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Practice Exams

Practice Problems by Chapter

Mechanism Flashcards

OChem1 Practice Exams

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Dr. Peter Norris, 2014

1

Chemistry 3719 Practice Exam A1

This exam is worth 100 points out of a total of 600 points for Chemistry 3719/3719L. You have 50 minutes to

complete the exam and you may use molecular models as needed. Good Luck.

1. (8 pts) Draw structures for all of the cycloalkane isomers with the formula C5H10, and then give each of your

structures an acceptable name. (Klein Chapter 4)

2

2. (9 pts) Give each of the following organic molecules an acceptable name. You may use either systematic or

common names for substituents. (Klein Chapter 4)

CH3

CH2CH3

a.

b.

c.

3. (8 pts) Fill in the ground state electron distribution for O (from the periodic table), and then construct hybrid

orbital pictures for the highlighted O atoms that describe the orbitals used for bonding in each case. (Klein

Chapter 1)

Explain briefly why hybrid orbitals are necessary here:

3

4. (12 pts) Give the products expected from each of the following acid-base reactions. Then label each acid

with an approximate pKa and indicate whether products, reactants, or neither are favoured in each case.

(Klein Chapter 3)

+ CH3CH2OKa.

b.

c.

CO2H

CCH

+ LiN[CH(CH3)2]

OH

+ CH3ONa

5. (8 pts) Give the three acids below an approximate pKa value and then explain why they have such different

acid strengths even though they each have an OH group as the acidic component. (Klein Chapter 3)

4

6. (6 pts) Complete the following structures by adding any needed formal charges. (Klein Chapter 2)

7. (6 pts) Draw a second resonance structure and an overall hybrid for each of the following species. (Klein

Chapter 2)

a.

b.

c.

H3C

O

O

CH3

O

CH2

H

C N N

H

8. (6 pts) For the three molecules below, discuss their relative solubilities in aqueous solution in terms of the

forces that are playing a role here. (Klein Chapter 1)

5

9. (10 pts) Draw two chair conformations for cis-1-t-butyl-3-methylcyclohexane that are related through a

ring-flip. Then circle which conformation you expect to be more stable and explain why. Then do the same

for the trans isomer and, finally, indicate whether the cis or trans isomer should be more stable overall and

explain your choice. (Klein Chapter 4)

10. (9 pts) Draw Newman depictions that correspond to the following conformations. (Klein Chapter 4)

a. The least stable conformation of 3-methylnonane along the C-4–C-5 bond.

b. The most stable conformation of 1,1-dibromo-6-chlorohexane along the C-3–C-4 bond.

c. A gauche conformation for 5-methyl-2-heptanol along the C-3–C-4 bond.

6

11. (9 pts) Rank the following species as indicated: (Klein Chapters 1-3)

a. Lowest to highest electronegativity (1 = lowest, 5 = highest)

b. Lowest to highest pKa (1 = lowest, 5 = highest)

c. Lowest to highest boiling point (1 = lowest, 5 = highest)

12. (9 pts) Within each pair of molecules below, indicate which is more stable and then give a brief explanation

for your choices in each case. (Klein Chapters 3-4)

a.

b.

c.

ONa ONa

vs.

Br

BrBr

Brvs.

OLi NHLi

vs.

1

Chemistry 3719 Practice Exam A1 Key



1. (8 pts) Draw structures for all of the cycloalkane isomers with the formula C5H10, and then give each of your

structures an acceptable name. (Klein, Chapter 4)

2


common names for substituents. (Klein Chapter 4)

CH3

CH2CH3

a.

b.

c.

5-ethyl-1-isopropyl-2-methyl-3-propylcyclohexane

5-isopropyl-2,6,7-trimethylnonane

bicyclo[4.3.1]decane

3. (8 pts) Fill in the ground state electron distribution for O (from the periodic table), and then construct hybrid

orbital pictures for the highlighted O atoms that describe the orbitals used for bonding in each case. (Klein

Chapter 1)

Explain briefly why hybrid orbitals are necessary here:

The most obvious reason is that the lone pairs in the ground state structure of Oxygen are not equivalent; the

hybridization model fixes that.

3

4. (12 pts) Give the products expected from each of the following acid-base reactions. Then label each acid

with an approximate pKa and indicate whether products, reactants, or neither are favoured in each case.

(Klein Chapter 3)

+ CH3CH2OKa.

b.

c.

CO2H

CC

H

+ LiN[CH(CH3)2]

OH

+ CH3ONa

+ CH3CH2OH

CO2K

CC

Li

+ HN[CH(CH3)2]

ONa

+ CH3OH

pKa ~ 5 pKa ~ 16products favoured

pKa ~ 26 pKa ~ 38products favoured

pKa ~ 16 pKa ~ 16neither favoured

5. (8 pts) Give the three acids below an approximate pKa value and then explain why they have such different

acid strengths even though they each have an OH group as the acidic component. (Klein Chapter 3)

The first molecule has pKa = 16 since it is an alcohol, the second has pKa ~5 since it is a

carboxylic acid, and the third has pKa = 10 since it is a phenol. The difference in pKa values is

down to the relative stabilities of the conjugate bases formed from each; the first compound

produces an anion in which the charge is localized on O and therefore unstable. The carboxylic

acid produces a conjugate base in which the charge is delocalized onto two O atoms, which is

quite stable. The third molecule also has a resonance-stabilized conjugate base, however here the

charge is delocalized into O and C, which isn’t as favourable as the carboxylate situation. The

more stable the conjugate base, the more acidic the associated acid.

4

6. (6 pts) Complete the following structures by adding any needed formal charges. (Klein Chapter 2)

7. (6 pts) Draw a second resonance structure and an overall hybrid for each of the following species. (Klein

Chapter 2)

8. (6 pts) For the three molecules below, discuss their relative solubilities in aqueous solution in terms of the

forces that are playing a role here. (Klein Chapter 1)

Molecule A will be the least soluble in water since although features a hydrogen-bonding

alcohol it also has a hyprophobic alkyl chain that will repel water. Molecule B has a

shorter alkyl chain so the OH group will have more of a role than in A and increase water

solubility. Molecule C should be the most soluble since it is ionic.

5

9. (10 pts) Draw two chair conformations for cis-1-t-butyl-3-methylcyclohexane that are related through a

ring-flip. Then circle which conformation you expect to be more stable and explain why. Then do the same

for the trans isomer and, finally, indicate whether the cis or trans isomer should be more stable overall and

explain your choice. (Klein Chapter 4)

10. (9 pts) Draw Newman depictions that correspond to the following conformations. (Klein Chapter 4)

a. The least stable conformation of 3-methylnonane along the C-4–C-5 bond.

b. The most stable conformation of 1,1-dibromo-6-chlorohexane along the C-3–C-4 bond.

c. A gauche conformation for 5-methyl-2-heptanol along the C-3–C-4 bond.

6

11. (9 pts) Rank the following species as indicated: (Klein Chapters 1-3)

a. Lowest to highest electronegativity (1 = lowest, 5 = highest)

3 5 4 1 2

b. Lowest to highest pKa (1 = lowest, 5 = highest)

2 4 1 3 5

c. Lowest to highest boiling point (1 = lowest, 5 = highest)

3 1 4 2 5

12. (9 pts) Within each pair of molecules below, indicate which is more stable and then give a brief explanation

for your choices in each case. (Klein Chapters 3-4)

1




1. (9 pts) For each of the following pairs of molecules, provide the configuration of each chiral centre and then

indicate whether the two molecules within a pair are enantiomers, diastereomers, or are identical. (Klein

Chapter 5)

a.

b.

c.

and

and

and

OH

Br

CH3

CH3

H Br

H OH

O CH3

CH3 O CH3

CH3

CH3

H OHCH2Cl

FHH3C

CH2Cl

OH

F

2


common names for substituents. (Klein Chapters 7-8)

a.

b.

c.

Br

OH

OH

F F

I

Cl

3. (8 pts) Provide a detailed mechanism for the following transformation that uses curved arrows to show the

breaking and forming of bonds. Explain why this product is the major product formed. (Klein Chapter 8)

3

4. (8 pts) The following solvolysis conditions give two major products in equal amounts but, when measured

for the mixture, [α]D ≠ 0. Provide the products and a mechanism for this process, as well as an explanation

for the optical rotation data. (Klein Chapter 7)

5. (8 pts) The following reactions have only slight differences; however the ratio of products formed in each

case is dramatically different. Provide a detailed mechanistic explanation for this data. (Klein Chapter 8)

6. (6 pts) Rank the following in decreasing order of reaction rate in SN1 with HBr (4 = fastest, 1 = slowest) and

then give a brief explanation for your choices. (Klein Chapter 7)

4

7. (10 pts) On the axes given below, draw a complete reaction profile for the following reaction that includes

diagrams of all intermediates and transition states involved. (Klein Chapters 6-7)

P.E.

reaction coordinate

NaCl

ClBr

8. (8 pts) Give the major products expected from each step in the following synthetic sequence. Be careful to

take into account any stereochemical issues along the way. (Klein Chapter 7)

5

9. (15 pts) Give the expected products, major and minor where applicable, in each of the following situations.

Be sure to take into account any changes in stereochemistry that may occur. (Klein Chapters 7-8)

1.

2. NaCN, DMF

H2SO4, heat

CH3OH

NaOCH3, DMF

NaN3, DMSO

a.

b.

c.

d.

e.

OH SO2ClH3C

pyridine

O

OH

O

H

BrH

CH2OH

CH3

H OH

H I

I

6

10. (8 pts) Give a mechanism for the following reaction using curved arrows to describe all bond-forming and

breaking events. Then draw a transition state for the R.D.S. and explain why this is the major product.

(Klein Chapter 8)

11. (6 pts) Convert the following stereochemical representations as directed. Working out R/S configurations

might help but are not required as part of the answer. (Klein Chapter 5)

12. (5 pts) Dehydration of a tertiary alcohol in the presence of catalytic acid through the E1 pathway is thermo-

dynamically unfavourable since sigma bonds are swapped for a pi bond. Explain how this process is made

viable such that the alkene is able to be formed and isolated in high yield. (Klein Chapter 8)

1




1. (9 pts) For each of the following pairs of molecules, provide the configuration of each chiral centre and then

indicate whether the two molecules within a pair are enantiomers, diastereomers, or are identical. (Klein

Chapter 5)

2


common names for substituents. (Klein Chapters 7-8)


breaking and forming of bonds. Explain why this product is the major product formed. (Klein Chapter 8)

The secondary carbocation that is formed undergoes rearrangement to produce a better tertiary cation that

no longer has the strained cyclobutyl ring involved. The tertiary carbocation is then deprotonated to give

the most highly substituted alkene which is stabilized by four electron-donating alkyl groups through

inductive effects.

3

4. (8 pts) The following solvolysis conditions give two major products in equal amounts but, when measured

for the mixture, [α]D ≠ 0. Provide the products and a mechanism for this process, as well as an explanation

for the optical rotation data. (Klein Chapter 7)

5. (8 pts) The following reactions have only slight differences; however the ratio of products formed in each

case is dramatically different. Provide a detailed mechanistic explanation for this data. (Klein Chapter 8)

6. (6 pts) Rank the following in decreasing order of reaction rate in SN1 with HBr (4 = fastest, 1 = slowest)

and then give a brief explanation for your choices.

The better the possible carbocation, the faster the SN1 reaction; 3° > 2° > 1° > CH3

4


diagrams of all intermediates and transition states involved. (Klein Chapter 7)

P.E.

reaction coordinate

NaCl

Cl

Cl

A B

C

A =

B =

C =

+

-

+

+

+

-

H

Cl

‡

‡

‡

Br

Br

Br

8. (8 pts) Give the major products expected from each step in the following synthetic sequence. Be careful to

take into account any stereochemical issues along the way. (Klein Chapter 7)

5


Be sure to take into account any changes in stereochemistry that may occur. (Klein Chapters 7-8)

1.

2. NaCN, DMF

H2SO4, heat

CH3OH

NaOCH3, DMF

NaN3, DMSO

a.

b.

c.

d.

e.

OH OTs CNSO2ClH3C

pyridine 1. 2.

O

OH

O

+

H

BrH

HOCH3

1. 2.

SN1 - racemic

SN2 - inversion

SN2 - inversion E2

CH2OH

CH3

H OH

H I

CH2OH

CH3

H OH

N3 H

SN2 - inversion

I

O

OCH3

O

OCH3

O

E1 - Zaitsev outcome

6

10. (8 pts) Give a mechanism for the following reaction using curved arrows to describe all bond-forming and

breaking events. Then draw a transition state for the R.D.S. and explain why this is the major product.

(Klein Chapter 8)

NaOCH2CH3

Br

H

Br

H

OCH2CH3

OCH2CH3

‡-

-

E2

The major product is the most highly-substituted and stabilized alkene (tetrasubstituted)

11. (6 pts) (6 pts) Convert the following stereochemical representations as directed. Working out R/S

configurations might help but are not required as part of the answer. (Klein Chapter 5)

12. (5 pts) Dehydration of a tertiary alcohol in the presence of catalytic acid through the E1 pathway is thermo-

dynamically unfavourable since sigma bonds are swapped for a pi bond. Explain how this process is made

viable such that the alkene is able to be formed and isolated in high yield. (Klein Chapter 8)

Run the reaction at high temperature to increase the contribution of entropy to the overall free energy of

the system (∆G = ∆H – T∆S) since the reaction is entropically favoured (1 mole of alcohol produces two

moles of product; alkene and water). High temperature also allows the more volatile alkene to be removed

by distillation which will then force the equilibrium towards product to compensate.

1




1. (9 pts) Within each of the following pairs of molecules, indicate which will react at a faster rate in the given

reaction; then give a brief explanation for your choice in each case. (Klein Chapters 7-9)

a.

b.

c.

CH3

H3C H

CH3or in electrophilic addition

Br

H H

Br

D Dor in E2 elimination

OH OHor in SN2 substitution

2

2. (9 pts) Give each of the following organic molecules an acceptable name using any systematic or common

names for substituents where appropriate. Be sure to include stereochemical designators where needed.

(Klein Chapteres 5-10)

a.

b.

c.

F

OH

Br

Cl

F

HO CH3

3. (9 pts) For the following conversion, provide a detailed mechanism for the first operation only (i.e. reaction

of the alkene with O3) that uses curved arrows to show the breaking and forming of bonds. (Klein Chapter

9)

3

4. (15 pts) Give the ultimate major product expected in each of the following situations. (Klein Chapters 9-11)

1. O3

2. Zn, H2O

Br2, heat

CH3CO3H

Br2 in H2O

H2, Pd

a.

b.

c.

d.

e.

