To determine which product the deduction has produced we will use melting point, IR, and TTL.
Key Experimental Details and Observations: To perform this experiment 1. 005 g of Bennie was placed in a 125 ml Erlenmeyer flask. We then added approximately 10. 0 ml Of 95% ethanol, a stir bar, and placed the mixture on a hot plate until the Bennie was dissolved. The solution was allowed to cool to room temperature and it began to rationalizes. Once cooled, we placed the substance on a stir plate and added 0.
03 g of sodium boyfriend over approximately t-von. ro minutes. By the end of the two minutes the substance went from the yellow color of the Bennie to n opaque pale yellow; likewise, the substance also began to admit heat once the sodium boyfriend was added.
The substance was allowed to stir for two more minutes and then we let it cool to room temperature for about ten minutes so that the precipitate could dissolve. However, while cooling bubble began to form at the top of our substance. After cooling, we added approximately 10. ml of hot water (-?76. 5 ; C) and allowed the solution to boil on a hot plate; adding the hot water also caused the solution to become a clear pale yellow solution. 15. 0 ml of water was slowly added to the mixture, white foamy precipitate began to form almost immediately on top of the solution after adding the addition water.
We then removed the stir bar and allowed the solution to cool to room temperature before placing it in an ice bath for approximately five minutes. To collect the product we used vacuum filtration and a Boucher funnel.After fully drying the product under heat lamps it was a white shiny flaky powder with a total mass of 0. 816 g.
Our next objective was to determine what product had been formed from the Bennie reduction. To determine this we first performed an IR and a melting point analysis. For the first trial of the melting point analysis our range was 130. 0 – 134. 0 ;C while the second was 131. 0 – 135.
0 ;C. Likewise, we performed thin layer chromatography (TTL) to help determine which product was formed.The developing solvent was a 2:1 mixture of hexane and ethyl, approximately 5 ml of the solution was poured into the watching glass. We used ethyl acetate to dissolve racemes benzene standard (spot one), mess- hydrogenation standard (spot two), our product (spot three), and a co-spot (spot 4).
We then spotted our plate with the four samples. Rest Its: Table 1: Literature melting point (co) Us absence Literature M. P.
(00 Bennie 94. 0 – 95. 0 co Racemes benzene 135. 0 – 137. 0 co Racemes hydrogenation 122. 0 – 123. 0 co Mess – hydrogenation 137. – 139.
0 CE Table 2: Experimental melting points (co), unknown compound Melting point Unknown substance, Trial 1 130. 5 – 136. 0 CE Trial 2 132. 0 – 137. 0 co Average 131 . 25 – 136.
5 co Table 3: Weight and yield Of product Theoretical yield 1. 024 g Weight of product 0. 816 g Yield of product 79.
67% Table 4: TTL Discussion and Conclusion: The first method used to determine the identity of the product was melting point. The first trial the range was 130. -? 136. 0 co and for the second 132.
0 137. 0 ‘C; the average was 131. 25 – 136. ‘C.
Moreover, the expected value for benzene is 135. 0 – 137. 0 co, for racemes hydrogenation 122. 0- 123. 0 co, and 137. 0 – 139. 0 ICQ for mess-hydrogenation.
In comparison, the product obtained from the reduction is depressed with an increased range indicating that it was not completely pure. However, the average range (1 31. 25 – 136. 5 co) is larger than what would be expected if the product was a racemes mixture of hydrogenation. Likewise, the second method used to determine the identity of the product as an Infrared spectroscopy (IR).On the IR there is a peak at 3376. 42 that may have been cause by an O-H stretch and a peak at 3062.
74 most likely caused by a C-H stretch aromatic compound. There is also a peak at 2900. 24 caused by a C-H stretch and peaks at 1496. 50 / 1452.
05 probably caused by a stretch on an aromatic compound. These structures are present on both hydrogenation and benzene; however, the lack of a carbonyl group (C=O) eliminates benzene as a possible product. Our final method used to determine the identity of the product was thin layer chromatography (TTL).TTL is a useful tool in determining the polarity of substances. Compounds that travel farther on the silica gel plate are less polar. The first spot on the plate was benzene, it is less polar than mess- hydrogenation so it traveled further and had a higher Revalue, 0. 4, than the other spots on the plate.
The second spot was standard mess-hydrogenation, it is less polar than benzene so its RFC value was only 0. 2. The third spot, the unknown product, had an identical RFC value of standard mess-hydrogenation indicating that the unknown product has the same polarity as mess- hydrogenation.Likewise, because the RFC was identical the unknown product must be fairly pure. Our last spot was a co-spot of all three substances. It was a much larger spot than the others starting by spot two and three and ending around the same area as the benzene spot. Based on the results of all three tests, the unknown products identity may be mess-hydrogenation. The melting point eliminated racemes hydrogenation based on the range, but still left the possibility of benzene or mess- hydrogenation.
However, the IR eliminated benzene based on the absence of a carbonyl group.The last test, TTL, further confirmed that the unknown reduce was not benzene and was most likely mess-hydrogenation supported on the matching RFC value of 0. 2. Since the mess-hydrogenation product formed both of the carbonyl groups where reduced. After the first carbonyl reduction a choral center formed. The R moisteners created an S choral center and the S mentioned created the R choral center.
The theoretical yield for the unknown product was 1. 024 g, but the observed yield was 0. 816 g; therefore, the percent yield was 79. 67%. The loss of product may be a result of loss during vacuum filtration or product left on the filter paper.