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Chemistry 303

fall, 2006

THIRD EXAMINATION

7:00 PM, DECEMBER 5th, 2006

Duration: 2.0 hr

Name____________________________________________________________

Lab TA___________________________________________________________

(if you do not know his/her name, give day of lab section—Not Shawn nor Bill)

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—

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.

USE THE ARROW FORMALISM CAREFULLY FOR ALL MECHANISMS

BE SURE TO INCLUDE ALL FORMAL CHARGES

IF YOU ARE ASKED TO WRITE THE PRODUCTS FROM A PROCESS INVOLVING RACEMIZATION, BE SURE TO

DRAW BOTH STEREOISOMERS TO MAKE IT CLEAR.

Write only in the space provided for each question. 

Score:

p2___________/10 p3___________/10 p4___________/14 p5___________/12

p6___________/08 p7___________/12 p8___________/15 p9 _________/16 p10 __________/10

Total: /107

There are 10 pages in this exam; please check now to be sure you have a complete set.

Please be aware that a small number of students will be taking the exam at different times up until the morning on Wednesday. It

would be well not to discuss the exam until after that time.

Pledge:_________________________________________________________________________________

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I. (10 pts).  Considering the following pair of reactions, please choose the one that you predict will be faster and draw the

 product(s) from that reaction ( 2 pt s)

ClINaI(1)

A.

(2)

ClBrNaI

ClBr

 

 A. (04 pts).  Name the mechanism and draw it, including the transition state for the rate-determining step.

The substrate for eq 1 is an allylic bromide and will react faster in both an SN1 (resonance delocalized cation) and SN2process (pi bond stabilization of the transition state. The conditions do not include a strong base, so E2 is not likely. The

solvent is fairly non-polar, so ionization is unlikely. The chloride is not underconsideration as a leaving group, since in (1)it is attached to an sp2 orbital (strong bond) and in 2, it is secondary (hindered). Chloride is intrinsically a less good

leaving group than Br because it forms a stronger bond to carbon and is a slightly less stabile anion (weaker conj base).

Most likely SN2.

ClBr

I

H H

Cl

BrI

!" !"

TS

ClI

+ Br

 

 B. (04 pts).  Then explain why your choice is the faster reaction, focusing on the single most important reason for thedifference and based on the TS picture.

H H

Cl

BrI!" !" In this TS, the pi orbitals can interact with the p orbital

at the carbon undergoing substitution, stabilizing this orbital andtherefore lowering the TS energy.

 

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II. (14 pts). Consider the molecule B. It is one of these versatile substrates that can do any of the mechanisms we have

talked about. Which mechanism occurs depends on base/nucleophile and solvent character.

H

OSO2Me

H

B

diastereoisomer of B

H

OSO2Me

H

or 

H

OSO2Me

H

 

A. (02 pts).  Draw next to B a diastereoisomer of B, showing clearly the arrangement at each stereogenic center (as in B).

B.  Consider an E2 mechanism.

(1) (04 pts).  Suggest specific conditions (specific base/nucleophile/solvent) from the following lists (one from each) which would

tend to favor E2 over the others. Explain your choices briefly.

Best base/nucleophile: NaOMe, LDA, NaBr, Na acetate, NaCN

Best solvent: water, CH2Cl2, isopropyl alcohol, CH3CN, hexane

Base is required for E2: best- LDA (very strong, hindered so not a good nucleophile)

second best-- NaOMeno others are acceptable—too weak

Should avoid a polar solvent in order to minimize the chance of competing SN1/E1.

Best—hexaneSecond best-- CH2Cl2 

No others are acceptable

(2) (04).  Draw the mechanism for the E2 reaction from B, including the transition state for the rate-determining step. Draw the

structure of the product carefully. Use the sawhorse projection carefully to illustrate.

PhMe

H

OSO2Me

H

B:

PhMe

H

OSO2Me

H

:B!"

!"

PhMe

H

seriousgaucherepulsion

Z-isomer

repulsion

 

Cont…

NLi = LDA

 

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(3) (04 pts). Structure B and its diastereoisomer need not react at the same rate. Which would have a higher rate in the E2

reaction? Explain using Newman projections.

PhMe

H

OSO2Me

H

B:

lessseriousgauche

PhMe

H

E-isomer

  lessrepulsion

diastereoisomer of B

The diastereoisomer of B has less gaucherepulsion in the required conformation forthe best E2 TS, therefore it proceeds fastercompared to the E2 for B.

C. Consider the SN1 reaction for B where NaCN provides the nucleophile, dissolved in a solvent.

(1) (04 pts).  Write the mechanism for the S  N 1 reaction of B in water/NaCN with heating to initiate reaction.

