Contents

Contents. 1

Schedule of Topics – Fall 2015. 3

Gen-Chem Review 1 – Bonding & Resonance. 5

Review of Lewis Dot Structures. 6

Resonance Structures (powerpoint). 10

Using Arrows in Resonance Structures. 11

Gen-Chem Review 2: Molecular Shapes and Polarity. 17

Gen-Chem Review: Intermolecular Forces. 18

Resonance Structures and Extended Molecular Orbitals. 20

Prepare for Unit 1 Test!. 29

Isomers,  Models, and Naming Exercise. 31

Conformations and Newman Projections. 34

Models: Conformations of Substituted Cyclohexanes. 36

Hydrocarbon Formulas and Hydrogen Deficiency. 39

Models: Introduction to Stereochemistry. 43

The R, S System for Naming Chiral Compounds. 46

Cis-Trans and E-Z naming of Alkenes. 48

More practice with Isomers & Chirality. 51

Prepare for Unit 2 test!. 55

Acid-Base Reactions (powerpoint). 56

Acid-Base Reactions, Arrows, and pKa Values. 59

Substitutions and Eliminations. 65

One-Step (bimolecular) Nucleophilic Substitution. 65

Coordination and Heterolysis (Karty 7.3). 70

One-Step Elimination Reactions. 72

Importance of Reaction Mechanism for SN2 Reactions: 78

Stereochemistry of SN1 Reactions: 84

Carbocation Rearrangement (Karty 7.7). 87

Factors Influencing Reaction Mechanisms. 91

Applications of Substitutions. 93

Reactions of Epoxides. 93

Stereochemistry and Regiochemistry of Bimolecular Eliminations. 94

Worksheet: E2 Eliminations. 95

Applications of Elimination. 99

Electrophilic Additions. 103

Carbenes, Oxidation, Ozonolysis, and Hydrogenation of Alkenes & Alkynes. 110

Catalytic Hydrogenation of Alkenes & Alkynes. 113

Addition of Halogens, Epoxidation: Triangular Additions. 115

Hydroboration-Oxidation. 119

Introduction to Infrared Spectroscopy. 123

Introduction to 1H NMR Spectroscopy. 127

Thinking about Molecule Identification – Spectroscopic Summary. 134

Radicals and Radical Reactions. 135

 


 

Schedule of Topics – Fall 2015

Date

Day

Topic

Karty Chap.

Week 1

1

Aug 24

Mon

Intro, syllabus,  expectations, Adapting from Gen-Chem

2

Aug 25

Tue

Gen-Chem Review 1: Atoms & Bonding Basics

1

3

Aug 26

Wed

Nomenclature Part 1: Simple Compounds

Nom-1

4

Aug 28

Fri

Gen-Chem Review 2: Geometry & Intermolecular Forces

2

Week 2

5

Aug 31

Mon

Gen-Chem Review 3: Molecular Orbitals & Bonding

3

6

Sep 01

Tue

Nomenclature Part 2: Unsaturated Hydrocarbons

Nom-2

7

Sep 02

Wed

Catch-Up & Review

8

Sep 04

Fri

Test #1, on Gen-Chem &  Nomenclature

1-3 & Noms

Week 3

Sep 07

Mon

Labor Day - Campus Closed

9

Sep 08

Tue

Isomers & Conformers

4.1-4.4

10

Sep 09

Wed

Conformations & Energies

11

Sep 11

Fri

Cyclohexane Conformations

4.5-4.9

Week 4

12

Sep 14

Mon

Constitutional Isomers & IHD

4.10-13

13

Sep 15

Tue

Finish  isomers,  begin stereoisomers

4, 5.1-5.2

14

Sep 16

Wed

Chirality and Models

5.2-5.4

15

Sep 18

Fri

The R/S and E/Z naming systems

Nom-3

Week 5

16

Sep 21

Mon

Diastereomers;  drawing Fischer & Newman diagrams

5.5-5.7

Sep 22

Tue

Student Success Day - No Class or Lab

17

Sep 23

Wed

Properties of Isomers

5.8-5.12

18

Sep 25

Fri

Catch-Up & Review

Week 6

19

Sep 28

Mon

Test #2, on Conformers & Isomers

4, 5

20

Sep 29

Tue

Acid-Base Reactions: Arrows & Mechanisms

6.1-6-3

21

Sep 30

Wed

Acidities of Different Molecules

6.4-6.6

22

Oct 02

Fri

Finish Acid-Base reactions

6

Week 7

23

Oct 05

Mon

Substitution Reactions: Some Elementary Steps

7.1-7.3

24

Oct 06

Tue

SN2 and SN1 - Introduction

8.1

25

Oct 07

Wed

One-Step Elimination Reactions - Introduction

7.4

26

Oct 09

Fri

Two-Step Elimination Reactions - Introduction

8.2

Week 8

27

Oct 12

Mon

Test #3, on Mechanism Basics

6, 7.1-7.4, 8.1-8.2

28

Oct 13

Tue

How Do We Know? Evidence For Reaction Mechanisms

8.3-8.4

29

Oct 14

Wed

Details of SN2 & E2: Stereochemistry & Regiochemistry

8.5, 9.10

Oct 16

Fri

No Classes - MEA weekend

Week 9

30

Oct 19

Mon

Carbocation Stability & Rearrangement

7.7, 7.8, 8.6

31

Oct 20

Tue

Finish Ch. 8

32

Oct 21

Wed

Comparing Substitutions & Eliminations

9.1-9.2

33

Oct 23

Fri

Factors Influencing Reaction Mechnisms

9.3-9.9

Week 10

34

Oct 26

Mon

Predicting Products

9.10-9.14

35

Oct 27

Tue

Finish Ch. 9

36

Oct 28

Wed

Catch-up & Review

37

Oct 30

Fri

Test #4, on Multi-Step Mechanisms

IC, 8, 9

Week 11

38

Nov 02

Mon

Return & go over T4, begin Synthesis

10.1

39

Nov 03

Tue

Applications of SN2 & SN1

10.1-10.6

40

Nov 04

Wed

Reactions of Epoxides

10.6-10.7

41

Nov 06

Fri

Applications of Elimination

10.8-10.10

Week 12

Nov 09

Mon

Electrophilic Additions & Carbocation Formation

11.1-11.3

42

Nov 10

Tue

Regiochemistry of Additions

11.3-11.8

43

Nov 11

Wed

Campus Closed: Veterans Day

44

Nov 13

Fri

Finish Electrophilic Additions, Reduction of Alkenes

11, 25.7

Week 13

45

Nov 16

Mon

Addition of Carbenes & Halogens

12.1-12.3

46

Nov 17

Tue

Oxymecuration & Epoxidation

12.4-12.5

47

Nov 18

Wed

Hydroboration-Oxidation

12.6-12.8

48

Nov 20

Fri

Catch-up & Review

Week 14

49

Nov 23

Mon

Test #5, on Synthesis & Alkene Reactions

10-12

50

Nov 24

Tue

Identification of Molecules: Infrared

15.4-15.7

51

Nov 25

Wed

Identification of Molecules: Intro to NMR

16.1-16.9

Nov 27

Fri

Thanksgiving Break

Week 15

52

Nov 30

Mon

Identification of Molecules: More NMR

16.4-16.12

53

Dec 01

Tue

Identification of Molecules: Combining IR & NMR

54

Dec 02

Wed

Radical Formation & Stability

25.1-25.4

55

Dec 04

Fri

Reactions of Radicals

25.4-25.6

Week 16

56

Dec 07

Mon

Radical Chain Reactions & Polymers

25

57

Dec 08

Tue

Catch-Up & Review

58

Dec 09

Wed

Test #6, on Spectroscopy & Radicals

15, 16, 25

59

Dec 11

Fri

Review for Final Exam

Finals Week

Dec 16

Wed

Final Exam- 7:45 - 9:45 AM - ACS Standardized Test

TBA

Chemistry Magic Show - Theater - evening

 

 

 

 

 

 

 


 

Gen-Chem Review 1 – Bonding & Resonance

 

 

 

 


Review of Lewis Dot Structures

Draw out the Lewis atom symbols for H, C, N, O, and F.

 

 

1.       How does the number of paired vs. un-paired electrons in the atom compare to the number of bonds each atom makes?


2.       Draw out the Lewis dot structures for these molecules:

a.       CF4

b.      H­2O

c.       NH3

3.       How do the Lewis structures and shapes of CH4, NH3, H2O, and HF compare to each other?

 

 

 

4.       Now draw CH3CN. (Note that one C is attached to 3 H and the other C is bonded to the N.) Does that also fit the pattern you saw in #3?

 

 

 

5.       Repeat #4 for acetic acid, CH3CO2H.

 

 

 

Does the typical bonding pattern hold for multiple bonds? How might you modify your statement in #3 to make it fit molecules with multiple bonds?

 

 

Now draw the two ions in ammonium hydroxide, NH4+ and OH-.

1.       Which atoms have a formal charge?

 

2.       Which atoms don’t follow the bonding pattern you saw in #3?

3.       What happens to the formal charge of an atom when it is bonded to a different number of atoms than is usual? Can you explain this in terms of electrons “owned” and “shared”?

 

 

 

Draw the Lewis dot structure of the acetate ion, CH3CO2-.

1.       Does each atom fit the rules you set out in the previous questions?

 

 

2.       Compare the acetate structure to the acetic acid structure you drew in #5.

 

3.       Focusing on the charged O atom in acetate, how is its bonding situation different from the O-H group in acetic acid? 

 

You should remember the concept of resonance structures from GenChem. Draw a pair of resonance structures, and the “average” resonance structure, for acetate ion.

 

 

1.       What is changed between the two resonance structures? What hasn’t changed?

 

 

2.       Now assign the VSEPR geometries for each carbon atom in CH3CN and CH3CO2-. (It might be helpful to recopy or re-draw the structures.)

 

 

3.       What do the two structures in #2 above have in common? Where are they different?

 

 

 

 

4.       Draw the Lewis structures, and assign the geometries, for CH3CH3, CH2CH2, and CHCH.

 

 

5.       Draw a Lewis structure for the complex molecule (CH3)2CCHCH2OCH3.

 

 

 

On a blank paper, list any of the ideas from this worksheet that you are still a little less than confident about. If there are any that you can’t figure out from reviewing the textbook, please ask about them tomorrow.

Future activities:

·         We will represent organic molecules using a highly simplified representation of the structure called a “skeleton structure”, including the three-dimensional geometry.

·         We will make a deeper exploration of resonance structures, which are a vitally important concept in O-Chem.

·         You will build models (starting this week!) of molecules to help you connect the 2-dimensional pencil-and-paper representation to the three-dimensional geometry.


 

Introduction to Skeleton Structures

The most convenient representation of an organic molecule is a very simplified representation called a skeleton structure. In it, most carbon and hydrogen atoms are omitted. All C-C bonds are represented by line segments, with a C atom assumed at each end unless another atom symbol is shown. Double bonds are represented by two parallel lines, triple bonds by three lines. The hydrogen atoms are assumed to be present in order to fill each C to its usual bonding pattern of four bonds, if no other atoms are shown.

The molecule C3H5Cl could be written out as a structural formula, CH2CHCH2Cl Notice that each C is followed immediately by what is attached to it: CH2 CH CH2Cl . The middle C is bonded to both the left and right, and it is given one more H in the formula. Since that is only 3 bonds, there must be a fourth which is a double bond. To decide whether the double bond goes to the left or to the right, note that the left C also has only 3 bonds detailed: two to H and the tirt to the nect C along in the chain. The right C (last C in the formula) has its chain bond and three others specified. It can’t be involved in a double bond because all four of its bonds are already accounted for. Thus, you can convert the structural formula into the Lewis structure at right.

It is time-consuming to write out all of the H atoms, so the more compact skeleton structure below is much quicker and (to the experienced eye) conveys all the same information.

1.       Confirm that adding the H atoms to each vertex of the skeleton structure to make a total of four bonds for each C gives the Lewis structure shown.

 

2.       Convert the skeleton structure of this molecule, a  flavor ingredient in Juicy Fruit gum, to a full Lewis dot structure.  Show that its formula is C7H12O2.

 

 



3.       Draw out the skeleton structures for these molecules:

a.       CH3CHClCH3

b.      CH3CH(OH)CH3

c.       CH­3COCH3

d.      CH3CH2CH2CH(OCH2CH3)CH2CH=CHCH2CH=CH2

 

 

Resonance Structures (powerpoint)


 

Using Arrows in Resonance Structures

Here are the resonance structures for nitromethane, a highly flammable compound used in “top fuel” drag racing (mixed with 10% methanol, CH3OH).

 

Review your skills from Gen-chem :

In resonance structures, __________ do not move. Resonance structures show changes in _______ bonds, not ________ bonds, and also in ______  _______.  Resonance structures (do/do not) change an atom’s hybridization or geometry.

Explore:

1.       What is the chemical formula of nitromethane? ___________________

2.       Identify any non-zero formal charges, and the overall charge, for Structure 1.

 

3.       Describe the geometry and hybridization around the N atom in Structure 1.

 

 

4.       Are the answers to questions 2 & 3 you gave about Structure 1 the same as you’d give for Structure 2? Explain.

 

 

5.       Now examine the bonds and lone pairs in each.

a.       The bond between N and the upper O in Structure 1 has  ______ s (sigma) and ____ π (pi) bonds.

b.      The bond between N and the lower O in Structure 1 has  ______ s (sigma) and ____ π (pi) bonds.

c.       The upper O in Structure 1 has _____ lone pair(s) and a _____ formal charge.

d.      The lower O in Structure 1 has _____ lone pair(s) and a _____ formal charge.

e.      The bond between N and the upper O in Structure 2 has  ______ s (sigma) and ____ π (pi) bonds.

f.        The bond between N and the lower O in Structure 2 has  ______ s (sigma) and ____ π (pi) bonds.

g.       The upper O in Structure 2 has _____ lone pair(s) and a _____ formal charge.

h.      The lower O in Structure 2 has _____ lone pair(s) and a _____ formal charge.

i.         How would you answer question 5 about the average structure?

 

Examine the arrows drawn near Structure 1:      

6.       Where does arrow a begin and end? What change does it make as the picture changes to Structure 2?

 

 

 

7.       Where does arrow b begin and end? What change does it make as the picture changes to Structure 2?

 

 

 

Summary Questions:

Curved arrows show movement of ____________________. Arrows can start or end at either a ______ bond or a _____ ______. If an arrow begins at an atom, that atom will __________ its formal charge, and if it ends at an atom then the formal charge will ___________.

Practice:

Draw in the arrows and lone pairs for the resonance structures of the propionate ion, (an anti-mold agent used in baked goods as the as calcium salt) below:

 

2,4-pentanedione is a relatively good acid, as organic molecules go. (Sec. 4.8). Draw two resonance structures, with curved arrows, showing how the negative charge can be moved to each of the O atoms.



