You should allow yourselfexactly 135 minutes to take this exam.

Last Name:

First Name:

Section TF:

Important Notes:

1. This exam consists of 12 problems on 12 pages, plus this cover sheet and two pages for scratch workat the end of the exam.

2. All answers must be written in the spaces provided. Do not write anything in red ink. Answerswritten on the back of a page will not be graded. Anything written on the scratch page will not be graded.

3. You are allowed to use a set of molecular models for this exam, but NO calculators.

SOLUTIONS

Name:11.

(a)

(c)

(b)

(d)conc. H2SO4

For each of the reactions shown below, predict the single major product and determine which mechanism,SN1, SN2, E1, or E2, is operative. Be sure to indicate stereochemistry if it is relevant.

MeI

Mechanism:

Mechanism:

Mechanism:

Mechanism:

E2

SN2

E1

ONa OMe

BrNaOH

OH

Cl

EtOHOEt

(+/-)

SN1

Name:22.

(a)

(c)

(b)

(d)

Fill in each box with the single major product of the indicated transformation. Be sure to indicatestereochemistry clearly if it is relevant.

I

OEt

NaOAc

AcOHOEt

OAc

(+/-)

Br1. Li0 (2 equiv.), Et2O

2. H2O

H2O

H+ cat.

1. HBr

2. SNaS

OH

(+/-)

3 Name:

3.

Desired Product

Starting Material

Provide a multistep synthesis for the desired product from homoallyl bromide, shown below. You may useany additional organic or inorganic reagents. The best answer will require four or fewer steps.

Br

homoallyl bromide

EtO

CN

(+/-)

NaOEtEtO

HBr

EtO

Br

(+/-)

NaCN

4 Name:

4. For each of the following species, provide a complete Lewis structure, showing all atoms, bonds, andlone pairs. Then, using skeletal structures, draw the best alternate resonance structure in the other box.Include all nonzero formal charges.a)

b)

c)

Complete Lewis Structure: Best Alternate Resonance Structure:

Complete Lewis Structure: Best Alternate Resonance Structure:

Complete Lewis Structure: Best Alternate Resonance Structure:

B

OCH3

O

H2C N N C N

H

H

N N NH2C

O

CCC

H

H HC

H

O

C CH3HC

C C

CB

C

OC

H

H H

H

H

H

H

HH H

H

B

OCH3

5 Name:

5. Consider the structure shown below.

a. Provide the best possible systematic name for this compound.

b. Use a Newman Projection to show the lowest energy conformation of the compound around the bondindicated in bold below. Draw your Newman projection from the vantage shown by the "eyeball" below. Ifthere is more than one conformation of lowest energy, draw all such conformations.

You may represent the rest of themolecule using the "R" notation:

R

Cl

3-chloro-5-methylhept-1-eneor

3-chloro-5-methyl-1-heptene

Cl

R

H3C H

CH3

HH

6 Name:

6. Consider the following compound, related to the chemotherapeutic drug epothilone B:

a) Circle all stereogenic carbons in the molecule above, and then assign R/S and E/Z configurationswherever necessary.

b) Identify each of the following pairs of structures as structural isomers, enantiomers, diastereomers,identical, or none. You do not have to explain your choice.

O O

O

O

H2N

O

OH

MeO OMe

O O

OHOH

and

structural enantiomers diastereomers identical noneisomersThe above molecules are:

O

O

O

O

and

structural enantiomers diastereomers identical noneisomersThe above molecules are:

OOO

O

O

O HH

and

HH

structural enantiomers diastereomers identical noneisomersThe above molecules are:

(E)(R)

(R)

(S)

(S)

7 Name:

7. Cordycepic acid, isolated from fungal growths on arthropods, is a component of traditional Chinese herbalremedies. Naturally isolated samples were shown to have the specific rotation [ ]D = +40.3o

a) Cordecypic acid was originally reported to have the structure shown below. Why must this structure be incorrect?

CO2H

HO

HOOH

OH

cordycepic acid

The specific rotation, [ ]D, of an ammonium salt of myoinositol-1-phosphate (structure shown below) is +4.5o.OH

HO

OH

OH

HO OP

O

OO

12

34

56

myoinositol-1-phosphate[ ]D = +4.5o

b) What is the specific rotation of the ammonium salt of myoinositol-2-phosphate? (Hint: Use the numberingscheme shown above, exchanging an OH group for a phosphate group as necessary.)

c) What is the specific rotation of the ammonium salt of myoinositol-3-phosphate?

Although the molecule has stereogenic centers, itcontains an internal mirror plane of symmetry asshown and is thus achiral and meso. This meansthat the optical rotation must be 0o and thestructure that was initially reported is wrong.

CO2H

OH

HO

HO

HO

O

HO

HO

HO

OH

OH

PO

OO

Myoinositol-2-phosphate contains aninternal mirror plane of symmetry asshown and is thus achiral and meso.This means that the optical rotation is0o.meso!

myoinositol-2-phosphate:

Myoinositol-3-phosphateis the enantiomer ofmyoinositol-1-phosphate,so it has the samemagnitude of opticalrotation, but the oppositesign, so [ ]D = _4.5o

OH

OH

OH

HO

OHOP

O

OO

OH

HO

OH

OH

HO OP

O

OO

myoinositol-1-phosphatemyoinositol-3-phosphate

enantiomers!

