1. Carbenes, :CH2 Preparation of simple carbenes 1. carbene 2.
Mechanism of the a elimination.

2. Reactions of Carbenes, :CH2 (not for synthesis)
Addition to double bond.
liquid
Insertion into C-H bond
Formation of ylide (later)

3. Simmons Smith Reaction (for synthesis, addition to alkenes to yield cyclopropanes)
CH2I Zn(Cu)  ICH2ZnI
Carbenoid, properties similar to carbenes.

4. Template for Reactions
Why stereospecific, why from same side as OH group?
Interaction with metal holds the carbenoid on the top side.

5. Electronic Structure
Electrons paired, singlet

6. Triplet and Singlet Methylene
Dominant form
in solution
Gas phase
Rotation can occur around this bond.

7. Aldehydes and Ketones
Chapter 16 7

8. Structure
Aldehydes
Ketone
Carbonyl group
sp2
2-pentanone
pentanal 8

9. Examples of Naming 9

10. Resonance
result 10

11. Extension of resonance11

12. Boiling points Water Solubility
For compounds of comparable molecular weight…
Alkanes, ethers < aldehydes, ketones < alcohols < carboxylic acids
Dipole-dipole
Hydrogen Bonding
Dispersion Forces
Water Solubility
Ketones and Aldehydes, like ethers, can function as hydrogen bond acceptors and smaller compounds have significant water solubility. 12

13. Recall Preparation from Alcohols
Can also be done using KMnO4 in base with heat or bleach in acid solution (HOCl).
Be sure you can balance this kind of reaction.
Use PCC to limit oxidation of primary alcohol to the aldehyde. Secondary are oxidized to ketone.
Primary alcohol
RCH2OH RCH=O
PCC
Secondary
PCC
R2CHOH R2C=O 13

14. Preparations, con’d Reaction of acid chloride and Gilman
But where do we get these?? 14

15. Note that we have two possible disconnects available15

16. Example: Prepare 2-butanone from ethyl alcohol
Requirement to start with ethanol suggests a disconnect into two carbon fragments.
Done! 16

17. Aldehydes from carboxylic acids
Reduction
And from alcohols, as before:
Oxidation 17

18. A Common Sequence
Observe these parts at this moment. 18

19. Reactions Addition of a nucleophile: Nucleophilic Addition
Good nucleophile, usually basic +
Attack of nucleophile occurs on both sides of carbonyl group.
Produces both configurations. +
Overall: H – Nu was added to carbonyl group double bond.
Notice that the CO bond order was reduced from 2 to 1. The addition reduced the bond order. We will use this idea later. 19

20. Reaction can also be done in acid environment.
Nucleophiles not expected to be as strong (why?) but the oxygen may become protonated making the carbonyl a better electrophile (why?).
Very electronegative, protonated oxygen. Pulls the pi electrons into itself strongly.
Problem: If there is too much acid present the nucleophile may become protonated, deactivating it 20

21. Addition of Grignard (Trumpets Please)
Recall the formation of a Grignard and its addition to an oxirane
Carbonyls may be added to in same way…
If a new chiral center is created both configurations will be produced. 21

22. Common Reactions of Grignards
ROHRXRH(D)
Both of these reactions extend carbon chain & keep -OH functionality at end of chain. Can extend further.
Examine reaction with ester further.
ROH + R’CH2OHRX + R’CO2HR2C(OH)R’
ROH + R’R”CHOH
RX + R’R”CO RR’C (OH)R”
ROHRXRCO2H
ROH + R’CH2OHRX + R’CHORCH(OH)R’ 22

23. Grignard Reacting with an Ester. Look for two kinds of reactions.
Substitution
Any alcohol will do here.
But where does an ester come from?
Acid chloride
Perhaps this carboxylic acid comes from the oxidation of a primary alcohol or reaction of a Grignard with CO2.
Addition 23

24. New bond. Disconnect site.
Synthetic Planning… Use of epoxides and carbonyls offer different disconnect sites.
Pattern
HO-C-C-R
epoxide
Nucleophile
New bond. Disconnect site.
New bond. Disconnect site
Pattern
HO-C-R
carbonyl
Want this to be the nucleophile (Grignard). 24