The extremely reactive diazomethane molecule is used to insert a –CH2– unit into reactive bonds. (Sec. 10.5). Fill in the missing lone pair(s) to make the formal charges correct as shown, then draw a resonance structure, showing the arrows.

 

 

Helpful hint: It doesn’t really matter which side (top or bottom) of a bond you draw the arrows. The important thing is that they start and end at the right places.

Part 2: Finding and Fixing Arrow Mistakes.

Using the curved-arrow notation for organic chemistry is a vital part of communicating your knowledge. Sentence wrong arrows grammar with confused using like is a. The exercises in this section should help you spot and fix common types of arrow mistakes.

Tip: If the skeleton structures confuse you, draw out one of them as a full Lewis structure.

Wrong Example 2a: 

Look back at your answer to the Summary Question on the previous page. How is WE2a inconsistent with that?

 

 

Wrong Example 2b:

This example forgot something.

(1)    What was left out?

(2)    How is the formula of the molecule on the right side of WE2b different from the left?

 

 

Wrong Example 2c:    

This example is only a little bit wrong. What’s the matter? Can you draw the structure that would result from the arrow as drawn (Hint: it has formal charges in three places!)

 

Draw in a correct arrow to fix examples 2a-2c:

 

 

 

 

 

 

Wrong Example 2d:

Look carefully at the far left carbon in both sides of WE2d. How did it change?

 

 

 

 

Arrows are also used in describing reactions. The next examples will extend your skills as you figure out what’s wrong with the arrows in this reaction of hydroxide with ammonium ion:

Wrong Example 2e:

Wrong example 2f:

(1)    Based on your knowledge of the conventions for arrows, what’s wrong with example 2e?



(2)    What’s wrong with example 2f?

 

(3)    Redraw the reaction with (two) correct arrows:

 

 

Summary Questions:

An arrow ending on a sigma bond shows that ______________________________________.

The atom at either end of an arrow can change its __________________ but not its _______________ or ______________. Arrows always __________  an electron-rich location and ___________ an electron-poor location. In resonance structures, arrows can show movement of only ______ bonds or lone pairs, but in reactions they can show movement of _____________________.


 

Gen-Chem Review 2: Molecular Shapes and Polarity

    


 

Gen-Chem Review: Intermolecular Forces

  

          


 

 

Resonance Structures and Extended Molecular Orbitals

Summary Tips:

Resonance structures only move lone pairs and π bonds; not σ bonds or atoms.

Resonance structures are a way of “freezing out” electron distributions within an extended pi bond made of more than two p orbitals lined up parallel to each other.

Resonance structures do not involve sp3 hybridized C atoms.

You should always be able to draw arrows showing how electrons move in going from one resonance structure to the next.

Resonance structures may alter formal charges, but the overall charge cannot change.

Model 1: Resonance Structures

 

Review: Re-draw the two resonance isomers as full Lewis structures, showing all C and H atoms.




In an earlier worksheet, you learned about the electron-arrow convention for showing how electrons move in a resonance structure. Review that knowledge by drawing arrows on your answer to #1, showing how the two resonance isomers are related.

What is the meaning of the dashed line in the average representation?

 

 

 

 

Based on the Lewis structure of the resonance isomer with the double bond on the right side, what should be the geometry and hybridization of each carbon atom? Is that the same for the isomer with the double bond on the left?

 

 

 

How is a “resonance isomer” similar to the definition of isomers you already learned? How is it different? Is “isomer” a good term for this?

 

 

 

 

 

 

Model 2: The extended π system

Another representation of the ion in Model 1 shows the three carbon atoms’ pz orbitals overlapping with each other to form an extended π system. (By convention, a unique axis is usually assigned to be the z axis, but it doesn’t matter to the molecule how we describe it!)

 

Do the hybridizations you found in Model 1, Question 4 fit the representation in Model 2?

Usually, we define a bond as a shared pair of electrons in overlapping orbitals. Does this model fit that definition?

 

 

In the C3H5+ ion, how many electrons should be assigned to the extended π bond above?


Does the extended π system give information about the average charges of the atoms?

 

 

Now consider the anion shown here:

What is the formula of the anion? ________________

Draw out a full structure, including all H atoms and lone pairs.




Draw a resonance structure that shows the negative charge on the oxygen atom.

 

 



Draw the average resonance structure, using the symbol ∂- to show the partial negative charges.

 

 

 

Do you think the values of the ∂- charges will be equal? Which will be the most negative?

 

 

 

Model 3: The LCAO-MO Model

A more sophisticated view of the extended π system in Model 2 uses linear algebra to come up with a new rule: the number of molecular orbitals (MOs) in a molecule must be equal to the number of atomic orbitals (AOs) available. We can treat the σ bonding MOs  separately from the π system, so for now only the three pz orbitals, and the three π molecular orbitals they generate, need to be considered.

Various mathematical rules help generate the molecular orbitals shown in the diagram. The slightly larger size of the center orbital signifies that it has a higher contribution to the shape of the MO.

In the diagram, vertical dotted lines represent nodes in the MO. The superscript on πnb stands for “non-bonding”, an intermediate level that does not contribute to either bonding or anti-bonding interactions.

Unlike the extended π model, however, only two electrons are allowed in each molecular orbital.

How many electrons are in the π bonding system according to the previous models?


Which orbital(s) is/are occupied by electrons in this model? Which is/are unoccupied?

 

 

Does this model help give information about the charge distribution? Does it agree with the resonance picture in Model 1, and/or the extended π system picture in Model 2?

 

 

 

 Where could an incoming negatively charged atom such as Cl- bring its pair of electrons to make a new bond?

 

 

 

Using the Models Together

Draw a basic resonance picture for the related anion, C3H5-. Include electron-moving arrows.




To your figure above, add the average resonance picture for C3H5-, including dashed lines and fractional charges.



How many electrons should be included in the  π bonding system for C3H5-?  ___



How would you change your answers to Questions 7, 8, and 9 in Model 2, above, for the C3H5- anion?







Consider the MO diagram in Question 3.  Which MOs should be occupied for C3H5-?

 

 

Adding the electron distribution for all occupied MOs in the anion, how does that compare to the average resonance picture? To the extended π picture?

 

 

 

 

 

Review Questions:

Here is an average structure for a very resonance-stabilized cation.

What is the formula of this cation?

 

How many MOs would be generated in the LCAO-MO model?  _________

Draw resonance structures showing which carbon atoms have a charge. Then draw δ+ symbols on the average structure above.

 

 

 

As a general rule, having more possible (legitimate) resonance structures implies greater stability. This is explained by the extended π model as giving the electrons more “room” to move; or by MO theory as forming lower energy molecular orbitals.  Which of the two cations in the exercise do you think is lower energy?


 

More fun with resonance structures:

  

This aromatic ring is so common, and so stable,  it is often symbolized by drawing the π bonds as a circle instead of individually. Draw electron arrows in the left representation.

For these ions, draw at least one or two more resonance structures, including arrows. Then draw an average (dotted-line) structure showing how the charge is distributed over the molecules. Include arrows.

 

 

 

 

 

As a general rule, having more possible (legitimate) resonance structures implies greater stability. This is explained by quantum theory as giving the electrons more “room” to move; or by MO theory as forming lower energy molecular orbitals.

Use resonance structures to decide which of these structures should be stabler.

 

                             or             

 

A fun example of resonance stabilization is found in squaric acid (really!) This organic acid is about 1000 times more acidic than carboxylic acids like formic or acetic acid.

Draw the hydrogen squarate ion, then the squarate (2-) ion, and some resonance structures.

 


NOM-2 Functional Groups, Bonding, and Hybridization

Complete the table below, following the example of an ester in the first row. Generic alkyl groups (or sometimes H) are given the symbols R, R′, R″, etc. and atoms other than C or H are called “heteroatoms”.

Functional Group

General Formula

Example with (at least) 4 C atoms

Skeleton Structure of example

Example’s name

Ester

RCO2R

CH3CO2CH2CH3

Ethyl acetate

 

 

CH3CH2CH(OH)CH3

 

 

 

ROR

 

 

 

 

 

 

 

cyclohexanone

 

 

CH3(CH2)3CHO

 

 

Carboxylic acid

 

 

 

 

Amine (1°)

 

 

 

 

 

 

 

 

Diethyl acetamide

 

R-CN

 

 

 

 

 

 

 

 

 

 

(CH3CH2)2NH

 

 

Aromatic ring

 

 

 

 

 

 

(CH3)2CHCH2CHBrCH3

 

 

 

RRC=CRR

 

 

 

 

 

Answer these questions about heteroatoms and bonding in functional groups.

These two groups have triple bonds:  ______________ and _____________

Aldehydes, ketones, esters, and acids all have: _________________________

Groups containing N include: ________________, ___________, and ____________.

Two groups containing C-O single bonds are _____________ and _______________.

Four groups with only sp3 hybridized C (plus heteroatoms) include: ___________________________________________________________________.

A group containing both N and O heteroatoms is: _______________

 


 

 

 

Prepare for Unit 1 Test!

 

 


 


Isomers,  Models, and Naming Exercise

Form into teams of three or four. Each team should have a model set.

1.    Introduction to model sets.

a.       The atoms are color coded, with most of the atoms drilled with holes to represent the typical bonding pattern. The color convention is black = C, white = H, red = O, blue = N (some with four holes to make ammonium compounds), yellow = S or P, green = halogens (some kits also have orange for Br and purple for I), 5-hole brown = any trigonal bipyramid atom, 6-hole silver = any octahedral atom.

b.      There are three bond pieces in the kits. The long thin grey bonds are flexible and used for double, triple, or strained bonds. The thicker grey bonds are for single bonds, and the short white ones are for building semi-space-filling models.

c.       There may also be an oval bond-removal tool with a keyhole-shaped opening to help take out the white bonds.

2.    Build a model of methane (CH­4). Remove one H and insert a CH2- to get C2H6. Since all four H atoms are identical, it doesn’t matter which one you remove.

3.    Choose one of the H atoms, remove it, and substitute a Cl.

a.       Are all the H atoms on C2H6 identical?

b.      Is it possible to make more than one isomer of C2H5Cl?


4.    Remove the Cl and replace it with a CH3, forming C3H8. Are there two possible isomers of C3H8?

5.    Repeat step 3 to look at the possible isomer(s) of C3H7Cl. Name each.


6.    Repeat step 4 to look at the possible isomers of C4H10.


7.    Build all possible isomers of C4H10, and set them side by side.

a.       Draw the structure of each isomer, and name it.

 




b.      For each isomer, count how many different H atoms there are. (If you’re not sure, put on a Cl atom to compare).


c.       How many total isomers of C4H9Cl should be possible?


 

 

d.      Challenge: There are actually two subtly different isomers which could be written CH3CHClCH2CH3. Try to build them both and compare. How are they the same or different? (We’ll study this more in chapter 7.)


 

8.       More isomers

a.       Is it possible to make more than one isomer of C2H4Cl2?


b.      How many isomers are possible for C3H8O? Build each, name it, and write the structural formula. Why are there more isomers of C3H8O than C3H7Cl?


9.       More challenges: How many isomers can you build (and name) with these formulas?

a.       C5H12 (try to find 3)



b.      C5H11Cl Hint: start with each of the C5H12 you made, and



c.       C5H10 (use bendy-bonds for double bonds or strained rings)


d.      C5H12O




e.      C5H10O

 




10.   Build something large using C, H, O, and Cl. Each member of the team should try to name it. Compare names and see if you can come up with a correct name.

 

 

 

 

 

 

Questions:

1.       Do you think the number of isomers of CnH(2n+2) increases linearly with n? Why or why not?

 

 

2.       Based on the building method from steps 3, 4, 5, 6; it seems that the number of alkanes (compounds with only C & H, and all C being sp3 hybridized) with n carbon atoms is the same as the number of  chloroalkanes with (n-1) carbons. Is this always true? Why or why not?

 


 

Conformations and Newman Projections

Organic molecules in liquid (or gas) phase are constantly twisting and moving. However, some arrangements are more stable than others, which has implications for how they will react.

Build a model of butane and hold it so you’re looking along the bond between C2 and C3.

The representation in skeleton structure form (far right) is called a Newman projection. The front carbon is designated by a point, the back by a circle.

Twist the C2-C3 bond by 180 degrees and draw another Newman projection.

Now twist 60 degrees at a time and draw a series of Newman pictures until you get back to the original Z-shaped picture.

 

 

 

 

 

 

 

 

The first diagram in your series has the two methyl groups in the “eclipsed” conformation. When groups are not directly in line with each other, the conformation is called “staggered”. The diagram above (and the final one in the series) is called “anti”. The conformation in which the two methyl groups are separated by 60° is called “gauche”.

 

 

Convert this picture to a skeleton structure with a wedge-dash to indicate geometry.

 Build a model to help you!

 

 

 

 

 

 

Build this, then draw it as a Newman projection, looking along the arrow.

 


 

Models: Conformations of Substituted Cyclohexanes

(Covers: Karty sections 4.5-4.9)

Work in teams of 2 or 3 students with at least one model kit each.

Note: When using the short bonds, try using the beige keyhole-shaped tool to remove the bonds. It works better than fingernails or teeth!

Monosubstituted Cycloalkanes

1.       Build a model of cyclohexane, including hydrogen atoms, using the short white bonds which give a closer bond length. This closer length gives a space-filling model.

2.       Twist your cyclohexane into the normal chair conformation; so that three H atoms on alternating C point up (axial), three on the other C point down (axial), and the other six are pointing equatorial.

3.       Build a methyl group, again using the short bonds, and substitute it in an equatorial position. Note how it doesn’t interact much with neighboring H atoms.

4.       Move the methyl group to an axial position. How much does it interfere with the other axial H atoms?

 

The energy difference between an axial methyl and an equatorial methyl group is small, but it is enough that gas-phase methylcyclohexane is mostly in the conformation where the methyl can be equatorial.

 

Disubstituted Cycloalkanes

5.       Now build cis-1,3-dimethylcyclohexane and look at the methyl-methyl interference when both methyl groups are in the axial position.

6.       Twist it so that both methyl groups are in the equatorial position. How does the interference compare?

7.       Repeat steps 5 & 6 with cis-1,4-dimethylcyclohexane.

8.       Repeat steps 5 & 6 with trans-1-4-dimethylcyclohexane.

 

Questions:

a.       Which conformation do you think is lower energy for cis-1,4-dimethylcyclohexane?

b.      Which conformation is lower for the trans isomer?

c.       Can you guess whether the cis or trans isomer of 1,3-dimethylcyclohexane should be stabler? (Hint: Think about all possible conformations.)

d.      What do you think would be the geometry effect of a tert-butyl group on a cyclohexane? 

Newman Projections of Cyclohexane

Here is a Newman projection of chair cyclohexane. Note that it is a “double” because of the roughly parallel C-C bonds. The Newman projection easily shows that the chair conformation has all-anti arrangements of the H atoms, and the unavoidable gauche of the -CH2- units at front and back.

