aldehydes ketones hacc

• Important in the synthesis of many pharmaceuticals

– The basis upon which much of the remaining concepts in this course will build

– The carbonyl group is common to both ketones and aldehydes

20.1 Ketones and Aldehydes

Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e 20 -1

1. Identify and name the parent chain:

– Be sure that the parent chain includes the carbonyl carbon.

– For aldehydes, replace the e with an al.

– For ketones, replace the e with an one.

20.2 Nomenclature of Aldehydes


1. Identify and name the parent chain:

– Numbering the carbonyl group of the aldehyde takes priority over other groups.

20.2 Nomenclature of Aldehydes


Klein, Organic Chemistry 1e

Section: 20.2 What is the correct IUPAC name for the

following?

H

Cl O

a. 1-aldo-3-chlorohex-4-ene

b. 4-chlorohept-5-enal

c. 3-chlorohex-4-enal

d. 1-aldo-4-chlorohept-5-enal


What is the correct IUPAC name for the following?

O

Br

a. (E)-7-bromohept-5-en-4-one

b. (Z)-7-bromohept-5-en-4-one

c. (E)-1-bromohept-2-en-4-one

d. (Z)-1-bromohept-2-en-4-one

20.3 Preparing Aldehydes and Ketones


• What makes the carbonyl carbon a good electrophile?

1. RESONANCE: There is a minor but significant contributor that includes a formal 1+ charge on the carbonyl carbon.

– What would the resonance hybrid look like for this carbonyl?


What makes the carbonyl carbon a good electrophile?

2. INDUCTION: The carbonyl carbon is directly

attached to a very electronegative oxygen atom.

3. STERICS: How does an sp2 carbon compare to an sp3?


• Consider the factors: resonance, induction, and sterics.

• Which should be MORE REACTIVE as an electrophile, aldehydes or ketones? Explain WHY.


• We want to analyze how nucleophiles attack carbonyls and why some nucleophiles react and others don’t.

– Example attack:

• If the nucleophile is weak, or if the attacking nucleophile is a good leaving group, the reverse reaction will dominate.

– Reverse reaction:


• Show the nucleophilic attack for other nucleophiles.

– consider OH–, CN–, H–, R–, H2O.

– When the nucleophile attacks, is the resulting intermediate relatively stable or unstable? WHY?

– If a nucleophile is also a GOOD LEAVING GROUP, is it likely to react with a carbonyl? Explain WHY.

– Compare attack on a carbonyl with attack on an alkyl halide.


• If the nucleophile is strong AND NOT a good leaving group, then the full ADDITION will occur (Mechanism 20.1).

– The intermediate carries a negative charge, so it will pick up a proton to become more stable.


• If the nucleophile is weak and reluctant to attack the carbonyl, HOW could we improve the electrophile?

• We can make the carbonyl more electrophilic:

– Adding an acid will help. HOW?

– Consider the factors that make it electrophilic in the first place (resonance, induction, and sterics).


• With a weak nucleophile, the presence of an acid will make the carbonyl more attractive to the nucleophile so the full ADDITION can occur (Mechanism 20.2).

– Why is acid generally not used with strong nucleophiles?


• Is water generally a strong or weak nucleophile?

• Show a generic mechanism for water attacking an aldehyde or ketone.

– Will the overall process be fast or slow?

– Will the overall process be product or reactant favored?

– Would the presence of an acid improve the reaction?

20.5 Water as a Nucleophile


– How do the following factors affect the equilibria: entropy, induction, sterics?


Acetone Formaldehyde

Hexafluoroacetone

• To avoid the unstable intermediate with two formal charges, the reaction can be catalyzed by a base (Mechanism 20.3).

– How does the base increase the rate of reaction? Will it make the reaction more product-favored?


• The reaction can also be catalyzed by an acid (Mechanism 20.4).

– How does the acid increase the rate of reaction? Will it make the reaction more product-favored?


• An alcohol acts as the nucleophile instead of water.

– Notice that the reaction is under equilibrium and that it is acid catalyzed.

– Analyze the complete mechanism (Mechanism 20.5) on the next slide.

– Analyze how the acid allows the reaction to proceed through lower energy intermediates.

20.5 Acetals – Formation


• After the hemiacetal is protonated in Mechanism 20.5, the water leaving group leaves. Instead, you might imagine an INTERMOLECULAR collision that causes the water to leave.

– Why is the INTERMOLECULAR step unlikely?



• 5 and 6-membered cyclic acetals are generally product favored

• How do entropy, induction, sterics, and Le Châtelier’s principle

affect the equilibrium?

• Acetals can be attached and removed fairly easily.

– Both the forward and reverse reactions are acid catalyzed.

– How does the presence of water affect which side the equilibrium will favor?

20.5 Acetals – Equilibrium Control


• We can use an acetal to selectively protect an aldehyde or ketone from reacting in the presence of other electrophiles.

• Fill in necessary reagents or intermediates.

20.5 Acetals – Protecting Groups


• Fill in necessary reagents or intermediates.

