1Chemistry 303

fall, 2005

THIRD EXAMINATION7:00 pm December 6th

Duration: 2 hr

Name________________________KEY_________________________

Lab TA__________________________________________________________(if you do not know his/her name, give day of lab section. NOT Brow or Clay)

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:

p2______/14 p3_______/13 p4_______/08 p5_______/11

p6______/12 p7_____/10 p8____/11 p9_____/06 p10_____/07 p11_____/10

Total: _________/102

There are 12 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 morning onWednesday. It would be well not to discuss the exam until after that time.

NOTE: You will be asked to write mechanisms; that requires you always use the arrow formalism.Please use it precisely, carefully, and correctly. Show charges carefully. Distinguish between anintermediate and a transition state. When you draw cyclohexanes in the chair form, be sure to indicatethe angle of the axial and equatorial substituents accurately.`

PLEDGE:________________________________________________________________

2I. (14 pts). Consider each of the following pairs of reactions.

a. Write the single most likely mechanism and product(s) for the faster reaction in each pair. Pay

attention to stereochemistry; e.g., if the product is a racemic mix, draw both isomers.

b. Give the name of the mechanism SN1 SN2 E1 E2.c. Give the single most important reason for the difference in reaction rate.

Br

Br

OMe

THF

Na CN

THF

Na CN

1.

2.

A.

HGood nucleophile, fairly non-polar solvent, secondary bromide substrate SN2 conditions

CN

OMe

+ NaBr

The primary alkyl bromide reacts faster in the SN2 mechanism due to less stericcrowding in the transition state.

________________________________________________________________________________________

H

AgNO3

CN

Br

AgNO3

CH3

Br

EtNH2

EtOH

B.

1.

2.

HR-configuration

R-configuration

Silver ion to force ionization. Moderate nucleophile present.rate determining step is ionization. Product determining stepis nucleophile addition, SN1. E1 gets partial credit but the presenceof a moderate nucleophile favors SN1.

The rate of ionization depends on the stabilization of the resultingcation. CN destabilizes the cation due to the inductive with-drawing effect.

H

CH3

OEtH

CH3

OEt+ racemic

racemic product mixture due to planar cation intermediate

H

CH3

H

CH3

O

H

Et

3II. ( 21 pts). Consider the following related reactions.

A. (5 pts). Using the R,S system, indicate the absolute configuration at each stereogenic center in A.

Name the most likely mechanism to operate on A under the conditions shown and draw it using the

arrow formalism, indicating the exact product(s) carefully.

For each product, indicate the configuration of each stereogenic center.

MeSNaCl

Me HEt

H

HH

H

A

THF solvent

MeH

Et

H

HH

H

SMe

Strong nucleophile, low polarity solvent,secondary halide = SN2

R-configuration

MeS

B. (8 pts). Name the most likely mechanism to operate on A under the conditions shown and draw it using the

arrow formalism, indicating the exact product(s) carefully.

For each product, indicate the configuration of each stereogenic center.

NOTE: NMR analysis of the product mixture shows peaks in the region 5-7 ppm.

Cl

Me HEt

H

HH

H

A

AgNO3?

THF

Forced ionization, low polarity solvent.NMR data suggest alkene product and confirm E1 mechanism

Me

HEt

H

HH

H

Me

HEt

H

HH

Me

HEt

H

HH

+

major

E1

4 C. (8 pts). For the slightly more complicated case, B is converted to C by a multi-step process.

Draw the mechanism carefully making clear the configuration of each carbon at each step.

Name each step [SN1, SN2, E1, E2]. HINT: Note carefully the location of the Me and Et groups.

This is tough to visualize and the product is deliberately drawn to not correlate directly with B.

Do some visual gymnastics.

Indicate the configuration of all stereogenic carbons in the product using the R,S nomenclature.

ClO

Me HEt

H

HMeSNa

B

THF solvent

O

HMe

Et

HH

SMeC

Cl

O

Me H

Et

H

H

MeS Cl

O

MeHEt

H

H

MeS

O

MeH

Et

H

H

MeS CSN2

SN2

S configuration

R configuration

S-configuration

MeS

5III. (23 pts). Consider the structure D (no stereochemistry intended).

A. (2 pt) How many stereogenic centers are there? ______

B. (2 pt) Assuming only structures in their lowest energy conformation,

how many stereoisomers are possible?

______

C. (7 pts) Using the chair representation:

a. Draw the most stable stereoisomer of D, in its lowest conformational form. Call it D-1.

b. Draw also its ring flip (other chair) isomer, call it D-2.

[Note Table 3.2 of substituent steric effects on the last page for your reference.]

c. Explain carefully the difference in stability between D-1 and D-2, by listing gauche, eclipsing,

and 1,3-diaxial interactions, for the Cl and CH3 shown on the structure. You need not be quantitative.

Cl

H

H

H3C

D-1

Cl

HCH3

HD-2

g

g

g

1,3-diaxial

g g

g

g = gauche interaction

g

In D-1, the Cl has three gauche interactions while the CH3 has one gauche.

In D-2, the Cl has two gauche interactions with the ring CH2 groups, and one gauche with a CH3

The CH3 has two gauche interactions with the ring CH2, and a 1,3-diaxial interaction with a CH3

A satisfactory answer would point out the gauche interactions approximately, and note the 1,3-diaxial

interaction carefully, and do not mis-identify interactions.

Cl

CH3

D

6

D. (6 pts). Using the chair representation:

a. Draw a diastereoisomer of D-1, in its lowest conformational form. Call it D-3.

b. Draw also the ring flip (other chair) isomer of D-3, call it D-4.

c. Explain carefully the difference in stability between D-3 and D-4, taking into account gauche,

eclipsing, and 1,3-diaxial interactions, for the Cl and CH3 where relevant. You need not be quantitative.