9.       Align your cyclohexane (you might want to rebuild it with stick bonds instead of short bonds) to look like the Newman projection.

10.   Twist the cyclohexane into a boat conformation and draw its Newman projection.

a.       Count the anti, gauche, and eclipsed interactions.

 

b.      Why do you think the chair is stabler than the boat?

 

 

11.   Build methylcyclohexane, twist it so the methyl is axial (chair), and draw it so that the methyl group is on one of the four C atoms that are part of the Newman projection. Count the gauche C-C interactions.

 

 

12.   Twist your methylcyclohexane so that the methyl is equatorial. Now how many gauche interactions are there?

 

 

13.   There are seven possible isomers of cis or trans dimethyl cyclohexane with the methyls in various positions. Can you draw all of them?  Which do you suppose are the most and least stable?

 

 

 

 

 

14.   Fill in methyl groups for a Newman projection of one of the most and of one of the least stable isomers, keeping the CH2 groups in front and back so that the methyls are directly on the Newman carbons.

 

 

 

15.   Count the anti and gauche C-C interactions in each.

 

 

 

 

16.   Speculate on the effect of larger alkyl groups such as ethyl, isopropyl, etc. on cyclohexane conformations.


 

Hydrocarbon Formulas and Hydrogen Deficiency

Formulas of Hydrocarbons

The simplest category of organic molecules is the hydrocarbons, those containing only C and H. Consider a series of molecules, CH4, C2H6, C3H8, C­4­H­10 with increasing length of carbon chains.

1.       Draw out their skeleton structures (except CH4, which has no skeleton structure!)

What pattern do you see in the structure series? What is the pattern in the formula?

 

 

2.       How many C and H atoms would you expect for the fifth in the series? The eighth? Draw their skeletons.

 

 

 

3.       Can you come up with a general rule predicting the number of hydrogen atoms for a hydrocarbon with n carbon atoms?

 

 

4.       Does this rule apply to branched hydrocarbons such as CH3CH2C(CH33 or  CH3CH2CH(CH3)CH2CH(CH2CH3)CH2CH3?

 

 

Unsaturation

The term unsaturation generally means the existence of double or triple bonds in a molecule. Each double bond replaces two H atoms on neighboring carbons with a second bond. –CH2–CH2– becomes –CH=CH–  and the second bond is called a π bond. Thus, for each π bond in an organic molecule the formula decreases by two H atoms. A ring structure could also be interpreted as taking H atoms off of non-adjacent carbon atoms and connecting them with a single bond. This has the same effect on the formula. Therefore, ring structures and double bonded structures could be isomers.

5.       How many C and H atoms are there in

 

6.       Draw a skeleton and a full structure for a 5-carbon ring and count the H atoms. Is it an isomer of the above molecule?

 

7.       What is the general rule for the formula of a hydrocarbon with a double bond or a ring?

 

8.       How many H atoms would there be in a 5-carbon molecule with two double bonds? With a triple bond? With both a ring and a double bond? Are these all isomers? Draw a skeleton structure for each.

 

 

 

 

Index of Hydrogen Deficiency

The term “Index of Hydrogen Deficiency” (IHD) means the number of H2 molecules that a particular organic structure has fewer than the maximum number. Compare the formulas for five-carbon molecules you found for questions 5, 8, and 11. The number of hydrogen atoms decreases by two (or by one H2) for every ring or double bond. 

You should have found earlier that a simple hydrocarbon fits the formula CnH(2n+2) whether it is linear or branched. A ring or a double bond will decrease that by two H, making CnH2n. This would have a IHD of one since it is one H2 less than fully saturated.

A triple bond would change –CH2–CH2– to –CºC– with four fewer H atoms (or two H2 molecules) for a IHD of 2 from the two π bonds.

Rule: For a hydrocarbon CnHx, the IHD is calculated as (2n+2-x)/2 since (2n+2) would be the fully saturated, non-ring hydrocarbon.

The degree of unsaturation is a quick way of predicting the formula by looking at the number of double bonds and rings present.

1.       Use IHD to predict the formula of each of the six-carbon hydrocarbons below and then verify by careful counting. 

2.       Complete the table:

Structure

# rings

# π bonds

IHD

Formula

a

 

 

 

 

b

 

 

 

 

c

 

 

 

 

d

 

 

 

 

e

 

 

 

 

f

 

 

 

 

g

 

 

 

 

 

3.       Find three isomer pairs: ____ and ____,        ____ and ____,       ____ and ____

 

 

4.       Draw another isomer with 6 C atoms and an IHD of two.

 

5.       Which structure has the highest IHD?

 

6.       What are the formulas of hydrocarbons with 7 C and an IHD of 0, 1, or 2?

 

7.       Draw two skeleton structures each for a 7-carbon molecule with IHD of 1 and 2.

 

 

 

Unsaturation and Heteroatoms

A heteroatom is any atom other than C or H. Different heteroatoms in the formula have different effects on the IHD.

Halogens. Since halogens (F, Cl, Br, I) almost always form one bond to C, they should be considered as replacing an H. Thus, add the number of halogens to the number of hydrogens when finding IHD.

Example: Find the IHD of C3H5Cl, pictured on page 1.

Oxygen can form either single bonds to C and H, or a C=O double bond. If it forms a C-O-H or C-O-C structure, it does not affect the formula. If it forms a C=O, the C is “missing” two H atoms and the double bond adds one to the IHD. Thus, you can ignore O atoms in determining the IHD. They either have no effect or the C=O π bond shows up because of the “missing” H atoms that might otherwise be present.

Example: find the IHD of isopentenyl acetate, page 1, and verify that this agrees with the given formula.

 

What is the IHD for C4H7BrO ?

 

Draw at least two isomers of C4H7BrO as skeleton structures.


 

Models: Introduction to Stereochemistry

Materials: Plastic molecular model sets with small mirrors, tennis ball, gloves

Part 1: Recognizing Chirality

1. Build a model of 1-bromo-1-chloroethane (using yellow for Br) with your model set. Include all H atoms.

a.       Draw a careful wedge-dash picture of it.

 

 

b.      Draw a Newman projection, looking at it with the methyl group toward you.

 

 

 

2.       Compare your model with your neighbor's. Are they exactly the same? If not, how are they different?

 

 

 

3.       Look at your model with a mirror. Is it identical to its mirror image? Is the mirror image identical to your neighbor’s model?

 

 

 

 

4.       Now, one of you should remove the Br and Cl and switch their places. Now compare your models. Are they mirror images? Are they identical? Are they different?

 

 

5.       Make your models so that they are mirror images. Draw a wedge-dash picture and a Newman picture for both isomers. Use a vertical dotted line to represent a mirror.

 

 

 

Information: In real molecules, this mirror-image property is called chirality. (adjective form: chiral) It is a form of stereoisomerism. The most common feature of chiral molecules is that one C atom has four different things bonded to it.

 

6.       Does the four-groups rule apply in this case? Which carbon is it, and what are the four groups?

 

 

 

Part 2: Representing Chirality

Here are four representations of 2-chlorobutane. Note that this is a “road-kill” view: it is flat and shows no indication of which stereoisomer it is. To represent the stereoisomer, we use the wedge-dash convention only on the carbon where it matters.

 

At right, above, is one of the two possible isomers. Build a model look to like this, including H atoms.  

7.       Draw in the H atom on carbon #2 using the wedge-dash convention.

8.       Draw a Newman projection looking along the C-2 to C-3 bond. (C-2 in front)

 

 

 

9.       Pair up two models. On one, swap the Cl and the H on C-2. Draw the Newman projection from the same viewpoint as before. Are the Newman projections also mirror images?

 

Important Rule: You can interconvert enantiomers by swapping any pair of atoms on the chiral C. When drawing enantiomers, switching a  to a  also changes the other atom attached to the chiral center C.

 

Verify this rule by, on the second model, switch the H and the CH3. This should result in both models being identical again.

 

The molecule pictured above is assigned the designation “R” for right-handed. The opposite configuration (which your models should now show) is called “S” from the Latin for left-handed. The next unit shows how that naming system works.

 

Significance: Molecules with different chirality have some properties that differ from each other, and some that are the same. In general, chiral isomers will behave identically when interacting with a non-chiral system, but differently when interacting with a chiral system.

Consider a left or right hand – they are identical mirror images, and both can hold a ball equally well, but only one can interact properly with a glove. Since almost all biological molecules have chirality, it can be very important for pharmaceutical molecules to have an appropriate “handedness” if they too are chiral.

 

 


 

The R, S System for Naming Chiral Compounds

 

 

 

 

 

 

 

When assigning stereochemistry, the first task is to identify any stereocenters (carbons with four different groups) and then rank each group according to a set of rules called the Cahn-Ingold-Prelog system, or CIP for short. The rules are fairly simple:

1.       Each atom is ranked according to its atomic number.

2.       Isotopes of atoms are ranked by mass

3.       If two atoms have the same atomic number, the atoms attached to them are ranked, and so on until a point of difference is found. 

4.       Double bonds count as two bonds to the same atom, with one bond on the further atom counting as back to the nearer one. For example, a –CH=O group has a C bonded to an H and twice to O, and the O is bonded back to the C.

5.       Tiebreakers: In assigning priority, move away from the stereo-center one atom at a time until there is a difference in the structures.

Practice with Cahn-Ingold-Prelog Rules

1.       Rank these groups:

¾     -H

¾     –OH

¾     -CH3

¾     –CH(CH3)2

¾     -Cl

¾     -NH2

¾     –CHCH2CH2CH3 (n-butyl)

¾     -CH(CH3)CH2CH3 (sec-butyl)

¾     –CH2CH(CH3)2 (isobutyl)

¾     -C(CH3)3 (tert-butyl)

 

2.       Draw out the vinyl group, R-HC=CH2, as it would be counted by the C-I-P rules.

3.       Where would a vinyl group rank in the list from above?

4.       Explain the naming of this compound as best you can.  

 

 

 

5.       Can you draw and name the other three possible stereoisomers of 1-bromo-3-chloro-2-methyl penta-1,3-diene?

 


Cis-Trans and E-Z naming of Alkenes

Here are two stereo-isomers of 2-butene:

They cannot be interconverted without breaking the π bond, which has a high energy barrier.  The isomers are designated the same way as disubstituted cycloalkanes: if the branches are on the same side, the alkene is is called cis and if they are on opposite sides it is called trans.

1.       Which is which above?

 

2.       Can 1-butene have cis/trans isomers? Why or why not?

3.       Draw trans-3-methyl 2-pentene.


4.       Add another carbon to the methyl group in #3 to make 3-ethyl-2-pentene and comment on whether it is cis or trans.

 

5.       Move the methyl group you added in #4 so as to make 2,3-dimethyl-2-pentene and decide if that is cis or trans.

 

 

6.       Can you generalize about what allows an alkene to have a cis or  trans isomer?


Careful measurement of reaction energies shows that trans alkenes are slightly stabler than cis. This effect is attributed to the eclipsed cis-alkyl groups on each end of the bond interfering with each other, called steric interference. The strain is about 1 kcal/mole (4.2 kJ/mole) for methyl groups and larger for bulkier groups.

1.       Build models of the cis- and trans- isomers of 2-butene. Fill in H and CH­3 groups on the Newman projections of each:

 

2.       Compare your model with that of someone who made the other isomer. Which do you think is more stable?

 

 

3.       Then draw cis- or trans- 2,2,5,5-tetramethyl-3-hexene (you might want to draw them as skeletons first!).

 

 

 

The cis-trans naming convention follows the path of the longest chain of carbon atoms. However, it is possible for the alkene to have stereoisomers that are ambiguous as to which should be called cis or trans. A slightly different system, the E-Z system, assigns a priority to the groups based on the atomic number of substituents. In this system, the pairs of substiutuents at each end are assigned higher or lower priority. If the higher-priority groups are on the same side (cis to each other) the alkene is labeled Z and if they are on opposite sides it is labeled E.

An example of where EZ gives different results than cis-trans happens if there are substituents with high atomic number off the main chain.

·         Draw both isomers of 3-chloro-pent-2-ene. Build one, and compare with a group which built the opposite. Assign cis-trans and E-Z to both. Explain why the E-Z notation differs from the cis-trans notation compared to 3-methyl-pent-2-ene. Does the same apply to 4-chloro-pent-2-ene?

 

 

·         Draw both (roadkill view) isomers of 4-chloro-3-methyl-pent-2-ene and assign E-Z and cis-trans.

 

 

 

 

·         4-chloro-3-methyl-pent-2-ene has a chiral center. Draw all four stereoisomers (wedge-dash) and assign the R and S.

 

 


 

More practice with Isomers & Chirality

1.       Here is 3-methyl-cyclohexanone as a road-kill view (no wedge-dash information).

a.       Can this have a cis-trans isomer?

b.      Mark the chiral carbon(s) in the structure with a star ☆.

c.       Can you tell whether the chirality is R or S?

 

2.       Here is a ball-and-stick picture of 3-methyl-cyclohexanone (the O is at far left.) You might want to build this with a model set now.

a.       Re-draw it as a skeleton structure, in the same perspective as the chair conformation shown.

 

 

 

b.      Re-draw it in the orientation shown in #1 above, but this time with wedge-dash stereochemistry as needed.

c.       Assign the Cahn-Ingold-Prelog (CIP) priority to the groups around the chiral carbon.

d.      Is this structure the R or the S isomer?

 

3.       Now, draw 3-methyl-cyclohexanone with the methyl group on the left (so the numbering goes counterclockwise instead of clockwise.) Include a wedge  or dash in the same direction, up above the paper or back below it, you drew for 2(b).

a.       Was that a mirror flip, or a rotation?

 

b.      Did moving the methyl from right to left change the (R,S) designation? Verify by assigning the configuration.


 

4.       Draw 1,3-dimethylcyclohexane, as a top-view hexagon,  in all four possible arrangements: both up, one up and one down, both down, or one down and one up.

Up, up

Up, down

Down, down

Down, up

a.       Identify all stereocenters with a star ☆.

b.      Assign each chiral center as R or S.  

c.       Which, if any, of these are chiral? Which are meso (internal mirror canceling chiral centers)? If there is an internal mirror, draw it in using a dashed line  - - - - -

 

 

 

d.      What to the meso ones have in common? What do the chiral ones have in common?

 

 

e.      What about 1,4-dimethylcyclohexane?


 

 

5.       You can use the pair-swap rule to help assign chirality from a Newman projection.

Here’s a chiral molecule.
                                                           

a.       What’s the molecular formula?

b.      What’s the name (leaving out R,S for now)?

c.        Is the chiral carbon in front or in back?


d.      On the center diagram, redraw as a wedge-dash picture with the groups in the same positions as the Newman projection.

e.      Now redraw but swap the front (chiral) H with the back group, so the H is in the back.

f.        Assign CIP priority numbers and do the “steering wheel” to decide if the far right picture is R or S.

g.       Was the original (center) structure R or S?