CH3

CH3

4

5. (9 pts) Provide the major product expected under the following reaction conditions, and then give a detailed

mechanism (using curved arrows) that describes its formation. (Klein Chapter 10)

6. (9 pts) Show the major product expected under the following reaction conditions, name that product, and

then give a detailed mechanism (using curved arrows) that describes its formation. (Klein Chapter 11)

5

7. (8 pts) The following conditions induce a free radical halogenation process at the allylic position with only

two products forming with a combined [α]D = 0; give a mechanistic explanation for this observation. (Klein

Chapter 11)

8. (8 pts) Provide the major product expected under the following conditions and then a detailed mechanism

for its formation that includes all steps that lead to this product. (Klein Chapter 11)

9. (6 pts) Provide the reagents required to complete the following transformations. (Klein Chapter 9)

6

10. (8 pts) The following hydroboration-oxidation sequence results in the alkylborane and alcohol shown below.

Explain how the regiochemical and stereochemical outcomes help determine the mechanism in each step.

(Klein Chapter 9)



P.E.

reaction coordinate

HBr

Br

1




1. (9 pts) Within each of the following pairs of molecules, indicate which will react at a faster rate in the given

reaction; then give a brief explanation for your choice in each case. (Klein Chapters 7-9)

2

2. (9 pts) Give each of the following organic molecules an acceptable name using any systematic or common

names for substituents where appropriate. Be sure to include stereochemical designators where needed.

(Klein Chapteres 5-10)

a.

b.

c.

F

OH

(3R,6S)-6-fluoro-7-methyloct-4-yn-3-ol

Br

Cl (R,Z)-2-bromo-5-chloro-6-methylhept-2-ene

F

HO CH3

(1S,3R)-3-fluoro-5-methylcyclohept-4-enol

3. (9 pts) For the following conversion, provide a detailed mechanism for the first operation only (i.e. reaction

of the alkene with O3) that uses curved arrows to show the breaking and forming of bonds. (Klein Chapter

9)

3

4. (15 pts) Give the ultimate major product expected in each of the following situations. (Klein Chapters 9-11)

1. O3

2. Zn, H2O

Br2, heat

CH3CO3H

Br2 in H2O

H2, Pd

a.

b.

c.

d.

e.

CH3

CH3

H

O

O

H

Br

O

Br

OH

CH3

CH3

H

H

4

5. (9 pts) Provide the major product expected under the following reaction conditions, and then give a detailed

mechanism (using curved arrows) that describes its formation. (Klein Chapter 10)

6. (9 pts) Show the major product expected under the following reaction conditions, name that product, and

then give a detailed mechanism (using curved arrows) that describes its formation. (Klein Chapter 11)

5


two products forming with a combined [α]D = 0; give a mechanistic explanation for this observation. (Klein

Chapter 11)

8. (8 pts) Provide the major product expected under the following conditions and then a detailed mechanism

for its formation that includes all steps that lead to this product. (Klein Chapter 11)

9. (6 pts) Provide the reagents required to complete the following transformations. (Klein Chapter 9)

6

10. (8 pts) The following hydroboration-oxidation sequence results in the alkylborane and alcohol shown below.

Explain how the regiochemical and stereochemical outcomes help determine the mechanism in each step.

(Klein Chapter 9)

The first step involves regioselective addition of the borane reagent such that the larger substituent

on the alkene avoids the large bicyclic framework on boron; the oxidation step results in

stereochemical retention, which is only possible with a concerted migration step.



P.E.

reaction coordinate

HBr

Br

Br

A B

C

A =

B =

C =

H Br+

-

+

+

+

-

H

Br

‡

‡

‡

1

Chemistry 3719 Practice Exam B1



1. (8 pts) Fill in the ground state electron distribution for N (from the periodic table), then construct hybrid

orbital pictures for the highlighted N atoms that describe the orbitals used by N for bonding in each case.

(Klein Chapter 1)

2

2. (20 pts) For each of the following acid-base reactions, draw the conjugate acid and conjugate base that are

expected to be formed. Label the acids on each side of the equations with approximate pKa values, and

indicate in each case if you expect the reaction to favour the left-hand side, the right-hand side, or neither.

(Klein Chapter 3)

+ KNH2

N

H

OH

+ NaOH

CF3CH2OK + CH3CH2OH

O

OHOLi

+

NLi CCH

+

a)

b)

c)

d)

e)

3

3. (15 pts) Provide acceptable names for the following molecules. Trivial or IUPAC names may be used as

appropriate. (Klein Chapter 4)

CH3

H CH2CH3

CH3

HH3C

CH2CH(CH3)2

H

H

a)

b)

c)

d)

e)

4

4. (6 pts) Draw all of the isomeric alkanes that have the molecular formula C5H12 and give each of them a

suitable name. (Klein Chapter 4)

5. (6 pts) Provide any missing formal charges that are necessary in the following molecules. (Klein Chapter

1)

a.

b.

c.

6. (9 pts) Draw a second resonance structure for each of the following and indicate which structure you expect

to be more significant in each system. Be sure to put lone pairs in the correct places. (Klein Chapter 2)

a.

b.

c.

H3C

O

NH

CH3

O

S

OH

5

7. (10 pts) For cis-1-t-butyl-2-methylcyclohexane and trans-1-t-butyl-2-methylcyclohexane, draw two ring-

flipped chair forms for each isomer, then indicate the preferred conformation for each of the two isomers,

and finally indicate which if the two isomers is more stable. Explain your choices in terms of the steric

interactions that both isomers will experience in their respective chair conformations. (Klein Chapter 4)

8. (10 pts) Draw structures that match the following situations: (Klein Chapter 4)

a) Newman projection of the least stable conformation of dodecane along the C-4 – C-5 bond axis.

b) Newman projection of the most stable conformation of 2-methylheptane along C-3 – C-4.

6

9. (8 pts) In the following questions, one of the two species is less stable than the other. Indicate which is less

stable in each case and explain the reason for the decreased stability. (Klein Chapter 4)

10. (8 pts) Indicate which of the following molecules should be soluble in water and which will not be soluble.

Briefly explain your choices. (Klein Chapter 1)

1

Chemistry 3719 Practice Exam B1 Key



1. (8 pts) Fill in the ground state electron distribution for N (from the periodic table), then construct hybrid

orbital pictures for the highlighted N atoms that describe the orbitals used by N for bonding in each case.

(Klein Chapter 1)

2

2. (20 pts) For each of the following acid-base reactions, draw the conjugate acid and conjugate base that are

expected to be formed. Label the acids on each side of the equations with approximate pKa values, and

indicate in each case if you expect the reaction to favor the left-hand side, the right-hand side, or neither.

(Klein Chapter 3)

+ KNH2

N

H

OH

+ NaOH

CF3CH2OK + CH3CH2OH

O

OHOLi

+

NLi CC

H

+

ONa

+ H2O

pKa ~ 10 pKa ~ 16R.H.S favored

CF3CH2OH + CH3CH2OK

pKa ~ 16 pKa < 16L.H.S favored

+ NH3

KN

pKa ~ 38 pKa ~ 38neither favored

NH CC

Li

+


O

OLiOH

+


a)

b)

c)

d)

e)

3

3. (15 pts) Provide acceptable names for the following molecules. Trivial or IUPAC names may be used as

appropriate. (Klein Chapter 4)

CH3

H CH2CH3

CH3

HH3C

2,3-dimethylpentane

CH2CH(CH3)2

H

H

trans-1-isobutyl-3-isopropylcyclohexanol

Bicyclo[4.2.1]nonane

3,4,6-trimethylnonane

1-(sec-butyl)-4-cyclobutyl-2-methylcyclopentane

a)

b)

c)

d)

e)

4

4. (6 pts) Draw all of the isomeric alkanes that have the molecular formula C5H12 and give each of them a

suitable name. (Klein Chapter 4)

5. (6 pts) Provide any missing formal charges that are necessary in the following molecules. (Klein Chapter

1)

a.

b.

NO

c.

6. (9 pts) Draw a second resonance structure for each of the following and indicate which structure you expect

to be more significant in each system. Be sure to put lone pairs in the correct places. (Klein Chapter 2)

5

7. (10 pts) For cis-1-t-butyl-2-methylcyclohexane and trans-1-t-butyl-2-methylcyclohexane, draw two ring-

flipped chair forms for each isomer, then indicate the preferred conformation for each of the two isomers,

and finally indicate which if the two isomers is more stable. Explain your choices in terms of the steric

interactions that both isomers will experience in their respective chair conformations. (Klein Chapter 4)

The trans isomer will be more stable since both of the large groups are able to be

equatorial thus avoiding any destabilizing 1,3-diaxial interactions. Both of the

conformations of the cis isomer will have a large group in an axial position.

8. (10 pts) Draw structures that match the following situations: (Klein Chapter 4)

a) Newman projection of the least stable conformation of dodecane along the C-4 – C-5 bond axis.

b) Newman projection of the most stable conformation of 2-methylheptane along C-3 – C-4.

6

9. (8 pts) In the following questions, one of the two species is less stable than the other. Indicate which is less

stable in each case and explain the reason for the decreased stability. (Klein Chapter 4)

10. (8 pts) Indicate which of the following molecules should be soluble in water and which will not be soluble.

Briefly explain your choices. (Klein Chapter 1)

Solubility is based on interactions between the solute and solvent and is generalized by

the term “like dissolve like” – here the OH groups are capable of hydrogen bonding with

water while the hydrophobic alkyl groups will repel. Large alkyl portions outweigh the

importance of the OH group and cause insolubility as listed; many OH groups (bottom

left) will enhance solubility.

1




1. (8 pts) Provide a detailed mechanism, using curved arrows to show the breaking and forming of bonds, that

accounts for the following conversion. (Klein Chapter 8)

2

2. (12 pts) Provide structures of the major and minor organic product(s) that are expected to be formed in each

of the following situations. (Klein Chapters 7-8)

NaOCH3

H3PO4, heat

KO-tBu

NaCl

CH3

NaN3, DMF

CH3

Br

Br

Br

CH3OH

OTs

Cl

OH

a)

b)

c)

d)

e)

f)

3

3. (9 pts) For each of the following pairs of molecules, identify any stereogenic centers, label them as having

the (R) or the (S) configuration, and then indicate the relationship between the two molecules; are they

enantiomers, diastereomers, or are they identical? (Klein Chapter 5)

4. (8 pts) Give all of the expected organic products from the following reaction and label those products as

being major, minor, or very minor. Then give an explanation for your assignments. (Klein Chapter 8)

a)

b)

c)

4

5. (12 pts) Give the major organic product formed under the following reaction conditions and then a detailed

mechanism, using curved arrows to show bonds being formed and broken, to describe the transformation.

(Klein Chapters 6-7)

Draw a reaction profile on the axes below for the formation of the organic product above that includes

structures of the reactant, any intermediate(s), and a transition state for the rate-determining step only.

Potential

energy

Reaction coordinate

5

6. (8 pts) Which isomer undergoes faster E2 reaction with base, cis-1-bromo-4-t-butyl cyclohexane or trans-1-

bromo-4-t-butylcyclohexane? Use diagrams to help with your explanation. (Klein Chapter 8)

7. (9 pts) Give each of the following molecules acceptable names; you may use trivial or IUPAC names for

substituents as needed. (Klein Chapters 7-8)

H

Br

I

Cl

a)

b)

c)

6

8. (8 pts) Give a complete mechanism that accounts for the following conversion. The optical rotation (i.e.

[α]D) of the starting material is +54°; what do you expect the rotation of the product to be, and why? (Klein

Chapters 5-7)

8. (8 pts) In the boxes provided, give a structure of a suitable organic starting material that would be needed to

produce the given product in each of the following cases. (Klein Chapters 7-8)

KOt-Bu

H2SO4, heat

NaN3 in DMF

H2O

CH2

N3

H3C OH

a)

b)

c)

d)

7

9. (18 pts) Answer each of the following questions by circling your choice and then giving a few words of

explanation for why you made that choice. (Klein Chapters 7-8)

a) Which of the following species is the better nucleophile? Why?

b) Which of the following compounds would react faster in a solvolysis reaction with CH3OH? Why?

c) Which of the following solvents would allow for faster SN2 reactions? Why?

d) Which of the following species is the better leaving group? Why?

e) Which of the following compounds would completely deprotonate ethanol? Why?

f) Which of the following compounds would react faster in SN2 reactions? Why?

1




1. (8 pts) Provide a detailed mechanism, using curved arrows to show the breaking and forming of bonds, that

accounts for the following conversion. (Klein Chapter 8)

Alkyl migration to convert a secondary carbocation

Into a more stable tertiary carbocation

2


of the following situations. (Klein Chapters 7-8)

NaOCH3

H3PO4, heat

KO-tBu

NaCl

CH3

NaN3, DMF

CH3

Br

H3C OCH3

(Zaitsev)

(racemic, via 3o carbocation)

(inversion through SN2)

(via 2o to 3o carbocation

rearrangment)

Br

+

Br

CH3OH

major minor

OTs N3

Cl

CH3 CH3

+ (Hoffman)

major minor

OH

(Zaitsev)+

major minor

Cl

a)

b)

c)

d)

e)

f)

3

3. (9 pts) For each of the following pairs of molecules, identify any stereogenic centers, label them as having

the (R) or the (S) configuration, and then indicate the relationship between the two molecules; are they

enantiomers, diastereomers, or are they identical? (Klein Chapter 5)

4. (8 pts) Give all of the expected organic products from the following reaction and label those products as

being major, minor, or very minor. Then give an explanation for your assignments. (Klein Chapter 8)

a)

b)

c)

4

5. (12 pts) Give the major organic product formed under the following reaction conditions and then a detailed

mechanism, using curved arrows to show bonds being formed and broken, to describe the transformation.


Draw a reaction profile on the axes below for the formation of the organic product above that includes

structures of the reactant, any intermediate(s), and a transition state for the rate-determining step only.

potential

energy

Reaction coordinate

OH2+

+

OH

OH2

R.D.S.

5

6. (8 pts) Which isomer undergoes faster E2 reaction with base, cis-1-bromo-4-t-butyl cyclohexane or trans-1-

bromo-4-t-butylcyclohexane? Use diagrams to help with your explanation. (Klein Chapter 8)

7. (9 pts) Give each of the following molecules acceptable names; you may use trivial or IUPAC names for

substituents as needed. (Klein Chapters 7-8)

H

Br

I

Cl

(E)-3-bromo-5-methylhept-2-ene

(S)-5-iodo-3,3-dimethylcyclopent-1-ene

1-chloro-2-propylcyclohex-1-ene

a)

b)

c)

6

8. (8 pts) Give a complete mechanism that accounts for the following conversion. The optical rotation (i.e.