25. Patterns to recognize: carbonyl vs oxirane
We can create the following fragments of target molecules by using an organometallic (carbon nucleophile)
The difference is the extra CH2 when using an oxirane. 25

26. Synthetic Planning… Three different disconnects possible
Give synthetic routes to
If none of the Rs are H then these three synthetic routes may be available. 26

27. Example: Synthesize from ethanol
Done
Preliminary Analysis
Hmmm, even number of carbons, at least that is good; ethanol is a two carbon molecule.
Now the problem is to divide it up into smaller fragments.
Ether linkage is easily constructed. Williamson.
Two butyl groups attached to the central 2 carbon fragment. Grignard + ester. 27

28. Bisulfite Addition Addition product.
Practical importance: liquid carbonyl compounds can be difficult to purify. The bisulfite addition products will be crystalline and may be recrystallized. 28

29. Addition of Organolithium Compounds to Carbonyls
Generally the reactions are the same as for Grignards but the lithium compounds are more reactive (and more difficult to handle).
bromocyclohexane
Decreased reactivity of electrophile due to steric hindrance to attack. So we used the alkyl lithium instead of a Grignard. 29

30. Nucleophiles derived from terminal alkynes
Can do all the reactions of an alkyne and an alcohol. But remember that we have two acidic groups: the more acidic OH and the less acidic terminal alkyne. We discussed this problem earlier.
For example, once formed, the new alkynyl alcohol can be hydrated in two ways, Markovnikov and anti Markovnikov.
Note that the regioselectivity used here is only effective if this alkyne is terminal. Otherwise get a mixture.
Carefully observe the structure of the products, the relationship of the OH and the carbonyl. 30

31. Addition of hydrogen cyanide
basic
Think of what the mechanism should be….
pH issue. Slightly basic media so that HCN has partially ionized to cyanide ion, the actual nucleophile.
Followed by protonation of the alkoxide ion (perhaps by unionized HCN). 31

32. Follow-up reactions on the cyanohydrins…
We saw this hydrogenation before. 32

33. hydrolysis of the nitrile group to a carboxylic acid.
Let’s see what we can do with the mechanism of the
hydrolysis of the nitrile group to a carboxylic acid.
Overall
The action is at the nitrile group, CN --> CO2H.
But how does a nitrile group behave? What could be happening?
We are breaking the CN bond; bond order goes from 3 to 0. Probably stepwise.
Chemically speaking: the nitrogen of the nitrile is basic (lone pair) and can be protonated. This makes it a better electrophile (cf. carbonyl). Multiple bond can undergo addition (cf. carbonyl) reducing bond order.
Goal: Break the C to N bonding and create C-O bonds.
Considerations: neither the electrophile (RCN) nor the nucleophile (water) is very reactive. Since we are in acid protonate the CN group to make it a better electrophile. Then attack it with the water nucleophile to add water. This results in reduction of C-N bond order and creation of C to O bonds . 33

34. Again, we are in acid environment. Let’s protonate something….
Protonate the multiple bonded N atom to make better electrophile and attack with the nucleophile, water.
What have done so far? Reduced the CN bond order from 3 to 2 and added one O to the C. Moving in the right direction!
Want to reduce the CN bond order to zero and introduce more O on the C. Keep going! To induce the water to attack again (adds another O) need to increase the reactivity of the electrophile. Protonate again!! On the O. 34

35. Initial equilibrium with acid
Now want to get rid of the NH2. We have all the O’s we need.
We know what we have to do. Have to get the N protonated to make it a good leaving group.
Done. 35

36. Wittig Reaction or combine them the other way… Example, synthesize
Substitution Elimination
Example, synthesize
or combine them the other way… 36

37. Wittig Reaction Mechanism
Acidic hydrogen
Nucleophilic substitution
Nucleophilic center
Phosphonium ylide
betaine 37

38. Friedel Crafts Acylation
And then all the reactions of ketones… 38

39. Formation of Hydrates, carbonyls and water.
Carbonyl side of equilibrium is usually favored. 39

40. Hemiacetals and Acetals, carbonyls and alcohols
Addition reaction.
(Unstable in Acid; Unstable in base)
(Unstable in Acid; Stable in base)
Substitution reaction 40