6.       Here’s a more complicated Newman projection:

a.       Which carbon(s) is/are chiral? (Think about having four different groups).

b.      Fill in the groups on  the “sawhorse” version of this molecule:


            

Make sure the same groups are anti to each other.


c.       Redraw as a more conventional wedge-dash picture with the main C chain (and the OH) in the plane of the paper.





d.      Assign any stereocenter(s) as R or S.

e.      Try to name the molecule. Hint: Multiple stereocenters are named with numbers such as (2R,4R).

 

 

 

 

 

 

The last page of the second exam was full of compounds that you now recognize to be chiral. This page is reproduced on the other side.  Build models of each compound, and try it identify all chiral centers. (You can also manipulate the models to work out the Newman or wedge-dash projections as asked on the test. Build that skill!)


 

 

1.       (13 points) Here is a skeleton structure of a molecule. .

a.       (10 points, 5 each) Draw two Newman projections, showing the structure as it would appear from the points of view at (a) and at (b). Be sure to label all substituents including H. 

 

 

b.      (3 pts.) What is the name of the molecule?

2.       (8 points total) Here is a Newman projection of a molecule.

a.       (4 pts.) Re-draw  this as a wedge-dash skeleton structure, with the longest C chain in the plane of the paper.

 

 

b.      (3 pts.) Name it correctly.

 

c.       (1 pt.) Circle the correct word.  The Br and the O atoms are:

eclipsed      anti      opposed      gauche    trans   cis      staggered     (none of those words)

3.       (6 points total ) Here is a chair-conformation substituted  cyclohexane. 

a.       (3 pts.) Re-draw it as a top view, so the ring is a hexagon, with correct wedge-dash notation.

 

b.      (3 pts.) Name it correctly.

 

 

Prepare for Unit 2 test!

 

Acid-Base Reactions (powerpoint)

   

 

 

 

 

 

 


 

Acid-Base Reactions, Arrows, and pKa Values

Related sections in Karty: 6.1-6.3

Acid-base reactions are a key part of organic chemistry and a frequent step as part of more complicated reactions. By definition, an acid base reaction involves moving an H+ from one substance (the acid) to another (the base).

1.    Here is a drawing of ethoxide (the conjugate base of ethanol) reacting with acetic acid:

a.       On the product side circle the new bond(s) formed in this reaction.

b.      On the reactant side, draw a slash through the bond(s) that break. 

c.       What happened to the third lone pair on the ethoxide O?


d.      What happened to the electron pair that connected the H to the O in acetic acid?


e.      Draw electron-pushing arrows showing what happens in the reaction.

 

 

2.    The pKa values for both neutral molecules are given.

a.       Which is more acidic?

b.      Which way does the equilibrium go?

c.       Which is the strongest base drawn above?

d.      Explain, using resonance structures, the relative stability of the conjugate bases.

 

 

 

 

 

You may remember from Gen Chem that reaction rates are a function of activation energy and orientation of molecules.  Why do you think acid-base reactions have very easy orientation requirements?

 

The range of pKa values seen in organic chemistry is much broader than that seen in general chem, because we are not restricted to water as a solvent. In water, the strongest possible acid is H3O+, because anything stronger will react with H2O. (Thus, aqueous HCl has very few hydrogen chloride molecules.) And the strongest possible base is OH- for similar reasons.

Reflect: If you used acetic acid as a reaction solvent, what are the strongest acid and base possible? Draw structures.



In organic chemistry, use of other solvents allows much stronger acids or bases to be used. For example, sodium amide (Na+ NH2-) is sometimes used as a moderately strong base. Hint: don’t confuse with RC(O)NR2, also spelled  amide. The amide in NaNH2 rhymes with chloride in NaCl, while amide RC(O)NR2 rhymes with “Sam hid”.

1.       Draw the Lewis dot structure of the amide ion, NH2-.

 

2.       What is the conjugate acid of amide?  (Its pKa is about 38).

3.       Do you think sodium amide would be stable in water solution? If not, why?

Draw a reaction, similar to the one above, showing sodium amide reacting with acetone. The pKa of acetone is about 20. Draw correct electron-pushing arrows. (Where is the H on acetone? Draw in the one that is removed, including its bonding electron pair.)

4.       Draw a resonance structure for the anion product you just made. 

5.       Which of the two resonance structures do you think is more significant, and why? 

 

 

 

Key ideas: The relative acidity of organic compounds can be explained by several effects, but each has to do with comparing the relative stabilities of the acid and of the conjugate base. 

1.                   H bonded to electronegative atoms. All other things being equal, hydrogens bonded to more electronegative atoms are more acidic.

2.                   Resonance Structures. If a negative charge can be spread over several atoms by resonance through π bonds, the resulting anion is more stable than one without resonance structures.

3.                   Inductive effect. If very electronegative atoms (or functional groups) are present in a molecule, the “pull” of electron density through σ bonds will stabilize the anion.

4.                   If internal hydrogen bonding is possible, this can stabilize either the acid (raising pKaor its conjugate base (lowering pKa)

5.                   If interactions with solvent can stabilize an anionic conjugate base, the related acid might be more acidic in that solvent than in another.

6.                   If like (negative) charges are near each other this can destabilize the base.

 

Practice Problems:

Apply Key Idea #1 to rank the relative acidity of the three types of hydrogen atoms in ethanolamine, HOCH2CH2NH2. (Can you predict any difference between the two different C-H groups’ acidity?)

 

 

Apply Key Idea #2 above to explain why nitroalkanes (RCH2NO2, pKa = 10)  are even more acidic than most alcohols (R-OH, pKa = 16). Use a picture in your explanation.

 

 

Apply Key Idea #3 above to predict the relative acidity for the series of chloroacetic acids with zero to three chlorine atoms, CHxCl(3-x)CO2H.

 

 

Resonance and Substitution Positions:

1.       Apply Key Idea #2 to predict which H+ is more acidic in each isomer, then #4 to compare the acidity of these two molecules: (Hint: Draw conjugate bases!)

 

 

 

 

 

 

2.       The pKa of phenol is 9.98, of p-aminophenol is 10.3 (5 times less acidic than phenol), and of p-nitrophenol is 7.1 (800 times more acidic). Use resonance structures to explain why these effects happen.

3.       Which Key Idea explains the fact that p–nitrophenol is about 20 times more acidic than m-nitrophenol?

 

 

 

 

 

 

The structure of picric acid (pKa = 0.19) is shown here. Why do you think its pKA is 10 units lower than phenol? Draw a dotted-line average resonance structure of the picrate anion in the box :

 

 

 

Base Strength:

Tables such as in the textbook usually list pKa of an acid. For bases, remember that conjugate base strength is inversely proportional to conjugate acid strength. Students often make the mistake of pairing “strong acid means weak conjugate base” and not remembering that it’s a very broad scale. So the conjugate acid of a really strong base like butyllithium, C4H9Li, is butane; not an acid at all.

Like relative acidity, relative base strength can be explained by comparing the stability of a base and its conjugate acid. Disruption of resonance as a lone pair is forced to become a bonding pair (rehybridizing out of a p orbital into a σ bond) is very important to explaining acid-base properties.

Exercise: Explain why aniline (C6H5NH2) is a weaker base than butylamine, C4H9NH2.

 

 

 

Challenge: Draw out the skeleton structures for these dicarboxylic acids:

Name

Structural formula

pKa(1)

H2A HA-

pKa(2)

HA-   A2-

Maleic acid

HO2CCH=CHCO2H
(carbon chain is cis or C-shaped)

1.9

6.3

Fumaric acid

HO2CCH=CHCO2H
(carbon chain is trans or zig-zag shaped)

3.1

4.6

Malonic acid

HO2CCH2CO­2H

2.8

5.7

Succinic acid

HO2CCH2CH2CO2H

4.2

5.6

Oxalic acid

HO2CCO2H

1.3

4.3

 

1.       How can you explain the fact that there is such a big gap from first to second pKa in maleic acid?

 

 

2.       Why is pK1 for maleic acid so much lower, and pK2 higher, compared to fumaric?

 

 

 

3.       Comparing fumaric to succinic acid, what is the effect of the π bond?

 

4.       Comparing malonic to succinic acid, can you explain why pKa (1) is different?

 

 

5.       Can you make any other observations about these acids?

 

 

 


 

Substitutions and Eliminations

One-Step (bimolecular) Nucleophilic Substitution

MODEL 1:  Here are some substitution reactions.

a.       N≡C- + CH3CH2CH2Cl → CH3CH2CH2C≡N + Cl-

b.     

c.       Br- + CH3-OH2+ → Br-CH3 + H2O

Explore:

1.       For each reaction, what gets substituted?

a.       Cl- gets substituted by NC-  

b.       

c.         

 

2.       Compare the reactions above to an acid-base reaction. 

a.       Complete the acid-base reaction that happens between cyanide and hydrogen chloride:

N≡C:-            +          H―Cl  ®

 

b.      Label each species in (2a) as acid, base, conjugate acid, and conjugate base.

 

c.       Compare the role of CN- in the model reaction (a) with its role in the acid-base reaction.

 

 

d.      Can an acid-base reaction be described as a substitution? Why or why not?

 

 

e.      What are some ways a substitution might be different from an acid-base reaction?

 

 

 

Vocabulary: In the example reactions, one reactant is a Lewis base. Reminder: A Bronsted base is a species that bonds to H+ ions, a Lewis base is a species that can bond to any positive (or partial positive) charge. In organic chemistry, the Lewis base is called a nucleophile (from “nucleus lover”, because it is attracted to positive charges).

3.       Which reactant is the nucleophile in each reaction? Draw their Lewis dot structures.

 

 

 

(a)                                      (b)                                (c)  

 

Note that each reaction also makes a non-carbon containing product. This product is called the leaving group, or L. Draw a  box  around the leaving group for each reaction in Model 1.

 

Explore:

4.       True or false: Nucleophiles are always negatively charged. ___________

 

5.       True or false: Leaving groups are always negatively charged. ___________

 

6.       What feature do all the nucleophiles have in common? (Hint: look at Lewis structures!)

 

 

7.       What pair of electrons forms the new bond Nu-C?

 

 

8.       What happens to the electron pair in the C-L bond?

 

 

Flashback:

9.       Assign the R, S stereochemistry to the reactant and product in reaction b.

 

Vocabulary: The opposite of a nucleophile is “electrophile”, meaning an “electron-lover”, a species that wants to bond to negative (or electron-rich) things. Carbon atoms attached to electronegative atoms can act as electrophiles because the polar bonds give a ∂+ charge to the carbon.

10.   Is an electrophile a Lewis acid, Lewis base, or neither? Compare your answer with others in your group.

 

 

 

11.   Draw the skeleton structure of 1-chloroethane and show the partial charges of the polar bond. Which carbon is an electrophile?

 

 

 

 

12.   Draw a CH3CH2NH2 (ethylamine) molecule next to the 1-chloroethane. These two substances react easily. Try to predict what might form from the reaction, assuming a 1:1 ratio.

a.       What could be the leaving group L in the reaction? ______________

b.      What is the nucleophile in the reaction? ________________

 

c.       Draw the structure of the organic product.

 

 

d.      Does the organic product have a positive, negative, or neutral charge? (Hint: Think about acid-base equilibrium!)

 

 

e.       What is the name of the organic product? ___________________________

 

Model 2: Electron-Arrow Mechanisms

Recall that we have used electron arrows to describe resonance structures:

Here is the electron-arrow mechanism for the acid-base reaction of acetylide anion with ethanol:

1.       Identify the electron pair from the reactants that becomes the new C-H bond in the products. Draw a circle around it.

2.       Where does the third lone pair (on O) in the products come from?


3.       What do the curved arrows represent…

a.       In the resonance structure?

 

 

b.      In the reaction?

 

 

c.       How do the curved arrows in the reaction diagram help show which bond(s) are breaking and/or forming?

 

 

4.       Fill in curved arrows to show the substitution reaction of acetylide with bromoethane.

 

Self-Check Reflections on #4: Asking yourself, and answering, these questions will help you make sure you truly understand the goals of this worksheet. You do not need to write down your answers.

Can you identify the nucleophile, electrophile, and leaving group in the reaction above? What happened to the carbon atom’s lone pair from the acetylide reactant? Is there an arrow to show that? Where did the fourth lone pair on bromide come from? What new bond formed? What bond broke? How many arrows did you draw? In what way(s) is/are your drawing similar to the acid-base drawing in Model 2?

 

 

Review:

5.       Try rewriting the three reactions from Model 1 using the electron-arrow mechanism. You might want to ask yourself the self-check questions above, too.

a.       N≡C- + CH3CH2CH2Cl → CH3CH2CH2C≡N + Cl-

 

b.     



c.       Br- + CH3-OH2+ → Br-CH3 + H2O

 

 


 

Coordination and Heterolysis (Karty 7.3)

It is possible for the two electron-pair movements in an SN2 reaction to happen sequentially instead of simultaneously. In a coordination step, an electron-rich species (Lewis base) uses its pair of electrons to form a bond to an electron-deficient species (Lewis acid). And in a heterolysis step, a bond breaks with the electron pair going to one atom while the other is left electron-deficient. The hetero in heterolysis means the bond breaks unevenly, whereas homolysis refers to a bond breaking equally, with one electron going to each of the atoms. The homolytic bond energy is usually used for energy calculations.

Explore: Why wouldn’t the bond energies be the same for homolysis vs. heterolysis? (think about NaF, for example) 

 

You may have seen coordination compounds of transition metals in General Chemistry. A common lab experiment involves measuring the equilibrium constant for Fe3+ +  SCN1- FeSCN2+.   

Explore: Fill in the blanks with “coordination” or “heterolysis”

In the reaction Fe3+ +  SCN1- FeSCN2+, the → is ____________ and the ← is  _____________

An example of an organic heterolysis reaction is (CH3)3C-Br → (CH3)3C+ +  Br-.

1.       Draw that reaction, using a curved arrow to show the electron movement.

 

 

The reverse reaction, coordination, can be illustrated by (CH3)3C+ +  CH3CO2- → (CH3)3C-OCOCH3.

2.       Draw that reaction, using a curved arrow to show the electron movement.

 

 

 

3.       What is the type of functional group in the product?

 

 

4.       When the two steps are combined, what is the overall net reaction?

 

Unimolecular Nucleophilic Substitutions – A Two-Step Process

Background:

Some nucleophilic substitution reactions have a rate law that is second order overall, because they depend on the concentrations of both nucleophile and substrate (the carbon electrophile). Those are the category called SN2 (bimolecular or 2nd-order.) Others show kinetics that are first order in substrate and zero order in nucleophile – the concentration of nucleophile does not influence the reaction rate! Those reactions are called SN1, for Substitution, Nucleophilic, 1st order (or unimolecular).

