reduction of 9-fluorenone/aldol

• Catalytic hydrogenation

the addition of a molecule of dihydrogen across a carbon-carbon or carbon-heteroatom double or triple bond in the presence of a metal catalyst such as platinum, palladium, nickel, or rhodium that is often adsorbed on an inert solid support like carbon

reduce the p-bonds in a variety of functional groups

• Hydride reducing agent

(LiAlH₄ / NaBH₄)

metal hydrides may be thus used to reduce a carbon-oxygen or a carbon-nitrogen double bond in the presence of a carbon-carbon double bond, a reaction that is not easily accomplished by catalytic hydrogenation.

produce H:^-

broad range of selectivities toward various unsaturated functional groups

• Racemic mixture

if the product carbon is a stereocenter, attack from both faces gives a 50:50 mix of enantiomers.

• Asymmetric synthesis

Reactions resulting in preferential formation of one enantiomer

• Hydride reducing agent strength
  
  

  • metal electronegativity difference

  • Valence shell electron configuration

  • Solvent issues

  • Disadvantage and advantage of using NaBH₄

electronegativity difference

• why can use methanol (B vs Al difference)

valence shell electron configuration

• same group = same reactivity

solvent issues

• must use aprotic, dry solvent

NaBH₄

• less reactive and thus more selective. It reduces aldehydes and ketones but not esters

• safer and easier to handle

• Sodium borohydride may be used in alcoholic and even aqueous solutions, because it reacts much more rapidly with the carbonyl group than with the solvent

• General reaction for flourenone

• IR analysis (will show picture of product) for fluorenone reduction

OH stretch 3200-3500

1450-1600 C=C stretches

• Melting Point Analysis

9-fluorenone (starting) MP is 80-83

9-Fluorenol (product) MP is 152-158

IR analysis (picture of starting) for fluorenone reduction

ketone stretch around 1650-1750

aromatic c-h stretch 2850-3000

aromatic c=c 1450-1600 ish

mechanism of reduction for fluorenone

• α-Hydrogen/Proton

Hydrogens on the carbon next to the carbonyl (the α-carbon) unacidic. removing one turns the a carbon into a nucleophile

• α-Carbon

carbon next to the carbonyl

pka 18-20

the one that gets deprotonated by a base (like OH) to form an enolate

• Enolate

deprotonated form of enol (c=c-o-)

• Mixed aldol condensation

when two diff carbonyl compounds are the reactants

• Why are α-protons more acidic than an alkane proton

• Resonance Stabilization: Deprotonation forms an enolate, delocalizing the negative charge onto oxygen (vs. localized on carbon in alkanes).

• Inductive Effect: The electronegative carbonyl carbon pulls electron density away, weakening the alpha C-H bond

they can form enolates which makes the conjugate base more stable

• Mixed aldol condensations

  • What are the problems

  • How do we solve them

  • Why don't we need to worry about the problems with our reaction

problems

• multiple enolates

• multiple electrophiles

• equilibrium must be on products side

solve

• form enolate and electrophile separate

• use powerful base (LDA) to form enolate (equilibrium problem)

why don’t worry?

• we can mix together and use weak base because only acetophenone has alpha proton (solves multiple enolates) and acetophenone will react with p-anisaldehyde has less steric hindrance

• General reaction and mechanism

• IR analysis aldol (starting) acetophenone

• IR analysis aldol (product)

carbonyl peak (ketone) - 1650-1750

carbon-hydrogen stretches - weak (around 3000)

• MP analysis (aldol)

product mp: 73-76

• IR analysis aldol (starting) p-anisaldehyde