Chapter 10 Summary of π Bond Chemistry to π C=C 13.01 Chapter 11 at π C=C to π C=C Chapter 1 LG at π C=C Chapter 13 at α position C= activates position ow does C= activation α position? Via or. E base E ow acidic is the α position? N pk a K a / deprotonates N deprotonates / deprotonates N deprotonates
α alogenation 13.0 Summary: Br / Mechanism: Br Br nly works w/ C= that have significant amount of enol : & Fails w/,, & (too little presesnt) Work around : ell - Vollhardt - Zelinski eaction 1. PBr 3 / Br. Mechanism: Br P Br Br Br Br acid bromide enol Br
α Bromination 13.03 Usefulness of ell-vollhardt-zelinski: Amino-Acid Synthesis 1. PBr 3 / Br. N 3 " " 1. PBr 3 / Br. 3. N 3 Iodoform reaction : Base - catalyzed α iodination Br methyl Mechanism: more than original ketone CI 3 CI 3 Iodoform (CI 3 ) is a solid that precipitates out of solution Iodoform test is an old qualitative test for
Carbon Nucleophiles: Direct alklylation with enolates 13.04 nly amounts of enolate formed. pk a (acetone) = ; pk a (water) = LDA enolate formed pk a (acetone) = ; pk a (amine) = LDA = L D A N nbuli N LDA is a sterically amide nucleophile, but base always used at temperatures Examples: Cl has limitations of S N : does not work at carbon does not work at subst. carbon
Aldol eaction 13.05 - not terribly useful reaction - gives new bond fct. groups Mechanism: Use two C= molecules, one as electrophile one as nucleophile aldehyde The Aldol eaction has two possible outcomes based on reaction conditions 1. base conc. & temp. T. base conc. & temp. (80 C - 100 C) 100 C ( ) α, β - unsaturated ketone via 100 C
Cross-Aldol eaction - Using two different C= 13.06 ow to control cross-aldol! Too many E & N 1. Use one component with no hydrogens has no 's only enolate can form @ high temp. Example: dehydrate see p. 13.05 for details. Intramolecular with 1,4 or 1,5 diketone Entropy favors rxn rather than rxn. nly works for 5- & 6-membered rings.
Aldol - Problem Solving 13.07 etro-synthesis 1. Identify & carbons. Cut between & carbons 3. Insert & convert to α β β-hydroxy carbonyl α 1. Identify & carbons. Cut between & carbons 3. Insert & β α,β-unsaturated carbonyl Me
eformatsky eaction 13.08 Consider of β-hydroxy ester Me Problem: This is a Aldol Both have s two enolates Two solutions to this dilemma 1. eformatsky eaction 1. PPBr 3 / Br. Me / 3. Zn 4. Mechanism PBr 3 Br Me / Zn ZnBr Works just like "Latent" enolate Formed in %. Use strong base to get 100% enolate 1. LDA. 3. LDA
Michael eaction - Introduction 13.09 eactions of aldol products:? Consider the resonance structures of enone: δ δ δ esonance structure imply that there are sites for nucleophile to attack: Why does it attack? Depends on the nature of the nucleophile Nuc Nuc nucleophiles add to carbon; ( nucleophiles have " " charge) alkyl lithium 3 C C=C bond Grignard 3 C nucleophiles add to carbon; ( nucleophiles have " " charge) cuprates 3 C Cu C 3 C= bond enolates
Michael eaction 13.10 Michael eaction = enolate addition to enone ow to recognize Michael product? Me / Me Look for Mechanism: β - dicarbonyl Me Me Typical Michael Donors ( Nucleophiles ) β- keto ketone β- keto ester β- dinitrile β- di ester nitroalkane N C C N 3 C N pk a ~ 9 ~ 10 ~ 1 ~ 13 ~ 10 If the donor is not acidic (aldehyde, ketone, & ester), use LDA LDA LDA -78 C -78 C pk a = ~ 19 pk a = ~ 5 Typical Michael Acceptors( Electrophiles ) C N N N α,β - unsaturated aldehyde α,β - unsaturated ester α,β - unsaturated nitrile α,β - unsaturated ketone α,β - unsaturated nitro cmpd. α,β - unsaturated amide
obinson Annulation 13.11 Annulation means formation obinson Annulation = reaction reaction Me / Me Mechanism: Me Me Me Me Michael Problem Solving Me Me Me Me Me Et Et Et Et C 3 N Et Et
Claisen eaction 13.1 Aldol = enolate Claisen = enolate Et Et 1,3 - ester = β - ester Compare to Aldol: Et Et β - ketone. utcome different because no good Note that Claisen product is acidic than reactant Therefore, product ( ) base catalyst Et Et Et Et Et Et Et Et pk a Therefore amounts of base required Dieckmann eaction = Claisen Me Me Me Me 1 3 4 5 6 7 Cyclic
Cross Claisen 13.13 Cross Claisen = Trouble Et 4 products ( what are they? ) Et Choose one ester that has no 's These esters can only act as formate benzoate carbonate oxalate Claisen Problem Solving Et Et Et Me Me Me Me Me Me Me Et Et Et Et Et etro-synthesis - Find carbons α and β to ester Break C α C β Insert " " & " " β α β α X Y Me Me Me X Y Et Et Et Et X Y Et Et Et X Me Me Me
Alkylation 13.14 Alkylation requires deprotonation For " " acidic α-carbons, use LDA Br Why use LDA? b/c would undergo to avoid Aldol make sure deprotonation, i.e. LDA For " " acidic α-carbons, use Et Cl Et Me Br Me ydrolysis & Decarboxylation β-keto ester 3 3 Δ Δ ketone β-di-ester carb. acid Mechanism = Saponification Decarboxylation Tautomerization β-keto ester 3 P.T. Δ C Likewise β-di-ester saponify 3 decarboxylate C tautomerize
More β - C= Ester ydrolysis & Decarboxylation 13.15 A complex example : Dieckmann Saponification Decarboxylation 1. Me / Me. 3 / Δ Me / Me Δ 3 Acetoacetic Ester Synthesis = Alkylation Saponification Decarboxylation 1. Et / Et. X acetoacetic ester 3. 3 / Δ Et Et Br 3 Δ Malonic Ester Synthesis = Alkylation Saponification Decarboxylation 1. Et / Et dimethyl malonate. X 3. 3 / Δ Et Et Br 3 Δ
686 kcal/mol otosynthesis 6 C 6 C 6 1 6 6 Glycolysis ATP ADP P 3 13.17 espiration 686 kcal/mol P 3 ADP ATP P P 3 3 3 P etro- Aldol 3 P P 3 P 4 3 P P 3 NAD NAD 3 P P 3 ADP ATP P 3 P 3 NAD NAD oxaloacetic acid Vitamin B 1 NAD NAD S S ATP ADP P 3 hydroxysuccinic acid citric acid P 3 fumaric acid aconitic acid FAD FAD succinic acid isocitric acid NAD Vitamin B 1 NAD C -oxoglutaric acid C NAD NAD