Iodoethane Lab Report

Water was added to the glue in a separate cup and another separate ratio of water was added to the sodium borate. The mixtures were mixed with glass stirring rods. In each trial, different mass ratios were used. Grams was the metric unit that was used in this experiment. After adding up the ratio of glue/water/sodium borate/water for one trial, forty grams should be the total mass. In trials

Based on the 1H NMR spectrum that was collected, a few things can be determined. Based on deshielding and electronegativity, the peak that occurs around 4.7ppm is associated with the O-ethylsaccharin product and the peak at 3.8 ppm is associated with the N-ethylsaccharin product. Based on the height ration, the N-ethylsaccharin product is the more prevalent result.

After 10 minutes the reaction liquid was separated from the solid using a vacuum filtration system and toluene. The product was stored and dried until week 2 of the experiment. The product was weighed to be 0.31 g. Percent yield was calculated to be 38.75%. IR spectra data was conducted for the two starting materials and of the product. Melting point determination was performed on the product and proton NMR spectrum was given. The IR spectrum revealed peaks at 1720 cm-1, which indicated the presence of a lactone group, and 1730 cm-1, representing a functional group of a carboxylic acid (C=O), and 3300cm-1, indicating the presence of an alcohol group (O-H). All three peaks correspond with the desired product. A second TLC using the same mobile and stationary phase as the first was performed and revealed Rf Values of 0.17 and 0.43for the product. The first value was unique to the product indicating that the Diels-Alder reaction was successful. The other Rf value of 0.43 matched that of maleic anhydride indicating some

The two remaining sp2 hybrid orbitals on oxygen are used to hold oxygen's lone pairs(bruice). O-ethylsaccharin is then less stable because the Ethyl group is attached to the Oxygen that used to be a carbonyl group, giving the Carbon an sp3 configuration (joined to two other carbons, the Oxygen with the Ethyl group and a Hydrogen). This put strain on the ring, and therefore is less stable.”(Richard y.a.). Upon mixing the reactants, sodium saccharin slightly dissolved producing a clear colorless liquid. When placed at 80°C hot bath, the solution completely dissolved and turned yellowish-green. Iodoethane is a clear and colorless liquid, but when exposed to light and air the Iodide dissociates from ethane and gives off a yellow color as a sign of decomposition. The solution was covered to prevent this from happening. But, as iodide dissociates from CH3CH2 that then gave off its yellowish color which shows SN2 reaction taking place. SN2 reaction happened fast The limiting Reagent is Iodoethane , as the alkylating agent; it was not used in excess and dictated how far the reaction went. The Na+ binds with I-(noted disappearance of the yellow color, as I- binds with Na+) then the ethane group bonds with either the Oxygen of saccharin or the Nitrogen of saccharin. The final product after vacuum filtration had some unreacted material, indicated by some yellow green solid formation.

We first, determined the percent recovery of both vanillin and unknown and then determined the melting range of both compounds. Determining the melting point range, verifies whether or not students purified the compounds properly. Pure compounds tend to have smaller and higher melting points than the original compound. Students worked individually when determining the melting range of vanillin—we were to determine the melting ranges of crude vanillin and the recrystallized vanillin made in the first lab. Moreover, students worked in pairs when determining the melting range of the unknown by using a commercial apparatus. For both solids, we watched the crystals melt. We recorded the temperature at which the solid first began melting and when the solid was completely

The wet, crude product was placed into the 50 mL Erlenmeyer flask. Small amounts of CaCl2 were added to dry the solution. The flask was sealed and the mixture was swirled and left to settle. Once

40 g of ice and approximately 30 ml of sulfuric acid is cautiously added to a 100 mL beaker respectively. Weigh 7.6 g of ammonium chloride and 14.0 g of ammonium bromide and place it in another beaker, crushing the lumps until a powdery mixture remains. The powdery mixture is then transferred to a 125 mL Erlenmeyer flask. Add the ammonium salts into the sulfuric acid mixture. Heat is applied to dissolve the salt. Once the

After the mixture finished refluxing, the flask was then cooled on ice. A sulfuric acid solution was then prepared by pouring 4.5 mL of concentrated H2SO4 over 50 grams of ice and then diluted to 75 mL by adding enough tap water to reach 75 mL. The sulfuric acid solution was then cooled on ice.

