CH 237 Lab 8 PDF

Title CH 237 Lab 8
Course Elem Organic Chem Lab
Institution University of Alabama
Pages 6
File Size 167.8 KB
File Type PDF
Total Downloads 97
Total Views 151

Summary

Download CH 237 Lab 8 PDF


Description

Robinson 1 Kaitlyn Robinson CH 237-102 TA: Dayne Fraser 30 July 2014 Experiment 8: Aldol Condensation of an Unknown Aldehyde and an Unknown Ketone INTRODUCTION: In this experiment, I performed a double aldol condensation reaction with an unknown aldehyde and an unknown ketone, and I analyzed the product in order to determine its identity. The purpose of this experiment was to utilize a variety of techniques in the laboratory for determining molecules. IR spectroscopy was used to determine which unknown was an aldehyde and which was a ketone. After performing the double aldol condensation reaction with one equivalent of the identified ketone and two equivalents of the identified aldehyde, 1H NMR was performed. The spectra were examined and used to identify the product of the condensation reaction. Melting point analysis was also performed on the product of the reaction. RESULTS: 1) XNA (cinnamaldehyde) and AHK (cyclohexanone) yielded a product of 2,6dicinnamylidene-cyclohexanone.

2) Cinnamaldehyde: (2mL)(1.05g/1.0mL)(1mol/132.16g)= 0.016mol Cyclohexanone: (0.5mL)(0.948g/1.0mL)(1mol/98.15 g)= 0.0048mol (limiting reagent)

Robinson 2 Theoretical Yield: (0.0048mol cyclohexanone)(326.43g 2,6-dicinnamylidenecyclohexanone/1mol 2,6-dicinnamylidene-cyclohexanone)= 1.57g 3) I isolated 1.27g of my 2,6-dicinnamylidene-cyclohexanone product. Percent yield: (1.27g actual yield/1.57g theoretical yield)*100%= 80.89% 4) The melting range of my recrystallized product was 168°C-176°C. DISCUSSION: When the unknown aldehyde and the unknown ketone were added together, a bight yellow viscous liquid was formed. As sodium hydroxide was added and the double aldol condensation reaction proceeded, the mixture turned cloudy and the bright yellow color changed into an orangeyellow color. The precipitate did not form until after heating the reaction mixture while stirring, at which point a frothy precipitate formed. After the vacuum filtration, the precipitate was composed of small grainy particles. The recrystallization formed miniature crystals, which were bright orange in color. To decide which solvent to use, a small amount of crude product was placed into 3 test tubes, and then ethanol, toluene, and a 9:1 mixture of ethanol and acetone were each added to separate test tubes. The toluene test tube dissolved the crude product immediately, so it was, therefore, too good of a solvent. The remaining two test tubes were then immersed into a hot water bath while stirring to attempt to dissolve the crude product. After a while, both test tubes dissolved the product. They were then placed in an ice water bath to induce crystallization. The product in both test tubes recrystallized, but the ethanol created the best crystals, so it was chosen as the reaction solvent. The melting range of the product was from 168°C to 176°C. This range is close to four different possible products. The product of p-tolualdehyde and acetone has a melting point of 175°C. The product of p-tolualdehyde and cyclohexanone has a melting point of 170°C. The

