In the lab we added bromine (Br2) to trans cinnamic acid, which formed the product 2,3-dibromo-3-phenylpropanoic acid. This product could be in either Erythro form or Threo form. We know what form was produced based on the melting point. A high melting point, between 200 and 202 °C, or 115-160 °C for an impure form, means it is an erythro type. If the melting point of the product was between 80-91 °C it was a threo form compound. There are 2 places or centers on the molecule where the bromine could bond, which makes 4 stereoisomers possible in the reaction. The product could attack either anti or syn, and could attack to either the top or bottom side. Anti means the bromines would attack to the opposite sides, so they aren’t in the way of each other. Syn means on the same side, so both bromine’s …show more content…
It prefers to add bottom anti so that when the Br adds from the backside it doesn’t interfere with the other bonded Br. The bromines are like big beach balls and don’t want to be near each other because they would be fighting for space. This is the steric argument. The first part of the bottom side anti addition of bromine to trans cinnamic acid is when the bromine adds to the left side, to the COOH side of the trans cinnamic acid molecule. Figure 1: Initial anti attack approach of bromine to the bottom side of the trans-cinnamic acid: The Br adds to the left side with the COOH so that when the carbocation forms, it goes to the left side with the phen where it can be neutralized by phen’s 6 pi electrons. There is a double bond between the 2 carbons in the trans cinnamic acid. Pi electrons from this double bond move down towards the Br and start to form a bond with the bromine. This is a bridgehead approach, so a bridgehead is initially formed between the Br and the two central carbons from the trans-cinnamic
The percent yield of products that was calculated for this reaction was about 81.2%, fairly less pure than the previous product but still decently pure. A carbon NMR and H NMR were produced and used to identify the inequivalent carbons and hydrogens of the product. There were 9 constitutionally inequivalent carbons and potentially 4,5, or 6 constitutionally inequivalent hydrogens. On the H NMR there are 5 peaks, but at a closer inspection of the product, it seems there is only 4 constitutionally inequivalent hydrogens because of the symmetry held by the product and of this H’s. However, expansion of the peaks around the aromatic region on the NMR show 3 peaks, which was suppose to be only 2 peaks. In between the peaks is a peak from the solvent, xylene, that was used, which may account to for this discrepancy in the NMR. Furthermore, the product may have not been fully dissolved or was contaminated, leading to distortion (a splitting) of the peaks. The 2 peaks further down the spectrum were distinguished from two H’s, HF and HE, based off of shielding affects. The HF was closer to the O, so it experienced more of an up field shift than HE. On the C NMR, there are 9 constitutionally inequivalent carbons. A CNMR Peak Position for Typical Functional Group table was consulted to assign the carbons to their corresponding peaks. The carbonyl carbon, C1, is the farthest up field, while the carbons on the benzene ring are in the 120-140 ppm region. The sp3 hybridized carbon, C2 and C3, are the lowest on the spectrum. This reaction verifies the statement, ”Measurements have shown that while naphthalene and benzene both are considered especially stable due to their aromaticity, benzene is significantly more stable than naphthalene.” As seen in the reaction, the benzene ring is left untouched and only the naphthalene is involved in the reaction with maleic
The goal of this two week lab was to examine the stereochemistry of the oxidation-reduction interconversion of 4-tert-butylcyclohexanol and 4-tert-butylcyclohexanone. The purpose of first week was to explore the oxidation of an alcohol to a ketone and see how the reduction of the ketone will affect the stereoselectivity. The purpose of first week is to oxidize the alcohol, 4-tert-butylcyclohexanol, to ketone just so that it can be reduced back into the alcohol to see how OH will react. The purpose of second week was to reduce 4-tert-butylcyclohexanol from first week and determine the effect of the product's diastereoselectivity by performing reduction procedures using sodium borohydride The chemicals for this lab are sodium hypochlorite, 4-tert-butylcyclohexanone