[α]D) of the starting material is +54°; what do you expect the rotation of the product to be, and why? (Klein

Chapters 5-7)

Br

CH3

CH3

H2O OH

CH3

H

CH3CH3

O

CH3

O

=

H

H

H

HOH2

OH2 OH2

The chiral starting bromide converts to an achiral product so [α]D should be 0°.

8. (8 pts) In the boxes provided, give a structure of a suitable organic starting material that would be needed to

produce the given product in each of the following cases. (Klein Chapters 7-8)

KOt-Bu

H2SO4, heat

NaN3 in DMF

H2O

CH2

N3

H3C OH

I

OH

E2

E1

Br

SN2

H3C Br

SN1

a)

b)

c)

d)

7

9. (18 pts) Answer each of the following questions by circling your choice and then giving a few words of

explanation for why you made that choice. (Klein Chapters 7-8)

a) Which of the following species is the better nucleophile? Why?

b) Which of the following compounds would react faster in a solvolysis reaction with CH3OH? Why?

c) Which of the following solvents would allow for faster SN2 reactions? Why?

d) Which of the following species is the better leaving group? Why?

e) Which of the following compounds would completely deprotonate ethanol? Why?

f) Which of the following compounds would react faster in SN2 reactions? Why?

1




1. (7 pts) Give a detailed mechanism that shows the main steps that lead to the two products in the following

reaction. Include structures of any intermediates that are formed. (Klein Chapter 11)

2

2. (7 pts) Give the expected major product and then provide a detailed mechanism, using curved arrows to

show the breaking and forming of bonds, for the following reaction. (Klein Chapter 10)

3. (7 pts) Give the expected major and minor products formed under the following conditions. Then give a

mechanism for the formation of both the major and minor products. (Klein Chapter 11)

3


of the following situations. When more than one reaction is involved, a product from each is required.

(Klein Chapters 9-11) Br2, heat

HBr

1. O3

NaOCH3

CH3

dilute H2SO4

H2, Pt

CH3

2. Zn, H2O

Br

CH3OH

a)

b)

c)

d)

e)

f)

4

5. (8 pts) In the boxes provided, give a structure of the organic starting material that would be needed to

provide the given product in each of the following cases. (Klein Chapters 10-11)

Br2, heat

CH3CO3H

Br

O

Br Br

excess HBr

Br

6. (9 pts) Give the major product formed from free radical bromination of the following molecules and then

give a name for each of those products. (Klein Chapter 11)

a) Methylcyclopentane

b) 2,3-Dimethylbutane

c) 2,2,4,4-Tetramethylpentane

a)

b)

c)

d)

5

7. (9 pts) Give each of the following compounds acceptable names. You may use either the IUPAC system or

the common names for substituents. Watch out for any stereochemical issues (i.e cis/trans or R/S). (Klein

Chapter 10)

8. (6 pts) In each of the series of molecules below, circle the most stabilized species, and then give a few

words of explanation for your choice. (Klein Chapters 3-11)

a)

b)

c)

a.

b.

c.

6

9. (7 pts) Give all of the possible monobrominated products formed under the following conditions. Then

provide a detailed mechanism, pushing curved arrows to describe all important bond-forming and bond-

breaking events, that accounts for the formation of the major product in the following reaction. (Klein

Chapter 11)

10. (7 pts) Provide a detailed mechanism for the following reaction that explains the bond-forming and bond-

breaking steps on the way to the product. (Klein Chapter 9)

7

11. (7 pts) Provide a detailed mechanism, pushing curved arrows to describe bonds forming and breaking, that

accounts for the formation of both products in the following reaction. (Klein Chapter 9)

12. (8 pts) Give the major and minor final products formed in each of the following situations, and then give an

explanation for the different outcomes. (Klein Chapter 9)

a)

b)

1




1. (7 pts) Give a detailed mechanism that shows the main steps that lead to the two products in the following

reaction. Include structures of any intermediates that are formed. (Klein Chapter 11)

2

2. (7 pts) Give the expected major product and then provide a detailed mechanism, using curved arrows to

show the breaking and forming of bonds, for the following reaction. (Klein Chapter 10)

3. (7 pts) Give the expected major and minor products formed under the following conditions. Then give a

mechanism for the formation of both the major and minor products. (Klein Chapter 11)

3


of the following situations. When more than one reaction is involved, a product from each is required.


Br2, heat

HBr

1. O3

NaOCH3

CH3

dilute H2SO4

H2, Pt

CH3

2. Zn, H2O

Br

CH3OH

Br

CH3

O

HO

H3C OH

(via 3o radical)

(via 3o carbocation)

H

H

(via syn addition)

CH3

O

O

O1. 2.

Br

(via 3o carbocation)

(more highly substitutedalkene formed - Zaitsev)

a)

b)

c)

d)

e)

f)

4

5. (8 pts) In the boxes provided, give a structure of the organic starting material that would be needed to

provide the given product in each of the following cases. (Klein Chapters 10-11)

Br2, heat

CH3CO3H

Br

O

Br Br

excess HBr

Br

H

Na

6. (9 pts) Give the major product formed from free radical bromination of the following molecules and then

give a name for each of those products. (Klein Chapter 11)

a) Methylcyclopentane

b) 2,3-Dimethylbutane

c) 2,2,4,4-Tetramethylpentane

a)

b)

c)

d)

5

7. (9 pts) Give each of the following compounds acceptable names. You may use either the IUPAC system or

the common names for substituents. Watch out for any stereochemical issues (i.e cis/trans or R/S). (Klein

Chapter 10)

Cl

OH

Br

(R,Z)-hept-4-en-1-yn-3-ol

(R)-3-bromo-3-ethynylcyclopent-1-ene

(R)-4-chlorohex-1-yne

8. (6 pts) In each of the series of molecules below, circle the most stabilized species, and then give a few

words of explanation for your choice. (Klein Chapters 3-11)

a)

b)

c)

a.

b.

c.

6

9. (7 pts) Give all of the possible monobrominated products formed under the following conditions. Then

provide a detailed mechanism, pushing curved arrows to describe all important bond-forming and bond-

breaking events, that accounts for the formation of the major product in the following reaction. (Klein

Chapter 11)

10. (7 pts) Provide a detailed mechanism for the following reaction that explains the bond-forming and bond-

breaking steps on the way to the product. (Klein Chapter 9)

7

11. (7 pts) Provide a detailed mechanism, pushing curved arrows to describe bonds forming and breaking, that

accounts for the formation of both products in the following reaction. (Klein Chapter 9)

12. (8 pts) Give the major and minor final products formed in each of the following situations, and then give an

explanation for the different outcomes. (Klein Chapter 9)

a)

The outcome here is governed mainly by the relative steric interactions in the transition states of

the hydroboration step. The large –BR2 group will avoid the more crowded end of the alkene; the

oxidation step retains the regiochemistry and stereochemistry (syn addition) from the first step.

b)

In the acid-catalyzed addition of H2O to this unsymmetrical alkene the product distribution

depends upon the type of carbocation that could be formed; the major product will be formed via

the tertiary carbocation, which is more stabilized by hyperconjugation than the secondary (the

Markovnikoff rule).

1

Chemistry 3719 Practice Exam C1



1. (6 pts) Label all of the sp2 hybridized atoms within the following molecules. (Klein Chapter 1)

a)

b)

O

H

O

NCH3

c) O

2

2. (15 pts) Draw the required structures in each of the following situations. (Klein Chapter 4)

a) The highest energy chair conformation of trans-1,2-dibromocyclohexane

b) A Newman projection of a gauche conformation of n-octane along the C-2 – C-3 bond axis.

c) The less stable isomer of 1,2-diisopropylcyclopropane.

d) A Newman projection of the most stable conformation of n-hexane along the C-3 – C-4 bond axis

e) A boat conformation of cyclohexane.

3

3. (15 pts) Give acceptable names for the following molecules. Trivial or IUPAC names may be used for

substituents. (Klein Chapter 4)

CH(CH3)2

C(CH3)3

a)

b)

c)

d)

e)

4

4. (15 pts) Give the products from each of the following acid-base reactions. Then give the acids on each side

of the equations approximate pKa values and indicate in each case if you expect the reaction to be

exothermic, endothermic, or approximately thermoneutral in each case. (Klein Chapter 3)

+ H2O

CH3CH2CH2OK+

+

+

+

C CCH3

NH

CH3CH2NHLi

Li

OH

H LiN(CH3)2

OLi OH

a)

b)

c)

d)

e)

5

5. (9 pts) Draw acceptable structures for each of the following molecules. (Klein Chapter 4)

a) trans-1-Ethyl-3-isopropylcyclopentane

b) Cyclopentylcyclohexane

c) 2,4,4-Trimethyl-3-octanol

6. (10 pts) Draw all of the isomers possible for C6H14 and then give each structure an acceptable name. (Klein

Chapter 4)

6

7. (8 pts) For the highlighted atom in the following molecules, fill in the ground state electronic configuration

on the left, and then draw a picture on the right of the hybridization model that best explains the bonding of

the highlighted atoms. (Klein Chapter 1)

8. (6 pts) Which of the two molecules below is the stronger acid? Give their approximate pKa values and

explain your choice in terms of the structures and relative stabilities of the respective conjugate bases.

(Klein Chapter 3)

O

OH OH

7

9. (6 pts) Draw a second resonance structure for the following anions and then an overall resonance hybrid for

each. (Klein Chapter 2)

10. (10 pts) Consider the cis- and trans- isomers 1-t-butyl-3-ethylcyclohexane. (Klein Chapter 4)

a. Draw the lowest energy ring conformation for both isomers and explain your choices.

b. Indicate which isomer is less stable and explain your reasoning using a diagram to highlight any

destabilizing interactions.

1

Chemistry 3719 Practice Exam C1 Key



1. (6 pts) Label all of the sp2 hybridized atoms within the following molecules. (Klein Chapter 1)

a)

b)

O

H

O

NCH3

c) O

2

2. (15 pts) Draw the required structures in each of the following situations. (Klein Chapter 4)

a) The highest energy chair conformation of trans-1,2-dibromocyclohexane

Br

Br

HH

b) A Newman projection of a gauche conformation of n-octane along the C-2 – C-3 bond axis.

c) The less stable isomer of 1,2-diisopropylcyclopropane.

d) A Newman projection of the most stable conformation of n-hexane along the C-3 – C-4 bond axis

CH2CH3

H H

CH2CH3

HH

e) A boat conformation of cyclohexane.

3

3. (15 pts) Give acceptable names for the following molecules. Trivial or IUPAC names may be used for

substituents. (Klein Chapter 4)

CH(CH3)2

C(CH3)3

6-ethyl-3,5,5-trimethylnonane

bicyclo[5.3.0]decane

6-isopropyl-2,5-dimethylundecane

trans-1-(tert-butyl)-3-isopropylcyclohexane

6-cyclopropyl-2,3-dimethylheptane

a)

b)

c)

d)

e)

4

4. (15 pts) Give the products from each of the following acid-base reactions. Then give the acids on each side

of the equations approximate pKa values and indicate in each case if you expect the reaction to be

exothermic, endothermic, or approximately thermoneutral in each case. (Klein Chapter 3)

+ H2O

CH3CH2CH2OK+

+

+

+

C CCH3

NH

CH3CH2NHLi

Li

OH

H LiN(CH3)2

OLi OH

+ LiOHH

base acid: pKa ~ 16 pKa ~ 50

exothermic

base acid: pKa ~ 16 pKa ~ 16

+

OH OLithermoneutral

+C CLiCH3 HN(CH3)2

baseacid: pKa ~ 25 pKa ~ 38

exothermic

+ NLi

CH3CH2NH2

acid: pKa ~ 38base pKa ~ 38

thermoneutral

CH3CH2CH2OH+

OK

acid: pKa ~ 10 base acid: pKa ~ 16

exothermic

a)

b)

c)

d)

e)

5

5. (9 pts) Draw acceptable structures for each of the following molecules. (Klein Chapter 4)

a) trans-1-Ethyl-3-isopropylcyclopentane

b) Cyclopentylcyclohexane

c) 2,4,4-Trimethyl-3-octanol

6. (10 pts) Draw all of the isomers possible for C6H14 and then give each structure an acceptable name. (Klein

Chapter 4)

6

7. (8 pts) For the highlighted atom in the following molecules, fill in the ground state electronic configuration

on the left, and then draw a picture on the right of the hybridization model that best explains the bonding of

the highlighted atoms. (Klein Chapter 1)

8. (6 pts) Which of the two molecules below is the stronger acid? Give their approximate pKa values and

explain your choice in terms of the structures and relative stabilities of the respective conjugate bases.

(Klein Chapter 3)

The electron-withdrawing C=O in the carboxylic acid makes the OH proton

more positive than in the alcohol, and the C=O also allows the charge in the anion to delocalize thus making the conjugate base more stable.

7

9. (6 pts) Draw a second resonance structure for the following anions and then an overall resonance hybrid for

each. (Klein Chapter 2)

10. (10 pts) Consider the cis- and trans- isomers 1-t-butyl-3-ethylcyclohexane. (Klein Chapter 4)

a. Draw the lowest energy ring conformation for both isomers and explain your choices.

This conformation of the cis isomer can have both substituents equatorial thus avoiding any destabilizing 1,3-diaxial interactions that would occur in the “ring-flipped” conformation. The trans isomer has to have one substituent axial and will prefer the conformation in which the smaller ethyl group is axial and not the larger t-butyl group, which would inevitably have significantly more destabilizing 1,3-interactions.

b. Indicate which isomer is less stable and explain your reasoning using a diagram to highlight any

destabilizing interactions.

The trans isomer is less stable here since in either ring conformation one of the bulky groups will have to be in an axial position thus destabilizing the system due to 1,3-diaxial interactions (below).

1




1) (10 pts) Give a complete mechanism for how both products are formed in the following reaction. Which

product do you expect to be major and which minor? Briefly explain your answer. (Klein Chapter 8)

2

2) (18 pts) Give organic product(s), labeling major and minor where applicable, for the following reactions.