20.5 Acetals – Protecting Groups


• Are amines stronger or weaker nucleophiles than water?

– If you want an amine to attack a carbonyl carbon, will a catalyst be necessary?

– Will an acid (H+) or a base (OH-) catalyst be most likely to work? WHY?

– What will the product most likely look like? Keep in mind that entropy disfavors processes in which two molecules combine to form one.

– Analyze the complete mechanism (Mechanism 20.6) on the next slide.

20.6 Primary Amine Nucleophiles


• The mechanism requires an acid catalyst. Note that the optimal pH to achieve a fast reaction is around 4 or 5.

– Why does the reaction slow down below pH 4?

– Why does the reaction slow down when the pH is greater than 5?


• A proton transfer alleviates the +1 charge in both mechanisms. The difference occurs in the LAST step.

– For 1° amines (Mechanism 20.6): the NITROGEN atom loses a proton directly.

– For 2° amines (Mechanism 20.7): a neighboring CARBON atom loses a proton.

20.6 Primary Amine Nucleophiles vs. Secondary Amine Nucleophiles



Section: 20.6

5. What is the product(s) of the following reaction?

A. B. C. D.

NH2

O

N NHN HN

[H+] -H2O


Section: 20.6

6. What is/are the product(s) of the following reaction?

[H+] -H2O

A. B. C. D.

NH

O

N NN N

• Reduction of a carbonyl to an alkane:

• Hydrazine attacks the carbonyl via Mechanism 20.6 to form the hydrazone, which is structurally similar to an imine.

• The second part of the mechanism is shown on the next slide (Mechanism 20.8).

20.7 Wolff-Kishner Reduction


• In general, carbanions are unstable and reluctant to form.


• What drives this reaction forward?

• Is OH- a catalyst in the mechanism?



Section: 20.6

7. What is the product(s) of the following reaction?

A. B. C. D.

O

NH2 OHN

1) NH2NH2, H+

2) KOH/H2O/heat

NH2

• Note the many similarities between the acid catalyzed mechanisms we have discussed:

– Carbonyl is protonated first: • Makes the carbonyl more electrophilic

• Avoids negative formal charge on the intermediate

– Avoid high energy intermediate with two formal charges

– Acid protonates leaving group so that it is stable and neutral upon leaving

– Last step of mechanism involves a proton transfer forming a neutral product

• Overall: under acidic conditions, reaction species should either be neutral or have a +1 formal charge.

20.8 Mechanism Strategies



8. List the steps in order for each of the following types of

mechanisms (steps may need to be repeated)?

I. Acetal formation

II. Imine formation

III. Enamine formation

A. Loss of Leaving group

B. Nucleophilic attack

C. Proton transfer

A. I = C, B, C, A, B, C; II = C, B, C, A, C; III = C, B, C, C, A, C

B. I = C, B, C, A, B, C; II = C, B, C, C, A, C; III = C, B, C, A, C

C. I = C, B, C, C, A, B, C; II = C, B, C, A, C; III = C, B, C, A, C

D. I = C, B, C, C, A, B, C; II = C, B, C, A, C; III = C, B, C, C, A, C

E. I = C, B, C, C, A, B, C; II = C, B, C, C, A, C; III = C, B, C, C, A, C

• Under acidic conditions, thiols react nearly the same as alcohols. Examples:

20.8 Mechanism Strategies – Sulfur Nucleophiles


• Conditions to convert a ketone into an alkane:

1. A thioacetal is formed via an acid catalyzed nucleophilic addition mechanism.

2. Raney Ni transfers H2 molecules to the thioacetal converting it into an alkane.

• Recall the Clemmenson (Section 19.6) reduction can also be used to promote this conversion.

20.8 Mechanism Strategies – Alternative to Wolff-Kishner


• To be a nucleophile, hydrogen must have a pair of electrons.

– H:1- is called hydride

• Reagents that produce hydride ions include LiAlH4 (LAH) and NaBH4. Hydrides will react readily with carbonyls.

20.9 Hydrogen Nucleophiles


• Identify the nucleophile.

– Will the reaction be more effective under acidic or under basic conditions? WHY?

– Show a complete mechanism (Mechanism 20.9).

– Analyze the reversibility (or irreversibility )of each step.

– Describe necessary experimental conditions. Why are there two steps in the reaction?


• Carbon doesn’t often act as a nucleophile. WHY? What ROLE does carbon most often play in mechanisms?

• To be a nucleophile, carbon must have a pair of electrons it can use to attract an electrophile:

1. A carbanion with a -1 charge and an available pair of electrons. However, carbanions are relatively unstable and reluctant to form.

2. A carbon attached to a very low electronegativity atom such as a Grignard.

20.10 Carbon Nucleophiles


• Identify the nucleophile.

– Will the reaction be more effective under acidic or under basic conditions? WHY?

– Show a complete mechanism (Mechanism 20.10). Three equivalents of the Grignard are necessary.

– Analyze the reversibility or irreversibility of each step.

– Describe necessary experimental conditions. Why are there two steps in the reaction?