The proposed mechanism for an SN1 reaction is:

1.       R-L R+  + L-               (slow, RLS)

2.       R+ + Nu- → R-Nu       (fast)

The positively charged intermediate is called a carbocation. R = alkyl or other C group.  The reaction coordinate diagram shows the R-L bond breaking step as the first transition state, the R+ intermediate as the valley, and the R-Nu bond forming step as the second transition state.

Since the rate of an SN1 reaction depends on breaking the R-L bond, the factors influencing reaction rate are: stability of R+, stability of the leaving group L-, and ability of the solvent to support ion separation.

Questions

1.       Draw a Lewis structure of (a) (CH3)3C-Br, (b) the (CH3)3C+ intermediate, and (c) the product of its SN1 reaction with H2O. (The H2O gives an H+ to another H2O ­molecule.)

 

 

 

 

 

2.       Describe the VSEPR geometry and hybridization of the (CH3)3C + intermediate.

 

 

3.       Use a model set to build a model of the intermediate. The center C atom can be represented by a 5-hole ball (tan or brown), but be careful to use the correct holes!

4.        Is there a preferred orientation for the carbocation to approach the nucleophile?

 

5.       What can you predict about geometry changing as the nucleophile bonds with the carbocation?

 

 

One-Step Elimination Reactions

The Bimolecular Elimination Reaction (Karty 7.4)

What happens if the incoming Lewis base (nucleophile) is also a good Brønsted base? In that case, it may be possible to remove an H+ from the substrate, instead of doing an SN2 reaction.

Note that in an elimination reaction, the H atom removed is on a carbon adjacent to the carbon with the leaving group.

 


 

Two-Step Elimination Reactions

Observation: Alkene side products are frequently observed in reactions that give SN1 substitution.

Questions:

1.       Write the formula of the reactant and (elimination) product:

a.       2-bromo-2-methyl butane __________

b.      2-methyl-but-2-ene ___________

What must be lost from the starting reactant to give an alkene product?

 

In the SN1 mechanism to give the methoxy product, what is the first step? Draw the reactant and how it forms the intermediate.

 

 

What second step is needed to make an alkene from the intermediate you drew in the answer to #2? ___________________________

Draw an arrow-pushing mechanism to show a methoxide (CH3O-) doing the second step, as you answered question 4.

 

 

 

Do the steps in your answers to #3 and #4 combine to answer  #2? __________

 

When a tertiary or resonance stabilized carbocation forms an alkene, the reaction is categorized as an E1, standing for first-order (or unimolecular) elimination. These reactions are first-order in substrate: the concentration of base does not matter in the rate law. Like SN1 reactions, the rate-limiting step is formation of the carbocation. And, like SN1 reactions, substrates that do not easily form carbocations are unlikely to have an E1 reaction.

 

 

Questions:

1.       Rank these compounds in order of their ability to have an E1 reaction, 1 being most reactive.

One of those compounds cannot do E1 even though it has a reasonably stable carbocation. Explain why.

 

Draw a complete electron-arrow mechanism for the E1 reaction of 2-iodopropane with sodium ethoxide, NaOCH2CH3.

 

 

 

 

 

Observation: When H+ can be removed from different carbons, the distribution of possible products generally favors alkenes with more carbons attached to the sp2 hybridized C atoms. Thus, in the reaction in the first example, more 2-methyl-2-butene is generated than 2-methyl-1-butene. Observations like this were first published in 1875 by the Russian chemist Alexander M. Zaitsev, so the trend is known as Zaitsev’s Rule.  Furthermore, trans alkenes are slightly stabler than cis alkenes.

Challenge: Draw 3-bromo-2,3-dimethyl pentane and four alkenes it could make by E1 elimination. Rank the likely abundance of each, in a product mixture, from 1 (most) to 4 (least).

 


 

Carbocation Rearrangement:

As is seen in SN1 reactions, carbocation rearrangements can sometimes generate unexpected products. These rearrangements can move a hydrogen atom or alkyl group from an adjacent carbon atom to a carbocation. It only helps if the newly formed cation is more stable.

(Final proton-removal step not shown!)

In the above reaction, the triflate leaving group breaks off to form a secondary carbocation. This carbocation can give an SN1 product or either of two alkenes. A hydride-shift carbocation rearrangement allows formation of a new alkene and a new SN1 product. The relative amounts of elimination, substitution, and rearrangement products depend on the reaction conditions.

Question: Which alkene above is predicted to be the major product by Zaitsev’s Rule?

Another common reaction is the elimination of water by acid catalyzed E1 reactions. The net reaction is HC-COH C=C + H2O. In the presence of strong acid, or if the product alkene is distilled away, the reversible reaction can be forced to near completion.

Question: Draw a mechanism for the acid-catalyzed dehydration of 2-methyl-2-butanol and predict the major and minor products by Zaitsev’s Rule.

 


 

Review Question:

Write an electron-arrow mechanism to explain the formation of these three products when 1-(1-iodoethyl)cyclohexene reacts with methanol. Hint: Resonance!

 

 

 

 

 

 

 

Challenge:

Refluxing cyclopentylmethanol in methanol with sulfuric acid generates several products. Try to write a mechanism that accounts for each. Use a C-C bond shift to form the C6 ring!

 


 

Summary of Multi-Step Reactions


 

 

Importance of Reaction Mechanism for SN2 Reactions:

Kinetics studies (Chem 1042) of nucleophilic substitution reactions show that there are two different rate laws: one that is first order in both reactants (second-order overall) and another that is first order in the electrophile and zero order in the nucleophile. This suggests two different mechanisms, which are called SN2 (for Substitution, Nucleophilic, 2nd order) and SN1 (for first order rate law). Each of the examples in the previous worksheet is SN2. We’ll look at the first-order SN1 later.

A reaction rate that is first order in each reactant suggests that the rate-limiting step involves a collision of the two reactants. The general reaction can be written as:

A lone pair from the nucleophile forms a new bond to the electrophilic carbon, while the polar C¾L bond breaks with its electrons going to L. The observed stereochemistry for this reaction is below:

With only one collision step in the reaction mechanism, the reaction coordinate diagram shows one energy peak. The arrangement of atoms at this peak is called the “transition state” (sometimes “activated complex”) and is shown by the notation [transition]. Even though this is shown as an energy maximum on the reaction coordinate diagram, it actually represents the “pass” the atoms must cross to travel between higher energy geometries.

If reactants collide with enough energy and the correct orientation to reach the transition state, they can go on to make products. Here are two possible orientations for the reactants as they approach the transition state, with partially formed or broken bonds shown by - - - -

 

Explore:

1.       Which transition state orientation agrees better with the observed stereochemistry?

 

Draw δ- symbols on the diagram above to show the distribution of negative charge as bonds are broken and formed. Which orientation minimizes charge repulsion between δ- charges?

 

Draw and build a model of S-2-bromopentane, and the product of its reaction with CH3O- . What is the stereochemistry of the 2-methoxypentane product, R or S

 

 

 

Draw in electron-pushing arrows to complete the reaction mechanism from #3. Pay attention to which electron pairs become new bonds and where the electron pairs from broken bonds go.

 

 

 

Suppose you could run a reaction of (2S, 3R)-2-bromo-3methylpentane with Br- as nucleophile. (This can be done if you use isotope-labeled 79Br and 81Br, but the reaction quickly randomizes as the leaving-group bromide can go be a nucleophile somewhere else.) Would the product of the reaction be an enantiomer or a diastereomer of the starting material? Draw careful pictures showing the stereochemistry.

 

 

 

 

 


 

 

In organic chemistry, reaction orientation involves conformation of the reactants as they approach each other. If a reactant has large, bulky parts of the molecule blocking access to the electrophilic carbon, the nucleophile may not be able to get to it easily. This means that fewer collisions will have a chance of successful reaction and therefore the rate will be slower. In addition, bulky groups will interfere with each other when forming the trigonal bipyramid transition state, making the activation energy higher. This effect is called “steric hindrance”, steric meaning bulky and hindrance meaning slowing or interfering.

The importance of steric hindrance can be seen by comparing the relative rates of reaction for the same leaving group and nucleophile with different alkyl groups. As the alkyl groups get larger, it becomes slightly more difficult for the nucleophile to reach the substrate in the transition state. The increased activation energy makes the reaction slower, by about a factor of 30-40 for each methyl group on the electrophilic C, and less for carbon atoms farther away. (See Table 9-5, page 489 in Karty).

Remember the vocabulary of primary, secondary, and tertiary alkyl groups. The relative rates for SN2 reactions are methyl > primary > secondary >> tertiary.

1.       Use the short white bonds to build semi-space-filling models of 1-chloropropane and 2-chloropropane. Which has more room for a nucleophile to form a new bond? (try CH3S-, using short bonds and a yellow ball for S)

 

 

Replace an -H by -CH3 on the electrophilic C to make 2-chloro 2-methyl propane. How crowded is that? What is the common name for this molecule?

 

 

 

 

Draw a skeleton structure of 1-chloro-4,4-diethylhexane.  Do you think that would have a significantly slower rate compared to 1-chlorohexane? How about compared to 3-chlorohexane?

 

 

 

Sulfonate esters, R-SO2-O-R’, make very good leaving groups. Three frequently used ones are para-toluenesulfonate (tosylate), trifluoromethanesulfonate (triflate), and methanesulfonate (mesylate)

1.       Draw a skeleton structure of sec-butyl triflate, including wedge/dash at the chiral center. What isomer did you draw, R or S? Now draw a complete curved-arrow mechanism for reaction of sec-butyl triflate with cyanide ion, :N≡C:- (the C is more nucleophilic).

 

 

 

 

 

 

The cis and trans isomers of 4-tert-butyl 1-chlorocyclohexane have dramatically different SN2 rates. Draw (or build models of) each isomer in the chair conformation, and try to explain which reacts faster and why. Recall that there is a strong conformation influence of the t-butyl group.

 


 

 

Intramolecular SN2 Reactions

 An SN2 reaction by definition is second order in nucleophile and electrophile. But, if a single molecule has both a nucleophile and electrophile, the reaction does not require a difficult collision and can instead be reached by a conformational change of one reactant; often a lower activation energy barrier.

Example: Mustard gas was used in the First World War to kill and injure many thousands of troops on both sides. Its formula is ClCH2CH2SCH2CH2Cl.

Questions:

1.       What kind of reaction is the biomolecule + mustard reaction?

What can you say about the stability of the cyclic sulfonium ion?

 

 

Your Turn:

Write an arrow-pushing reaction mechanism to show the intramolecular SN2 that happens after 5-bromo-1-pentanol reacts with a base. This is an example of a reaction called the Williamson Ether Synthesis, a generally useful reaction for making ethers.

 

 

 

This kind of cyclization is important in making a 3-membered cyclic ether called an “epoxide”. Draw reaction that shows formation of an epoxide from R,R-3-chloro-2-butanol. Do you think stereochemistry might be important?

 

Vocabulary Summary:

Nucleophile: Electron-rich reactant; a Lewis base.

Electrophile: Electron-poor reactant

Leaving group: Atom, ion, or molecule that breaks away as a product

Inversion of configuration: Change in stereochemistry of a chiral reactant in SN2 reactions

Steric Hindrance: effect of slowing reactions by difficulty in getting to a reactive conformation

Substituent: A functional or alkyl group; “the t-butyl substituent” or “an alcohol substituent”

Substrate: A molecule on which the reactions are done; in this case the electrophilic C.


 

 

Stereochemistry of SN1 Reactions:

The observed products in SN1 reactions of chiral molecules typically are racemic mixtures.

1.       Assign the R,S stereochemistry of the three chiral centers  in the above reaction.

Draw out a full mechanism for this reaction, showing the carbocation intermediate, and leading to formation of one of the two products. Include electron arrows!

 

 

 

 

 

 

 

 

Explain why the product is a racemic mixture.

 

This reaction is helped by a resonance stabilization of the intermediate. Can you draw it?

 

 

 

Relative Rates of S­N1 Reactions:

Hyperconjugation

The observed rate of SN1 reactions shows that reactions at tertiary carbons proceed much faster than reactions at secondary carbons. Primary and methyl carbons are nearly inert to S­N1 reactions. The explanation for this effect has to do with an orbital overlap effect called hyperconjugation, in which electron density from a neighboring C-H or C-C bond can stabilize an empty p orbital. The figure at right shows hyperconjugation of an empty p orbital with a C-C σ bond (to the left) and with a C-H σ bond (front right). Notice that the empty p orbital cannot get involved with C-H bonds in the node plane, since it has no overlap with them.  Hyperconjugation is not quite the same as a full π bond, or a resonance structure, since it involves an sp3 hybrid orbital and somewhat less overlap.

 

Draw all five isomers of C4H9Br (including a pair of stereoisomers). Then draw the carbocations that form when the Br- is  removed from each isomer.

isomer

 

 

 

 

 

carbocation

 

 

 

 

 

Write 1°, 2°, and 3° in the boxes in the second row to show which carbocation(s) above are primary, secondary, or tertiary.

 

Which two carbocations are identical?

 

 

Rank the isomers as (1) fastest,  (2) medium, (3) slowest to SN1 reactions, writing the number in the boxes in the first row.

 

Rank the isomers as (A) fastest, (B) medium, or (C) slowest to SN2 reactions. Use the ideas of steric hindrance from earlier.

 

Comment on the relationship between SN1 rates and SN2 rates for alkyl halides.

 

 

Resonance Stabilization

Resonance stabilization of the carbocation increases the reaction rate. The ability to spread the positive charge over more C atoms’ p orbitals has a significant stabilizing effect, making it much easier to form the carbocation in the rate-limiting step. When a nucleophile adds to the carbocation, it can potentially add to any site that has significant positive charge from a resonance structure.

1.       Draw two resonance structures of the carbocation formed from the methylcyclopentenyl methanesulfonate shown.

 

 

 

 

 

When (4R)-4methylcyclopent-2-en-1-yl cation reacts with water, it forms two pairs of diastereomers which are constitutional isomers of each other (four isomers total.) Can you draw all four methylcyclopentenol isomers?

 

 

 

 

Draw pictures of both compounds, then explain why 1-phenyl-1-bromobutane is much more reactive than 2-bromopentane in SN1 reactions.


 

Carbocation Rearrangement (Karty 7.7)

In SN1 reactions, side products are sometimes observed.  For example, reaction of 1-cyclohexyl 2,2-dimethylpropyl triflate in methanol shows several substitution products:

1.       Give three reasons you’d expect this to be an S1 reaction instead of SN2.

a.         

b.        

c.         

Draw the carbocation that forms initially in this reaction in the box:

Toward the right side of the blank below, draw the carbocation that leads to Product 2 (take off a CH3O-). What has to change from the ion in #2 to get to this carbocation?

 

 

 

 

Toward the right side of the blank below, draw the carbocation that leads to Product 3. What has to change from the ion in #2 to get to this carbocation? (this is trickier!)