KCN, DMF

NaOCH3

HBr

HCl

KOt-Bu

H2SO4, H3C

OH

OOH

CH3

Cl

CH3OH,

OH

Br

OSO2CF3

H3C

a)

b)

c)

d)

e)

f)

3

3) (10 pts) Give the major organic product expected below, then a draw the reaction profile (potential energy

vs. reaction coordinate) for the conversion that includes pictures of the transition state(s) involved. (Klein

Chapters 6-7)

4) (9 pts) Give acceptable names for each of the following molecules. (Klein Chapters 5-8)

OH

Br

OH

Cl

OH

a)

b)

c)

Potential

energy

Reaction coordinate

4

5) (8 pts) Give the major organic products formed under the following conditions and a detailed mechanism

that explains their formation. Will the product mixture have an optical rotation or will [α]D = 0? (Klein

Chapter 7)

6) (9 pts) Give structures for each of the following molecules. (Klein Chapters 7-8)

a) (E)-4-Methyl-3-hexen-2-ol

b) 1-Bromo-5-tert-butylcyclohex-1-ene

c) trans-2-Isopropenylcyclopentanol

5

7) (12 pts) Tamiflu®

has been developed by Roche Pharmaceuticals to alleviate the symptoms caused by the

influenza virus. The 14-step synthesis of Tamiflu®

begins with the chiral natural product shikimic acid

(below). Identify all of the stereocenters in both shikimic acid and Tamiflu®

and label them as having the

(R) or (S) configuration. (Klein Chapter 5)

shikimic acid

HO

HO

HO

O

OH

H2N

HN

O

O

O

O

Roche

Pharmaceuticals14 steps

Tamiflu

8) (8 pts) Working out the configurations of the chiral centers (i.e. R or S) will help in the following questions.

(Klein Chapter 5)

a. Convert the following “wedge-dash” depiction into a Fischer projection.

b. What is the relationship between the following molecules (same, enantiomers, diastereomers)?

6

9) (10 pts) Provide explanations, using Newman projections, to explain the following product outcomes. (Klein

Chapter 8)

10) (6 pts) In each of the following cases, circle your choice and then give a few words of explanation for why

you made that choice. (Klein Chapter 7)

a. Reacts faster in SN2 reactions with NaN3 as the nucleophile?

b. Better solvent in SN1 reactions?

c. Better nucleophile in SN2 reactions?

a)

b)

1




1) (10 pts) Give a complete mechanism for how both products are formed in the following reaction. Which

product do you expect to be major and which minor? Briefly explain your answer. (Klein Chapter 8)

OH

+

H2SO4

heat

H OH2

OH2H

H

H2O

H2O

-H2O

-H3O

-H3O

Rearrangement occurs where the more stable tetrasubstituted alkene is preferred over trisubstituted alternative; sp

2 hybrid C in alkenes is stabilized by alkyl groups through

induction, thus the more alkyl groups the better the alkene.

2

2) (18 pts) Give organic product(s), labeling major and minor where applicable, for the following reactions.


KCN, DMF

NaOCH3

HBr

HCl

KOt-Bu

H2SO4, H3C

OH

OOH

CH3

Cl

CH3OH,

OH

Br

OSO2CF3

H3C

H3C H3C

H

+

H

major minor

CN

only inverted

product via SN2

OCl

CH3O

CH3

Cl+

racemic mixture via SN1

+

major minorE2 - Zaitsev

E1 - Zaitsev

Br only product

via SN2

H3C

+

H3C

major minor

E2 - Hofmann

a)

b)

c)

d)

e)

f)

3

3) (10 pts) Give the major organic product expected below, then a draw the reaction profile (potential energy

vs. reaction coordinate) for the conversion that includes pictures of the transition state(s) involved. (Klein

Chapters 6-7)

HClHO Cl(+ H2O)

HO

+ HCl

H2O

Cl

O

H

Cl

H

+

-

OH H

Cl

+

-

H

H

Potentialenergy

Reaction coordinate

SN2

4) (9 pts) Give acceptable names for each of the following molecules. (Klein Chapters 5-8)

a)

b)

c)

4

5) (8 pts) Give the major organic products formed under the following conditions and a detailed mechanism

that explains their formation. Will the product mixture have an optical rotation or will [α]D = 0? (Klein

Chapter 7)

In this SN1 reaction the leaving group breaks away to leave a flat carbocation that has three different alkyl groups attached. When the bromide nucleophile attacks it may do so from either face of the cation to produce enantiomeric products in approximately equal amounts. The optical rotation of this racemic mixture will therefore be zero as the rotations of the enantiomers will cancel.

6) (9 pts) Give structures for each of the following molecules. (Klein Chapters 7-8)

a) (E)-4-Methyl-3-hexen-2-ol

b) 1-Bromo-5-tert-butylcyclohex-1-ene

c) trans-2-Isopropenylcyclopentanol

5

7) (12 pts) Tamiflu®

has been developed by Roche Pharmaceuticals to alleviate the symptoms caused by the

influenza virus. The 14-step synthesis of Tamiflu®

begins with the chiral natural product shikimic acid

(below). Identify all of the stereocenters in both shikimic acid and Tamiflu®

and label them as having the

(R) or (S) configuration. (Klein Chapter 5)

shikimic acid

(R)

HO

(S)

HO

(R)HO

O

OH

H2N

HN

O

O

O

O

Roche Pharmaceuticals

14 steps

Tamiflu

(S)

(R)

(R)

8) (8 pts) Working out the configurations of the chiral centers (i.e. R or S) will help in the following questions.

(Klein Chapter 5)

a. Convert the following “wedge-dash” depiction into a Fischer projection.

b. What is the relationship between the following molecules (same, enantiomers, diastereomers)?

6

9) (10 pts) Provide explanations, using Newman projections, to explain the following product outcomes. (Klein

Chapter 8)

10) (6 pts) In each of the following cases, circle your choice and then give a few words of explanation for why

you made that choice. (Klein Chapter 7) F07 Exam 3

a. Reacts faster in SN2 reactions with NaN3 as the nucleophile?

b. Better solvent in SN1 reactions?

c. Better nucleophile in SN2 reactions?

a)

b)

1




1. (9 pts) Give the major and minor alkyl halide products expected from the following reaction and then show

a mechanism (using curved arrows) that describes how the major product is formed during the reaction.

(Klein Chapter 11)

2

2. (18 pts) Give the expected major products from each of the following reactions. When more than one set of

reagents are given the major product from each step is required. Where stereochemistry is an issue, label the

products as being either single stereoisomers, enantiomers, or diastereomers. (Klein Chapters 7-10)

1. O3

2. Zn, H2O

DMSO, 50 oC

1. H-B(C6H11)2

NaCN

2. NaOH, H2O2

Cl

Br2

CH3CO3H

CCH

H1. NaNH2

2. CH3CH2I

a.

b.

c.

d.

e.

f.

3

3. (8 pts) Give the major and minor product(s) from the following reaction and then a complete mechanism

(using curved arrows) for the formation of the major product. Explain briefly why this product is major.

(Klein Chapter 9)

4. (12 pts) Draw acceptable structures for each of the following molecules. (Klein Chapters 7-10)

a. (4S,5S)-5,7-dimethyl-1-octyn-4-ol

b. (2R,3S,Z)-2-bromo-5-fluoro-3-methyl-4-heptene

c. (1R,2S)-1-bromo-2-isopropylcyclohexane

4

5. (12 pts) Provide the major product from each step of the following reaction sequence: (Klein Chapter 10)

6. (8 pts) Give an explanation for the different product outcomes in the following reactions. (Klein Chapter

11)

5

7. (8 pts) Draw the organic product produced under the following ozonolysis conditions; then draw a detailed

mechanism for its formation. (Klein Chapter 9)

8. (8 pts) In each of the following reactions give a mechanistic explanation that accounts for the observed

regiochemistry and/or stereochemistry in each product. (Klein Chapter 9)

a. Explain the observed stereochemistry and regiochemistry in the following hydroboration:

b. Explain the observed stereochemistry in the following addition of Br2:

6

9. (9 pts) Provide a complete mechanism for the following conversion. (Klein Chapter 9)

10. (8 pts) Give the expected product from the following portion of the hydroboration-oxidation sequence and a

mechanism for its formation. Be careful to account for any important stereochemical issues. (Klein Chapter

9)

1




1. (9 pts) Give the major and minor alkyl halide products expected from the following reaction and then show

a mechanism (using curved arrows) that describes how the major product is formed during the reaction.

(Klein Chapter 11)

2

2. (18 pts) Give the expected major products from each of the following reactions. When more than one set of

reagents are given the major product from each step is required. Where stereochemistry is an issue, label the

products as being either single stereoisomers, enantiomers, or diastereomers. (Klein Chapters 7-10)

1. O3

2. Zn, H2O

DMSO, 50 oC

1. H-B(C6H11)2

NaCN

2. NaOH, H2O2

Cl

Br2

CH3CO3H

CC

HH

1. NaNH2

2. CH3CH2I

OO

O

1.O

O2. +

CN

SN2 reaction only gives the one

(inverted) product

B(C6H11)2H1.

OHH2.

Br

Br

Br

Br

+

Enantiomers formed (50/50)

(R,R) (S,S)

O O+

Diastereomers formed (~50/50)

CCH

1. CCCH2CH3H

2.

a.

b.

c.

d.

e.

f.

3

3. (8 pts) Give the major and minor product(s) from the following reaction and then a complete mechanism

(using curved arrows) for the formation of both products. Explain briefly why one product is major. (Klein

Chapter 9)

The major product in this electrophilic addition reaction is produced via the more stabilized tertiary carbocation, which is preferred over the alternative secondary carbocation, due to increased hyperconjugation;

the tertiary carbocation has more adjacent (beta) σ bonds, which are able to share electron density with the electron-poor positive carbon.

4. (12 pts) Draw acceptable structures for each of the following molecules. (Klein Chapters 7-10)

a. (4S,5S)-5,7-dimethyl-1-octyn-4-ol

b. (2R,3S,Z)-2-bromo-5-fluoro-3-methyl-4-heptene

c. (1R,2S)-1-bromo-2-isopropylcyclohexane

Br

4

5. (12 pts) Provide the major product from each step of the following reaction sequence: (Klein Chapter 10)

Br

Br Na

O

1. Br2 in CCl42. excess NaNH2

then H+ quench

3. CH3CH2CH2CH2Li4. CH3CH2CH2Br5. Na, liq. NH3

6. CH3CO3H

1. 2. 3.

4. 5. 6.

6. (8 pts) Give an explanation for the different product outcomes in the following reactions. (Klein Chapter

11)

The difference in the product distributions is due to the difference in the stability of the Br and the Cl radicals. The Br radical is more stable and is therefore able to be more selective when it comes to abstracting a hydrogen atom to produce a carbon radical. In this case the more stabilized 2

o radical is formed preferably leading to

the 2o

alkyl halide being the vast majority of the product. The Cl radical is much more reactive, and therefore less selective, resulting in significantly more of the 1

o

alkyl halide product being formed with Cl2 as the halogen radical source.

5

7. (8 pts) Draw the organic product produced under the following ozonolysis conditions; then draw a detailed

mechanism for its formation. (Klein Chapter 9)

8. (8 pts) In each of the following reactions give a mechanistic explanation that accounts for the observed

regiochemistry and/or stereochemistry in each product. (Klein Chapter 9)

a. Explain the observed stereochemistry and regiochemistry in the following hydroboration:

This reaction involves a syn addition of the BH3 molecule hence the shown cis stereochemical outcome. The regiochemical preference is explained by the smaller H bonding to the less accessible carbon of the alkene and the larger BH2 group adding to the less hindered carbon.

b. Explain the observed stereochemistry in the following addition of Br2:

This addition reaction involves the formation of a carbocation that is stabilized by donation of electron density from the first Br that has added (bromonium ion). The large Br then blocks one face of the molecule such that the bromide ion may only attack from the opposite face, hence the trans stereochemistry.

6

9. (9 pts) Provide a complete mechanism for the following conversion.

10. (8 pts) Give the expected product from the following portion of the hydroboration-oxidation sequence and a

mechanism for its formation. Be careful to account for any important stereochemical issues. (Klein Chapter

9)

The migration step is concerted and the initial stereochemistry of the addition (syn) is retained to give the product shown.

1

Chemistry 3719, Fall 2013 Exam 1 Student Name: “Y” Number:

This exam is worth 100 points out of a total of 700 points for Chemistry 3719/3719L. You have 50 minutes to complete the exam. Good Luck!

1. (8 pts) For the highlighted atom in the following molecules, fill in the ground state electronic configuration from the periodic table on the left, and then draw a picture on the right of the hybridization model that best explains the atomic orbitals used in bonding by the highlighted atoms.

Energy hybridize

Energy hybridizeNCH

OH

2

2. (10 pts) Provide the required values for each of the following species as indicated.

a. Give each of the following atoms an approximate electronegativity value:

b. Give each of the following acids an approximate pKa value:

3. (9 pts) In each of the following pairs of molecules one of the two is more stable than the other. Circle the more stable molecule and then briefly explain your choices in the space on the right.

a.

b.

c.

vs.

vs.

Li OLi

CH2 O

vs.

FOLi

OF

FH

OLi

OH

H

3

4. (9 pts) Provide each of the following alkane derivatives with an acceptable name. You may use either the common or IUPAC names for substituents.

a.

b.

c.

5. (9 pts) For the following systems, draw all of the other important resonance structures, including any lone

pairs, and then a resonance hybrid for each molecule.

O

a.

O

b.

c.

H

4

6. (12 pts) Give the products expected from each of the following acid-base reactions. Then label each acid with an approximate pKa and indicate whether deprotonation will be complete or incomplete in each case.

a.

b.

c.

OH+

OH

+ KOCH2CH3

NaO

+NLi

7. (8 pts) The following molecule has been designed as a synthetic intermediate in the formation of glutathione drug analogs. Identify each of the eight (8) functional groups contained within the molecule.

5

8. (9 pts) In each of the pairs of compounds below, indicate whether the structures are related as stereoisomers, constitutional isomers, or are the same compound. Then circle the more stable of the two structures and give a few words of explanation for your choices.

CH3

CH3

CH3H3C

a.

and

b.

c.

H

H3CH

CH3

H

Hand

H3C

HH

CH3

H

H

and

9. (8 pts) For each of the following structures, add any missing lone pairs that are needed and label any formal charges that are necessary.

Oa. b.

H O N

O

O

c. d.O

OO

O

6

10. (9 pts) Draw Newman depictions that correspond to the following conformations. a. The most stable conformation of 2-methyloctane along the C-4–C-5 bond.

b. The least stable conformation of 2,2-dimethylhexane along the C-3–C-4 bond.

c. A gauche conformation for hexylcyclopropane along the C-3–C-4 bond of the hexyl chain.