20.10 Grignard Example



Section: 20.10

10. What is the product of the following reaction?

A. B. C. D.

O1) CH3CH2CH3MgBr

2) H2O

-OOH HO

• The cyanide ion can act as a nucleophile.

• Disadvantage: EXTREME toxicity and volatility of hydrogen cyanide.

20.10 Cyanohydrin


• Advantage: synthetic utility

20.10 Cyanohydrin



Section: 20.10


A. B. C. D.

O

1) KCN, HCl

2) LAH

3) H2O

HO CN HOHO

NH2 OH

OCN

• Like the Grignard and the cyanohydrin, the Wittig reaction can be very synthetically useful. What do these three reactions have in common?

– Similar to the Grignard, one carbon is a nucleophile and the other is an electrophile.

20.10 Wittig Reaction


• The ylide carries a formal negative charge on a carbon.

• In general, carbons are not good at stabilizing a negative charge.

– What factors allow the ylide to stabilize its formal negative charge?

– Why is the charged resonance contributor the major contributor?

20.10 Wittig Reaction – Wittig Reagent or Ylide


• The Wittig mechanism (Mechanism 20.12):

– Which of the steps in the reaction is mostly likely the slowest? WHY?

– The formation of the especially stable triphenylphoshine oxide drives the equilibrium forward.

20.10 Wittig Reaction


• To make an ylide, you start with an alkyl halide and triphenylphosphine.

• Example:

• The first step is a simple substitution. The second step is a proton transfer.

20.10 Wittig Reaction – Formation of an ylide


• Is the base used in the second step strong or weak? Why is such a base used?

20.10 Wittig Reaction – Formation of an Ylide


• Overall, the Wittig reaction allows two molecular segments to be connected through a C=C.

• Example:

– Note how the colored segments are connected.

20.10 Wittig Reaction – Overall


• Overall, the Wittig reaction allows two molecular segments to be connected through a C=C.

• Use a retrosynthetic analysis to determine a different set of reactants that could be used to make the target.

20.10 Wittig Reaction – Overall



Section: 20.10


A. B. C. D.

O(C6H5)3P-CH2

• An oxygen is inserted between a carbonyl carbon and neighboring group.

• Mechanism 20.13 shows the movement of electrons.

20.11 Baeyer-Villiger


• Which step in the equilibrium is most likely the slowest? WHY?

• Note the last step is not reversible. WHY?

20.11 Baeyer-Villiger


• If the carbonyl is asymmetrical, use the following chart to determine which group migrates most readily.

• Predict the product of the reaction, and give a complete mechanism.

20.11 Baeyer-Villiger Example



Section: 20.11


A. B. C. D.

O

OO

O O

MCPBA

O

OO O

• Recall the questions we ask to aid our analysis-

1. Is there a change in the carbon skeleton?

2. Is there a change in the functional group?

• Changes to the carbon skeleton:

C–C bond formation

• Name each reaction.

20.12 Synthetic Strategies


• Recall the questions we ask to aid our analysis

1. Is there a change in the carbon skeleton?

2. Is there a change in the functional group?

• Changes to the carbon skeleton: C–C bond cleavage

• Name the reaction.

20.12 Synthetic Strategies



Alternate Answer:

Section: 20.12

14. What is the correct order of reagents to achieve the following

synthesis.

HINT: If the reactant has two isolated pi bonds when treated with a Wittig reagent, a

mixture of products will form and the synthesis will fail (same problem if there are two

carbonyl groups when performing a reaction that involves a carbonyl group).

O

N

a. HOCH2CH2OH/H+

b. HSCH2CH2SH/H+

c. CH3NH2/H+

d. (CH3)2NH/H+

e. 1) LAH 2) H2O

f. 1) CH3MgBr 2) H2O

g. PCC, CH2Cl2

h. 1) KMnO4, NaOH, heat 2) H3O+

i. 1) HCN/KCN 2) LAH 3) H2O

j. CH3CH=PPh3

k. Dilute H2SO4

l. 1) Hg(OAc)2, H2O 2) NaBH4

m. MCPBA

n. TMSCl, Et3N

o. TBAF

A. l, g, c

B. l, g, d

C. l, n, j, o, c

D. j, l, g, c

E. j, l, g, d

• STRONG peak for the C=O stretch:

typical carbonyl typical conjugated carbonyl

– Aldehydes also give WEAK peaks around 2700–2800 cm-1 for the C–H stretch.

20.13 Spectroscopic Analysis – Infrared Spectroscopy


• Protons neighboring a carbonyl are weakly deshielded by the oxygen.

• Aldehyde protons are strongly deshielded, usually appearing around 9 or 10 ppm.

– Why is the aldehyde proton shifted so far downfield?

• In the 13C NMR, the carbonyl carbon generally appears around 200 ppm.

20.13 Spectroscopic Analysis – NMR Spectroscopy


• Predict 1H NMR shifts, splitting, and integration and 13C shifts for the following molecule.

20.13 Spectroscopic Analysis – NMR Spectroscopy


aldehydes ketones hacc

Documents