2-ethyl-1,3-hexanediol. The molecular weight of this compound is 146.2g/mol. It is converted into 2-ethyl-1-hydroxyhexan-3-one. This compounds molecular weight is 144.2g/mol. This gives a theoretical yield of .63 grams. My actual yield was .42 grams. Therefore, my percent yield was 67%. This was one of my highest yields yet. I felt that this was a good yield because part of this experiment is an equilibrium reaction. Hypochlorite must be used in excess to push the reaction to the right. Also, there were better ways to do this experiment where higher yields could have been produced. For example PCC could have been used. However, because of its toxic properties, its use is restricted. The purpose of this experiment was to determine which of the 3 compounds was formed from the starting material. The third compound was the oxidation of both alcohols. This could not have been my product because of the results of my IR. I had a broad large absorption is the range of 3200 to 3500 wavenumbers. This indicates the presence of an alcohol. If my compound had been fully oxidized then there would be no such alcohol present. Also, because of my IR, I know that my compound was one of the other 2 compounds because of the strong sharp absorption at 1705 wavenumbers. This indicates the presence of a carbonyl. Also, my 2,4-DNP test was positive. Therefore I had to prove which of the two compounds my final product was. The first was the oxidation of the primary alcohol, forming an aldehyde and a secondary alcohol. This could not have been my product because the Tollen’s test. My test was negative indicating no such aldehyde. Also, the textbook states that aldehydes show 2 characteristic absorption’s in the range of 2720-2820 wavenumbers. No such absorption’s were present in my sample. Therefore my final product was the oxidation of the secondary alcohol. My final product had a primary alcohol and a secondary ketone
The sequence of BRCA1 protein shows that there are quite a few of cysteine residues. These cysteine residues form disulfide bonds which help stabilize the secondary structure of the protein. The secondary structure of BRCA1 protein indicates that there are alpha helices and beta turns which are connected by loops and turns. The 3D structure of BRCA1/BARD1 RING-domain heterodimer is shown in figure 1.
... which is catalyzed by b-ketoacyl CoA thiolase. The products are acetyl-CoA and a long chain fatty acyl CoA that is 2 carbons shorter than the original fatty acyl CoA. One complete round of β-oxidation cleaves 2 carbons from the fatty acid chain, and the process continues until the entire fatty acid chain is broken down into acetyl propinoyl CoA. For example, an 18 carbon chain fatty acid would need to go through 9 rounds of β-oxidation in order to be completely metabolized.
Ensure gloves are worn at all times when handling strong acids and bases within the experiment of the preparation of benzocaine. 4-aminobenzoic acid (3.0g, 0.022 moles) was suspended into a dry round-bottomed flask (100cm3) followed by methylated sprits (20 cm3). Taking extra care the concentrated sulphuric acid of (3.0 cm3, 0.031 moles) was added. Immediately after the condenser was fitted on, and the components in the flask were swirled gently to mix components. It should be ensured that the reactants of the concentrated sulphuric acid and the 4-aminobenzoic acid were not clustered in the ground glass joint between the condenser itself and the flask. In order to heat the mixture to a boiling point, a heating mantle was used and then further left for gently refluxing for a constituent time of forty minutes. After the duration of the consistent forty minutes the rou...
The competing enantioselective conversion method uses each enantiomer of a kinetic resolution reagent, in this case R-HBTM and S-HBTM, in separate and parallel reactions, where the stereochemistry of the secondary alcohol is determined by the rate of the reactions. When using the CEC method, the enantiomer of the secondary alcohol will react with one enantiomer of the HBTM acyl-transfer catalyst faster than with the other HBTM enantiomer. The mnemonic that identifies the absolute configuration of the secondary alcohol is as follows: if the reaction is faster with the S-HBTM, then the secondary alcohol has the R-configuration. In contrast, if the reaction is faster with the R-HBTM, then the secondary alcohol has the S-configuration. Thin layer chromatography will be used to discover which enantiomer of HBTM reacts faster with the unknown secondary alcohol. The fast reaction corresponds to a higher Rf spot (the ester) with a greater density and a slower reaction corresponds to a lower Rf spot with high de...
