Aldol condensation - lab PDF

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Aldol Condensation Objective: To perform a double aldol condensation, isolate the dehydration product, recrystallize from a suitable solvent and analyze by IR and NMR spectroscopy. Introduction: In this lab we perform the aldol condensation. It builds new carbon carbon bonds with the help of reactivity of the carbonyl functionality; dipolar. Oxygen is more electronegative than carbon, hence there is polarization of the carbon-oxygen pi bond. When we look at the structure of the carbonyl compound, the δ+ charge on the carbonyl carbon affects the hydrogens near it. The alpha hydrogens become more acidic because of the carbonyl functionality. This is due to the δ+ charge and the conjugate base after deprotonation will be stabilized by resonance to the enolate anion. If we react a base with acetaldehyde, it will result in the acetaldehyde enolate which will attack the carbonyl carbon of another molecule of acetaldehyde. It forms the product aldehyde-alcohol derivative which is called aldol. Equation:

Safety considerations: Aldehydes and ketones are usually skin and eye irritants. Aqueous sodium hydroxide solutions are corrosive and cause burns. Rinse with water if it comes in contact with skin.

Procedure and observations: Procedure

Observations

Using a pre-set pump dispenser, measure 0.8mL of the aldehyde derivative in a preweighed 25mL Erlenmeyer flask. Weigh the flask. Add 0.2 mL of the ketone derivative in the Erlenmeyer, and weigh again. Add 4.0mL of 95% ethanol and 3mL of 2M sodium hydroxide solution. Stir the solution for 15minutes with a magnetic stirrer.

Longer if precipitate is still forming.

Heat the mixture in a hot water bath (~80°C) for 10–15 minutes.

Solution is cloudy - heat the reaction.

Cool the flask to room temperature. Cool the flask in ice water. Obtain two clean test tubes - 8 mL of 95% ethanol to the first tube, 4 mL of a prepared solution of 4% acetic acid in ethanol in the second. Chill the content of these two tubes in the ice bath. Collect the product using vacuum filtration. Rinse the product with half of the chilled ethanol, then with the acetic acid/ethanol solution, and finally with chilled ethanol. Recrystallize the product from a suitable solvent. Filter the crystals and allow them to dry on top of the filter. Selecting a recrystallization solvent: Place about 50 mg of the sample in a test tube.

Solvents available: toluene, 95% ethanol, acetone, and hexanes, deionized water.

Procedure

Observations

Add about 0.5mL of a solvent at room temperature, Stir, if most of the solid dissolves, then product is too soluble in this solvent. Repeat with a less polar solvent. If none/few of the crystals dissolve at room temperature, heat the test tube in a hot water bath. If the crystals don’t dissolve in refluxing solvent, repeat with a more polar solvent.

Don’t add more than 1.5 mL of solvent.

If the crystals dissolved, allow the solution to cool to room temperature and then place it in an ice bath.

Lots of crystals - good solvent.

Data collection/calculations: (Chosen aldehyde): para-tolualdehyde (Chosen Ketone): Cyclohexanone 0.174 g: Cyclohexanone 0.186 mg = 0.000186 g: product crude product before recrystallization:" 164 - 167°C after recrystallization:" 168 – 169°C 111 degrees celsius: product melting point Theoretical yield: 0.000186g of para-tolualdehyde x 1 mol/120.15 g = 0.00000154g 0.17g of cyclehexanone x 1mol/ 98.14 = 0.0017 mol of cyclehexanone.

IR spectrum: Crude Aldol

IR spectrum: Aldol Recrystallized

The table below shows the melting points of Double Aldol Condensation Product.

Discussion: In this lab experiment, we learnt double aldol condensation using para-tolualdehyde and Cyclohexanone. The dehydrated product was isolated and recrystallized using a suitable solvent and then we obtained the IR spectroscopy shown above. We collected 0.000186 g of product. The melting point obtained was 111 degrees celsius and after recrystallization was " 168 – 169°C which matches the literature melting point table given above. Our IR spectrum showed C=C peak at about 1590 cm-1 and a C=O peak at 1660 cm-1. In conclusion, our lab was successful. The aim was to perform an aldol condensation. The dehydrated product was then isolated, and recrystalised in a solvent. Error in this experiment may have occurred if there was insufficient amount of ruction time between the ketone and the aldehyde. There may be other side products. Other sources of error could be if we did not allow our product to cool in an ice bath for long. If not enough ethanol was used in washing the filtered product, there would have been other crystals left bheind. References: Department of Chemistry. (2015-2017). CHM 205/206 Lab Manual. Miami, Florida: University of Miami at Coral Gables. Padias, Anne B “Making the connections: A how to guide for organic chemistry lab techniques”. New York: Hayden McNeil, 2007.

Post lab questions: 1.

2. Yes, Yes Yes No, no reaction.

No, no reaction. Yes

3. Le chatelier’s principle states if one side equilibrium is altered then the equilibrium shifts to counteract the change which means it restores the equilibrium. If the concentrations of both the aldehyde and ketone increases, this shifts the equilibrium of the product throughout the experiment allowing more product to form. In an aldol condensation removing water shifts the reaction towards the right. 4. Ketone enclaves are good nucleophile, However in aldol reactions ketones are not very successful. They are less reactive than aldehydes due to steric and electronic effects, Using excess of a less hindered reagent will drive the reaction which allows for a reasonable yield of the product.

5.

6. If we mix acetone and butane in the presence of a base it leads to the formation of an enolate. The base removes the alpha hydrogen. The possible products are

7. A...