Lab Report #8

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University of Alabama *

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Chemistry

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Apr 3, 2024

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Peyton Yarbrough Lab Report #8 Ryan Clemens December 5, 2007 Aldol Condensation Using an Unknown Aldehyde and Unknown Ketone RESULTS 1. Structure of product and starting materials: Aldehyde: 135 Ketone: 322 2. Theoretical yield of aldol product: 0.5 mL C 8 H 8 O 2 x 0.95g/1mL x 1mol/84.12g x 202.27g/1mol = 1.14g 3. Isolated and percent yield of product: 1.04g 1.04g/1.14g = 91% 4. Melting range of product: 236-238°C 5. Attached 6. Attached DISCUSSION 1. When I mixed the unknown ketone and the unknown aldehyde with the ethanol and sodium hydroxide solution a precipitate formed immediately and was completely done with minimal mixing. The precipitate was more yellow than orange and very thick and spongy. After I had separated the precipitate by vacuum filtration I used toluene as my solvent. The mixture was still cloudy and not completely dissolved when I put it in the ice bath despite using an extended amount of solvent. After recrystallization I obtained a large amount of bright yellow precipitate and washed it with toluene.
2. I chose toluene as my solvent. I tried everything else and it dissolved the unknown compound the most upon heating. After using an excessive amount of toluene the mixture still remained a cloudy upon heating but after the ice bath I still retrieved a large amount of product. 3. My range was much larger than what was expected. I initially thought that I had a mixture of 4- methylbenzaldehyde and cyclopentanone, but I think I washed the final product with too much toluene and it altered the melting range drastically. 4. The IR spectra for my aldehyde had a very sharp peak at 1704 cm -1 indicating a carbonyl group, a strong peak at 640 cm -1 and 769 cm -1 indicating an aromatic ring, and a strong sharp peak at 1169 cm -1 indicating a ether C-O bond leading me to assume my aldehyde was 4-methoxybenzaldehyde. My IR spectra for my ketone had very strong peaks at 2966 cm -1 and 1747 cm -1 indicating sp 3 hybridized carbons and cyclopentanone. 5. My 1 H NMR spectra consisted of 5 peaks which was consistent with my product. It consisted of 2 doublets on the aromatic ring, and 3 singlets. The line of symmetry prevented more peaks. PROBLEMS 1. MECHANISM is ATTACHED 2. . 3. The IR spectra for the product would of had to have a peak for the aromatic rings that were attached to the initial ketone. The ketone IR spectrum does not contain peaks for aromatic rings because none are present.
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