Robinson 3 product of cinnamaldehyde and cyclohexanone has a melting point of 180°C. The product of cinnamaldehyde and 4-methylcyclohexanone has a melting point between 163°C and 164°C. The other possible products had melting points that were either too high or too low to be considered. After analyzing the IR and NMR spectra, I determined that cinnamaldehyde and cyclohexanone were the starting materials, meaning that a pure product would have a melting point of 180°C. However, the measured melting range of my product was a little lower than that, indicating that there were still impurities in the recrystallized product. The unknown XNA was the aldehyde cinnamaldehyde and the unknown AHK was the ketone cyclohexanone. The IR spectrum of AHK has large peak at around 1750 cm-1, which indicates the C=O ketone bond on the cyclic hexane. The IR spectrum of the product of the aldol condensation reaction would be different from that of the starting ketone. In the IR spectrum of the ketone, there are peaks for C-H stretching between 3000 cm-1 and 2800 cm-1. In the IR spectrum of the product, there would instead be peaks representing alkene double bonds at a lower frequency, between 1680 cm-1 and 1600 cm-1. The IR spectrum of XNA has an aldehyde carbon-hydrogen peak at 2814 cm-1 and aromatic peaks located around 3000 cm-1. It also has two pi-bond peaks at 1627 cm-1 and 1577 cm-1, which would not be present in the IR spectra of any of the other possible aldehydes because cinnamaldehyde is the only possible reactant with two pi-bonds outside of the aromatic ring. In addition, my reaction mixture had a distinct smell of cinnamon, confirming my conclusion that cinnamaldehyde was my unknown aldehyde. The 1H NMR spectrum of my product indicates that my product is both symmetrical and highly conjugated. There are 22 hydrogens present in the structure divided into 7 signals on the 1H NMR spectrum. The product formed from cinnamaldehyde and cyclohexanone is the only mixture that produces a product with 22 hydrogens. The two pi-bonds on each side of the ketone of my product are unique to my structure, and the signals for these would not be seen on other spectrums.

Robinson 4 The H1 NMR spectrum is consistent only with this product because there are no alkane hydrogens present on the spectrum, meaning that my product did not contain any extra substituents on the benzene rings, and this rules out any product that would be produced from 4-methylbenzaldehyde or 4-methoxybenzaldehyde. Additionally, the hydrogens not connected to the aromatic ring, which would be the hydrogens from the starting ketone, are arranged in a cyclic fashion, ruling out acetone. Finally, the only combination that matches the spectrum according to the number of hydrogens would rule out any combination that includes benzaldehyde, cyclopentanone, or 4methylcyclohexanone, leaving cinnamaldehyde and cyclohexanone as the only possibilities for my reactants. There are 5 peaks in the aromatic region on the 1H NMR spectrum for my product, there is a triplet at around 7.5 ppm with an integration of 6 hydrogens, a triplet that occurs between 7.3 ppm and 7.4 ppm showing an integration of 4 hydrogens, a triplet at around 7.3 ppm with an integration of 2 hydrogens, a quadruplet occurring between 7.1 ppm and 7.2 ppm that shows an integration of 2 hydrogens, and a doublet between 6.9 ppm and 7 ppm with an integration of 2 hydrogens. The rest of the 1H NMR spectrum displays 2 peaks: a triplet at 2.8 ppm that shows an integration of 4 hydrogens and a pentet occurring at 1.5 ppm with an integration of 2 hydrogens. The peak near 0 is indicative of the solvent used and was not considered in the 1H NMR spectrum analysis of my product. CONCLUSION: Based on the IR analysis of the two starting materials of this reaction, I can conclude that the aldehyde used (XNA) was cinnamaldehyde, and the ketone used (AHK) was cyclohexanone. From the analysis of the 1H NMR spectrum as well as the melting range of my product, I determined that my product was 2,6-dicinnamylidene-cyclohexanone. This confirms my conclusion regarding the starting materials used, because they react to produce this product.

Robinson 5 QUESTIONS/PROBLEMS: The amount of toluene necessary to recrystallize 3.2 grams of compound A is 16 mL:

( 3.2 g compound A ) ×

(1.0 mL toluene ) =16 mL toluene ( 0.2 g compound A )

The maximum amount of compound A that could be recovered if the saturated solution was allowed to cool to 0°C is 2.4 grams:

(

)

( 3.2 g compound A ) − 16 mL toluene ×0.05 g =( 3.2 g compound A )−(0.8 g toluene ) mL ¿ 2.4 g compound A The amount of compound A that would be recovered if you accidentally used twice as much toluene as was necessary is 1.6 grams.

[

(

( 3.2 g compound A ) − (16 mL toluene × 2 )∗ 0.05

g solubility mL

(

¿(3.2 g compound A )− 32mL toluene ×0.05 g solubility mL ¿ ( 3.2 g compound A )− ( 1.6 g toluene)=1.6 g compound A

)

)]

Robinson 6 MECHANISM:

Hydroxide does act as a true catalyst in this reaction. For each equivalent of hydroxide that is consumed in this mechanism, there is a hydroxide molecule produced....


Similar Free PDFs