PhMe

H

CN

H

PhMe

H

H

CN

CN

PhMe

H

CN

H

+PhMe

H

OSO2Me

H

 

(2) (06 pts).  Is the reaction likely to be faster or slower if CH 2Cl2 is used as solvent in place of water

(same temperature, etc)? Draw two reaction coordinate diagrams to show the relative effect of changing the solvent.

The rate determining step is the ionization of the leaving group to give the intermediate cation. This involves very strong

increase in charge as one proceeds to the intermediate, and the TS is strongly stabilized by polar solvents relative to theeffect on the reactants.

cation

TS

non-polar solvent

extent of reaction

E cation

TS

polar solvent

extent of reaction

E!G#!G#

 

Cont…

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(3). (04 pts).  Like many S  N 1 reactions, this one is complicated by rearrangement before addition of the nucleophile

(substitution with rearrangement). Consider the possible rearrangement products and draw here the mechanism for the process that

is likely to be most favorable. Explain your choice. Show the product(s) clearly,

PhMe

H

H

PhMe

H

OSO2Me

H

Me

H

H

CN

NC

PhMe

H

H

PhMe

H

H

CN

NC+

Me

H

H

etc

tertiary benzylic cation,stabilized by resonancedelocalization

The same cation as before can allow migration of a H to give a very stable tertiary cation, with resonance delocalization.

Migration of the Me group would give a secondary, resonance stabilized cation, not quite as favorable.

__________________________________________________________________________________________________________

III. (08 pts).  Draw the single most likely product for the following reaction, consistent with the conditions and data. Then

write a detailed mechanism and label the rate-determining step. Show all intermediates. Name the mechanism.

Br

NH2

heatQ; C12H15N + HBr

THF solvent

OTHF

 

NH2   NH2N

H   H

N

HQ

RDS

SN1 mechanism

Selected 1H NMR spectral data for Q:

two singlets at ! 1.8 (3H), 1.85 (3H)

one alkene H (! 5.4, d, J=7Hz)

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IV. (08 pts).  One can imagine cyclohexanes with heteroatoms in the ring (e.g., S and T). The usual chair

conformational analysis is used to consider conformational isomers.

tBu

O

O

tBu

O

O

OH

R   S   T  

A. (04 pts).  As you know, the axial form of R is disfavored by more than 5 kcal/mol. But in S, the preference for equatorial

over axial is only 1.4 kcal/mol. Please rationalize this difference by explaining why axial R is so unfavorable and then why axial S is

much less so. Draw the relevant structures carefully in the chair form.

HH

unavoidablegaucheinteractions

OO

reducedgaucheinteractions  

B. (04 pts).  Perhaps surprisingly at first glance, in T the axial conformer is actually favored compared to equatorial.

 Explain. 

O

O

OH

axial stabilized by H-bonding.Not available in the equatorial

 

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V. (12 pts).  Consider the trisubstituted cyclohexanes, H and J 

and the E2 reaction in the presence of strong base.

A. (04 pts).  Draw here the two chair forms of H  (H-1 and H-2) and the two

chair forms of J (J-1 and J-2). 

Br

Cl

Me

Hb

Hc

Ha

HdBr

ClMe

Hc

Ha

H-2

Br

Cl

Hb

Hc

Ha

HdBr

Cl

Hc

Ha

J-1   J-2

Me

Me

H-1

 

B. (08 pts). Which of the four isomers above will react faster in the E2 process?   Explain your choice carefully.  Several different

elimination products are possible.  Draw carefully the single most likely product from the E2 reaction and explain why it is formed

 preferentially. Draw the mechanism carefully. 

An axial leaving group is required for the anti-parallel TS preferred for E2, so only H-2 and J-2 are candidates.Br is a better leaving group than Cl (weaker bond and more stable anion). So Ha is the preferred proton to pick off.

J-1 is the low energy state for that molecule and an increase of energy is necessary to move to the all axial version J-2

before the elimination can begin. On the other hand, H-2 has only two axial substituents and both have relatively small

repulsive values (0.5 compared to 1.7 for Me). Relatively little energy is necessary to achieve the axial Br conformation,

H-2.

This added energy to access J-2 means the overall activation energy for E2 from J is higher than for H.

Br

Cl

Hc

Ha

J-2

Me

:B

Cl

Me

Cl

Me

 

Br

ClH

Br

ClJ  

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VI.  (15 pts).  Note the formation of X under the conditions specified.

OBF3

CH3OH solventX

IR: 3500 cm-1 (only significant functional group absorption)

UV: no significant absorption

C5H10O2  C-13 NMR: 5 peaks

H NMR: ! 2.0 (broad s, 1H), 3.5 (s, 3H), 4.35 (d, 2H, J=7 Hz), 4.40 (d, 2H, J=7 Hz)

  5.45 (1H, d of t, J=12, 7 Hz), 5.50 (1H, d of t, J=12,7 Hz

A. (02 pts). What is the structure of the product, X? [You may purchase this structure at a penalty of 7 pts] 

B. (08 pts).  Write a careful mechanism for the formation of X

showing all intermediates (not transition states).

O BF3   O

BF3

O

BF3

O

BF3

Me-O-H

O

BF3

O

H

O

BF3

O

-H+

H+

HO

O

+ BF3

 

C. Support your structure for X  by analyzing the spectral data:

1. (02 pts).  How are the IR data consistent with the structure? Give one point. 

The peak at 3500 cm-1 is consistent with the presence of the –OH group

2. (03 pts)  Give one key element of the H NMR data which supports your structure clearly.

presence of broad peak at 2.0 is not very definitive. Same for singlet at 3.5 for the –Ome

better:

Presence of two alkene H (5.45, 5.50)

Coupling pattern for the alkene H showing –CH2- adjacent

Coupling pattern for the –CH2- showing one H adjacent

Two peaks at appropriate chemical shift to be inbetween an O and a double bond (4.35, 4.40)

OH

eO  OHMeOor

OH

MeO  OHMeO

or 

 

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VII. (16 pts). Consider the molecule E. It undergoes reaction in methyl alcohol solvent with silver nitrate added to give two

isomeric products, each of them a pure single compound.

OCH3H

Cl H

AgNO3

 CH3OH(solvent)

F + G C8H18O2a

b

[!]D = +4.5o

[!]D = 0o

E

[!]D = +10.0o

A. (02 pts).  In the reactant, E:

What is the absolute configuration at carbon (a)?  R  S cannot tell  (circle single best answer) 

What is the absolute configuration at carbon (b)?  R  S cannot tell  (circle single best answer) 

B. (02 pts).  The rotation for E of +10.0o in a solution requires that: (circle all correct answers)

(1) E has one at least one stereogenic center with the R configuration.

(2) The solution of E caused rotation of polarized light to the right

(3) The concentration of E in the solution is 0.1 M

(4) The sample vial containing E was spinnng at +10 rpm

C. (04 pts).   Draw the exact structure for F and for G, showing clearly the configuration at each stereogenic carbon. Explain how

 your structures are consistent with the [! 

] values ( F has a positive value and E has a value of zero). 

OCH3H

H3CO   HF

OCH3H

H   OCH3G  

F is chiral, capable of having a non-superimposable mirror image and therefore is capable of rotating the plane of plane

polarized light. It is present as one enantiomer, so there should be a net rotation.

G has a plane of symmetry (through the central carbon-carbon bond). The stereogenic centers at the end are mirror images

of each other. The mirror image is the same structure. This is a MESO isomer, and does not rotate polarized light.

D. (06 pts).  Write a mechanism for the formation of F and G which makes clear how each forms under these conditions. 

OCH3H

H3CO H

OCH3H

H OCH3

OCH3H

Cl HAg+

OCH3H

H

H H

OCH3H

H3CO HF

OCH3H

H OCH3G

HOMe

addition to eitherlobe of the empty p  orbital

 

E. (02 pts).  How are F and G related?  Enantiomers Diastereoisomers Stereoisomers  None of these

(circle all correct choices)

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VIII. (10 pts). The reaction of an amine with nitrous acid gives the diazonium ion group, which very easily loses N2 as a

leaving group in a spontaneous ionization process. In the case shown here, the elements of an E1 elimination appear [N2 and H+

],

along with Q [C6H10O]. However, Q is not the simple E1 product.

NH2

OH

CH3

HONO

N

OH

CH3

N

Q + N2  + H+

PN  

A. (05 pts).  Draw the structure of the simple E1 product one might expect from P and write the (hypothetical) mechanism of

its formation. Give two elements of the spectral data that are clearly inconsistent with the structure. 

OH

CH3

N

OH

CH3

N

OH

CH3

H

H

-N2

-H+

 

B. (05 pts).  Draw the structure of Q showing clearly the configuration at the stereogenic center(s), and write the mechanism

of its formation. Show all intermediates; you need not represent transition states. Give two elements of the spectral data that are

clearly consistent with your structure.

O

CH3

N

OH

CH3

N

OH

H

-N2

-H+

H

Me migration fromtop face of five-membered ring

Me

O

MeH H H

-H+

O

MeH

resonance structurestabilized by additionalbond

 

The product structure is consistent with the presence of a C=O in a 5-membered ring (1751)with a Me doublet (adjacent H)

with 5-non equivalent C and a C=O in the C13 NMR

Spectral data for Q:

IR: 1751 cm-1 (s)

1H NMR: ! 1.1 ppm, 3H (d, J=7Hz)

! 1.5-1.7 ppm, 4H (multiplet)

! 2.2 ppm, 2H (t, J=7Hz)! 2.5 ppm, 1H (sextet, J=7Hz)

13C NMR:

peaks at 16, 18, 19, 30, 33, 210