41. Formation of Hemiacetals, catalyzed by either acid or base
Formation of Hemiacetals, catalyzed by either acid or base. Let’s do it in Base first.
But first let’s take stock.
We have an addition reaction.
Just mixing a carbonyl and an alcohol do not cause a reaction.
One of them must be made a better reactant.
Carbonyl can be made into a better electrophile by protonating in acid.
Alcohol can become a better nucleophile in base by ionization.
Use Base to set-up good necleophile.
Poor nucleophile
Good nucleophile
An addition of the alcohol to the carbonyl has taken place. Same mechanism as discussed earlier.
hemiacetal 41

42. Alternatively, hemiacetal formation in Acid
Protonation of carbonyl (making the oxygen more electronegative)
Attack of the (poor) nucleophile on (good) electrophile.
Deprotonation
Overall, we have added the alcohol to the carbonyl. 42

43. Hemiacetal to Acetal, Acid Only
Protonate the hemiacetal, setting up leaving group.
Departure of leaving group.
Attack of nucleophile
Substitution reaction, cf SN1.
Deprotonation 43

44. Equilibria
Generally, the hemiacetals and acetals are only a minor component of an equilibrium mixture. In order to favor formation of acetals the carbonyl compound and alcohol is reacted with acid in the absence of water. Dry HCl) The acetals or hemiacetals maybe converted back to the carbonyl compound by treatment with water and acid.
An exception is when a cyclic hemiacetal can be formed (5 or 6 membered rings). 44

45. Hemiacetal of D-Glucose
The alcohol
The carbonyl
Try following the stereochemistry here for yourself
The hemiacetal can form with two different configurations at the carbon of the carbonyl group. The carbon is called the anomeric carbon and the two configurations are called the two anomers. The two anomers are interconverted via the open chain form. 45

46. Stabilities of the Anomers…
Here note the alternating up-down relationships.
More stable b form, with the OH of the anomeric carbon is equatorial
Less stable a form.
Here see the cis relationship of these two OH groups, one must be axial. 46

47. Acetals as Protecting Groups
Synthetic Problem, do a retrosynthetic analysis
Target molecule N
Form this bond by reacting a nucleophile with an electrophile. Choose Nucleophile and Electrophile centers.
Grignard would react with this carbonyl.
The nucleophile could take the form of an organolithium or a Grignard reagent. The electrophile would be a carbonyl.
Br-Mg
Do you see the problem with the approach?? 47

48. Use Protecting Group for the carbonyl… Acetals are stable (unreactive) in neutral and basic solutions.
Create acetal as protecting group.
protect
Now create Grignard and then react Grignard with the aldehyde to create desired bond.
react
Remove protecting group.
deprotect
Same overall steps as when we used silyl ethers: protect, react, deprotect. 48

49. Tetrahydropyranyl ethers (acetals) as protecting groups for alcohols.
Recall that the key step in forming the acetal was creating the carbocation as shown…
There are other ways to create carbocations……
Recall that we can create carbocations in several ways:
1. As shown above by a group leaving.
2. By addition of H+ to a C=C double bond as shown next.
This resonance stabilized carbocation then reacts with an alcohol molecule to yield the acetal.
An acid
This cation can now react with an alcohol to yield an acetal. The alcohol becomes part of an acetal and is protected. 49

50. Provide a mechanism for the following conversion
Sample Problem
Provide a mechanism for the following conversion
First examination:
have acid present and will probably protonate
Forming an acetal. Keep those mechanistic steps in mind.
Ok, what to protonate? Several oxygens and the double bond. Protonation of an alcohol can set-up a better leaving group. Protonation of a carbonyl can create a better electrophile.
We do not have a carbonyl but can get a similar species as before. 50

51. Strongly electrophilic center, now can do addition to the C=O
The protonation of the C=C
Now do addition, join the molecules
Product
Now must open 5 membered ring here. Need to set-up leaving group. 51

52. Followed by new ring closure.
Leaving group leaves….
Followed by new ring closure.
Done. Wow! 52

53. Sulfur Analogs
Consider formation of acetal
Sulfur Analog 53

54. The aldehyde hydrogen has been made acidic
Why acidic?
Sulfur, like phosphorus, has 3d orbitals capable of accepting electrons: violating octet rule. 54 54

