Aldol Condensation + Michael Addition

OCHEM II Lab Final

What is the first step?

• aldol condensation of acetone and p-anisaldehyde

Condensation reaction

• combines two molecules while removing a small molecule such as water or an alcohol

  • First part:

    • removal of a proton from the a-carbon of one carbonyl compound molecules creates a nucleophile

    • nucleophile attacks the electrophilic carbon of a second molecule

  • Second part: dehydration

    • heat aldol with acid or base

    • can occur w/o heating under conditions

      • dianisalacetone (product) is very stable and relatively easy to produce because the double bonds formed by dehydration are conjugated with the carbonyl group and with the benzene rings

Second Step: Michael Addition

• base catalyzed addition of a compound containing an active a-hydrogen to the b-carbon of an a,b - unsaturated carbonyl compound

• dianisalacetone with dimethylmalonate under catalysis by methoxide ion

  • dimethyl-2,6-bis(p-methoxyphenyl)-4-oxo-cyclohexane-1,1-dicarboxylate

• Ethylene ketal formation may also be used to temporarily block a carbonyl group while a reaction is conducted at another site on a molecule

• after second site reaction is complete, the ethylene ketal-blocking group is removed by reaction with acid and water

Reactions

Precautions for ethyl ether?

• keep away from flames

• gloves + lab coat

Precautions for 25% sodium methoxide?

• keep away from flames

• wear gloves

Theoretical yield

Overall Theoretical Yield

Compare the NMR spectra for each stage of the synthesis sequence. Explain how the spectra support the identity of your product in each case.

• The NMR spectra for the aldol condensation sequence shows major peaks around 6.9-7 ppm which represent the aromatic protons that are two sets of doublets

• Peaks around 3.8 ppm show the methoxy protons, and vinyl protons are shown around 6-7 ppm.

• The NMR spectra for the Michael product shows major peaks that support the identity of the product.

  • The double peaks around 7.0 ppm represent the hydrogens right near the carbonyl in the top of the structure

  • The singlet peak at 3.755 ppm represents the hydrogens on the methoxy group

  • The singlet at 3.380 ppm showcases the hydrogens on the methyl ester groups. 

What are the steps for this experiment?

1. p-anisaldehyde + acetone + ethanol in flaks A

2. 10% NaOH in flask B

3. pour A into B and stir for 45 minutes

4. cool mixture in ice bath for 5 min and collect precipitate via vacuum filtration

5. rinse crystals with water and filter

6. wash crystals on funnel with acetic acid, stir gently, and filter

7. water rinse again and filter for 2 minutes

8. recrystallize the product with ethanol and dry for 10 minutes and record weight ( aldol product : dianisalacetone)

9. add aldol product + methanol + dimethyl malonate + 25% sodium methoxide to rbf with boiling chip

10. gently reflux for 15 min

11. cool flask to room temp + ice bath for 5 min

12. vacuum filter + wash with cold methanol + dry for 10 minutes , mp weight

Why do we add NaOH and stir the reaction for 45 minutes?

• acts as base to deprotonate acetone to create nucleophile

• allows enough time for double condensation to occur

• acetic acid wash neutralizes any residual NaOH left on the crystals, preventing the base from interfering with the product during drying

Why do we use sodium methoxide for the Michael reaction?

• deprotonate dimethyl malonate to create Michael donor

• 15 min reflux to do intramolecular claisen like condensation

• methanol prevents transesterification side reactions that would happen if used a different alcohol