Lab Report 8

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Feb 20, 2024

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Barberito 1 Max Barberito 12174194 CH-237-109 April 17 2023 Aldol Condensation Using an Unknown Aldehyde and Ketone Introduction The Aldol Dye Co. makes dyes using the double aldol condensation of a ketone with two equivalents of an aldehyde. The products of these reactions, conjugated dienones, usually have a yellow to orange color depending on the structures of the starting material. A recent shipment of aldehydes and ketones was just delivered, but while the materials were being shipped, the labels on the containers wore off and became unreadable. To determine the identity of the unknowns, first, scientists have been asked to take IR spectra of the starting materials to figure out which are the ketones and which are the aldehydes. Second, the aldehydes and ketones undergo an aldol condensation. The products are then characterized using H-NMR. The possible aldehydes are benzaldehyde, 4-methylbenzaldehyde (p-tolualdehyde), 4-methoxybenzaldehyde (p- anisaldehyde), and E-3-phenyl-2-propenal (cinnamaldehyde). The possible ketones are acetone, cyclopentanone, cyclohexanone, and 4-methylcyclopentanone.
Barberito 2 Results Starting materials Product
Barberito 3 Theoretical Yield .5 mL cyclopentanone x .951 g/mL cyclopentanone / 84.12 g/mol cyclopentanone x 1 cyclopentanone/ 1 (2E,5E)-2,5-dibenzylidenecyclopentanone = .0057 moles (2E,5E)-2,5- dibenzylidenecyclopentanone OR 1.47 g (2E,5E)-2,5-dibenzylidenecyclopentanone Actual Yield 1 g (2E,5E)-2,5-dibenzylidenecyclopentanone Percent yield 1g / 1.47 g * 100% = 68% Melting Range 163°C – 168°C Discussion To determine the identities of the unknowns, the experimenters conducted an aldol condensation reaction. First, 2 milliliters of an unknown aldehyde and .5 milliliters of an unknown were added to an Erlenmeyer flask. Second, 10 milliliters of ethanol and 7.5 milliliters of sodium hydroxide were added. This mixture was then stirred for around 15 minutes. A precipitate formed. The precipitate that formed was a yellow to orange color. This precipitate was then vacuum filtered off. This product was then cleaned with ethanol and acetic acid. Two small portions of the crude product were then dissolved in ethanol and toluene to determine the best solvent. The best solvent for the recrystallization process was toluene. When the small portion of the crude product was added to the ethanol, it dissolved very quickly, meaning it had a high solubility. Having a high solubility in the solvent used in a recrystallization process is
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Barberito 4 detrimental to the experiment because it will result in a lower yield. When the small portion of the crude product was added to the toluene, it didn’t dissolve nearly as quickly. Some of the powder had not dissolved after the initial addition to the solvent. All the powder dissolved in the toluene after a small additional portion was added and the mixture was heated. The crude product was then all dissolved in toluene. This mixture was then placed in an ice bath, and the recrystallized product formed. The melting point of the recrystallized product received from the reaction was between 163°C and 168°C. The possible product of this reaction that has a melting point similar to that is the product between cinnamaldehyde and 4-methyl cyclohexanone. The tested melting point of the product we received is 189°C. Possible reasons that our product didn’t match the tested values could be contamination errors with the recrystallization process. The IR spectra are attached below. Based on the spectra, the vial labeled VRA was the aldehyde, and the vial labeled EQK was the ketone. Both contain a peak around 1710 - 1720 cm -1 . This signal appears for a carbonyl bond, which both aldehydes and ketones possess. However, on the spectra labeled VRA, there is a medium strength signal around the 2700 – 2800 cm -1 , which represents the sp 2 C-H bond characteristic of aldehydes. The IR spectra of the aldol product and the starting ketone would have been different because the aldol product would have a C=C signal around 1625 cm -1 , but the ketone wouldn’t. Also, the C-H bending signal on the aldol products IR spectra would shift right to a lower value than the ketone. The NMR of the product determined that the structure was (2E,5E)-2,5- dibenzylidenecyclopentanone. The characteristic traits of (2E,5E)-2,5-
Barberito 5 dibenzylidenecyclopentanone in the NMR spectrum are the bunched up aromatic signals, with an outlying signal around 3 ppm. The signal at 3 ppm has an integration of 4, matching the 4 hydrogens that are on the beta and gamma carbons in cyclopentanone. The hydrogens on the benzene rings of (2E,5E)-2,5-dibenzylidenecyclopentanone are symmetrical in each of the ortho, para, and meta positions. The NMR of (2E,5E)-2,5-dibenzylidenecyclopentanone is also missing the signals of a methyl or ether substitution on the benzene rings, eliminating some products. Mechanism
Barberito 6 The hydroxide is used in a 1 equivalent ratio for each aldol reaction. In the double aldol reaction, there would need to be 2 equivalents to bring the reaction to completion.
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Barberito 7 Questions/Problems Compound A has a solubility of 0.25 g/mL in boiling toluene, and a solubility of 0.05 g/mL at 0ºC. a. How much toluene would be necessary to recrystallize 4.0 g of compound A? a. 4.0 g / 0.25 g/mL = 16mL toluene b. What would be the maximum amount of compound A that could be recovered if the saturated solution thus obtained was allowed to cool to 0ºC? a. .05g/mL * 16 mL = .8 g, so .8g would be dissolved. 4g-.8g = 3.2g would be recovered c. How much of compound A would be recovered if, instead, you accidentally used twice as much toluene as was necessary? a. .05g/mL * 32mL = 1.6g, so 1.6g would be dissolved. 4g-1.6g=2.4g would be recovered