What’s That Smell?
Due Date/Submission Date: 02/17/2014 I. Purpose
To synthesize a fragrance used commercially in two steps by using a chemoselective reduction reaction, and a polymer-supported reaction.
II. Reaction Equations and Mechanisms
III. Calculations
Part 1
Theoretical Yield
1) (1.62 g ethyl vanillin x 154.16 g ethyl vanillyl alcohol) / (166.18 g ethyl vanillin) = 1.503 g ethyl vanillyl alcohol 2) (0.3 g NaBH4 x 154.16 g ethyl vanillyl alcohol) / (37.83 g NaBH4) = 1.223 g ethyl vanillyl alcohol
Percent Yield = (1.03 g / 1.223 g) x 100 = 84.2%
Melting Point Range ( ̊C): 83.9, 84.4, 85.0
Melting Point Average ( ̊C): 84.43
Rf Values
Starting material = 4.0/5.5 = 0.73
Final product =
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While we refluxed the solution there was no color change, the solution stayed clear. After gravity filtration the solution stayed clear. After roto vap, solution was a milky brown color.
IR spec data showed a weak OH peak as well as strong C—O stretches around 1200 cm-1.
V. Discussion Questions: 1. Our percent yield for alcohol was 84.2% which is average. We rushed through our vacuum filtration and probably did not let the solid dry long enough and might have not transferred all of the solid to the vacuum filtration from the beaker. 2. The TLC results showed two different spots that traveled different distances on the TLC plate, one for ethyl vanillin and one for ethyl vanillin alcohol which proves there is no evidence of the starting material in our final product. The product (ethyl vanillin alcohol) was more polar and interacted more with the solvent than our starting material (ethyl vanillin). This increase in polarity is due to the extra alcohol group in the ethyl vanillin alcohol and the smaller Rf value also indicates it is more polar and pure than the ethyl vanillin. 3. The IR spectrum of the starting material shows a medium/strong C-O bond at around 1500cm-1, also the starting material shows a strong C-H bond at around 3000cm-1 and another medium C-H bond at 2865cm-1 indicating an aldehyde group whereas the product does not. The IR spectrum of the product shows a two weak broad O-H peaks at around
d.) The volume of Alcohol with the mass of 20 grams would be an estimated amount of 23.0 milliliters.
9. The accepted value for the density of water is 1 g/mL and the accepted density for isopropyl alcohol is 0.786 g/mL. Determine the percent error between your calculated densities and the accepted values for both water and isopropyl alcohol. Record the percent error in Data Table 4.
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.
reaction. The purpose was to learn what a SN1 reaction was with combining 2,5-dimethyl-2,5-hexanediol and HCl and to learn how to use an infrared spectrum for analyzing the reaction. The infrared spectrum determined the alcohol content
Many techniques and skills were developed in this lab. Among them were dehydration, isolation, drying, and distillation. We used all of these techniques to get the product we were looking for. In addition to these experimental techniques we also verified our product via spectroscopy which is a new technique. Using IR spectroscopy we were able to
Possible reasoning for a lower yield could be loss of product during separation process, particularly leaving some product behind in Erlenmeyer flask. Another possible explanation could be the wash of product with not enough cold water, which increases the solubility of the product, thus lowering the yield. Also the product was lost during purification process, recrystallization. Solid could be dissolved below the boiling point of the solution, thus required more solvent, resulting in a lower
=0.313g O_2 9.103×〖10 〗^(-3) mol O_2×(32g O_2)/(1 mol O_2 )= 0.291g O_2 percent yield=(0.291g O_2)/(0.313g O_2 ) ×100%=93.0% Run 2: 0.9g KClO_3×(1mol KClO_3)/(122.5 g KClO_3 )×(3 mol KClO_3)/(2 mol O_2 ) ×(32g O_2)/(1 mol O_2 )
The plate is sprayed with 5 ml of 1% ethanolic vanillin solution, followed by 3 ml concentrated net, then evaluated in vis. Colours are intensified by heating for 5 min at 100°C. Vanillin-phosphoric acid reagent (VP) (a) Dissolve I g vanillin in 100 ml of 50% phosphoric acid.(b) Two parts 24% phosphoric acid and eight parts 2% ethanolic vanillic acid. After spraying with either (a) or (b), the plate is heated for 10 min at 100°C, and evaluated in vis.
mass of acetic acid=mass of vingear in bottle×.07 Calculated the moles of CH3COOH in soda bottle in mol.mass of 〖CH〗_3 COOH×(1 mole)/(60.04g)
The product attained was a white, dry solid. The small amount of product lost during the second recrystallization was most likely do to impurities, which were filtered away with the methanol. Impurities that contributed to the low percent yield could be due to side reactions such as methyl o-nitrobenzoate and methyl p-nitrobenzoate. Although the percent yield attained was low, the product attained was fairly pure due to similarity in melting point and IR spectrum compared to standardly accepted values for methyl m-nitrobenzoate.
It is not uncommon to have a percent yield less than 100% as was observed in our experiment. Imperfect percent yield can be interpreted that the condition were not optimal and could be improved. Despite careful measurements some component will still be lost during transferring between containers. One way to get a better percent yield is by minimizing transfers. A low percent yield could be a result of evaporation
4. Calculate the number of moles of NaOH used in reactions one and two. Show work here and record your answer in Data Table 2.
To characterize the synthesized product using its boiling point, results of simple chemical tests and derivatization reactions, along with the determination of the melting points of the hydrazones and comparison of the hydrazones using their RGB values.
1. A 40-mol-% aqueous ethanol solution is to be distilled into two streams: a distillate with 98% ethanol and a side stream with 60%. 98% of the alcohol in the feed is to be recovered in these two products; only 2% goes into the bottoms. If the bottoms product contains 2% ethanol, calculate the amount of the different streams per 1000 kmol of feed stream.
The fragrance market is the major part of the care market. During the 90’s, this market has known a considerable growth. Currently, it might be entering a maturity phase. The perfume market is highly competitive and there are a lot of fragrance houses which are competing for sales. The total global market is over 25 billion Dollars.