 

 

 

 

The changes in #3 and #4 are examples of carbocation rearrangements. A group, and its σ bonding electrons, can shift from one position to an adjacent position. This typically can only happen if the shift results in increased stability of the carbocation.

 

Are the carbocations in #3 and #4 stabler than what you drew in #2? Why?


 

For #3 and #4 above, copy the carbocation in #2 as a reactant(on the left side of the blank), and show an electron-moving arrow that shows the methyl (#3) or hydrogen (#4) migration.


 

 

 

   

 


 

Factors Influencing Reaction Mechanisms

  

  

  

  

  


 

Applications of Substitutions

Reactions of Epoxides


 

Stereochemistry and Regiochemistry of Bimolecular Eliminations

  

  

  

  

 

Worksheet: E2 Eliminations

 

 


General Reaction:

 

 

Part 1: Regiochemistry of E2 Eliminations

Observation: Zaitsev’s Rule (eliminations tend to make the most substituted alkenes) applies to E2 as well as E1 eliminations.

 

1.       Draw each of these substrates and then any alkenes which could form from them by E2 elimination. (hint: One is a trick!)

a.       3-chloro-2,2-dimethylpentane

 

 

 

b.      1-chloro-2,3-dimethylpentane

 

 

 

c.       1-chloro-2,2-dimethylpentane

 

 

 

 

d.      3-chloro-2,3-dimethylpentane

 

 

 

 

What bromoalkane would be best to make 3-methyl-1-butene by E2 elimination?

 

 

 

 

Draw a bromoalkane that would give 3-methyl-1-butene as only a minor product. What would its major product be?

 

 

 

 

 

 

Part 2: Conformational Requirements for E2 Reactions

As you saw in the PowerPoint, the preferred conformation for E2 eliminations has the leaving group anti to the β-H that is being removed.

 

1.       Draw both a wedge-dash structure and a Newman projection → of 2-bromopropane, including the H atoms on carbon #1. Circle the H that is anti to the Br in both drawings.

 

 

 

Build a model of chlorocyclohexane, showing the β-hydrogens. Which H is anti and could be eliminated? What if you flip the chair conformation so the chlorine atom is axial instead of equatorial?

 

 

 

 

 

2.       Based on your model, which conformation of chlorocyclohexane is more reactive to E2?

 

 

3.       Compare the cis and trans isomers of 1-chloro-4-tert-butyl-cyclohexane. Knowing that the tert-butyl group is bulky enough to stay equatorial, which do you think would be more reactive to E2 elimination?

 

 

 

4.       Here is a wedge-dash structure of 2-bromo-3-methylpentane.

a.       Build a model, including the H atoms on C2 and C3.

 

b.      Assign the R,S configuration to both chiral C atoms.

c.       Draw a Newman projection of this molecule, looking from C2 to C3 and with the Br anti to the β-H



d.      When this molecule is dissolved in acetone and reacted with NaOH, it forms 3-methyl-2-pentene by a (Zaitsev-rule) E2 elimination. Which isomer, E or Z?

 

 

 

e.      There are three other stereoisomers of 2-bromo-3-methylpentane (an enantiomer and a pair of diastereomers). Draw Newman projections to decide whether they will make the cis or trans alkene.

 

 


 

 

Applications of Elimination

   

  


 

Review & Practice Problems

1.       Suggest reagents to do the following transformations, using reactions in your toolbox.

 

 

 

2.       Starting from any of the above (since they could all be made from cyclohexylmethanol), how might you make these? (More than one step might be needed)

 

 

 

3.       Use retrosynthesis to plan how to make this target compound, using cyclopentene and any other hydrocarbon with four or fewer C atoms. (plus reagents and solvents as desired.)

(a)    What bonds need to be formed to make the target?

(b)   What functional groups need to be changed?

(c)    What reagents are needed to do the changes?

(d)   Plan a sequence of reactions.

(e)   Look for possible problems with unwanted side reactions.

 

4.       Predict the major and minor products for these reactions.

5.       Suppose the minor products 1e and 1g were what you were trying to make. Suggest reactants and bases to make those a major (or at least more significant) product.

6.       Suggest a way to make compounds 1h and 1i exclusively. (You might want to refer to an earlier section!)

7.       What alcohols would you choose to make these alkenes by E1?


 

8.       Giant Roadmap:


 

Electrophilic Additions

Part 1:  Microscopic Reversibility and Alkene Additions

You've studied acid-promoted eliminations of alcohols, with a carbocation intermediate. The reverse reaction can also happen. The "principle of microscopic reversibility" suggests that every step in a reaction mechanism can happen backwards as well as forwards, that there is no magic direction a reaction has to happen. In practice, if a reaction is very exothermic, the reverse reaction's activation energy will be prohibitively large.

Reversible, modest energy barrier lets equilibrium happen easily and with both reactants and products present.

Energy barrier for reverse reaction is very large, little reactant at equilibrium

Huge energy difference means this reaction is effectively irreversible.

 

Key Idea: Each individual mechanism step is (potentially) reversible, and running the steps for an E1 elimination in the reverse order results in addition to an alkene.

Review: List the basic mechanism steps for the E1 reaction.

 

 

 

 

 

 

 

 

Review: Draw the mechanism for E1 elimination of 2-chloro-2-methylpropane to make 2-methylpropene. Use a generic base, B-.

 

 

 

 

1.       What bond breaks in the first step?

In the second step, one bond breaks and two bonds form. Which are they?



Compare the relative energies of the reactants (t-BuCl + B-); the intermediates (t-Bu+, Cl-, and B-); and the products (2-methylpropene, Cl-, and HB).


Which step is faster, first or second?


Draw an energy barrier diagram for that reaction in the box. Draw the curve so that each step gets a hump and each intermediate gets a valley, and the energies of the humps and valleys agree with your answers to questions 3 & 4.

Now, apply the principle of microscopic reversibility, but assume the base B-  is replaced by the strong acid HCl. Redraw the reaction, showing the same steps as before but in the reverse order and the reverse direction.

 

 

 

 

 

1.       What has changed in the reaction?

 

What energy features of reactants and/or products have changed to make the (formerly spontaneous) E1 reaction into a favorable addition?

 

 

How would that change the energies on a reaction coordinate diagram?

 

 

Representing Addition Mechanisms

"Bouncing arrows" are a modification of the electron-pushing arrows that help clarify where the  bonding electro pair goes as it reacts with the H+. Here is a bouncing-arrow picture of addition of HBr to C2(CH3)4.

That is the basic mechanism for polar (electrophilic) addition to an alkene.

Now draw a bouncing-arrow mechanism for adding H2O to trans-2-butene, catalyzed by H3O+ (the final step is a protonated alcohol giving an H+ to a water molecule, to generate 2-butanol).

 

 

 

 

1.       Which carbon atom became the carbocation?

 

Which carbon got the H, and which got the OH?

 

Is the product chiral? If so, is it formed as one enantiomer or as a racemic mixture?

 

 

How might the reaction be different if the reactant was cis instead of trans? Explain.

 

 

Model 2: Carbocation Stability

Recall from the chapter about elimination reactions that there are different types of carbocations: tertiary, secondary, primary, and resonance-stabilized.

Draw four isomers of the C4H9+ carbocation and classify them as primary, secondary, or tertiary. Label them as most stable, medium, or least stable.

 

 

 

 

 

 

Observation: When protonating an alkene, the H+ typically goes to the carbon that leaves the most stable carbocation.

Use a bouncing-arrow mechanism to show the protonation of 1-butene to make the more stable carbocation, followed by reaction of this carbocation with bromide ion.

 

 

 

 

 

 

 

 

Challenge: Explain why it would be very difficult to make isobutanol by acid-catalyzed hydration of an alkene.

Model 2: Markovnikov’s Rule.

Because the acid protonates the alkene to give the stabler carbocation, the product is easily predicted.  The trend is referred to as "Markovnikov's Rule, anfter the chemist who first discovered it.

1.       Put a * at the C atoms that become carbocations formed when these alkenes are protonated by a strong acid:

2.       Which, if any, will make a chiral compound when the carbocation reacts with a Lewis base (a nucleophile such as Cl– or H2O)?

3.       Which, if any, might have a carbocation rearrangement? Draw the rearranged carbocation in the box:

4.       Suggest reactions to make these alkyl halides and alcohols by Markovnikov addition to alkenes:

Challenge: There are two possible ways to make this ether; by the Williamson (SN2) method, or by alkene addition. Suggest reagents for both reactions.

 

Model 2: Resonance Stabilization

If the carbocation formed is resonance-stabilized, the reaction is even more favorable:

Observation: Compounds capable of forming stabler carbocations tend to react faster.

1.       Which of the above will react more quickly? More slowly?

2.       Which alkene(s) of the three above could make a racemic mixture of products?

a.       An achiral carbon atom that can become chiral in a reaction is called “prochiral”. Mark  any prochiral carbon atoms above with a «.

3.       Which could make more than one regioisomer, within Markovnikov’s guideline? Explain, drawing the intermediate.

4.       Which of the following would you expect to react with HBr faster?

Explain, and draw a mechanism showing the intermediate and the major product.

 

 

 

5.       Draw a complete mechanism for the reaction of 2-methyl-1-butene with D2O/D2SO4. (D = 2H, an isotope with identical reactivity.)

 

 

 

 

 

 


 

Carbenes, Oxidation, Ozonolysis, and Hydrogenation of Alkenes & Alkynes

Carbene: A very reactive species, CR2, that  combines electrophilic and nucleophilic behavior on the same atom. A carbene can be seen as an sp2  hybridized carbon atom, with a nucleophilic lone pair in an sp2 hybrid orbital and a vacant (electrophilic) p orbital. Carbenes add easily to alkenes to make cyclopropanes.

Based on the drawing above, is the addition syn or anti?

 

Draw electron arrows to show the bonds forming.

 

Preparation of Carbenes

Three good routes to carbenes are:

1. Elimination of N2 from diazomethane (explosive!)

2. Dichlorocarbene is made from CHCl3 and NaOH

3. A carbene-like species, I-CR2-Zn-I,  can be formed by reduction of CR2I 2 with Zn/Cu alloy.

 

 

Draw trans-1-ethyl-2-methylcyclopropane? What alkene and carbine could you use to make it?

Epoxidation & Hydroxylation

Recall that a peroxycarboxylic acid adds an O to an alkene, and the ring-opening reactions are generally anti addition.

Reaction of alkenes with transition metal oxides like OsO4 and MnO4- give syn di-hydroxylation. Osmium tetroxide gives better yields than permanganate, but it is expensive and much more dangerous.

The two reactions therefore can give different stereochemistry for forming diols from alkenes.

Practice: Draw the product of the reaction of the epoxide shown with OH-. Is it an enantiomer, diastereomer, or conformer of the diol from OsO4?

 

 

Draw the intermediates and products of reacting cyclopentene with MCPBA then OH-, or with KMnO4- in base.

Ozonolysis

Ozonolysis is a further oxidation of alkenes. The two-step ozonolysis process converts an alkene into two carbonyl groups (ketones or aldehydes). Don't worry about the mechanism yet!

 

 

 

 

What substance gives benzaldehyde as its only ozonolysis product?


 

Catalytic Hydrogenation of Alkenes & Alkynes

 

The generally useful reaction of hydrogen gas with alkenes or alkynes is very common. It is catalyzed by transition metals such as Pd, Pt, or Rh; generally supported on powdered carbon or ceramic beads. The hydrogen can be supplied by a tank or generated in the reaction vessel (for example from NaBH4 + acetic acid)

The reaction is quite selective under mild conditions, reacting only with alkenes & alkynes, not with aromatics, carbonyls, or nitriles. (These can be forced under higher pressures).

Alkynes typically react with 2 equivalents of H2, forming an alkane, unless a deactivated catalyst called Lindlar's Catalyst is used. This is a lead-doped palladium catalyst that forms cis alkenes.

Careful reaction of an alkyne with sodium in liquid ammonia reduces an alkyne in two steps with net  trans addition:

This is a good complement to the Lindlar catalyst!

 

Addition of Halogens, Epoxidation: Triangular Additions

Model 1:  Additions via cyclic intermediates

Additions of halogens and some other compounds show different results compared to acid-catalyzed additions. Reaction of alkenes with bromine shows a uniform pattern:

Review: Describe the stereochemical relationships between the dibromocyclohexane isomers shown. Can you assign all the R,S designations?

For each entry in the table below, choose identical, enantiomers, or diastereomers

A & B are:

A & C are:

A & D are:

B & C are:

B & D are:

C & D are:

 

The reaction is a stereospecific anti addition: the two bromine atoms always add to opposite sides of the double bond. A similar pattern is seen with chlorine addition.

The mechanism proposed to explain this reaction suggests a cyclic bromonium intermediate:

The alkene's π electron pair acts as a nucleophile to the electrophilic Br2, and displaces a Br- ion. The Br- ion then can react with either of the two C atoms in the ring via a back-side, SN2-like reaction.

Explore: Recall the “o-chem definition” of oxidation numbers; where C-C bonds add 0 to oxidation state, C-H bonds add -1 since H = +1, and C-X bonds add +1 since the X usually has oxidation state -1.

1.       Is this a redox reaction? If so, identify the oxidizing & reducing agents.

2.       Is addition of H2O to an alkene redox? Why or why not?

Review: Compare this mechanism with the formation of an epoxide followed by reaction with OH-.

Evidence for the cyclic bromonium  mechanism comes from additional observations: doing the reaction in the presence of Cl- ion gives some bromo-chloro addition, and reaction of aqueous halogens gives halohydrins, which have a halogen on one C and an -OH on the adjacent one. The stereochemistry is always anti.

Draw the complete electron-arrow mechanism for the formation of the chlorohydrin at right. The final step is proton transfer to H2O.

 

 

 

 

 

When the alkene reacting is asymmetrical, the incoming nucleophile tends to add to the side that would make the stabler carbocation; but the reaction is still anti. Notice the difference: the place the nucleophile reacts is referred to as "regiochemistry" (think about what region of the molecule reacts), while the spatial orientation is "stereochemistry". This is analogous to the observation about acid-catalyzed additions to epoxides.

What products would you expect from these reactions? 

 


 

Model 2: Mercury-Catalyzed Addition to Alkenes

In acid-catalyzed addition to alkenes, the relatively high temperatures and the possibility of carbocation rearrangements can give unwanted side reactions. To avoid this, a reaction involving (highly toxic) mercury(II) salts can be used. These compounds also add via cyclic intermediates:

The ring-opening regiochemistry and stereochemistry (inversion) are similar to that observed for epoxide ring-opening and halogen additions. In a final step, the mercury(II) is reduced to metallic mercury and replaced by an H from the hydride donor NaBH4, forming an alcohol. Yields are typically high but the environmental hazards of Hg compounds means the reaction is avoided if an acid-catalyzed addition works adequately and carbocation rearrangements are not a problem.