11. (9 pts) For each of the three molecules below, indicate whether you expect them to be a solid, liquid, or gas

at room temperature and whether you expect them to dissolve in water or not. Explain your choices briefly.

HOOH

OHO

HO

O

OH

OH

HO

HOO

sucroseanthracene

OH

1-octanol

1

Chemistry 3719, Fall 2013 Exam 1 – Key Student Name: “Y” Number:


1. (8 pts) For the highlighted atom in the following molecules, fill in the ground state electronic configuration from the periodic table on the left, and then draw a picture on the right of the hybridization model that best explains the atomic orbitals used in bonding by the highlighted atoms.

Energy hybridize

Energy hybridizeNCH

OH sp3

lone pairs to bonds

sp

lone pair to bond

to bonds

p

2

2. (10 pts) Provide the required values for each of the following species as indicated.

a. Give each of the following atoms an approximate electronegativity value:

b. Give each of the following acids an approximate pKa value:

3. (9 pts) In each of the following pairs of molecules one of the two is more stable than the other. Circle the

more stable molecule and then briefly explain your choices in the space on the right.

a.

b.

c.

vs.

vs.

Li OLi

CH2 O

vs.

FOLi

OF

FH

OLi

OH

H

A negative charge on a highly electronegativeatom such as O is more stable than on C

The lone pair on the right will delocalize suchthat some of the charge will be on O and not C

The highly electronegative F atoms will helpto spread negative charge through induction

3

4. (9 pts) Provide each of the following alkane derivatives with an acceptable name. You may use either the common or IUPAC names for substituents.

a.

b.

c.

bicyclo[3.3.3]undecane

2-isobutyl-5-isopropyl-1,1-dimethylcycloheptane

3,5-diethyl-6,7,7-trimethyldecane

5. (9 pts) For the following systems, draw all of the other important resonance structures, including any lone

pairs, and then a resonance hybrid for each molecule.

O

a.

O

b.

c.

H

O

OH

O

OH

O-

-

-

OH+

+

+

+

+

+

4

6. (12 pts) Give the products expected from each of the following acid-base reactions. Then label each acid with an approximate pKa and indicate whether deprotonation will be complete or incomplete in each case.

a.

b.

c.

OH+

OH

+ KOCH2CH3

NaO

+NLi

ONa+

OK

+ HOCH2CH3

HO

+NH

Li

pKa = 10 pKa = 16

K ~ 106

complete

deprotonation

pKa = 16 pKa = 16

K ~ 1incomplete

deprotonation

pKa = 25 pKa = 38

K ~ 1013

complete

deprotonation

7. (8 pts) The following molecule has been designed as a synthetic intermediate in the formation of glutathione

drug analogs. Identify each of the eight (8) functional groups contained within the molecule.

5

8. (9 pts) In each of the pairs of compounds below, indicate whether the structures are related as stereoisomers, constitutional isomers, or are the same compound. Then circle the more stable of the two structures and give a few words of explanation for your choices.

CH3

CH3

CH3H3C

a.

and

b.

c.

H

H3CH

CH3

H

Hand

H3C

HH

CH3

H

H

and

Same compound; right-hand ismore stable since both groupsare equatorial thus avoiding 1,3-diaxial interactions

Stereoisomers; having both CH3

groups on opposite sides of themolecule reduces steric strain

Constitutional isomers; right-handis more stable since it will haveless ring strain (no cyclopropane)

9. (8 pts) For each of the following structures, add any missing lone pairs that are needed and label any formal

charges that are necessary.

Oa. b.

H O N

O

O

c. d.O

OO

O

6

10. (9 pts) Draw Newman depictions that correspond to the following conformations. a. The most stable conformation of 2-methyloctane along the C-4–C-5 bond.

b. The least stable conformation of 2,2-dimethylhexane along the C-3–C-4 bond.

c. A gauche conformation for hexylcyclopropane along the C-3–C-4 bond of the hexyl chain.

11. (9 pts) For each of the three molecules below, indicate whether you expect them to be a solid, liquid, or gas

at room temperature and whether you expect them to dissolve in water or not. Explain your choices briefly.

HOOH

OHO

HO

O

OH

OH

HO

HOO

sucroseanthracene

OH

1-octanol

Anthracene is a solid since it has the formula C14H10 and there will be significant intermolecular dispersion forces holding molecules to each other. This compound is very non-polar so it would not be expected to be soluble in water. Sucrose has many OH groups which will allow H-bonding between molecules and cause this to be a solid. The OH groups also make the compound polar and, combined with the H-bonding, will make this material very soluble in water. 1-Octanol is a liquid since the greasy chain will only allow for weak intermolecular forces but the OH group will give some H-bonding. The non-polar alkyl chain will outweigh the OH group so this alcohol would not be expected to be soluble in water.

1



1. (10 pts) Draw two chair conformations for cis-1-t-butyl-4-ethylcyclohexane that are related through a ring-flip. Then circle which conformation you expect to be more stable and explain why. Then do the same for the trans isomer and, finally, indicate whether the cis or trans isomer should be more stable overall and explain your choice.

2

2. (15 pts) Give the expected products, major and minor where applicable, in each of the following situations. Be sure to take into account any changes in stereochemistry that may occur. If more than one set of reagents is employed, product(s) from each step is/are required.

1.

2. NaSH, DMSO

NaCN, DMSO

CH3OH

NaNH2, DMSO

a.

b.

c.

d.

e.

SO2ClH3C

pyridine

O

Br

O ICH3

OH

CH3Br

KOC(CH3)3

Br

OH

H

3

3. (9 pts) Provide each of the following molecules with an acceptable name. You may use either the common or IUPAC names for substituents and be careful to include any stereochemical descriptors where needed.

a.

b.

c.

Cl

F

CH3

CH2CH3

H Br

F H


diagrams of all intermediates and transition states involved.

4

5. (9 pts) In the boxes provided, identify the type of arrow-pushing process in each step within the following reaction mechanisms (i.e. nucleophilic attack, leaving group leaving, proton transfer, or rearrangement).

a.

b.

c.

HHH H

Br

H Br Br

OH OH2H OH2 H

H2O- H2O

Br

H2O- Br

OH H

H2OOH

6. (9 pts) The following solvolysis conditions give two major products in equal amounts and, when measured

for the mixture, []D = 0. Provide the products and a mechanism for this process, as well as an explanation for the optical rotation data.

H2O

O

Cl

5

7. (9 pts) For each of the following pairs of molecules, provide the configuration of any chiral centers and then indicate whether the two molecules within each pair are enantiomers, diastereomers, or are identical.

a.

b.

c.

and

and

and

CH3

H ClCH3

BrH H3CCH3

Cl

Br

O

OH

Br

O

OH

Br

8. (9 pts) The following substitution reaction begins with an enantiomerically pure starting material (i.e. only the R isomer) which has []D = +53° and, after the reaction, the isolated product has []D = -35°. Explain how this data helps you to decide which mechanism (SN1 or SN2) is operating here and suggest a simple experiment that would back up your choice. Push arrows to show the mechanism and give a product.

Br

:P(CH2CH2CH2CH3)3

ether

6

9. (10 pts) The molecule below is Haterumalide ND, a potent anti-tumor drug. Identify 5 (five) chiral centers within the compound and label them as being either the R or S configuration.

O

CH3

OAc

O

Cl

HO

O

H

H

CH3

CO2H

OH

10. (10 pts) Provide a detailed mechanism for the following transformation that uses curved arrows to show the breaking and forming of bonds. Identify each type of arrow-pushing process that you use and explain why this bromide is the major product formed.

1

Chemistry 3719, Fall 2013 Exam 2 - Key Student Name:

“Y” Number:


complete the exam. Good Luck!

1

1. (10 pts) Draw two chair conformations for cis-1-t-butyl-4-ethylcyclohexane that are related through a ring-

flip. Then circle which conformation you expect to be more stable and explain why. Then do the same for

the trans isomer and, finally, indicate whether the cis or trans isomer should be more stable overall and

explain your choice.

2


Be sure to take into account any changes in stereochemistry that may occur. If more than one set of reagents

is employed, product(s) from each step is/are required.

1.

2. NaSH, DMSO

NaCN, DMSO

CH3OH

NaNH2, DMSO

a.

b.

c.

d.

e.

SO2ClH3C

pyridine

O

Br

O I

CH3

OH

CH3Br

CH3CN

SN2 with inversion

KOC(CH3)3

major

(large base is used)

O

OTs1.

O

SH2.

O OCH3

CH3O OCH3

CH3

+

SN1 - Racemix mixture formed (~50:50)

Intramolecular SN2Br

OHO

H H

+

SN2 with inversion

3

3. (9 pts) Provide each of the following molecules with an acceptable name. You may use either the common

or IUPAC names for substituents and be careful to include any stereochemical descriptors where needed.

a.

b.

c.

Cl

(2S,5S)-2-chloro-5-ethyloctane

F

(R,E)-5-fluoro-6-methylhept-3-ene

CH3

CH2CH3

H Br

F H

(2S,3S)-2-bromo-3-fluoropentane


diagrams of all intermediates and transition states involved.

4

5. (9 pts) In the boxes provided, identify the type of arrow-pushing process in each step within the following

reaction mechanisms (i.e. nucleophilic attack, leaving group leaving, proton transfer, or rearrangement).

a.

b.

c.

HHH H

Br

H Br Br

OH OH2

H OH2 HH2O- H2O

Br

H2O- Br

OH H

H2O

OH

proton transfer Nu: attackrearrangement

proton transfer L.G. leaves proton transfer

L.G. leaves Nu: attack proton transfer

6. (9 pts) The following solvolysis conditions give two major products in equal amounts and, when measured

for the mixture, [α]D = 0. Provide the products and a mechanism for this process, as well as an explanation

for the optical rotation data.

H2O

O

Cl

O

OH

O

OH

+

O O

O

O

O

+

H H H H

H2O H2O

OH2

enantiomers formed in equal amounts so

optical rotations cancel and [ ]D = 0

flat carbocation that is attackedequally from both sides

Leaving group leaves

Nu: attacks

Proton transfer

SN1!

5

7. (9 pts) For each of the following pairs of molecules, provide the configuration of any chiral centers and then

indicate whether the two molecules within each pair are enantiomers, diastereomers, or are identical.

a.

b.

c.

and

and

and

CH3

H ClCH3

BrHH3C (R)

(R) CH3

Cl

Br

(S)(S)

O

OH

Br

(R)(S)

O

OH

Br

diastereomers

enantiomers

identical

(S, S)

8. (9 pts) The following substitution reaction begins with an enantiomerically pure starting material (i.e. only

the R isomer) which has [α]D = +53° and, after the reaction, the isolated product has [α]D = -35°. Explain

how this data helps you to decide which mechanism (SN1 or SN2) is operating here and suggest a simple

experiment that would back up your choice. Push arrows to show the mechanism and give a product.

The fact that the chiral starting material produces a chiral product is only reconcilable

with the SN2 mechanism; the SN1 mechanism would produce a racemic mixture of

products for which [α]D = 0 because the rotations of the equal amounts of the R and S

enantiomers would cancel.

A simple experiment to support the SN2 assignment would be to add more nucleophile;

the reaction would go faster if this is actually a second order reaction in which the

nucleophile is involved in achieving the transition state in the rate-determining step.

6

9. (10 pts) The molecule below is Haterumalide ND a potent anti-tumor drug. Identify 5 chiral centers within

the compound and label them as being either the R or S configuration.


breaking and forming of bonds. Identify each type of arrow-pushing process that you use and explain why

this bromide is the major product formed.

Protonation by HBr produces an excellent leaving group which breaks away to leave a

secondary carbocation; this then undergoes alkyl rearrangement, with expansion of the

strained cyclobutyl ring, to produce a more stable cyclopentyl tertiary carbocation which is

finally trapped by the nucleophile to give the tertiary bromide.

1



1. (10 pts) Provide a detailed mechanism for the formation of the major product in the following conversion, using curved arrows to describe the making and breaking of bonds, and then explain the regiochemical and stereochemical outcomes in terms of the reactive intermediate(s) involved.

2

2. (15 pts) Give the expected major products in each of the following situations. Be sure to take into account any changes in stereochemistry that may occur. If more than one set of reagents is employed, product(s) from each step is/are required.

1.

2. NaOH, H2O2

1. excess NaNH2

1. CH3CO3H

1. Br2, heat

a.

b.

c.

d.

e.

1. NaNH2, THF

Br Br2. H2O

2. CH3CH2CH2Br

BH

CH3

CH32. H+/H2O

O

CH3

2. KOt-Bu, THF

3

3. (9 pts) Provide each of the following molecules with an acceptable name. You may use either the common or IUPAC names for substituents and be careful to include any stereochemical descriptors where needed.

a.

b.

c.

F

4. (10 pts) On the axes given below, draw a complete reaction profile for the following addition reaction that

includes diagrams of all intermediates and transition states involved. Then indicate which step is rate-determining and whether that step is unimolecular or bimolecular.

4

5. (9 pts) Provide a complete mechanism for the following conversion that features curved arrows to describe the making and breaking of bonds. Why is a ketone formed and not an aldehyde in this reaction?


two alkyl halide products being formed with a combined []D = 0. Give a complete mechanism for the formation of the alkyl bromide products and an explanation for the optical rotation data.

5

7. (9 pts) For each of the following pairs of structures, circle the one that should be more stable. Then write a few words of explanation for your choices.

a.

b.

c.

and

and

and

CH2 CH2

Hg

OAc

Hg

OAc

8. (9 pts) Give a detailed mechanism for the following ozonolysis reaction that uses curved arrows to show the making and breaking of all bonds. What is the driving force (i.e. thermodynamic motivation) for this kind of process that results in both carbon-carbon sigma and pi bonds being broken?

6

9. (12 pts) Provide a sequence of reagents and conditions that would allow for the following conversions. Show the expected major products from each step of your syntheses.

?H3C

H3C

H3C

H3C CH3

?

10. (8 pts) Explain why the following reactions give such different outcomes in terms of the percentages of products formed. Discuss the radicals formed and the transition states for H-abstraction in each case.

Cl2, h

Br2, h

Cl

Br

Br

Cl+

+

72% 28%

100 % 0%

1

Chemistry 3719, Fall 2013 Exam 3 - Key Student Name:

“Y” Number:


complete the exam. Good Luck!

1. (10 pts) Provide a detailed mechanism for the formation of the major product in the following conversion,

using curved arrows to describe the making and breaking of bonds, and then explain the regiochemical and

stereochemical outcomes in terms of the reactive intermediate(s) involved.