55. Recall early steps from the Wittig reaction discussed earlier
This hydrogen is acidic.
Why acidic? The P is positive and can accept charge from the negative carbon into the 3d orbitals 55

56. Some Synthetic Applications56

57. Umpolung – reversed polarity
What we have done in these synthetic schemes is to reverse the polarity of the carbonyl group; change it from an electrophile into a nucleophile.
electrophile
nucleophilic
Can you think of two other examples of Umpolung we have seen? 57

58. Nitrogen Nucleophiles58

59. Mechanism of Schiff Base formation
Attack of nucleophile on the carbonyl
Followed by transfer of proton from weak acid to strong base.
Protonation of –OH to establish leaving group.
Leaving group departs, double bond forms. 59

60. Hydrazine derivatives60

61. Note which nitrogen is nucleophilic
Nucleophilic nitrogen
Favored by resonance
Less steric hinderance 61

62. Reductive Amination
Pattern:
R2C=O + H2N-R’   R2CH-NH-R’ 62

63. Enamines
Recall primary amines react with carbonyl compounds to give Schiff bases (imines), RN=CR2.
Primary amine
But secondary amines react to give enamines
See if you can write the mechanism for the reaction.
Secondary Amine 63

64. Acidity of a Hydrogens a hydrogens are weakly acidic
Weaker acid than alcohols but stronger than terminal alkynes.
Learn this table…. 64

65. Keto-Enol Tautomerism
(Note: we saw tautomerism before in the hydration of alkynes.)
Fundamental process
Mechanism in base:
Negative carbon, a carbanion, basic, nucleophilic carbon.
Additional resonance form, stabilizing anion, reducing basicity and nucleophilicity.
Protonation to yield enol form. 65

66. Details… Base strength
Alkoxides will not cause appreciable ionization of simple carbonyl compounds to enolate.
Strong bases (KH or NaNH2) will cause complete ionization to enolate.
Double activation (1,3 dicarbonyl compounds) will be much more acidic.
For some 1,3 dicarbonyl compounds the enol form may be more stable than the keto form. 66

67. More details…
Nucleophilic carbon
nucleophilicity
Some examples: 67

68. Some reactions related to acidity of a hydrogens
Racemization
Exchange 68

69. Oxidation: Aldehyde  Carboxylic
Recall from the discussion of alcohols.
Milder oxidizing reagents can also be used
Tollens Reagent test for aldehydes 69

70. “Drastic Oxidation” of Ketones
Obtain four different products in this case. 70

71. Reductions: two electron
NaBH4
Or LiAlH4 71

72. Reductions: Four Electron
Clemmenson
Wolf-Kishner 72

73. Mechanism of Wolf-Kishner, C=O  CH2
Recall reaction of primary amine and carbonyl to give Schiff base. Here is the formation of the Schiff base. We expect this to happen.
These hydrogens are weakly acidic, just as the hydrogens a to a carbonyl are acidic.
Weakly acidic hydrogen removed. Resonance occurs. Same as keto/enol tautomerism.
Protonation (like forming the enol)
Perform an elimination reaction to form N2. 73

74. Haloform Reaction, overall
The last step which produces the haloform, HCX3 only occurs if there is an a methyl group, a methyl directly attached to the carbonyl.
a methyl
If done with iodine then the formation of iodoform, HCI3, a bright yellow precipitate, is a test for an a methyl group (iodoform test). 74

75. Steps of Haloform Reaction
The first reaction:
All three H’s replaced by
This must happen
stepwise, like this:
Pause for a sec: We have had three mechanistic discussions of how elemental halogen, X2, reacts with a hydrocarbon to yield a new C-X bond. Do you recall them?
Radical Reaction: R. + X-X  R-X X. (initiation required)
Addition to double bond: C=C + X-X  Br- (alkene acts as nucleophile, ions)
Nucleophilic enolate anion: 75

76. Mechanism of Haloform Reaction-1
Using the last of the three possibilities
One H has been replaced by halogen.
Repeat twice again to yield
Where are we? The halogens have been introduced. First reaction completed.
But now we need a substitution reaction. We have to replace the CBr3 group with OH. 76