Question: In the absence of C+ rearrangements, does mercury-catalyzed addition give the same Markovnikov regiochemistry as acid-catalyzed addition? Can you suggest an explanation for this?

Review: Draw the mechanism that gives the product on the left. Use the “bouncing arrow” notation for the hydride shift required, to help you think about how the electron shift goes.

Practice Problems

 

Choose an alkene and appropriate conditions to make each of these products:

Fill in products that form when 3-methyl-1-pentene is reacted with each of these reagents:

 


 

Hydroboration-Oxidation

Observation: Boron is less electronegative than hydrogen.

1.       Draw a Lewis dot structure of borane, BH3, then determine its likely VSEPR geometry and hybridization.

 

 

2.       Describe its bond dipoles and determine its overall molecular dipole.

3.       What other unusual feature does borane have?

BH3 is a good Lewis acid: its electron deficient boron atom wants to bond to any pair of electrons it can find. Commercially, BH3 is available as a solution in tetrahydrofuran, with which it forms a Lewis acid-Lewis base complex.

When BH3 reacts with alkenes, a B-H bond adds across the C=C π bond and forms B-C and  C-H σ bonds. As with polar additions such as HBr, the more positive end of the added molecule goes to the less substituted carbon; the more negative end goes to the side that could form the stablest carbocation. Steric bulk of the substituted end also helps the relatively small H atom go to the more substituted side. A detailed mechanistic explanation for this is in the textbook, figure 11.8 on page 428.

This addition is a syn addition: the B and the H add to the same side of the C-C bond.

Draw 1-ethylcyclohexene, and then the product of adding deuteroborane, BD3 to the C=C π bond.

 

 

 

 

 

Following the reaction of borane with an alkene, the C-B bond can be broken by oxidation with H2O2 in aqueous NaOH. This generates an alcohol functional group at the site where the boron had been, with the same stereochemistry. The overall reaction, occurring in 2 steps, is:

With BH3, there are three reactive B-H bonds, so there are usually 3 alkenes reacting with each boron atom before oxidation. Draw the intermediate for the above reaction, before the oxidation step.

 

 

 

What alkenes would you use to make these alcohols by hydroboration/oxidation?

Draw the products of acid-catalyzed hydration (or oxymercuration/reduction) for the above alkenes. Note that the alkenes can give different products with H+ or Hg2+ catalysis – compare with a study-buddy to think about both (one is stereochemistry, the other is a C+ rearrangement).

 

 

You may have noticed that there are two different boron-hydrogen compounds used. Describe the two and how their uses are similar or different.

 

 

Alkynes

In general, alkyne reactions are very similar to those of alkenes. Alkynes are often more reactive, so in some cases a reaction with a 1:1 mole ratio can produce a substituted alkene.

Finally, "color in" a wedge-dash for the last picture above. Circle the two bromines added first. Be careful, both additions are anti.

Hydroboration-Oxidation of Alkynes

Alkynes have two π bonds, and can therefore react with two equivalents of borane. Since BH3•THF has three B-H bonding sites, a large and difficult oligomer can result, and that is difficult to oxidize with the H2O2 step. To keep this situation from being a problem, a special bulky monofunctional borane is used: disiamylborane, short for di(sec-isoamyl)borane.  This bulky borane can only react in a 1:1 ratio with an alkyne.

Below left, draw the intermediate formed between disiamylborane and 1-butyne.

What would this intermediate make after treatment with H2O2 and NaOH? Draw it above, at right.

The final, isolated product has a rearrangement called a "tautomerization" to form a carbonyl compound. The first step in that reaction is deprotonation of the most acidic H to give a resonance-stabilized intermediate. Re-protonating (at a different location) gives a carbonyl compound. Draw the mechanism with correct electron arrows. The acid-catalyzed version of this mechanism is on page 425.


Introduction to Infrared Spectroscopy

            

 

Interpreting an Infrared Spectrum

Look for the peaks that can tell you what functional groups might be there or are probably absent. You will probably want to refer to the textbook or the table in the Mohrig lab techniques book. (A table like this might be provided on tests, but you should learn the most important peaks.)

Example: This is a spectrum of C7H6O. Before examining peaks, what can you say about the degree of unsaturation (rings and/or π bonds)?

Ä      What peaks do you see in the hydrogen region, n > 2800? Is it likely to be an alcohol?

 

Ä      Could it be an amine, nitrile, or carboxylic acid? (Without examining the IR, why not?)

 

 

Ä      What do you see in the triple-bond region, 2000 – 2800? Could there be an alkyne?

 

 

Ä      There are two important peaks in the double bond region, 1500-2000.

o   Around 1700, a strong peak says there must be a __________ bond.

o   What does the sharp peak around 1600 suggest?

Ä      Based on the functional groups you could identify, can you suggest a structure?

 

 

Questions IR Can (and can’t) Answer

Interpreting IR spectra, as you’ve seen in the lab, is more often about checking compounds whose identity you suspect than completely unknown materials.  Answer the following questions about what IR features (presence or absence) to check the results of possible lab experiments. Follow the format of the example…

  1. How pure was the ester product formed from an alcohol and a carboxylic acid?
    Look for the absence of any band from –OH above 3100 cm-1, so no alcohol or acid remains.
  2. Was the product of a reaction an aldehyde or a ketone?

 

  1. Is an alkyne terminal (C-CºC-H) or internal (C-CºC-C)?
  2. Is an ester phenyl acetate or methyl benzoate?

  3. Did the reaction convert a ketone to an alcohol?

Other questions are somewhat more difficult to answer but can sometimes be resolved by careful analysis if the relevant peaks are clearly defined. Notable examples are substituton patterns of aromatic rings and alkenes. But often, IR just does not provide useful information. For example, even an expert would have a hard time telling the IR of 2-pentanone from 3-pentanone. 

http://www.uwplatt.edu/~sundin/ir/_one_one.jpg

 

 


 

 

Introduction to 1H NMR Spectroscopy

The Technique:

NMR stands for Nuclear Magnetic Resonance. In NMR spectroscopy, a sample is put in a strong magnetic field, then hit with a radio-frequency energy pulse. This causes the nuclei in the molecules to change their quantum spin states, and as those spin states “relax” to their normal levels, they emit energy. This energy is then detected, decoded, and plotted. Nuclei commonly scanned this way include 1H, 13C (about 1.1% of all C atoms), 19F, and 31P. Many other atoms can also be analyzed with specialized probes. 2H, called deuterium, is invisible to NMR, so samples are usually dissolved in deuterated solvents such as D2O, acetone-D6, CDCl3, or C6D6.

Information Provided:

The NMR spectrum is a graph of signal intensity vs. frequency. The frequency axis is always scaled in PPM δ, or parts per million deviation from a standard. Most organic H atoms show resonance within 10-12 ppm of the hydrogen standard, which is defined as 0 PPM. By convention, 0 is at the right end of the axis and the higher ppm extends to the left.

Four main types of information can be derived from a proton NMR spectrum:

1.       The number of types of H atoms. Each chemically different H atom will show up at a different position, though sometimes these positions overlap a bit.

2.       The relative number of H atoms of each type. The spectrum’s integral (area under the peaks) is proportional to the number of H atoms generating that peak.

3.       The peaks are “split” by neighboring H atoms in a way that allows us to calculate the number of neighbors. Additional information can be derived from the amount of splitting.

4.       The position of each peak in the spectrum depends on the chemical environment of the hydrogen atoms. H atoms near more electronegative atoms, or near π bonds, tend to be shifted to higher ppm positions. This is called “downfield” or “deshielded”.

Interpretation:

Analyzing a given spectrum for those four properties is a matter of practice and a systematic approach. It will help you to remember the acronym tHINC: types of H, Integration, Neighbors, Chemical shift.

Types of Hydrogen:

Each chemically distinct H in a molecule gives rise to a peak in the NMR spectrum. You can usually tell how many different kinds of H atoms there are by counting the peak groups. This fails in two cases: for longer alkyl groups, the very similar positions tend to overlap each other; and for aromatic rings the exact assignment can be difficult.

Integration:

The integrated area of a peak is proportional to the number of H atoms that generate that peak. Peak areas can be measured in percentage, or if you know the formula then it’s easy to multiply percent by total number of H to get the number of H atoms that give rise to each peak. For example, (CH3CH2)2O has ten H atoms in two groups, which have a 3:2 peak area ratio.

Neighbors & Splitting:

The splitting by neighboring H atoms is a bit confusing. Rather than go into a technical explanation, let me use an analogy. Suppose you and a friend have $10 in pennies, and you play a game. The game is that each of you takes turns putting a penny in a cup, shake it, and dump it out on the table. You get a penny that is heads, your friend gets it if it is tails. After playing the game with one penny, you have a 50% chance of $9.99 or a 50% chance of $10.01, but you can’t have $10.00 exactly. If you play with two pennies, you have a 25% chance that you’d get $10.02, a 50% chance of $10.00, and a 25% chance of losing both pennies to get $9.98.

Now suppose you played the game many times, each time starting at $10.00, and graphed the results. The graph would look like this:

Challenge: What would the expected winnings be if you played the game with three pennies?

A similar statistical event happens with 1H nuclei. Each H atom has a (nearly) equal chance of having its chemical shift either raised or lowered slightly by nearby H atoms. The resulting patterns can be decoded to find out how many H atoms are on the next C atom over from the H atoms responsible for that peak. The relative intensities in the pattern can be modeled by considering the pennies game, above, or by using a mathematical tool called Pascal’s Triangle.

The peak patterns are called singlet (no splitting, one peak, meaning no neighbors), doublet (two equal peaks), triplet, quartet, and so on.

Chemical Shift

The exact position of each peak group is determined by its chemical environment. Tables of chemical shift ranges are handy (end of Mohrig book, or page 554 in the Hornback textbook), but there are some groups you should learn. Alkyl peaks tent to have low δ values, around 0.5 to 2 ppm. Nearby electronegative atoms will increase the shift. Aromatic ring hydrogen atoms are typically at around 7 to 8 ppm.

Putting it All Together

Once you’ve worked through the tHINC process for a spectrum, you will probably have a collection of facts like “there are three H atoms with one neighboring H and an electronegative atom nearby” and “There is an aromatic ring with four H atoms”. Do you know something about the formula of the molecule (is there a halogen or an oxygen? Is there additional information such as an infrared spectrum?  Is the DU at least 4 if you think there’s an aromatic ring?), you have to connect each chunk into a reasonable structure.

The following examples should help you connect the structure-spectrum assignment.

In small-group this Friday & Monday, you’ll get a chance to play with an NMR simulator to generate your own spectra.

Example 1:

This is the simulated spectrum of chloroethane, CH3CH2Cl. The integrated peak ratios show 40% of the signal intensity due to the peak group at 3.42 ppm and 60% due to the peak group at 1.49 ppm.

1.       Draw out the Lewis structure of chloroethane including all hydrogen atoms. How many chemically distinct H atoms are there? Circle each group of distinct H atoms.

 

 

2.       Based on the integration values, which peak groups relate to which H atoms?

 

 

3.       Explain the splitting pattern. Why do the split peaks have the intensity patterns they have? 

 

 

 

4.       Explain the relative positions of the peaks.


 

 

Example 2: 

Identify this compound, C7H14O. The IR shows a peak at about 1720 cm-1 which might be useful to your identification. The dotted lines represent 10% intervals in the integration.

 

a. What is the significance of the IR peak at 1720 cm-1?

 

b. Figure out the number of H atoms corresponding to each signal in the NMR spectrum.

 

 

c. Explain the splitting pattern of the peaks at 2.5 and 1.2 ppm.

 

d. Draw a structure and assign the peaks.

 

Example 3:

The formula of this compound is C9H12O.

Some of the peaks are blown up to help you see the splitting pattern.

1.       How many types of H are there? _____________

2.       Convert the integration percents to the number of H atoms.


3.       How many neighbors does the peak at 1.4 ppm show?

 

4.       How many lines are in the peak at 4.7 ppm? What does that say about neighbors?


5.       The peaks at 7 and 7.4 are characteristic of what functional group?


6.       Compare your answers for questions 2 & 3 to come up with a possible sub-structure.



7.       What is the DU of the formula?


8.       How many of the C atoms are accounted for by what you’ve considered so far? How about the H atoms?

 

9.       Use the electronegativity and chemical shift idea to figure out where the O goes.

10.   Try to draw a structure consistent with everything you’ve figured out.

   


 

Thinking about Molecule Identification – Spectroscopic Summary

IR

·         Plot: Y-axis is % T, X-axis (usually) wavenumber in cm-1

·         Energy range: IR light is ~ 101 kJ/mole, about 0.02 to 0.1 x bond energy

·         Based on vibrations

·         Bond strength & atom masses

·         Vibrations change molecule polarity, relative intensity related to dipole

·         Tells functional groups by characteristic peak positions

·         Generally  can’t assign all peaks

·         Similar molecules nearly indistinguishable (i.e. 3-hexanone vs. 3-pentanone)

·         Computers can match structures by statistics esp. in fingerprint region

1H NMR

·         Plot: x-axis is ppm (parts per million), Y-axis is signal intensity

·         Energy range: Very low (radio waves)

·         Based on magnetic properties of spinning nuclei

·         Instrumentation is expensive

·         All peaks can be assigned for simpler molecules

·         Tells structure information

o   # of Types of H by peak groups

o   Relative numbers of H by integration

o   Splitting/neighbors – Pascal’s Triangle

o   Chemical Shift

·         Functional groups inferred from shift

·         Coupling of one H to neighbors can be traced


 

Radicals and Radical Reactions

A “radical” is the chemical term for a molecule with an unpaired, or odd, electron. (There are some instances of diradical molecules, typically short-lived excited states, but we won’t worry about those.) By definition, a radical does not fill the octet rule.  Because of this, radicals are extremely reactive and only a few odd-electron molecules are stable enough to isolate. Typically, radicals are intermediates in chemical reactions.

The term “radical” dates back to the 19th century, before a thorough understanding of bonding, when any piece of a molecule would be termed a radical.

Ethyl chloride, for example, may be converted into ethyl alcohol, the latter may be transformed into ethyl iodide, and this again may be converted into butane, but during all these interactions the group C2H5- remains unchanged, and behaves, in fact, as if it were a single atom,

C2H5.Cl +  H-OH = C2H5-OH + HCl

C2H5-OH + HI =  C2H5.I  + HgO

2 C2H5-I + 2Na = C2H-C2H5 + 2NaI.

Numerous examples of a similar kind might be quoted; amongst others, the changes by which the five compounds, CH3-Cl, CH3.OH CH3.O-CHg, CH3-I, and CH3-CH3, may be successively transformed one into the other.

Groups of atoms, such as C2H5- and CH3-, which act like single atoms, and which enter unchanged into a number of compounds, are termed radicles, or sometimes compound radicles.

