, EtOH

or EtO-Br

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VI. (15 pts).  Consider the selective formation of L from H under the conditions shown.

a. NaHCO3COOH

LH

b. I2

undefined configuration

Et3NM C9H12O2

[13C NMR: ! 23, 25, 29, 31, 42,

82, 122, 132, 176 ppm]

O

O

I

+ Et3NH I

 

A. (01 pts). What is the absolute configuration of H ?  R or S circle one

B. (06 pts). Draw a mechanism for the reaction that accounts for formation of L as a single diastereomer.

Show clearly the stereochemistry of the substituents.

COOH

NaHCO3

CO

O

I--I

C O

O

I

Iodonium formation on theleast hindered face of the pi bond

C O

O

IL  

C. (02 pts).  Draw L in its lowest-energy

conformation using the chair version. 

D. (06 pts).  Draw the mechanism for the conversion

of L to M, showing clearly the structure of  M.

 Explain why this particular alkene is formed in terms

of the mechanism. Use the appropriate chair

representation of L.

I

O

O Et3N:C O

O

M M

alternate chair for L withonly good transition state for E2 (anti conformation)

L

("normal" chemical shifts in the

  13C NMR spectrum

C O

OAlternate isomer is not consistent

with a good anti mechanism, norwith the 13C NMR data:

a

b

(a) is moved downfield by O; (b) is movedupfield by O resonance donation.

O

I

H

O

O

O

I

O

O

 L

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IX. (08 pts). A. (04 pts).  Draw here all four stereoisomers of 1,2,4-trimethylcyclopentane and

label them 1, 2, 3 and 4. Then answer the following questions using your labels. If the answer is "none", write

"none". 

1   3   431   422

B. (01 pts). Give one pair of enantiomers.

C. (01 pts). Give one pair of diastereomers.

D. (01 pts). Give one meso structure.

E. (01 pts). List all chiral structures.

_______________________________________________________________________________________

End Lecture Exam Portion

2,3

1,2  or 1,3 or 1,4 or 2,4 or 3,4

1 or 4

2 and 3

1,2,4-trimethylcyclopentane

  (no stereochemistry  intended)

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X. (14 pts).  Laboratory:  Hugh B. DeMann, former orgo lab student, attempted to prepare 3-

methyl-2-butenyl acetate (2) from 1-chloro-3-methyl-2-butene (1) and acetic acid, as shown below.

H3C

H3C

Cl H3C

H3C

OCCH3

O

!

1 2

+ 3

CH3CO2H

 

To his chagrin, the reaction also produced isomer 3. The 1H NMR spectrum of isomer 3 is summarized below:

Isomer 3  1H NMR (CDCl3, 400 MHz): " 1.52 (s, 6H), 1.99 (s, 3H), 5.07 (dd, J  = 10.9 Hz, J  = 0.9 Hz, 1H)

5.17 (dd, J  = 17.5 Hz, J  = 0.9 Hz, 1H), 6.08 (dd, J  = 17.5 Hz, J  = 10.9 Hz, 1H)

 Hint:  read both parts of the question before analyzing the data. The mechanism and the spectral data shouldwork together to help you.

A. (11 pts).  Deduce the structure of isomer 3. Draw the structure of isomer 3 and label each unique

hydrogen on your proposed structure. Under the structure, indicate your chemical shift assignments for the

resonances at !  5.07, 5.17, and 6.08 ppm. Also, carefully explain the splitting patterns and make coupling

constant assignments for these resonances.O

CH3

CH3

Hz

Hx

Hy

CH3CO

3

 

" 1.52 (s, 6H) C(CH3)2 

1.99 (s, 3H) CH3CO2 5.07 (dd, J XZ = 10.9 Hz, J XY = 0.9 Hz, 1H) HX 5.17 (dd, J YZ = 17.5 Hz, J XY = 0.9 Hz, 1H) HY 6.08 (dd, J YZ = 17.5 Hz, J XZ = 10.9 Hz, 1H) HZ

HX at " 5.07 ppm is first split into a doublet by the cis HZ (J XZ = 10.9 Hz) and then into a doublet of

doublets by the geminal HY (J XY = 0.9 Hz).

HY at " 5.17 ppm is first split into a doublet by the trans HZ (J YZ = 17.5 Hz) and then into a doublet of

doublets by the geminal HX (J XY = 0.9 Hz).

HZ at " 6.08 ppm is first split into a doublet by the trans HY (J YZ = 17.5 Hz) and then into a doublet of

doublets by the cis HX (J XZ = 10.9 Hz).

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B. (3 pts). Tell Hugh (and us) why he should not have been surprised to obtain two isomeric products

 from this reaction.  (Hint: Formal mechanisms are not required here, but a brief mechanistic analysis of the

reaction seems warranted.)

The absence of a strong nucleophile suggests an SN1 mechanism. Ionization of 1  affords a

resonance-stabilized allylic carbocation, which can be captured by acetic acid at either of the carbonssharing the positive charge. Deprotonation affords the observed acetates 2 and 3.

H3C

H3C

Cl H3C CH2

H3C

H3C   CH2

H3C

H3C OH

O

1

- Cl-

SN1+

+

:        :

(a)   (b)

 

(a)

H3C

H3C

OCCH3

O

(-H+)   (-H+)

2

(b)

OCH3

CH3

CH3CO

3