22 ChapterL Carbonyl Condensation Reactions

1) NaOH, H2O (PART 1 OF ALDOL RXN)

(Bases = NaOH NaOEt and NaOMe are also acceptable)

2 Aldehyde → Beta Hydroxy Aldehyde

2 Ketone → Beta Hydroxy Ketone

Aldol Reaction (Aldehyde + Alcohol)

1b) ∆ (PART 2 of ALDOL RXN, if there is HEAT)

(Bases = NaOH NaOEt and NaOMe are also acceptable)

2 Aldehyde → Beta Hydroxy Aldehyde → Alpha Beta Unsaturated Aldehyde

2 Ketone → Beta Hydroxy Ketone → Alpha Beta Unsaturated Ketone

Aldol Condensation

E1cB = Elimination Unimolecular Conjugate Base

E1 = Enolate Intermediate lone pair makes a C double bond C AND OH leaves as L.G. E1 because eliminated comes from UNIMOLECULAR ENOLATE

cB = When NaOH or other base DEPROTONATES the Aldol Addition to form an ENOLATE which is the CONJUGATE BASE of the Aldol

1c) Same Reagents 1a + 1b COMBINED

Cross Aldol Reaction

4 Different Products

1) A + B

2) B + A

3) A + A

4) B + B

1d)
LDA

Direct Aldol Reactions

1) A + B ONLYYY

Reacting 1 Aldehyde with with LDA (VERY STRONG BASE) to get to 100% conversion of aldehyde to an ENOLATE

Add 2nd Aldehyde to Enolate

1) Control Over which Enlate Forms

2) Can Stop at Aldol Product (No Condensation)

Aldol Condensation only possible when…

1e) Reagents:

1a) NaOH, H2O (PART 1 OF ALDOL RXN)

(Bases = NaOH NaOEt and NaOMe are also acceptable)

1b) ∆ (PART 2 of ALDOL RXN, if there is HEAT)

Intermolecular Aldol Reaction

2) Reagent: Claisen Reaction

1) NaOEt, HOEt

2) H3O+

2 SAME Esters → Beta-Keto Ester

Enolate → Adding 2nd Ester → Kick Out L.G → Neutral → Deprotoanted → Back to Neutral

2b) Reagent: Cross Claisen Reaction

1) NaOEt, HOEt

2) H3O+

2 DIFFERENT Esters → Beta-Keto Ester (1 Ester has NO alpha Hydrogens)

-1 Ester No Alpha Hydrogens → ELECTROPHILE

-1 Ester Has Alpha Hydrogens → NUCLEOPHILE

1 Ketone and 1 Ester

-Ketone → NUCLEOPHILE

-Ester → NO Alpha Hydrogens → ELECTROPHILE

Ethyl Chloroformate or Diethyl Carbonates as ELECTROPHILES

Cross Claisen Reaction

-Make up to 4 products

-Best by converting 1 ester to enolate using LDA and then add 2nd ester

OR

-If One ester has 0 alpha hydrogens, it can’t be an enolate and it can only be attacked

3) Dieckmann Condensation

1) NaOEt, HOEt

2) H3O+ (mild acid)

Intromolecular + Claisen Condensation Reaction

Enolate → Intramolecular Claisen → L.G. → Neutral → Deprotonated → Neutral

4) Michael Reaction

1) NaOET, HOEt

2) H2O

Alpha/Beta Unssaturated Carbonyl (Electrophile) + Michael Donor (Nucleophile)

→ 1,5 Di-Carbonyl

Enolate → Michael Donor Attacks the Alpha Beta Unsaturated Carbonyl → Form Alkoxide and ALkene → Reform Carbonyl and have Alkene Attack the H from H2O → 1,5 Dicarbonyl

5) Robinson Annulation
1) NaOEt, HOEt

2) H2O

Michael Reaction + Intramolecular Aldol Reaction

Part 1: Michael Reaction

Enolate → Addition attacking Alkene and Forming Alkoxide → Water forms back Carbonyl and Breaks Alkene (1,5 Dicarbonyl)

Part 2: Intramoleclar Aldol Reaction

1,5 Dicarbonyl → Closest H near both Carbonyls Deprotonated → Attacks Carbon of Carbonyl O → H2O forms back the OH → E1cB to kick out OH and form Alkene