The purpose of this lab was to perform an electro-philic aromatic substitution and determine the identity of the major product. TLC was used to detect unre-acted starting material or isomeric products present in the reaction mixture.
In a small reaction tube, the tetraphenylcyclopentadienone (0.110 g, 0.28 mmol) was added into the dimethyl acetylene dicarboxylate (0.1 mL) and nitrobenzene (1 mL) along with a boiling stick. The color of the mixed solution was purple. The solution was then heated to reflux until it turned into a tan color. After the color change has occurred, ethanol (3 mL) was stirred into the small reaction tube. After that, the small reaction tube was placed in an ice bath until the solid was formed at the bottom of the tube. Then, the solution with the precipitate was filtered through vacuum filtration and washed with ethanol. The precipitate then was dried and weighed. The final product was dimethyl tertraphenylpthalate (0.086 g, 0.172mmol, 61.42%).
In parts A and B, the results were the presence of the lead cation and the barium cation. Lead was identified in part a during step six when the solution turned cloudy and very yellow, indicating the presence of a lead precipitate. During part B, step 13, the solution turned foggy and yellow then settled into a white precipitate, indicating the presence of barium. These results were reasonable. The unknown solution was number 4. During part C, the BaCl2 solution emitted a yellow color when it was placed into the bunsen burner flame. KCl was lavender, NaCl was orange, CaCl2 was also orange, LiCl was magenta, and CuCl2 was green. These results were also expected and coincide well with the given cation flame color table.
Benzyl bromide, an unknown nucleophile and sodium hydroxide was synthesized to form a benzyl ether product. This product was purified and analyzed to find the unknown in the compound. A condenser and heat reflux was used to prevent reagents from escaping. Then the solid product was vacuum filtered.
The most widely used method for synthesizing citric acid was discovered by an American chemist named James Currie in 1917 in order to contend with the lack of citrus imports caused by the First World War. This method requires the fungus culture “Aspergillus niger”. This mould is fairly common in nature but special strains are used by the major citric acid producing companies to increase the fungus’s production of citric acid. This method involves substances that contain glucose such as simple sugar or molasses being fed to the fungus. The fungus uses the glucose as food and produces citric acid and carbon dioxide as waste products. When the peak concentration of citric acid is reached, the mould would be filtered out. A solution of base, such as calcium hydroxide, would then be added to the solution to create a salt, in the case of calcium hydroxide, the salt, calcium citrate, will be produced. The formula for calcium citrate is Ca3(C6H5O7)2. The equation for the reaction of calcium hydroxide and citric acid is (3Ca(OH)2 + 2H3C6H5O7 -> Ca3(C6H5O7)2 + 3H2). After the salt is filtered out, the citric acid is regenerated from the salt by adding an acid. Sulphuric acid is commonly used in this process. The equation for the reformation of citric acid from the salt, calcium citrate, is (3H2SO4 + Ca3(C6H5O7)2 -> H3C6H5O7 + 3CaSO4). The additional product, calcium sulphate is a precipitate and can be filtered out leaving an isolated citric acid as the only remaining product. The two widely used methods of fermentation are submerged and surface fermentation. They are relatively similar with the only difference being in surface fermentation the fungus would grow on top of the glucose that is being fed to it. Submerged fermen...
Determining the Activation Energy of the Reaction Between Bromide Ion and Bromate Ion in Acid Solution
The light provides the energy to break the bond between the two bromine atoms, creating the bromine radicals. The energy required to break the bond is known as the bond dissociation energy. The bromine radicals can then attack the hydrocarbons, creating carbon radicals and HBr. These carbon radicals in turn attack the bromine that is not in radical form, which results in the creation of the R-Br products and regeneration of bromine radicals. This allows the reaction to continue though propagation. Different hydrogen atoms demonstrate different reactivity depending on the stability of the different transition states that are formed with the carbon radical
Transition: Substitution of Purine by Purine (A to G or G to A) and pyrimidine by pyrimidine (T to C or C to T).