The regiochemical outcome is a consequence of a bromine radical abstracting H on a tertiary carbon to

produce a tertiary radical (stabilized by hyperconjugation). The stereochemical outcome is the consequence

of adding a Br radical to either face of a planar (prochiral) alkene, which produces a planar (prochiral) radical

that then abstracts H from HBr. Overall two new stereocentres and 4 stereoisomers are formed.

2

2. (15 pts) Give the expected major products in each of the following situations. Be sure to take into account

any changes in stereochemistry that may occur. If more than one set of reagents is employed, product(s)

from each step is/are required.

1.

2. NaOH, H2O2

1. excess NaNH2

1. CH3CO3H

1. Br2, heat

a.

b.

c.

d.

e.

1. NaNH2, THF

Br Br2. H2O

2. CH3CH2CH2Br

BH

CH3

CH32. H+/H2O

O

CH3

2. KOt-Bu, THF

1. 2.

H

1. 2.

1.B

H

(+/-)

2.OH

H

(+/-)

1. CH3

CH3

O2.

CH3

CH3

OH

OH

1. 2.

meso (+/-)

(+/-)

O

CH3Br

O

CH2

O

CH3

or

3

3. (9 pts) Provide each of the following molecules with an acceptable name. You may use either the common

or IUPAC names for substituents and be careful to include any stereochemical descriptors where needed.

a.

b.

c.

F

(R)-5-fluorohex-1-yne

(Z)-cyclooct-1-en-3-yne

(1R,2S)-1-ethynyl-2-vinylcyclohexane

4. (10 pts) On the axes given below, draw a complete reaction profile for the following addition reaction that

includes diagrams of all intermediates and transition states involved. Then indicate which step is rate-

determining and whether that step is unimolecular or bimolecular.

4

5. (9 pts) Provide a complete mechanism for the following conversion that features curved arrows to describe

the making and breaking of bonds. Why is a ketone formed and not an aldehyde in this reaction?

A ketone is formed because the first addition of H+ to the alkyne will form a

secondary-like carbocation instead of a less stable primary-like carbocation.


two alkyl halide products being formed with a combined [α]D = 0. Give a complete mechanism for the

formation of the alkyl bromide products and an explanation for the optical rotation data.

During the first H-abstraction step a planar allylic radical is formed which then reacts with either Br2

or Br radical from either face to produce a racemic mixture of enantiomeric products. Due to the

symmetry within the molecule reaction at either C-1 or C-3 of the allylic radical will give the same

products regardless of where reaction takes place.

5

7. (9 pts) For each of the following pairs of structures, circle the one that should be more stable. Then write a

few words of explanation for your choices.

a.

b.

c.

and

and

and

CH2 CH2

Hg

OAc

Hg

OAc

Benzylic radical will be stabilizedby resonance delocalization

Trisubstituted alkene features three

electron-donating alkyl groups

that help stabilize sp2 carbon atoms

Mercurinium ion has all atoms withfull complement of electrons; the carbocation has a 6-electron C atom

8. (9 pts) Give a detailed mechanism for the following ozonolysis reaction that uses curved arrows to show

the making and breaking of all bonds. What is the driving force (i.e. thermodynamic motivation) for this

kind of process that results in both carbon-carbon sigma and pi bonds being broken?

Ozone is a very reactive molecule that features weak O-O bonds that get the chance to break

during the ozonolysis process. In the first step the weaker O-O and C-C pi bonds are swapped for

sigma bonds to form the molozonide, and in the second step one of the weak O-O bonds is broken,

along with the C-C sigma bond. The resultant recombination produces the ozonide shown above.

6

9. (12 pts) Provide a sequence of reagents and conditions that would allow for the following conversions.

Show the expected major products from each step of your syntheses.

?H3C

H3C

H3C

H3C CH3

?

NaNH2

H3C

H3C

H3C

H3C

CH3

CH3Br

excess H2, Pt

Br2, RT

xs. NaNH2

H3O+ quench

Br

Br

10. (8 pts) Explain why the following reactions give such different outcomes in terms of the percentages of

products formed. Discuss the radicals formed and the transition states for H-abstraction in each case.

In both reactions there is competition between the formation of tertiary and primary radicals with the

more stable tertiary radical being favoured in each case. The difference in product distributions is

explained by the greater reactivity, and therefore lower selectivity, of the Cl radicals. The H-

abstraction step for chlorination is exothermic and fast, so the transition state is early and has little

radical character. For bromination the T.S. is more like the radical so the system can sense which

type of radical is formed with tertiary being greatly favoured.

OChem1 Practice Problems

___________________________________________________________

___________________________________________________________

Dr. Peter Norris, 2014

Dr. Peter Norris OChem 1

1

Klein Chapter 1 Problems : Review of General Chemistry

1. Draw viable structures for molecules with the following molecular formulae. Remember that each atom has

a “standard” valence when forming neutral molecules, for example halogens and H are monovalent, O is

divalent, N is trivalent, and C is tetravalent. Each of the molecules has only single bonds between atoms.

2. Draw Lewis formula structures for molecules with the following molecular formulae. Show all single bonds

as single lines and any lone pairs where appropriate.

3. Draw structures for the following molecules that include any formal charges and lone pairs where needed.

Each of the molecules has either single bonds or double bonds between atoms and there may be ionic

bonds involved in some of the structures.

4. Indicate the hybridization of each of the C, N, and O atoms in the following molecules.

5. Draw a structural formula for each of the following molecules and then, using the δ+/δ- convention, label

any dipoles that are present each molecule.


1

Klein Chapter 1 Problems : Review of Gen Chem – Answers

1. “Draw viable structures for molecules with the following molecular formulae. Remember that each atom has

a “standard” valence when forming neutral molecules, for example halogens and H are monovalent, O is

divalent, N is trivalent, and C is tetravalent. Each of the molecules has only single bonds between atoms.”

CH3Cla. C2H5Fb. C2H6Oc. C3H7Brd.

C2H4Cl2e. C2H3F3f. C3Br8g. C2H4BrClh.

C

H

H

H

Cl C

H

H

H

C

H

H

F C

H

H

H

C

H

H

OH

C

H

H

H

O C

H

H

H

or

C

H

H

H

C

H

H

C

H

H

Br

or

C

H

H

H

C

H

Br

C

H

H

H

C

H

H

H

C

H

Cl

Cl

C

H

Cl

H

C

H

H

Cl

or

C

H

H

H

C

F

F

F

C

H

F

H

C

F

H

F

or

C

Br

Br

Br

C

Br

Br

C

Br

Br

Br C

H

H

H

C

H

Cl

Br

C

H

Cl

H

C

H

H

Br

or

2. “Draw Lewis formula structures for molecules with the following molecular formulae. Show all single bonds

as single lines and any lone pairs where appropriate.”


2

3. “Draw structures for the following molecules that include any formal charges and lone pairs where needed.

Each of the molecules has either single bonds or double bonds between atoms and there may be ionic bonds

involved in some of the structures.”

4. “Indicate the hybridization of each of the C, N, and O atoms in the following molecules.”

5. “Draw a structural formula for each of the following molecules and then, using the δ+/δ- convention, label

any dipoles that are present each molecule.”


1

Klein Chapter 2 Problems : Molecular Representations

1. Turn each of the following “bond-line” structures into more expanded representations in which each of the

C, N, O, and H atoms are shown.

2. From Table 2.1 in the Klein text, identify all of the functional groups present in the following molecules.

3. Add any missing lone pairs to the following line structures.

4. Draw a second resonance form for each of the following structures.


1

Klein Chapter 2 Problems : Molecular Representations – Answers

1. Turn each of the following “bond‐line” structures into more expanded representations in which each of the

C, N, O, and H atoms are shown.

2. From Table 2.1 in the Klein text, identify all of the functional groups present in the following molecules.

3. Add any missing lone pairs to the following line structures.


2

4. Draw a second resonance form for each of the following structures.


1

Klein Chapter 3 : Acids and Bases Worksheet

Concepts

The interaction of a protic acid with a base is the first actual chemical reaction that we study in 3719. We use it to introduce

some of the fundamental concepts of mechanism, i.e. how to describe the bond-breaking and bond-forming events that

occur as a starting material (left-hand side of the equation) is converted to product (right-hand side of the equation). Since

we use acids and bases on many occasions as reagents throughout 3719 and 3720, it is important that you know what

constitutes an acid or a base and what happens when the two interact.

Definitions: Acids donate protons (Bronsted definition) or accept electrons (Lewis definition)

Bases accept protons (Bronsted definition) or donate electrons (Lewis definition)

General Chem:

In this case HCl is the acid (proton donor), water is the base (electron donor, proton acceptor) and the green arrows show

the proton being transferred to generate the conjugate acid (H3O+) and the conjugate base (Cl

-). The strength of the acid in

water is measured by the dissociation constant (Ka); the larger the Ka, the stronger the acid. In other words, the stronger

the acid, the more the above reaction goes to the right. Since the values of Ka can range from very large to very small, we

use the pKa scale to give manageable numbers (pKa = -log10Ka). The pKa range for acids used in Organic Chemistry are in the

range –10 (very strong acid) to +50 (very weak acid).

pKa values from class (Table 3.1 in Klein text):

HI, HBr, HCl, H2SO4 –5 to –10 extremely strong acids

CH3CO2H 4.7 weak acid H2O, ROH 16-18 weaker acids NH3 38 very weak acid CH4 50 not at all acidic

For the most part in 3719 and 3720 we will not be using water as the base; rather we will use some organic or inorganic

base which has been chosen to deprotonate the acid to a particular extent. Since we know something about relative acid

strength from pKa values, we also know a lot about the relative (conjugate) base strengths:

The task in Organic Chemistry is to decide what happens in terms of the equilibrium position when a particular acid is mixed

with a particular base. If you understand this, then you can decide which acid or base to use in particular circumstances.

The problems on the next page will give you practice with these ideas.


2

1. For each of the following mixtures, complete the equation, then identify the acid and base on the left and the

conjugate acid and conjugate base on the right. Comparing acid and base strengths (from pKa values) decide whether

the reaction is a) likely to happen at all, b) whether equilibrium will be established, and c) if equilibrium is established

which side is favoured?

2. The equilibrium idea means that if the reaction is reversible there will be four species in solution at one time, the acid,

the base, the conjugate acid, and the conjugate base. Sometimes this is what is required, but at other times we need to

choose bases that will completely deprotonate every molecule of acid, i.e. send the reaction completely to the right.

These bases will include CH3CH2CH2CH2Li, NaNH2, and LiN(i-Pr)2. Weaker bases will include NaOH, NaOCH3, KOtBu, and

NaOCH2CH3. For each of these bases, give the products formed when they react with H2O, then use pKa values to get an

idea of the relative base strengths of these compounds.

3. In OChem 2 we will study reactions based on the deprotonation of ketones such as acetone, (CH3)2C=O, which has a

pKa of 19. Given the bases LiN(i-Pr)2 and NaOCH3, decide which will be useful to completely deprotonate acetone, and

which will be useful for setting up an equilibrium. Explain your choices.

4. Give the products from the following acid-base reactions and identify the acid and base on the left side, as well as the

conjugate acid and conjugate base on the right side of the equation.

O

OH+ NaOH

O + NaNH2

OH

+ CH3LiOH

5. For each of the reactions in question 1 and question 4, which will have an equilibrium constant (K) greater than 1, close

to 1, or less than 1? Explain your answers.

6. Draw the structure of the conjugate base that will be formed when 1 mole of HOCH2CH2CO2H reacts with 1 mole of

CH3CH2CH2CH2Li.

7. The careful choice of an appropriate solvent will play a major role in whether organic reactions will be successful. For

example, reagents such as CH3Li and NaNH2 will be incompatible with solvents such as water and ethanol. Why is this?


Klein Chapter 3 : Acids and Bases Worksheet – Answers

1. “For each of the following mixtures, complete the equation, then identify the acid and base on the left and the conjugate acid and conjugate base on the right. Comparing acid and base strengths (from pKa values) decide whether the reaction is a) likely to happen at all, b) whether equilibrium will be established, and c) if equilibrium is established which side is favoured?”

For this type of question you have to know the approximate pKa values discussed in class and have an idea of what they mean in terms of relative acid strength and also relative base strength. a.

Here we have two acids of similar strengths (1‐butanol on the left and H2O on the right) and therefore two bases of similar strengths. In this situation, the forward reaction is favoured to about the same extent as the reverse reaction. Therefore a) the reaction left to right is likely to happen, b) equilibrium will be established, and c) the equilibrium lies approximately in the middle (similar acid strengths, similar base strengths on both sides of the equation). b.

In this example we have acids of very different strength, the carboxylic acid on the left (pKa~5) is very much stronger than the ammonia on the right (pKa~38) therefore the left to right reaction is very much favoured. The reverse reaction however is unlikely to occur since the base on the right is far too weak to deprotonate such a weak acid as NH3. Therefore a) the reaction left to right will happen, b) equilibrium will not be established, and c) the right‐hand side is completely favoured. c.

Here we have a situation in which the reaction will go to the right. The CH3CH2CH2CH3 is such a weak acid (pKa~50) that it is impossible to deprotonate. A 28 unit pKa difference between acids means that this is a completely irreversible reaction. Therefore a) the reaction from left to right will occur to give the products shown above, b) there will be no equilibrium established, and c) the right‐hand side would be completely favoured in this case.

2. “The equilibrium idea means that if the reaction is reversible there will be four species in solution at one time, the acid,

the base, the conjugate acid, and the conjugate base. Sometimes this is what is required, but at other times we need to


choose bases that will completely deprotonate every molecule of acid, i.e. send the reaction completely to the right. These bases will include CH3CH2CH2CH2Li, NaNH2, and LiN(i‐Pr)2. Weaker bases will include NaOH, NaOCH3, KOtBu, and NaOCH2CH3. For each of these bases, give the products formed when they react with H2O, then use pKa values to get an idea of the relative base strengths of these compounds.”

+ H2O + LiOH

very strongbase

acid C.A. C.B.

pKa ~16 pKa ~50

CH3CH2CH2CH2Li CH3CH2CH2CH3 completely to the right

+ H2O + NaOH

very strongbase

acid C.A. C.B.

pKa ~16 pKa ~38

NaNH2 NH3 completely to the right

+ H2O + NaOH

very strongbase

acid C.A. C.B.

pKa ~16 pKa ~38

LiN(i-Pr)2 HN(i-Pr)2 completely to the right

+ H2O + NaOH

strongbase

acid C.A. C.B.

pKa ~16 pKa ~16

NaOH H2O equilibr ium, roughlyin the middle

+ H2O + NaOH

strongbase

acid C.A. C.B.

pKa ~16 pKa ~16

NaOCH3 HOCH3 equilibr ium, roughlyin the middle

+ H2O + NaOH

strongbase

acid C.A. C.B.

pKa ~16 pKa ~18

KOt-Bu HOt-Bu equilibr ium, slightlyto the right

+ H2O + NaOH

strongbase

acid C.A. C.B.

pKa ~16 pKa ~16

NaOCH2CH3 HOCH2CH3 equilibr ium, roughlyin the middle


3. “In OChem 2 we will study reactions based on the deprotonation of ketones such as acetone, (CH3)2C=O, which has a pKa

of 19. Given the bases LiN(i‐Pr)2 and NaOCH3, decide which will be useful to completely deprotonate acetone, and which will be useful for setting up an equilibrium. Explain your choices.” In this question you have to set up the equation and then decide whether the base employed is indeed strong enough to completely deprotonate the acid, i.e. send the reaction completely to the right.

Here (above) we have a very powerful base (very weak conjugate acid) and this reaction will proceed all the way over to the right, i.e. all of the acid molecules will be deprotonated.

In this case (above) we are using a weaker base and even though the left to right reaction is possible, the right to left is also now possible. Equilibrium will be established here which will favour the side that contains the weaker conjugate base, in this case the left hand side. The main consequence of using this type of base here is that we now have some of all four species in solution at once.

4. “Give the products from the following acid‐base reactions and identify the acid and base on the left side, as well as the conjugate acid and conjugate base on the right side of the equation.”


5. “For each of the reactions in question 1 and question 4, which will have an equilibrium constant (K) greater than 1, close

to 1, or less than 1? Explain your answers.” 1a) K ~ 1 since acids and bases on each side are of similar strengths; 1b) No equilibrium since the acid and very strong base on the left will send the reaction completely to the right, K > 1; 1c) No equilibrium since the acid and very strong base on the left will send the reaction completely to the right, K > 1; 4a) K > 1 since the stronger acid and base on the left will favour the reaction to the right; 4b) No equilibrium since the reaction will go completely to the right, K > 1; 4c) No equilibrium since the reaction will go completely to the right, K > 1.

6. “Draw the structure of the conjugate base that will be formed when 1 mole of HOCH2CH2CO2H reacts with 1 mole of CH3CH2CH2CH2Li.”

O

OHHO + CH3CH2CH2CH2Li

O

OLiHO + CH3CH2CH2CH3

pKa ~16 pKa ~5

The more acidic carboxylic acid proton (pKa ~5) will be removed before the less acidic alcohol proton (pKa ~16)

7. “The careful choice of an appropriate solvent will play a major role in whether organic reactions will be successful. For example, reagents such as CH3Li and NaNH2 will be incompatible with solvents such as water and ethanol. Why is this?” These reagents are very powerful bases so they will be rapidly protonated by the solvents in question and destroyed:


1

Klein Chapter 4 Problems : Alkanes & Cycloalkanes

1. Provide accurate names for each of the following alkanes. You may use either IUPAC or trivial names for any

substituents within these molecules.

2. Draw Newman projections for each of the following situations.

a. The highest energy conformation of 3-methylnonane along the C-4 – C-5 bond axis

b. The lowest energy conformation of trans-1-isopropyl-2-propylcyclohexane along the C-1 – C-2 axis

c. The lowest energy conformation of 7-ethyl-2,3,8-trimethyldecane along the C-5 – C-6 bond axis

3. Draw appropriate diagrams for each of the following situations and explain your choices.

a. The most stable chair conformation of 1,3-di-tert-butylcylcohexane

b. The less stable chair conformation of trans-1-isopropyl-4-methylcyclohexane

c. The less stable isomer of 1,2-di-isopropylcyclobutane

4. Indicate which species/conformation is more stable in each of the following situations and give a few words

of explanation for your choices.


1

Klein Chapter 4 Problems : Alkanes & Cycloalkanes – Answers

1. Provide accurate names for each of the following alkanes. You may use either IUPAC or trivial names for any

substituents within these molecules.

2. Draw Newman projections for each of the following situations.

a. The highest energy conformation of 3-methylnonane along the C-4 – C-5 bond axis

b. The lowest energy conformation of trans-1-isopropyl-2-propylcyclohexane along the C-1 – C-2 axis

c. The lowest energy conformation of 7-ethyl-2,3,8-trimethyldecane along the C-5 – C-6 bond axis


2

3. Draw appropriate diagrams for each of the following situations and explain your choices.

a. The most stable chair conformation of 1,3-di-tert-butylcylcohexane

b. The less stable chair conformation of trans-1-isopropyl-4-methylcyclohexane

c. The less stable isomer of 1,2-di-isopropylcyclobutane

4. Indicate which species/conformation is more stable in each of the following situations and give a few words

of explanation for your choices.

a. b.

c. d.

or

CH2CH3

CH(CH3)2

CH2CH3

CH(CH3)2

or

H

H CH3

CH3

CH3H

CH3

H H

CH3

CH3H

or or

Having the two large groups on opposite sides of the ring will avoid destabilizing interactions

Both of the large substituents will be able to be in axial positions therebyavoiding 1,3-diaxial problems

The right-hand conformation has only

one CH3-CH3 gauche interaction as

opposed to two on the left

Both of the large substituents will beable to be as far away as possible from each other when in the anti orientation


1

Klein Chapter 5 Problems : Stereoisomerism

1. Identify any chiral carbons in the following molecules and, using the Cahn-Ingold-Prelog rules, label their

configuration as either R or S.

2. Within each of the following pairs of molecules, identify the relationship between them as being either

enantiomers, diastereomers, or identical.

3. Convert each of the following structures as directed.

4. Identify all the chiral carbon atoms in cholesterol (below) and label them as either the R or S configuration.


1

Klein Chapter 5 Problems : Stereoisomerism – Answers

1. Identify any chiral carbons in the following molecules and, using the Cahn-Ingold-Prelog rules, label their

configuration as either R or S.

2. Within each of the following pairs of molecules, identify the relationship between them as being either

enantiomers, diastereomers, or identical.

3. Convert each of the following structures as directed.


2

4. Identify all the chiral carbon atoms in cholesterol (below) and label them as either the R or S configuration.


1

Klein Chapter 6 Problems : Reactivity & Mechanisms

1. In each of the following situations, indicate whether the reactions are favoured or disfavoured in terms of

enthalpic and entropic factors in the Gibbs free energy equation (Table 6.1 might help here).

2. Draw a mechanistic interpretation (using curved arrow(s) to show bonds forming and breaking), and then an

approximate transition state (including partial electron-densities) for each of the following events.

3. Draw curved arrows to describe each of the steps in the following reaction mechanism, and then label those

steps as being one of the four types discussed in class, i.e. nucleophilic attack, loss of a leaving group, proton

transfer, or rearrangement. Then draw a transition state for each step within the mechanism, and finally

indicate which step is rate-determining and the molecularity (unimolecular, bimolecular, etc.) of that step.


1

Klein Chapter 6 Problems : Reactivity & Mechanisms – Answers

1. In each of the following situations, indicate whether the reactions are favoured or disfavoured in terms of

enthalpic and entropic factors in the Gibbs free energy equation (Table 6.1 might help here).

2. Draw a mechanistic interpretation (using curved arrow(s) to show bonds forming and breaking), and then an

approximate transition state (including partial electron-densities) for each of the following events.


2

3. Draw curved arrows to describe each of the steps in the following reaction mechanism, and then label those

steps as being one of the four types discussed in class, i.e. nucleophilic attack, loss of a leaving group, proton

transfer, or rearrangement. Then draw a transition state for each step within the mechanism, and finally

indicate which step is rate-determining and the molecularity (unimolecular, bimolecular, etc.) of that step.

OHCl

HCl

OH2

H Cl

H2O

HH

Cl

A. proton transfer

B. loss ofleaving group

C. rearrangement

D. nucleophilicattack

carbocation formation is rate-determiningand this step is unimolecular

OH

H Cl

Transition states:

+

-

A. B.

OH

+

+

C. D.

H

+

+

H

Cl

-

+


1

Klein Chapter 7 Problems : Substitution Reactions

1. Provide each of the following alkyl halides with acceptable names. You may use either the functional class or

substitutive nomenclature as appropriate. Be sure to assign the configurations of any chiral centers.

2. Draw detailed mechanisms for the following reactions that use “curved arrows” to show the breaking and

forming of bonds and identify the mechanism that is operating.

3. Provide the products expected to be formed under each of the following sets of conditions.


1

Klein Chapter 7 Problems : Substitution Reactions – Answers

1. Provide each of the following alkyl halides with acceptable names. You may use either the functional class or

substitutive nomenclature as appropriate. Be sure to assign the configurations of any chiral centers.



a. b.

c. d.

H3C Br H3C OH

H2O

+ HBr

INaCN

DMF

CN

CH3

OHHCl

CH3Cl

S

CH3Br

DMSO SCH3

CH3 H3C O

H

HOH2

BrBr

SN1

stepwise via

3o cation CN

SN2 - inversion of stereochemistry

during concerted substitution

CBr

H

H

H

SN2 - backside attack on simple

methyl electrophile

CH3

OH2

H3C H CH3

Br

SN1

stepwise with

rearrangement


2



1

Klein Chapter 8 Problems : Elimination Reactions

1. Provide each of the following alkene-containing molecules with acceptable names. Be careful to assign the

configurations of any chiral centers (R or S) and double bonds (E or Z).





1

Klein Chapter 8 Problems : Elimination Reactions – Answers

1. Provide each of the following alkene-containing molecules with acceptable names. Be careful to assign the

configurations of any chiral centers (R or S) and double bonds (E or Z).



a.

b.

H3C Br CH3

OCH3+ HOCH3

+ NaBr

H2SO4

heat

OH

+ H2O

H

E2 with small base favours the trisubstituted (Zaitsev) product

H

OH2

H

H2O

- H2O

E1 with acid favours the trisubstituted (Zaitsev) product


2

c.

d.

H3PO4

KOtBu

THF

Cl

+ HOtBu+ KCl

OH

heat+ H2O

H

OtBu

E2 with large base favours the less-substituted (Hofmann) product

H

OH2

H

H2O

E1 with migration to the tetrasubstituted (Zaitsev) product

2o cation

3o cation



1

Klein Chapter 9 Problems : Addition Reactions



2. Provide the major and minor products expected to be formed under each of the following sets of conditions.

3. Provide the reagents and conditions required to facilitate the following synthetic conversions.


1

Klein Chapter 9 Problems : Addition Reactions – Answers



a.

b.

c.

H3C H3C

dil. H2SO4

HBr

OH

racemic

Br

achiral

CH3CO2H

racemic

H

OH

2. NaOH,

H2O2

H B1.

H

H

O

H

H

H

O

H

H

O

H

H

Electrophilic addition of water with rearrangement

2o cation

prochiral 3o cation

H

H Br Br

Electrophilic addition of HBr(hydrohalogenation) via themore stable tertiary cation

H

B

H3C H

B

OOH

H3C H

B

O OH

H3C H

O B

H OH

Hydroboration-oxidation viaconcerted addition followedby concerted rearrangement

syn add'n

migration withretention


2


3. Provide the reagents and conditions required to facilitate the following synthetic conversions.


1

Klein Chapter 10 Problems : Alkynes




3. Provide complete names for the following alkyne-containing molecules, including any R/S or E/Z descriptors.


1

Klein Chapter 10 Problems : Alkynes – Answers



a.

b.

c.

excess NaNH2

1. NaNH2

Br Br

2. CH3CH2Br

THF

H

O

H2SO4, H2O

NH2

H

Br

H

concerted "E2"

mixture of E/Z alkenes

concerted "E2"

NH2

NH2

Br

deprotonation

SN2

acetylide anion

H

H

HH2O

O

H

H

H HO

H

H

HO

H

H

H

H

O

H

H

H

H

protonation

trap cation

deprotonation protonation

deprotonation

H2OH

H2O

Enol

Ketone

alkyne hydrolysisincluding enol-ketotautomerism


2


3. Provide complete names for the following alkyne-containing molecules, including any R/S or E/Z descriptors.


1

Klein Chapter 11 Problems : Radical Reactions


forming of bonds and any important resonance structures. Then identify the mechanism that is operating.

2. Provide the major products expected to be formed under each of the following sets of conditions.

3. Draw all of the possible resonance structures for the following radicals.


1

Klein Chapter 11 Problems : Radical Reactions – Answers


forming of bonds and any important resonance structures. Then identify the mechanism that is operating.

a.

Br2, h

H3C H3C BrH

Br Br

Br

Br2

CH3 CH3

or

Br BrBr

radical abstraction

termination propagation

b.

NBS, heat

H2C CH2BrH

N Br

O

O

Br

CH2

CH2 CH2 CH2

Br Br or

CH2Br

termination

propagation

initiation

initiation

delocalized benzylic radical


2

c.

d.

HBr, H2O2

Br

H2O2, heat

OH

HO OH

OHBr H

Br

BrH Br

radicaladdition

3o radical

H abstraction

anti-Markovnikoff product

HO OHO

H

OH OH

etc.

initiation

propagation

propagation


3

2. Provide the major products expected to be formed under each of the following sets of conditions.

3. Draw all of the possible resonance structures for the following radicals.

a.

b.

c.

d.

CH2 CH2 CH2

CH2CH2 CH2


1

Klein Chapter 12 Problems : Synthesis

1. Provide the major products formed from each step in the following reaction sequences. Be careful to include

any changes in stereochemistry that may occur.

2. Give a sequence of reagents that would affect the following synthetic conversions. Each problem requires at

least two synthetic steps.

3. In each of the following problems, draw a starting material that could be used as a suitable precursor in the

conversion to the given product.


1

Klein Chapter 12 Problems : Synthesis – Answers

1. Provide the major products formed from each step in the following reaction sequences. Be careful to include

any changes in stereochemistry that may occur.

a.

b.

c.

d.

1. NBS, heatCH3

2. NaOCH3, THF

3. HBr, H2O2

4. KOtBu, THF

1. Br2, CCl42. excess NaNH2

3. H3O+ (quench)

4. NaNH2

5. CH3CH2Br

CH3

OH

1. HBr

2. NaOCH3,

3. O3

4. Zn, H2O

1. Br2, h

2. KOCH2CH3

3. H-BR2

4. NaOH, H2O2

H

H3C Br

1.

CH3

2.

H3C H

3.

H3C H

4.Br

racemic

Br

Br1. 2. 3.

H

2. 3.

Br

1.

racemic

2.

