1

Chemistry 303

fall, 2006

FINAL EXAMINATION

9:00 AM, JANUARY 24TH, 2007

Duration: 3.0 hr

NOTE: an additional 30 minutes will be allowed for completion of the course evaluation by those who have not done it

Name____________________KEY________________________________________

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.

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

USE THE ARROW FORMALISM CAREFULLY FOR ALL MECHANISMS

BE SURE TO INCLUDE ALL FORMAL CHARGES

WRITE SOMETHING

Write only in the space provided for each question.

Score:

p2___________/15 p3___________/15 p4___________/18 p5___________/14

p6___________/16 p7___________/18 p8___________/18 p9 __________/17

p10 __________/12 p11___________/12 p12___________/15 p13___________/20 p14__________/10 pts

Total: /200

There are 14 pages in this exam; please check now to be sure you have a complete set.

Pledge:_________________________________________________________________________________

2

I. (08 pts). Considering the following pair of reactions, note that reaction (a) is much faster than (b) under

these conditions. Write the most likely mechanism for reaction (1) and explain carefully why it proceeds faster than reaction (b).

NC

O

CH2Cl2 solvent

Cl

H3C

HBr

HBr

CH2Cl2 solvent

(a)

(b)

A. H+

NC

O

H3C

H

NC O

H3C

H

Br

NC

The lone pair on the ether can be protonated to activate the oxygen as a leaving group. In reaction (b), the chloride is a good leaving group for spontaneous ionization, but is not easily activated by protonation (lone pairs held tightly). In a more polar solvent, reaction (b) might proceed rapidly spontaneously, but in the non-polar solvent, the acid activation is necessary.

Br

Not catalytic in acid

__________________________________________________________________________________________

II. (22 pts). Consider reactions 1, 2, and 3 (next page)

A. The ratio Y:Z for reaction (1) is 95:5.

H2SO4(catalytic)

H2O solvent

OH

or

OH

X Y Z

(1)

i. (02 pts) What do you predict for the molecular rotation of Z? (+) (-) zero cannot tell (circle best answer)

ii. (05 pts) Write the best mechanism for this reaction and explain

carefully why Y is preferred. Show the catalytic role of the acid.

H2SO4(catalytic)

H2O solvent

OH

or

OH

X Y Z

(1)

H+

H H

resonance stabilized cation

H2O

H

OH

H

Y is the preferred product because the reaction goes through the best cation, a secondary benzylic cation with resonance stabilization. The cation leading to Z is simply secondary

-H+

Proton in for the first step and out in the last step: catalytic in H+

cont…

3

B. (07 pts). The ratio Y:Z for reaction (2) is also 95:5 when R = tBu in the boron reagent. This might seem surprising

until you think carefully about the parameters which influence the regioselectivity [I urge you to do so…].

a. R2BH

b. NaOH, H2O2

OH

or

OH

X Y Z

(2)

Write the best mechanism for step (a) of this process and explain why Y is favored.

a. R2BH

b. NaOH, H2O2

OH

or

OH

X Y Z

(2)

BH

RR

R2BH

no mechanism required

Two factors influence the regio selectivity, and they work in opposition in this case. The polar picture, with H being delived to the best cation site would lead to Z, while the steric factors (large tBu group) favor H addition to the benzylic position. Hard to predict, but the given result allows one to conclude that the steric effects dominate.

C. (08 pts). For this modified substrate (X’) in eq 3, the ratio of Y’:Z’ is now 25:75. R is still tert-butyl.

a. R2BH

b. NaOH, H2O2

OH

or

OH

X'Y' Z'

(3)

Me2NMe2N Me2N

i. (04 pts) Explain carefully the origin of the contrasting regioselectivity between reactions (2) and (3).

ii. (04 pts) Which would be the faster reaction, step (a) of reaction (2) or step (a) of reaction (3)?

a. R2BH

b. NaOH, H2O2

OH

or

OH

X'Y' Z'

(3)

Me2NMe2N Me2N

BR2H

Me2N

H BR2

!+

!-

New and stable resonance structure favors the polar factors to determine the regioselectivity. Steric factors are the same, but the now the polar factors become dominant and lead to the opposite regioselectivity.

BR2

Me2N

Me2N Me2N

Me2N

The Me2N group pushes electron density into the ring and to the alkene. Favors donation of electron density from C2 and build-up of (+) charge at C1.

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21

ii. The dimethylamino group pumps electron density into the arene ring and then into the alkene, as indicated by the

resonance structure shown in the mechanism. More electron rich alkene reacts more favorably (faster) with the electron deficient

boron reagent.