8 Name:

8. The compound below, called dihydropyran, reacts with methanol in the presence of strong mineral acid toform Compound A. The other possible product, Compound B, is not formed at all.

Resonance Explanation: Frontier Molecular Orbital Explanation:

b) Explain why in this case Compound A is the only product actually formed. You must provide both aresonance explanation and an explanation in terms of the molecular orbitals involved in the intermediates.

a) Draw a complete curved-arrow mechanism for the putative formation of Compound B.

O

dihydropyran

MeOH

H+ cat. O OOMe

OMe

Compound A Compound B(not formed)

O H O

Me

HO

O

H

Me

O

O

H

Me

O

H

Me

O

OMe

Compound A results f rom reaction via a resonance-stabilized carbocation, and lowering the energy of theintermediate thus results in the preferred reactionpathway.

O O

The other possible carbocation intermediate lacksany such stabilization and thus is a disfavoredpathway leading to no product.

Compound A results f rom reaction through acarbocation intermediate that is stabilized by a f illed-empty orbital interaction with the adjacent oxygen lonepairs:

O C

nO 2pC

The carbocation intermediate leading to Compound Bcan only be stabilized by hyperconjugation with theadjacent CH bonds, which are signif icantly poorerdonors than lone pairs and thus cannot suf f icientlystabilize the carbocation to favor this pathway.

9 Name:

9. It is observed that Compounds A and B below react with potassium iodide quite readily by the samemechanism to yield the corresponding alkyl iodides. However, Compound A reacts many times faster thanCompound B.

a) Provide a curved-arrow mechanism for the reaction of Compound A with potassium iodide.

c) Explain briefly why Compound A reacts so rapidly by this mechanism. For full credit, you will need to illustrateyour explanation using "cartoon" orbital sketches. (Hint: consider the transition state for the reaction!)

OCl

KIO

I

Compound A

ClKI

I

Compound B

many times faster than

b) What will happen to the rate of reaction in both cases if the concentration of KI is tripled?

OCl

:I

OI

This is an SN2 reaction, which is f irst-order in both reactants, so the rate of the reaction will tr iple.

In the SN2 transition state for reaction of Compound A, the electrophilic carbon approaches an sp2 hybridization,and thus possesses a vacant 2p orbital. Since overlap of f illed and empty orbitals is stabilizing, over lap of theadjacent f illed CO orbital with the 2p orbital on carbon lowers the energy of the transition state relative to that ofCompound B, thus speeding the reaction.

C

CI Cl

H H

O

10 Name:

10. You likely recall that anions of the type OH- and OR- are extremely poor leaving groups. Contrary to thisgeneralization, the following reaction proceeds rapidly and in excellent yield.

a) Draw a curved-arrow mechanism to account for the above reaction.

b) Briefly explain why the reaction above proceeds so readily. For full credit, your answer should include clearlydrawn resonance structures.

O

O2N

NaN3

O2N

ON3+

O

O2N N N N O2N

ON3+

Normally good leaving groups are conjugate bases of strong acids, which means that they are stable and weakbases. The leaving group here, although technically an alkoxide like EtO- or MeO-, is exceptionally resonance-stabilized (see structures below), so its conjugate acid is actually quite strong, making it a good leaving group.

O

N N

O O

NO

O

O

O

O

O

O

NO

O

O

NO

O

Name:11

11. Consider the following questions.

c) The reaction shown in part (b) above takes place in two steps. For each step, identify the donor and acceptororbitals by name (e.g. *C-H).

Step 1: Step 2:

b) Molecule A reacts with an aldehyde to yield theproduct shown below. In the box, add curved arrowsto show how you get from reactants to products ineach step. Draw lone pairs when necessary.

a) In the box provided at right, provide acomplete energy-level diagram for themolecular orbitals of the molecule A, below.

A:

a) Energy-level diagram for the molecularorbitals of molecule A:

Donor:

Acceptor:

Donor:

Donor:

d) Please draw the shapes of the HOMO and LUMO of the aldehyde in the above reaction.

HOMO: LUMO:

b)O

H CH3

SCH3H3C

(step 1)

(step 2)

CH2S

H3C

H3C

CH2S

H3C

H3C

CH3

SH3C CH3

O

O

CH3

____ ____

____ ____ ____

____ ____ ____

____

____ ____

____

____

____ ____

____ ____ ____

____ ____ ____

____

____ ____

*C-H

*CH

*CS+

*C-S+

nC-

nS+

C-H

CH

C-S+

CS+

nC-

*CO

nO

*CS+

OH3C

HO

HH3C

12 Name:

12. Provide a complete curved-arrow mechanism for the following transformation.

OH

HOO

conc. H2SO4

OH

HO

H+

OH

H2O

OH

OHO

HO

:sol

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