E/Z mixture

3.

racemic

BR2H

4.

racemic

OHH

CH3

Br

1.

CH3

2.H3C

3.

O

OO

H3C4.

O

O

H

H

2. Give a sequence of reagents that would affect the following synthetic conversions. Each problem requires at

least two synthetic steps.


2

3. In each of the following problems, draw a starting material that could be used as a suitable precursor in the

conversion to the given product.

OC

he

m 1

Me

ch

an

ism

Fla

sh

ca

rd

s

Dr. P

ete

r No

rris, 2

01

3

Ch

em

ica

l ch

an

ge

invo

lve

s b

on

ds fo

rmin

g a

nd

bre

akin

g;

a m

ech

an

ism

de

scrib

es th

ose

ch

an

ge

s u

sin

g c

urv

ed

arro

ws

to d

escrib

e th

e e

lectro

ns in

vo

lve

d

Th

ere

are

two

ma

in ty

pe

s o

f cu

rve

d a

rrow

to d

escrib

e e

ithe

r

2-e

lectro

n o

r 1-e

lectro

n p

roce

sse

s

Ab

ove

, the

X-Y

bo

nd

is fo

rmin

g a

nd

the

Y-Z

bo

nd

is b

rea

kin

g

Th

ere

are

~1

00

me

ch

an

ism

s in

OC

he

m1

an

d O

Ch

em

2

Me

ch

an

ism

Ba

sic

s

�L

on

e p

air d

on

or +

Lo

ne

pa

ir acce

pto

r = A

cid

-Ba

se

rea

ctio

n

�R

ate

-de

term

inin

g s

tep

is b

imo

lec

ula

r(o

nly

on

e s

tep

invo

lve

d)

�A

cid

-Ba

se

rea

ctio

ns a

re g

en

era

lly v

ery

fast (p

roto

n, H

, is a

cce

ssib

le)

�A

cid

-Ba

se

rea

ctio

ns a

pp

ea

r as c

om

po

ne

nts

of o

the

r me

ch

an

ism

s

HI, H

Cl, H

NO

3 , H3 P

O4

pK

a-1

0 to

-5

Super s

trong a

cid

sH

3 O+

pK

a–

1.7

RC

O2 H

pK

a~

5acid

sP

hO

HpK

a~

10

get

H2 O

, RO

HpK

a~

16

weaker

RC

CH

(alk

yn

es)

pK

a~

26

RN

H2

pK

a~

36

Extre

mely

weak a

cid

RC

H3

pK

a~

60

Not a

cid

ic a

t all

Ch

.3 A

cid

-Ba

se

Re

ac

tion

s: L

on

e-P

air

Do

no

rs

& A

cc

ep

tors

�C

hira

l 2°

ca

rbo

n w

ith le

avin

g g

rou

p a

ttach

ed

; stro

ng

nu

cle

op

hile

�R

ate

-de

term

inin

g s

tep

is b

imo

lec

ula

r(n

o c

arb

oca

tion

form

ed

)

�P

roce

ed

s w

ith “b

acksid

e a

ttack” a

nd

“ste

reo

ch

em

ica

lin

ve

rsio

n”

�T

ran

sitio

n s

tate

is d

escrib

ed

as b

ein

g trig

on

alb

ipyra

mid

alsh

ap

e

Ch

.7 S

tere

oc

he

mic

al C

ha

ng

e in

the

SN2

Re

ac

tion

reaction coordinate

MeCH(B

r)Et

+ X:

MeCH(X

)Et

+ Br:

�1

°a

lco

ho

l (or C

H3 O

H) +

H-X

→ a

lkyl h

alid

e =

SN2

rea

ctio

n

�R

ate

-de

term

inin

g s

tep

is b

imo

lec

ula

r(n

o c

arb

oca

tion

form

ed

)

�R

ea

ctio

n s

low

ed

by s

teric

cro

wd

ing

(CH

3 >

1°

> 2

°>

3°)

�C

on

ce

rted

pro

ce

ss, n

o re

activ

e in

term

ed

iate

invo

lve

d

reaction coordinate

ROH

+HX

RX+

H2 O

Ch

.7 P

rim

ary A

lco

ho

ls (&

CH

3 OH

) with

H-C

l/H-B

r/H

-I –S

N2

�3

°m

ole

cu

le (w

ith le

avin

g g

rou

p) +

nu

cle

op

hile

= S

N1

rea

ctio

n

�R

ate

-de

term

inin

g s

tep

is u

nim

ole

cu

lar

= C

AR

BO

CA

TIO

N F

OR

ME

D

�C

arb

oca

tion

is p

lan

ar s

o is

atta

cke

d fro

m b

oth

sid

es to

giv

e 2

pro

du

cts

�C

hira

l sta

rting

ma

teria

l giv

es ra

ce

mic

mix

ture

of e

na

ntio

me

ricp

rod

ucts

reaction coordinate

ROH

+HX

ROH2

+X

R++ O

H2

+X

RX+

H2 O

Ch

.7 S

tere

oc

he

mic

al C

ha

ng

e in

the

SN1

Re

ac

tion

�3

°a

lco

ho

l + H

-X →

alk

yl h

alid

e =

SN1

rea

ctio

n

�R

ate

-de

term

inin

g s

tep

is u

nim

ole

cu

lar

= C

AR

BO

CA

TIO

N F

OR

ME

D

�C

arb

oca

tion

sta

biliz

ed

by h

yp

erc

on

jug

atio

n(3

°>

2°

> 1

°>

CH

3 )

�S

tep

wis

e p

roce

ss, re

activ

e in

term

ed

iate

(ca

rbo

ca

tion

) invo

lve

d

reaction coordinate

ROH

+HX

ROH2

+X

R++ O

H2

+X

RX+

H2 O

Ch

.7 R

ea

ctio

n o

f a T

ertia

ry A

lco

ho

l with

H-C

l/H-B

r/H

-I = S

N1

�3

°o

r 2°

alc

oh

ol +

H2 S

O4

or H

3 PO

4→

alk

en

e =

E1

rea

ctio

n

�R

ate

-de

term

inin

g s

tep

is u

nim

ole

cu

lar

= C

AR

BO

CA

TIO

N F

OR

ME

D

�C

arb

oca

tion

sta

biliz

ed

by h

yp

erc

on

jug

atio

n(3

°>

2°

> 1

°>

CH

3 )

�P

rod

uct d

istrib

utio

n is

ba

se

d o

n re

lativ

e a

lke

ne

sta

bility

(Za

itse

vru

le)

H3 CC

H3 CH

3 C

OH

H3 CC

H3 C

CH2

H3 CC

H3 CH

3 C

OH

H

C CH3

H3 C

CH2

(+ H

2 O)

HOH2

(- H2 O

)

H2 S

O4or H

3 PO

4

heat

OH

H3 O

+regenerated

HH

reaction coordinate

potential energy

RCH2 C

R2 O

H+

H3 O

+RCH2 C

R2 O

H2

+H2 O

RCH2 C

R2

+H2 O

RCH=CR

2

Ch

.8 R

ea

ctio

n o

f 3°/2

°A

lco

ho

l with

H2 S

O4 /H

3 PO

4–

E1

�3

°, 2°, o

r 1°

alk

yl h

alid

e +

ba

se

→ a

lke

ne

= E

2 re

actio

n

�R

ate

-de

term

inin

g s

tep

is b

imo

lec

ula

r=

no

inte

rme

dia

te fo

rme

d

�Z

aits

ev

ou

tco

me

ba

se

d o

n a

lke

ne

sta

bility

(su

bstitu

tion

pa

ttern

)

�U

se

ful, p

red

icta

ble

pro

ce

ss s

ince

no

inte

rme

dia

tes a

re fo

rme

d

H

X

BH

X

B+ B-H

+X

Ch

.8 R

ea

ctio

n o

f 3°/2

°/1

°A

lkyl H

alid

e w

ith B

as

e –

E2

�A

lke

ne

+ H

-X (X

= C

l, Br, I) g

ive

s a

lkyl h

alid

e a

dd

ition

pro

du

ct(s

)

�T

wo

-ste

p p

roce

ss: s

low

er s

tep

is c

arb

oca

tion

form

atio

n

�O

utc

om

e b

ase

d o

n c

arb

oca

tion

sta

bility

(Ma

rko

vn

iko

ffru

le)

�C

arb

oca

tion

form

ed

so

rea

rran

ge

me

nts

are

a p

ossib

ility

H

H3 C

CH3

HH

Br

H

H3 C

H

CH3

Br

+

H

H

H3 C

H

CH3

H

Br

reaction coordinate

RHC=CR2

+HX

RH2 C

-CR2

+X

RHC-C

HR

2

+X

RH2 C

-CXR2

RHXC-C

HR

2

Ch

.9 E

lec

tro

ph

ilic A

dd

ition

of H

-X to

Alk

en

es

�A

lke

ne

+ d

ilute

H2 S

O4

(H3 O

+) giv

es a

lco

ho

l ad

ditio

n p

rod

uct(s

)

�T

hre

e-s

tep

pro

ce

ss, th

e firs

t be

ing

slo

w fo

rma

tion

of c

arb

oca

tion

�M

ajo

r pro

du

ct fo

rme

d v

iam

ore

sta

biliz

ed

ca

rbo

ca

tion

(Ma

rko

vn

iko

ff)

�C

arb

oca

tion

(s) g

en

era

ted

so

rea

rran

ge

me

nts

will b

e p

ossib

le

H

H3 C

CH3

HH

OH2

H

H3 C

H

CH3

H2 O

+

H

H

H3 C

H

CH3

H

O

+

H

H

H

H3 C

H

CH3

H

HO

H2 O

reaction coordinate

RHC=CR2

+H3 O

+

RH2 C

-CR2

+H2 O

RHC-C

HR2

+H2 O

RH2 C

-C(O

H)R

2RH(H

O)C

-CHR

2

Ch

.9 A

cid

-ca

taly

ze

d H

yd

ra

tion

of A

lke

ne

s

�A

lke

ne

+ H

-BR

2g

ive

s a

dd

ition

ba

se

d o

n s

teric

sa

nd

ele

ctro

nic

s

�F

irst s

tep

is c

on

ce

rted

sy

na

dd

ition

; no

inte

rme

dia

te(s

) form

ed

�S

eco

nd

(oxid

atio

n) s

tep

reta

ins o

rigin

al C

-BR

2ste

reo

ch

em

istry

�O

ve

rall o

utc

om

e is

op

po

site

to th

at o

bta

ine

d u

sin

g d

ilute

H2 S

O4

Ch

.9 A

dd

ition

-Ox

ida

tion

Hyd

ra

tion

of A

lke

ne

s

�O

nly

an

tip

rod

uct in

dic

ate

s th

at th

is is

no

t a s

yn

ad

ditio

n p

ath

wa

y

�O

ne

pro

du

ct o

nly

su

gg

ests

a m

od

ified

ca

rbo

ca

tion

inte

rme

dia

te

�F

orm

atio

n o

f the

bro

mo

niu

mio

ne

xp

lain

s s

tere

och

em

ica

lo

utc

om

e

�S

imila

r rea

ctio

n w

ith B

r2 /H

2 O g

ive

s o

nly

an

tia

dd

ition

of “B

rOH

”

Atta

ck fro

m o

pposite

sid

e p

refe

rred

Ch

.9 A

dd

ition

of H

alo

ge

ns

to A

lke

ne

s

�R

em

ark

ab

le p

roce

ss th

at b

rea

ks

bo

thth

e πππ π

an

d σσσ σ

bo

nd

sin

the

alk

en

e

�F

irst s

tep

is c

on

ce

rted

sy

na

dd

ition

; no

inte

rme

dia

te(s

) form

ed

�S

ub

se

qu

en

t ste

ps in

vo

lve

bre

akin

g o

f we

ak O

-O b

on

ds a

nd

the

C-C

bo

nd

�M

alo

zo

nid

ep

rod

uct is

the

n re

du

ce

d to

giv

e th

e c

arb

on

yl p

rod

ucts

alkene

ketones and/or a

ldehydes

RR

HHC

CO

RR

HH

1. O

3in CH3 O

H

2. Z

n, H

2 OO

Ch

.9 O

zo

no

lys

is o

f Alk

en

es

�S

tep

wis

e a

cid

-a

nd

Le

wis

acid

-ca

taly

ze

d a

dd

ition

of w

ate

r to a

n a

lkyn

e

�F

orm

al p

rod

uct o

f the

ad

ditio

n is

the

en

ol, w

hic

h is

ofte

n n

ot is

ola

ted

�T

au

tom

eris

mm

ost o

ften

the

n le

ad

s to

the

mo

re s

tab

le k

eto

ne

pro

du

ct

�P

roce

ss o

ccu

rs th

rou

gh

the

mo

re s

tab

le c

arb

oca

tion

(Ma

rko

vn

iko

ff)

Ch

.10

Hyd

ro

lys

is o

f Alk

yn

es

�A

lka

ne

+ C

l2 /Br2

an

d h

ea

t/ligh

t = R

ad

ica

l ha

log

en

atio

n re

actio

n

�N

on

-po

lar m

ech

an

ism

with

ho

mo

lytic

bo

nd

-bre

akin

g a

nd

form

ing

�S

ele

ctiv

ityo

bse

rve

d w

ith a

bstra

ctio

n o

f H (3

°>

2°

> 1

°C

-H b

on

d)

�B

rom

ina

tion

mo

re s

ele

ctiv

e th

an

ch

lorin

atio

n (B

r rad

ica

l se

lectiv

e)

Ch

.11

Cl/B

r S

ub

stitu

tion

on

Alk

an

es

–F

re

e R

ad

ica

ls

�A

lke

ne

+ H

-Br a

nd

pe

roxid

e g

ive

s a

lkyl h

alid

e a

dd

ition

pro

du

ct(s

)

�R

ad

ica

l pro

ce

ss w

ith u

su

al s

tep

s (in

itiatio

n, p

rop

ag

atio

n, te

rmin

atio

n)

�O

utc

om

e is

ba

se

d o

n re

lativ

e ra

dic

al s

tab

ility (M

ark

ov

nik

off

pro

ce

ss

)

�N

o re

arra

ng

em

en

ts o

bse

rve

d w

ith fre

e ra

dic

al in

term

ed

iate

s h

ere

Ch

.11

Ra

dic

al A

dd

ition

of H

-Br to

Alk

en

es

Documents

OChem1 Course Pack - hyperconjugation.comhyperconjugation.com/3719files/OChem1CoursePack.pdf · OChem1 Course Pack ... (8 pts) Draw structures for all of the cycloalkane isomers with