For this experiment, unknown sample 10 was given with the empirical formula of C4H10O. Several scans were conducted in order to discover the unknown alcohol sample. After obtaining the 1H NMR spectra of sample 10, 1H NMR signals and 13C NMR were used to determine the structure of the alcohol. The results of the 1H NMR displayed the total signals, chemical shift, multiplicity and the integration of sample 10. The chemical shift tells the electron environment near the 1H nucleus*. When near a electron withdrawing group 1H nucleus is deshielded and is shifted downfield towards a higher ppm which gives evidence to atom "A" being near an electronegative atom with an peak observed of 5.00. In addition, according to the 1H NMR chemical shift table a ppm of 2-5 represents a R-OH which coin side with atom "A" near an oxygen. Furthermore, Signal splitting otherwise known as multiplicity tells us the number of nuclei or hydrogen neighbors are on a particular molecule. Signals "B" the spectra illustrated

A mixture of thiamine hydrochloride (3.5 g, 0.01 moles), and deionized water (10 mL, 0.555 moles) was added into a 100 mL round bottom flask, and swirled until the thiamine hydrochloride had dissolved. Ethanol (30 mL, 0.521 moles) and 2 M sodium hydroxide (10 mL, 0.27 moles) were added to the flask, swirled in, and cooled in an ice bath for 4 minutes. Upon removal from the ice bath, the flask was gently heated by reflux for 30 minutes.

In order to determine the major product of the alkylation, an NMR test was run to figure

Analyzing the 13C NMR spectrum of the unknown sample #46, seven distinct signals were studied shown in table (2). Signal 1 had an observed peak value of 21.225 ppm and a calculated peak value11 of 20.8 ppm. Comparing with the literature ppm value5 range 5-30 ppm, signal 1 corresponds with CH3-C=O group. This signal was the second to the most upfield and shielded in the 13C NMR spectrum comparing with signal 7 because of the carbon high electron density. However, it was more downfield than signal 7 due to the presence of the oxygen in the group. Signal 2 had an observed peak value of 171.629 ppm and a calculated peak value10 of 171.0 ppm. Comparing with the literature value5 range 165-180 ppm, signal 2 corresponds with an ester group. Signal 2 is the most downfield and deshielded because the carbonyl group is in between two electronegative oxygens, therefore it could be concluded that there was an ester in the given compound.

This data would indicate that the ester was converted into the targeted alcohol. In order to confirm this assumption H-NMR was employed. Since both compounds contain a benzene ring and at least one methyl group, the peaks around the 7ppm and 1.5ppm regions would not be enough to distinguish between the two compounds (6, 7). Instead the signals at 5.9ppm and 4.8ppm were looked at to distinguish between the two compounds. Since esters contain multiple electronegative oxygens the proton attached to the carbon adjacent to the ester experiences a greater magnetic field and thus a more downfield signal at 5.9ppm (6, 13), whereas the single oxygen of the alcohol induces a slightly weaker upfield signal at 4.8ppm (7, 13). Additionally, for the NMR spectra with the signal at 4.8ppm, there was a single short peak at 2.7ppm that is a unique signature to the presence of an alcohol (13). Lastly TLC chromatography was used while monitoring the progress of the lipase reaction. The ester is slightly less polar than the alcohol because the ester lacks the ability to be a hydrogen bond donor making it more nonpolar by comparison. As a result while monitoring TLC the ester had traveled further up the polar stationary phase, silica plate, while the alcohol traveled slower the plate (11). This same logic was applied to explain why it took the alcohol longer to elute from column

The beaker was slowly heated on a hot plate with low stirring until most of the stilbene was dissolved. 0.4 g of pyridinium tribromide was measured and added to the beaker after 5 minutes of heating. Small amounts of ethanol were used to clean the sides of the beaker. The beaker was heated for an additional 10 minutes on low temperature. An ice bath was prepared. The beaker was removed from the hot plate and left to cool to room temperature. Once at room temperature, the beaker was placed in the ice bath for 15 minutes. The solid product was collected through vacuum filtration and the product was weighed and a melting point was taken. Waste was disposed of in the correct waste bins and lab bench was cleaned

Once the range is obtained, remove the thermometer and beaker. Pour out the hot water and allow both the thermometer and the beaker to cool. Once cool, repeat the previous steps with the second compound, tetradecanol. At the conclusion of the second experiment, repeat the steps for a third and final time using an equal mixture of both compounds. Be sure to record the melting point range for all three experiments.