chem 51lc ch 21

haloform reaction because the ketone has a COCH₃ group.
The CH₃ next to the carbonyl gets iodinated 3 times, then the CI₃ group leaves as HCI₃, and the carbonyl becomes a carboxylate.

For an unsymmetrical ketone, how do I choose kinetic vs thermodynamic enolate?

Unsymmetrical ketones have two different alpha sides, so two enolates can form.

Kinetic enolate:
Remove H from the less hindered alpha carbon.
Forms faster.
Reagents: LDA, -78°C, bulky strong base

Thermodynamic enolate:
Remove H to give the more substituted alkene/enolate.
More stable.
Reagents: small base/protic solvent/heat, like NaOEt/EtOH or acid/base reversible conditions.

Memory:
LDA, -78 cold = less substituted enolate.
Heat/reversible = more substituted enolate.

acid halogenation at the alpha carbon

NOT NAOH (base) because Acid stops at one X. Base can keep going.

• carbonyl forms enol, then enol attacks X2

• product has one halogen on the alpha carbon

  

• excess X2 and NaOH

• X2 can be Cl2 Br2 or I2

• haloform reaction

• works when the starting material is a methyl ketone

• base removes H, enolate forms, enolate attacks X2, repeat until all 3 alpha H are replaced

• final products are carboxylate and HCX3

excess X2 and NaOH

• X2 can be Cl2 Br2 or I2

• halogenation in base

• deprotnation with base forms enolate

• then enolate attacks X2

• repeats

• replaces alpha hydrogens with halogens

• product is a dihalide

• 

  
  Li2CO3 LiBr DMF
  
  
  
  Find where Br was added → put the double bond between that α-carbon and the next β-carbon that has H.

• reaction type: elimination from an alpha halo ketone

• The reagents remove H and Br from the α and β carbons, respectively, to form a new π bond

• uses Li2​CO3​ as a base in the presence of LiBr and the polar aprotic solvent DMF to facilitate the formation of the enone

• forms a double bond next to the carbonyl

• product is an alpha beta unsaturated ketone

• 

• α-substitution of an α-halo ketone using a neutral thiol (HSCH3​) as the nucleophile

• Br is the leaving group

• SCH3 replaces Br on the alpha carbon

• SN2: inversion of configuration. the Bromine starts on a wedge, and the resulting Sulfur group is on a dash.

• carbonyl alpha alkylation

• starting carbonyl must have alpha H

• base removes alpha H to make enolate

• enolate attacks alkyl halide by SN2

• product has new carbon carbon bond at alpha carbon

• best with methyl or primary alkyl halides

• malonic ester synthesis

• malonic ester reacts at the middle alpha carbon

• base removes alpha H to make enolate

• enolate attacks alkyl halide by SN2

• acid heat removes one carbonyl as CO2

• final product is substituted carboxylic acid

• double malonic ester alkylation

• same middle alpha carbon gets alkylated twice

• repeat enolate formation then SN2 alkylation

• acid heat hydrolyzes and decarboxylates

• final product is carboxylic acid with two added carbon groups

• use when final product is a substituted carboxylic acid

• acetoacetic ester synthesis

• acetoacetic ester reacts at the carbon between two carbonyls

• base removes alpha H to make enolate

• enolate attacks alkyl halide by SN2

• acid heat removes ester side as CO2

• final product is alkylated methyl ketone

• double acetoacetic ester alkylation

• same alpha carbon gets alkylated twice

• repeat enolate formation then SN2 alkylation

• acid heat hydrolyzes and decarboxylates

• final product is methyl ketone with two added carbon groups

• use when final product is a substituted ketone, not carboxylic acid

show Decarboxylation mechanism

excess X₂ / OH⁻ with ketones, how do I know product?

Look at the α-carbon next to the C=O.

• If it is a methyl ketone: R–CO–CH₃
  → it gets halogenated 3 times to R–CO–CX₃
  → then cleaves into RCOO⁻ + CHX₃
  → haloform reactionNo cleavage unless it can become CX3

• If it is not a methyl ketone: R–CO–CH₂R
  → it only replaces the available α-H’s with halogens
  → gives α,α-dihalo ketone if there are 2 α-H’s
  → no cleavage because it cannot form CX₃

what does t-BuOK / t-BuOH do in enolate

t-BuOK / t-BuOH = thermodynamic enolate conditions

• more substituted C=C/enolate = more stable

• usually removes the α-H that leads to the more substituted enolate

what does H2O and pyridine do

Pyridine is a weak base.

When placed in water, it extracts a proton H+ from the h2o molecule
pyridine helps remove extra H⁺ (by N lone pair) so the reaction does not become too acidic

reagent socl2:

convert carboxylic acids, acids into acid chlorides and alcohols into alkyl chlorides
use SOCl₂ when they want to turn a bad leaving group OH into a better leaving group Cl.

LDA

elimination with base , substit with nucleophile

br2, ch3coh

li2co3: removes a beta H next to the carbonyl.

libr: helps promote the elimination; it is not adding Br.

dmf: dissolves the reagents and helps the elimination happen

ester enolate alkylation

nitrile alkylation

How do I choose the alkyl halide for acetoacetic ester synthesis?

Acetoacetic ester synthesis makes a methyl ketone:

CH3COCH2R

Break the bond between the alpha carbon and the R group.
The R group comes from the alkyl halide.

Reagents:

1. NaOEt/EtOH

2. R-X

3. H3O+, heat

Memory:
Acetoacetic ester gives CH3COCH2R. R comes from the alkyl halide.