77. Mechanism of Haloform - 2
This is a substitution step; OH- replaces the CX3 and then ionizes to become the carboxylate anion.
Here’s how:
Attack of hydroxide nucleophile. Formation of tetrahedral intermediate. Anticipate the attack…
Reform the carbonyl double bond. CX3- is ejected. The halogens stabilize the negative carbon.
Neutralization. 77

78. Cannizaro Reaction Overall:
Restriction: no a hydrogens in the aldehydes.
a hydrogens
No a hydrogens
Why the restriction? The a hydrogens are acidic leading to ionization. 78

79. Mechanism
What can happen? Reactants are the aldehyde and concentrated hydroxide.
Hydroxide ion can act both as
Base, but remember we have no acidic hydrogens (no a hydrogens).
Nucleophile, attacking carbonyl group.
Attack of nucleophilic HO-
Acid-base
Re-establish C=O and eject H- which is immediately received by second RCHO 79

80. Experimental Evidence
These are the hydrogens introduced by the reaction. They originate in the aldeyde and do not come from the aqueous hydroxide solution. 80

81. Kinetic vs Thermodynamic Contol of a Reaction
Examine Addition of HBr to 1,3 butadiene 81

82. But which is the dominant product?
Mechanism of reaction.
Allylic resonance
But which is the dominant product? 82

83. Nature of the product mixture depends on the temperature.
Product mixture at -80 deg % %
Product mixture at + 40 deg % %
Goal of discussion: how can temperature control the product mixture? 83

84. When two or more products may be formed in a reaction A  X or A  B
Thermodynamic Control: Most stable product dominates
Kinetic Control: Product formed fastest dominates
Thermodynamic control assumes the establishing of equilibrium conditions and the most stable product dominates.
Kinetic Control assumes that equilibrium is not established. Once product is made it no longer changes.
Equilibrium is more rapidly established at high temperature. Thermodynamic control should prevail at high temperature where equilibrium is established.
Kinetic Control may prevail at low temperature where reverse reactions are very slow. 84

85. Nature of the product mixture depends on the temperature.
Product mixture at -80 deg % %
Product mixture at + 40 deg % %
More stable product
Thermodynamic Control
Kinetic Control
Product formed most quickly, lowest Ea 85

86. Formation of the allylic carbocation.
Can react to yield 1,2 product or 1,4 product. 86

87. Most of the carbocation reacts to give the 1,2 product because of the smaller Ea leading to the 1,2 product. This is true at all temperatures.
At low temperatures the reverse reactions do not occur and the product mixture is determined by the rates of forward reactions. No equilibrium. 87

88. Most of the carbocation reacts to give the 1,2 product because of the smaller Ea leading to the 1,2 product. This is true at all temperatures.
At higher temperatures the reverse reactions occur leading from the 1,2 or 1,4 product to the carbocation. Note that the 1,2 product is more easily converted back to the carbocation than is the 1,4. Now the 1,4 product is dominant. 88

89. Diels Alder Reaction/Symmetry Controlled Reactions
Quick Review of formation of chemical bond.
Electron donor
Electron acceptor
Note the overlap of the hybrid (donor) and the s orbital which allows bond formation.
For this arrangement there is no overlap. No donation of electrons; no bond formation.

90. Diels Alder Reaction of butadiene and ethylene to yield cyclohexene.
We will analyze in terms of the pi electrons of the two systems interacting. The pi electrons from the highest occupied pi orbital of one molecule will donate into an lowest energy pi empty of the other. Works in both directions: A donates into B, B donates into A.
B HOMO donates into A LUMO
Note the overlap leading to bond formation
LUMO
acceptor
LUMO
acceptor
A HOMO donates into B LUMO
HOMO
donor
HOMO
donor
Note the overlap leading to bond formation B A

91. Try it in another reaction: ethylene + ethylene  cyclobutane
LUMO
LUMO
Equal bonding and antibonding interaction, no overlap, no bond formation, no reaction
HOMO
HOMO

92. Reaction Problem 92

93. Synthesis problem 93

94. Mechanism Problem
Give the mechanism for the following reaction. Show all
important resonance structures. Use curved arrow notation. 94