4

III. (18 pts). Consider the stereoisomer V, where one stereocenter is labeled with deuterium.

A. (02 pts). What is the absolute configuration at C-3? R S cannot tell (circle one)

B. (16 pts). Treatment of V with sodium ethoxide in ethanol gave three alkenes. Draw the mechanism

for formation of each one, and indicate clearly the location of deuterium (if any) in each. Draw the structures carefully.

Br H

H D

V

Br

HH

D

:B

E2H

D

Z isomer, with one D

Br

DH

H

:B

E2 HH

E isomer, with no D

Br

DH

H

:B

E2H

H

H

DAlternate E2, one D in product

Br H

H D

V

23

5

IV. (14 pts). A. Note the relative rates for reaction of D with water

R

OSO2MeH2O

[products]

D

R rate

CH3- 2.0CH3CH2- 4.0(CH3)2CH- 10.0(CH3)3C- 180.0

i. (02 pts) Draw D in its most stable conformation with R = tert-butyl.

ii. (04 pts) Write the product and the mechanism for the most likely substitution process with R = (CH3)3C-.

iii. (04 pts) Use your mechanism to rationalize why the substrate with the t-butyl group reacts 90 times faster than the one with the

Me group. Explain carefully in terms of the transition state for the rate-determining step. R

OSO2MeH2O

[products]

D

R rateCH3- 2.0CH3CH2- 4.0(CH3)2CH- 10.0(CH3)3C- 180.0

OSO2Me

equatorial tert-butyl is favoredover any other substituent

SN1 H2O O

H

HO

H

+ H+

The rate determining step in the SN1 process involves a carbon atom going from having four substituents to having three substituents; there is relief of strain in this step. The larger tert-butyl group leads to a larger relief of strain (relatively higher energy reactant) compared to the methyl case.

iv. (04 pts) Write the product and the mechanism for the most likely elimination process with R = (CH3)3C-. The observed product

shows a triplet (J = 5 Hz) with area 1H at ! 5.35 ppm, and no other peaks above 2 ppm.

OSO2Me

SN1

H HH

H

triplet

cont…

6

B. (16 pts) Now, note the related reaction from C. The rate of this reaction (substitution + elimination) shows a much

bigger dependence on the substituent R, with the t-butyl group giving an enormous acceleration.

R

OSO2Me H2O[products]

C

R rateCH3- 2.0

CH3CH2- 15.4

(CH3)2CH- 67.0(CH3)3C- 45,000

i. (10 pts) Compared to the reaction in part A, why is this process so much more sensitive to the change from isopropyl to tert-

butyl? Explain carefully.

OSO2Me H2O

C

HH

OH

H

simple E1

strained double bond;no alkene H showing

rearrangement

HH H

observed elimination product

H

H

J = 2 Hz

The tert-butyl is now locked into the axial position and feels two gauche interactions. In particular, one of the methyl groups on the tert-butyl has to be in the repulsive interaction; it cannot rotate into a less repulsive conformation. The other substituents, such as isopropyl, also have the axial gauche interactions, but there is at least one accessible conformation with just a H pointing into the gauche interaction. So the steric effect for tert-butyl is strongly amplified in this system.

OSO2Me

HH

vs OSO2Me

H

HH

ii. (06 pts). Note that for the elimination product in this case, with R = tert-butyl, the 1H NMR spectrum shows two alkene H,

both doublets with J = 2 Hz and nearly the same chemical shift. Draw the structure of the elimination product and show the

mechanism of its formation.

See above

7

V. (18 pts). Consider the formation of cation H as the exclusive product from each of three different substrates, under the same

reaction conditions. The conditions are special: SbF5 is a powerful Lewis acid with an affinity for Cl greater than silver ion; SO2 is an

exceedingly weak nucleophile and allows the observation of free cations. The 13C NMR was used to establish the structure of H.

Cl

Cl

Clcation H

conditions for each reaction: SbF5 in SO2 solvent, -60o

13C NMR for cation H: 4 signals (s, 2 t, quartet)T

U V

A. (06 pts). What is the structure of cation H?

Explain how your structure is consistent with the NMR data.

By symmetry, cation H has 4 non-equivalent carbons: two –CH2- types (triplet)

one with no H (s, cation carbon)

one methyl group (quartet)

B. (06 pts). What is the mechanism of formation of cation H from T? If you are not sure about your mechanism, explain your

reservations or alternatives.

Cl

T

SbF5

H+ SbF5Cl

Full credit requires showing a concerted (one step)for ionization and h migration to give the tertiary cation. The alternate two step mechanism via a primary cation is certain to be of higher energyH

C. (06 pts). What is the mechanism of formation of cation H from V? If you are not sure about your mechanism, explain your

reservations or alternatives.

Cl

V

SbF5

+ SbClF5

CH3

H

HCH2

H

c

d

In this case, initial ionization leads to a secondary cation, c. Carbon migration is possible, as shown, to give an

unfavorable primary cation (d) and then a hydride shift to give the stable tertiary cation, H. While the primary cation is so

high energy that we tend to discount mechanisms that require them, here the system has no other choice in order to

eventually equilibration to the most stable cation, H.

Structure H:

8

VI. (18 pts). Note the series of bromides, a-e. When mixed with silver nitrate in water, substitution occurs to give the

corresponding alcohols (OH substitution for Br). Note also the very different rates for this process (monitoring the rate of

disappearance of reactant bromide).

O

OO

Br Br Br Br

O

Br

a b c d e

rel rate: 1.0 0.01 0.1 50,000 5,000

A. (06 pts) Why is (b) much less reactive than (a)? Explain with a general mechanism.

Br H

The general mechanism is ionization, SN2. Secondary leaving group and activated by silver ion. The rate determining step is the ionization step. The question is how do the oxygens in the ring affect the transition state for this step, and that translates into how do the oxygens affect the stability of the resulting cation.

H2O

H

O

H

H

H

OH + H+

Ag+

RDS

O

Br

In this substrate, the oxygen is close enough to exert electron removal by the inductive effect, due to its high electronegativity. That destabilizes the cation intermediate and slows the rate.

B. (06 pts) Why is (e) much more reactive than a, b, or c? Explain with reference to the general mechanism.

O

Br

O

H

O

H

Ag+

C. (06 pts) Why is (d) still more reactive? Explain with reference to the general mechanism.

H2O

O

H

O

H

H

H

OH + H+

O

Br

O

H

This intermediate forms instead of the high energysimple cation; involves two favorable 5-membered rings and a new sigma bond C-O. The other substrates can form parallel intermediates but each would involve a highly strained ring.

Ag+

Cation has resonance stabilization which is

estimated to be strong because it creates an

additional bond for the structure. This effect dominates over the destabilizing inductive effect.

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VII. (17 pts). Consider the reaction of molecule K under the conditions specified to replace hydrogen by bromide (HBr

byproduct). Reactions of this type are typically quite unselective and give a mixture of monobromide products.

Br2, h!monobromide substitution products + HBr

K A. (10 pts). Draw the major product (call it M) you expect from this reaction and the mechanism by which it forms. This is a

multi-step process; draw each step carefully. Identify the product-determining step. Use the appropriate arrows appropriately.

Br2, h! 2 Br•

Br• +

HH

+ H-Br

H

Br--Br

H

+ Br•

Br

H product determining step.

M

B. (07 pts). Also draw the product (N) expected to be the second most abundant and explain why it is not as favored as M, but

favored over other alternatives. Include in your answer a brief general discussion of what parameter(s) influence the selectivity in

this bromination reaction.

H

HH

H

H

There are five different C-H bond types (assuming the benzene C-H are all approximately the same).

Each has a different bond energy (shown on structure), depending on hybridization, number of substituents, etc. These can be found in the BDE table in the data sheets.

103

9395

85

110

In a radical reaction, the only thing that counts is the bond energy (no polar factors, etc). This depends onhybridization, resonance stabilization, etc. Product is determined by which bond is weaker in the product-determining step.

H

HBr

H

H

N

The tertiary C-H bond is the next weaker, compared to secondary, methyl, and aryl.

10

VIII. (24 pts). Organic synthesis is what most organic chemists do, and you already have the tools to plan some simple

syntheses (I hope). It requires that you recognize the standard reactions (one or more) that connect a starting material with a synthesis

target. It sometimes helps to work backward from the target to the starting material.

Please suggest how you might carry out the following conversions by writing a sequence of steps showing reagents and

reactants required. Shorter plans with fewer potential side reactions or byproducts are better.

Be specific about the reagents and products for each step; you need not write mechanisms.

Name the mechanism involved in each step (SN1, SN2, E1, E2, electrophilic addition, oxidation, radical, etc) wherever possible

Possibly useful reagents: HBr, BH3, H2O, RO-OR, RCO3H, LiNR2, NaOH, CrO3, OsO4, NaH, pyridine, HCl, NaI, SOCl2,

Pbr3, MeSO2Cl. You may use any reactants you wish, such as acetylene, EtI, etc. Just specific clearly for each step.

OH CN

?racemic

A. (12 pts)

OH

Many possibilities.

CN

HOH

Bonds that need to be made to get to the product.CN is readily available as a nucelophile in NaCN, so an SN2 reaction is a good idea. Need a leaving group at C-1

123

The substitution at C2 and C3 looks like electrophilic addition of H-OH to the pi bond. Now

decide whether you need Markovnikoff addition (H+,H2O) or anti-Markovnikoff (R2BH/H2O2)

Since OH is on the best cation site, Markovnifoff addition of H2O should give the right selectivity

OH

SOCl2

Cl

NC-

CN

chlorinationby SN2 SN2

cat H2SO4

H2O

Electrophilic add to alkene

CN

HOH

OR:

OHOH

Reversing the sequence gets you into some trouble:

cat H2SO4

H2O

Electrophilic add to alkene

HO H

need to make this into a leaving group selectively,in the presence of another -OH. Possible, buttricky.

Short sequences with reliable (i.e., not too creative…) procedures will get full credit.

cont…

11

?B. (12 pts) O

OHNote: one diastereoisomer (racemic) is desired, as drawn

O

HOPossible bond formations to get the product. But the formation of the bond to the-OH must occur from the side opposite the Me group

RCO3HO

reagent approaches from the less hindered face, away from theMe group

Electrophilic addition to the alkene

CH3CH2O

SN2

O

HO

addition to the less hindered end, typical of SN2

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IX. (15 pts). Note the following reaction promoted by base to equilibrate S with its isomer T.

Treatment of T with CrO3 gives U.

O

OHbase

!1.2 (3H, s) 1.4 (3H, s) 4.5 (1H, d, J=2 Hz) 9.5 (1H, d, J=2 Hz)

T UCrO3

S

1H NMR

13C NMR: 5 different signals in broad band decoupled spectrum

IR: 1740 cm-1 (strong)

i. 07 pts) Write the structure of T and the mechanism by which it forms from S. (Note: you may want to consider part ii

simultaneously)

O

O

:B-

H

O

O

SN2

O

O H+

OH

O

H

T

ii. (08 pts) Write the structure of U and the mechanism by which it forms from T. Explain how your structure is consistent with the

1H NMR data and the IR data. Note: you can buy the structure of U at a penalty of 5 pts.

OH

O

H+ Cr

O

O O

O

O

H

H Cr

OO

O O

O

H

Cr

OO

O

-H+

H

H

O

O

CH3

H3C

H

H

-H+

Cr

HO OH

O

+

U

!1.2 (3H, s) 1.4 (3H, s) 4.5 (1H, d, J=2 Hz) 9.5 (1H, d, J=2 Hz)

O

O

CH3

H3C

HH

U

Methyl groups are in different environments due positioning above and below the ring; the assignment shown is arbitrary. The methine attached to the ring is perturbed by the O, shows up at 4.5. The aldehyde appears at the position typical of -CHO groups.

The IR shows a typical aldehyde C=O stretch.

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X. (20 pts). Note that isomers T and P react under the same conditions to give different products.

Br

OHC8H14O + NaBr + H2

NaH

T

THFsolvent

O

Br

OH NaH

P

THFsolvent

C8H14O + NaBr + H2

U

Q

THF

(1)

(2)

A. (02 pts). How are P and T related? a. enantiomers b. diastereoisomers c. stereoisomers d.

conformational isomers

(circle all correct answers)

B. (08 pts). i. Draw T in both chair forms. ii. Circle the more stable chair.

iii. Estimate the difference in energy between the two chair forms using the Table of substituent effects on the Data Handout.

T

Br OH

O

H

Br

diequatorial, zero torsional strain

diaxial: -CH2Br ca 1.8 + CH2OH ca 1.8 + 1,3-diaxial ca 3-4 = 6.6-7.6 kcal/mol

C. The chemist who was carrying out these reactions obtained pure samples of U and Q, but realized he got them mixed up while

labeling the containers. He noted that one sample showed moderate peaks in the IR at 3500 and 1680 cm-1 while the other one showed

neither of these (and no other characteristic functional group peaks other than C-H stretch) but he was stumped to come up with exact

structures. He turned to his labmate and she used mechanistic reasoning to suggest the exact structures of U and Q.

i. (02 pts) What is the structure for U?

ii. (05 pts) Draw the mechanism of formation of U from T.

iii. (03 pts) How is your structure consistent with the IR information?

TNa+ H-

Br OH

O

H

Br OBr O

SN2+ Br-

U

The IR shows no significant functional group stretches; structure U has no functional groups that would give such peaks.

cont.

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D. (10 pts).

i. (02 pts) What is the structure for Q?

ii. (05 pts)Draw the mechanism of formation of Q from P.

iii. (03 pts) How is your structure consistent with the IR data?

Na+ H-OH

O

H

HO HO

E2+ Br-

Q

Br

Br

H

Br HH

P

Neither conformation of P is favorable for the intramolecular SN2; would give a highly strained bicyclic structure.

At the same time, one conformation, with the axial -CH2O- is ideally poised for an intramolecular E2.

The IR shows a peak for 1,1-disubstituted double bond at 1680 cm-1, and peak for the -OH stretch at 3500 cm-1.

________________________________________________________________________________________________________

That’s it!

Have a good break; see you in Chem 304 for bigger and more biological things…

MFS