Cis arrangement of Cl and CH3

Cl

H

H

H3C

D-3

Cl

H

CH3

H D-4

g

gg

gg

g

g

1,3-diaxialg = gauche interaction

g

g

D-4 is less stable than D-3 because it has a 1,3-diaxial interaction; also more gauche interactions.

Again, the grading was gentle on exactly what gauche interactions are present, but the 1,3-diaxial needed to be

specified carefully.

E. (6 pts). Which isomer (D-1, D-2, D-3, D-4) will react fastest in an E2 reaction?

Draw the mechanism here and use it to explain your choice carefully. Draw the product(s).

Specify in your answer a specific base you would choose to favor the E2. Explain your choices.

Cl

H

H

H3C

D-3

HH3C

+ MeOH + Cl

MeO

We want a strong base to initiate E2; alkoxide is a good choice or the R2N- anions.

D-3 will react faster because it is the only isomer with the Cl and H lined up (1,2-diaxial). antiparallel.

7 IV. (10 pts). Consider the reaction of X and Y with dimethylamine to give the product shown. Note

the kinetic expression for each reaction.

A. (4 pts). Write the mechanism for the reaction of X under these conditions.

Clearly show the transition state for the rate-determining step.

Show partial charges where relevant.

Br NHMe2 rate = k [X][NHMe2]NHMe2

X Br!

HH

BrMe2HN"+ "-

B. (6 pts). Write the mechanism for the reaction of Y under these conditions.

Explain the difference in the kinetic expressions.

Why is a different mechanism followed by Y?

Br

NHMe2 rate = k [Y]NHMe2

Y Br!

Br

NHMe2

This is an SN1 reaction, with ionization being the rate-determining step.This is a unimolecular reaction, with the rate expression shown. The reaction above in (A) is an SN2 process, with a bimolecular step and a rate expressionincluding two concentrations.

Y cannot undergo an SN2 process as the leaving group is attached to a tertiary carbon, too hindered for SN2. However, it can ionize easily to give a tertiary, resonance stabilized cation.

8 V. (11 pts). A. (7 pts). This question is from the Chem 301 Make-up exam

“First, please make an excellent, fully three-dimensional drawing of the following molecule (E) in its energy

minimum form.” [chair form]

“Next, explain in detail how F is formed from E.” [write mechanism]

NH2

H

H

E

H2N

F

I

NH2

I I

H2N

SN2 not favorabledue to the geometryof this chair form

Very favorableSN2 in the diaxial form

H2N

F

(actually the enantiomerof F; my mistake but probably of no consequence)

B (4 pts). “Finally, explain why the related compound G does not react as does E to give an analogous

product.”

H2N

H

H

H

H

G

II

H2N

H

H

As for E, this isomer cannotdo a good SN2. It also cannotflip to the other chair becausethe other ring "locks" it.

9 VI. (13 pts). Consider the following related reactions.

Perhaps surprisingly, reaction (2) is much faster than (1).

OHCH3

OS

O O

NaH

OCH3

OS

O O

Na+ H2

OH O S

O O

Ph

NaH

O

O S

O O

Ph

Na+ + H2

1.

2.

A. (6 pts). Write the mechanism for reaction (2), showing exactly the role of the NaH.

Name the mechanism: SN1 SN2 E1 E2

O-H O S

O O

Ph

NaH

O

O S

O O

Ph

Na+ + H22.

Na H

O O S

O O

Ph

SN2

10B. (5 pts). Explain in detail why reaction (1) is much slower. Consider transition state structure.

What mechanism is being followed: SN1 SN2 E1 E2

OHCH3

OS

O O

NaH

OCH3

OS

O O

Na+ H2

1.

OCH3

OS

O O

O

C

OS

O O

H

HH

!"

!"

Transition state requires a 180o bond angle

at the carbon undergoing substitution. This cannot

be achieved in this intramolecular case, with a 6-membered

ring transition state.

C. (2 pts bonus). In fact, it was noted that the rate-determining step for reaction (1) is bimolecular while the

RDS for reaction (2) is unimolecular. Explain.

OHCH3

OS

O O

NaH

OCH3

OS

O O

Na+ H2

1.

OCH3

OS

O O

OH3C

OS

OO Intermolecular version is the only reasonable way to achieve an SN2.

The intermolecular version is a normal, bimolecular SN2. Reaction 2 is an intramolecular SN2 which is

unimolecular.

11

VIII. (10 pts). Consider reactions (1) and (2). Reaction (1) is much faster than reaction (2).

Br

AgNO3

H2OOH Br

AgNO3

H2OOH

1. 2.

A. (4 pts). Write a careful mechanism for reaction (2), showing all intermediates (not transition states).

Identify the rate-determining step.

Br

AgNO3

H2OOH

2.

H2O

O H

H

a

b

c

Step (a) is the rate determining step. The forced ionization

by Ag+ is effective with the secondary bromide.

B. (6 pts). Write a careful mechanism for reaction (1), showing all intermediates (not transition states).

Identify the rate-determining step. Explain carefully why (1) is faster than (2).

Br

1.

AgNO3

This ionization would give a secondary cation, just likethe one shown in Rxn 2. Shouldtherefore have essentiallythe same rate.

Br

Alternative:

Ag

Formation of secondary cation hasextra driving force of ring opening of thecyclopropane, release of strain energy.

H2O

O HH

O

H

a

b

c

Step a is the rate determining step, and is much faster than the RDS in rxn 2 because of the extra strain energy

release.

12

Glossary:

Me = methyl Et = ethyl Ph = phenyl

O

THF