               --- W. H. Perkin & F. Stanley Kipping, “Organic Chemistry”, J. B. Lippincott Company, Philadelphia, 1894, p. 115 (http://archive.org/details/cu31924031195583)

Notation for Radicals and Radical Reactions

Just as a pair of electrons is given two dots, the radical is given a symbol of one dot. It is very important to draw that clearly and carefully so there is no ambiguity of where the radical is. Omitting the dot on a skeleton structure would mean it was simply an implicit (unseen) H atom. Note that this radical has no charge: since, under the rules of formal charges, a carbon atom has four electrons and in the radical it “owns” a single electron instead of half a pair, it still has a formal charge of zero.

In mass spectroscopy, radicals, cations, and radical cations are important in considering the mass fragmentation patterns. . The radical cations generated in MS come from electron beams violently knocking electrons out of molecules, either σ or π bonds. These can then break into fragments, and the cation fragments are measured. Neutral radicals can’t be observed in normal mass spectrometry. For example, ethylbenzene shows a fragmentation in which a methyl radical breaks off to leave the fairly stable benzyl cation as the most prominent peak. (The parent ion peak, a radical cation, is about 28% of the height of the base peak.)

In this example, the radical cation symbol is shown localized over the dashed CH2---CH3 bond to show that the radical and cation are in the same place:

Mechanism Arrows:

Reaction mechanism arrows for radical reactions use a single-barbed arrow to show that only one electron is moving. In the fragmentation reaction above, a single-barb arrow shows that the one electron in the (weakened) CH2-CH3 σ bond is moving to the methyl group, forming the methyl radical and leaving the benzyl cation behind. Watch for those single-barbed arrows in the mechanisms to follow!

Radical Sources

There are two important ways to generate radicals, heat or photolysis. Either way, a compound with a relatively weak bond is broken up to generate radicals. Commonly used radical reactant sources are peroxides such as benzoyl peroxide (also an ingredient in acne medications!) or di-tert-butyl peroxide, azobis(isobutyronitrile) or AIBN, N-bromosuccinimide as a source of Br radicals, or halogens such as bromine or chlorine.

Example:

Practice:

Using the barbed-arrow notation, draw the fragmentation of AIBN to generate two radicals and N2 gas, :NN:, from the middle nitrogens. You’ll need four barbed arrows!

 

 

Radical Stability & Geometry

Like carbocations, radicals on carbon atoms are electron-deficient; and like carbocations they can be stabilized by resonance or by hyperconjugation. Thus, a methyl radical is the highest energy, followed by primary, secondary, and tertiary radicals. Resonance also stabilizes radicals, thus a resonance-stabilized primary radical is even more stable than a tertiary radical. Also like carbocations, radicals in sp2 orbitals (vinyl or phenyl) are less stable.

These stability trends are the same as seen in bond dissociation energies, the energy required for homolytic (lit. “same-breaking”) bond cleavage. The reaction is  Since the H· radical has the same energy in each case, the energy difference must come from relative radical stability.

The carbon radical geometry is not 100% clear, since they cannot be isolated to study. However, their reactions are such that they could be planar or rapidly inverting pyramidal. If the radical carbon is sp2 hybridized, with the odd electron in a p orbital, it would be planar. On the other hand, simple VSEPR suggests that perhaps sp3 hybridization might be appropriate. If the radical is sp3 hybridized, the geometry inverts back and forth so rapidly that radical reactions generally lose stereochemistry, resulting in racemic products if the radical center was chiral. Experimentally, studies of methyl radicals trapped in frozen argon suggest the radical is no more than 5° out of planar, if it isn’t perfectly planar.

Radical Mechanisms

There are several mechanism steps that are unique to radical reactions. The simplest is radical coupling, where two radicals meet up and share their electrons to form a new σ bond.

R·                  ·R      →     R―R

Practice: Draw the barbed arrows for the above coupling.

Another radical step that is seen is disproportionation of radicals to an alkene and an alkyl.

This reaction is used in oil refineries to “crack” large alkanes from crude oil into smaller molecules that have greater value as fuels and chemical feedstocks. Since there are many sizes of hydrocarbon in the crude oil, and many possible places for the molecules to break, and many ways the resulting radicals could meet to disproportionate, the reaction generates a large stew of products that is then separated by distillation.

Both of the above reactions require two radicals to collide. In most reactions, however, the concentration of radicals is extremely small. In that case, it is far more likely for the radical to hit a normal, non-radical molecule instead. When that happens, the radical can “steal” an atom from another molecule, in a step called abstraction.

 

If there is an alkene present, the radical can add to the π bond:

Note that abstraction and addition result in a new radical! Since that radical can participate in another reaction, the possibility of a repeated cycle, called a chain reaction, occurs.

Radical Chain Reactions

A chain reaction consists of three phases: initiation, propagation, and termination. As the names suggest, in the initiation phase, a radical is formed. During the propagation phase, a radical reaction happens which gives a new radical as a product. This step might involve more than one reaction, but it can happen over and over again, thousands of times.  The propagation stage is where the main transformation happens. Finally, a termination step can happen. This step uses up the radicals as two of them combine in a coupling or a disproportionation; or if a radical is somehow deactivated by a different reaction. (Contaminants on glassware can be a problem here!)

Polymerization

Radical polymerization is when an alkene does repeated radical additions that generate longer and longer chains. Many types of plastics are, or were formerly, generated by radical polymerization. (Organometallic catalysis has provided more controlled polymerization that can result in higher quality plastics.) Any plastic with the name poly _________ene can, in principle, be made via radical polymerization. Here is an example of the polymerization of methyl methacrylate, which makes the strong clear plastic called “acrylic”.

Initiation Steps:  A small amount of peroxide initiator is mixed with an alkene. Heat or light causes it to break into radicals.

 

A peroxy radical then kicks off the chain by adding to an alkene monomer:

Propagation Steps: The resulting radical then adds to another alkene monomer, which lengthens the growing polymer chain by one unit.

The chain grows to hundreds of monomer units long by repetition of this step.

Termination Step: Two radicals finally might meet up to stop the reaction, either by coupling or by disproportionation.

Halogenation

One important radical chain reaction is replacement of a C-H bond with a C-X (X = Cl or Br); with H-X as byproduct. The net reaction is C-H + X-X à C-X + X-H. This is one of the few organic reactions that alkanes can do, aside from their use as fuels.

Initiation Step:  A chlorine molecule is broken by heat or by photon energy to give two radicals:

Propagation Step 1:  A Cl atom abstracts an H from an alkane, making HCl and an alkyl radical.

Propagation Step 2:  The alkyl radical abstracts a Cl atom from another chlorine molecule, generating an alkyl halide and a new chlorine radical.

This new chlorine radical can then go and find another alkane and continue the reaction with propagation step 1.

Termination: Each radical chain eventually terminates. At some point, two radicals might collide and join together again in a coupling, or disproportionate. The amount of side product from the termination step is usually extremely low, since there can be thousands of propagation steps before a termination step.

 

Practice Problems:

1.       Draw out the initiation and propagation step(s) for the chain reaction mechanism for making 2-bromopropane by radical halogenation.

 

 

2.       Draw three coupling reactions and an abstraction reaction that could be possible termination steps for bromination of propane.
 

 

 

 

3.       Comparing chlorination to bromination, would you expect the ratio of 1-chloropropane/2-chloropropane to be higher or lower than the 8%/92% for bromination? Why?

 

 

4.       Consider the possible monobromination products that could be made from 2-methylbutane.

a.       How many different types of H atoms are there on 2-methylbutane?
 

 

b.      Draw all possible bromination products. Are any of them chiral?

 

 

 

 

 

 

c.       Name all of them, including (R,S) as needed.

d.      Rank them in order of most to least likely to form (1 = most likely).

 

5.       Monobromination of this aromatic compound gives only one significant product even though there are four types of H atoms. Which is it, and why is that product so dominant?

 

 

 

 

 

Radical Allylic Bromination and Hydrobromination:

Usually, reaction of Br2 with an alkene results in anti addition, as seen in Ch. 11. Refresher: Fill in the cyclic bromonium intermediate.

However, a radical mechanism can go via the resonance stabilized allyl radical to brominate by replacing a C-H bond with a C-Br bond, one position over from the alkene group. The source of bromine for this reaction is N-bromosuccinimide, often abbreviated NBS.

Allylic bromination using NBS avoids both addition of Br2 to the double bond, and the hazards of handling elemental bromine. NBS apparently produces small concentrations of Br2 when it reacts with HBr (generated by radical reactions starting with the AIBN or peroxide initiator) in the reaction mixture. However, the kinetics of ionic bromination requires larger concentrations of Br2. The experimental rate law is rate = k[alkene][Br2]2, and the proposed active brominating agent is the cation Br3+ from the equilibrium 2 Br2  Br3+ + Br-. (http://classes.yale.edu/chem220/STUDYAIDS/bromination/brominationcyclohexene.html, accessed 9 December 2013).

Questions:

1.       There are four exchangeable H atoms on the ring. What kinds of product isomers could form?

 

 

 

 

2.        Refresh your memory: what’s AIBN? Draw it.

 

 

 

 

3.       Draw the radical intermediate that forms in this reaction, including resonance structures.

 

 

 

4.       Assuming the active form of bromine is Br2 from the NBS, draw out full arrow mechanisms for both propagation steps. Start with a bromine atom and cyclohexene.

 

 

 

 

 

 

Radical Hydrobromination

Addition of hydrogen bromide (HBr) can also go via a radical mechanism instead of an ionic mechanism. The observed product is the opposite regiochemistry of what would be expected by an ionic (Markovnikov) addition.

 

Refresher: Draw the cationic intermediate for Markovnikov addition in the {brackets}.

 

Radical bromination is a chain reaction. The initiation step can be started by a peroxide or by photolysis of the H-Br bond.

For the propagation steps, both an H atom and a Br atom must add to the alkene, and there must be some kind of alkyl radical intermediate.

 

 

 

 

1.       Review: Which radical is more stable, primary or secondary?

 

2.       Draw an arrow mechanism showing how that radical is formed from an atom radical and 1-butene. This is the first propagation step.

 

 

3.       Use the bond energy table (below) to estimate the energy of the first propagation step. 

 

 

4.       React the radical you drew above with another H-Br to form 1-bromobutane and a new radical, again being careful with electron arrows.  This is the second propagation step.

 

 

 

5.       Use the bond energy table to estimate the energy of the second propagation step. 

 

 

6.       How does the anti-Markovnikov regiochemistry of the addition support the idea that a Br radical adds to the alkene instead of an H radical?

 

 

7.       Compare the energies for these reactions. Which reaction is more likely? 

 

 

8.       What do you think would happen if an H radical added to the alkene as a propagation step? (Hint: Draw the likely resulting radical and what it would do next. Is this a sustainable cycle?) 

 

Radical Energies & Selectivity of Reactions

Radical reactions are driven mostly by bond energy. Breaking a bond is always hard to do (endothermic), forming a bond is exothermic. The slight differences in radical stability due to hybridization, hyperconjugation, and resonance structures have a large impact on rates and selectivity of radical reactions. Note that the order of stability for radicals is similar to the order of stability for carbocations, and for similar reasons!

This table shows some typical bond-breaking reaction energies. (Data from Table 2.1, page 33, and 21.1, page 920, in the Hornback textbook; and Table 8.9, page 389 in Kotz, Treichel, Townsend’s General Chemistry, 7th edition)

Reaction

Radical Formed

Energy (kJ/mole)

C6H5-H C6H5· + ·H

Phenyl

460

H2C=CH-H H2C=CH· + ·H

Vinyl

452

H3C-H H3C· + ·H

Methyl

435

CH3CH2-H CH3CH2· + ·H

Primary

410

(CH3)2CH-H (CH3)2CH· + ·H

Secondary

397

(CH3)3C-H (CH3)3C· + ·H

Tertiary

385

CH2=CH-CH2-H CH2=CH-CH2· + ·H

Allyl

372

C6H5CH2-H C6H5CH2· + ·H

Benzyl

356

H2 2 H·

Hydrogen

436

F2 2 F·

Fluorine

155

Cl2 2 Cl·

Chlorine

242

Br2 2 Br·

Bromine

193

I2 2 I·

Iodine

151

H-F H· + ·F

 

565

H-Cl H· + ·Cl

 

432

H-Br H· + ·Br

 

366

H-I H· + ·I

 

299

C-F C· + ·F

“average” C

485

C-Cl C· + ·Cl

“average” C

330

C-Br C· + ·Br

“average” C

276

C-I C· + ·I

“average” C

217

C-C C· + ·C

“average” C

339

C= C C: + :C

“average” C

606

C=C ·C—C·

(calc.)

267

Application

1.       Use the bond energies given to calculate the net energy for the reactions:

a.       CH4 + Cl2 CH3Cl + HCl

b.      (CH3)2CH­2 + Cl2 (CH3)2CHCl + HCl

c.       CH3CH­2CH3 + Cl2 CH3 CH2CH2Cl+ HCl

2.       How many H atoms could be replaced in reaction (b)? In (c)?

3.       How can you account for the observation that chlorination of butane gives about 78% 2-chlorobutane to about 22% 1-chlorobutane?

 

 

As a generalization, it is commonly assumed that the activation energy of a reaction (which influences the rate) is related to the other energies in the reaction.  If a reaction is endothermic, the activation energy is always more than the (positive) ∆H, and the transition state will be similar to the products of the reaction. If it is very exothermic, the activation energy is (usually) fairly low, with the transition state looking more like the reactants. This is the “Hammond Postulate” about the nature of transition states. A result of this idea is that, for endothermic reactions, the relative energy of different products has a very significant effect on the rates.

 

Draw in the barbed arrows and calculate the energies for these steps in the reaction of CH3CH2CH3 with Br2. Which step controls the product distribution?

 

vs.


 

Radical Reactions as Cycles

Students are often confused about radical chain reactions. Remember that a chain reaction has hundreds or thousands of cycles between the initiation and termination steps.

The general cycle of a chain reaction has an initiation step, in which the radicals are generated, followed by a series of propagation steps, which happen hundreds or thousands of times. Very rarely, a termination step removes an intermediate and breaks the chain.

You can easily convert a set of propagation steps written in the usual straight-line format to a cycle to emphasize the repeating process of the chain reaction.

Here is a cycle illustrating the auto-oxidation of an alkene (Hornback section 21.8, page 935).

In Step A, a peroxide radical (top) abstracts an H from cyclohexene to form peroxycyclohexene (product) and the allylic cyclohexenyl radical (bottom). In Step B, the cyclohexenyl radical adds to an O2 molecule to form a new peroxide radical.

 

Write out the two propagation steps for the radical reaction CH4 + Cl2 CH3Cl + HCl.

Initiation: Cl-Cl   2 Cl·

1.       Cl· + CH­4

2.        

Now draw them as a cycle: