When reacting with different solvents, the relative ratio of β-bromostyrene isomers can give a hint about the mechanisms the reaction underwent. By an atom attaching to either a cis or trans side of a molecule, the production of different isomers occurs. Using stereochemistry, there are clues that can be gathered about the mechanism used. In order for the elimination reaction to occur there are two options, either E2 or E1 pathway. Both of these pathways have characteristics that must be present for the reaction to occur. The E2 reaction need the molecule to have anti-coplanar stereochemistry, while the E1 reaction needs a strong leaving group, which will create a tertiary or secondary carbocation. The E2 reaction also needs a strong base in …show more content…
The first method is using the integrated NMR doublets and dividing them by the values of the integrated ether quartet (divided by two) and the value of the integrated doublets again. Another method is using GC. In this method, the area of the peak for trans peak needs to be divided by the total area of the peaks for both trans and cis peaks (Experiment VII). The reaction with 2-butanone produced a one and five percent trans isomer fr0m GC and NMR respectively. Water as a solvent produced 79 and 80 percent trans isomer from GC and NMR (Kimmell). Both relative percentages feel within the accepted range of similarity (5-10%). The class data showed similar trends to the experimental data of relative percent mentioned above. The trans isomer relative percent for the solvent of 2-butanone is six percent and the trans isomer relative percent of the water acting as a solvent is 79 percent. (Compiled data). There were a few miscellaneous missing data points and most of the results were within the five to ten percent error margin. The class set of data held true to the personal experimental values achieved through the NMR characterization for the 2-butanone and the GC and NMR characterization of the reaction using the water. In terms of reaction pathways, it seems the reaction with water as a solvent acted in the E1 pathway and the 2-butanone reaction followed the E2 pathway. This can be determined because the E1 pathway produces more trans isomers than the E2 pathway does, which are congruent to our findings. The E2 reaction in figure 1 and 2 shows how the cis isomer is formed by decarboxylation. The figures of 3 and 4 show the trans isomer formation of β-bromostyrene via the E1 pathway. The solvents used played a role in which mechanistic pathways the reaction underwent. The solvent effect caused a difference of 75 percent in the stereochemistry of the molecules.
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 ratios for NaBH4, MPV, and L-selectride are 24.2:75.8, 43.6:56.3, 91.3:.86 respectively. According to analysis of the 1H-NMR spectrum, it is shown that the trans product formed over the cis. The mechanism for L-selectride is very similar to that of NaBH4, but NaBH4 primarily yields the more trans isomer whereas the L-selectride primarily the cis isomer. The reason for this is because in NaBH4, the hydride is not being blocked when convert to OH so it’s free to do a top attack to make a lot more of the the trans isomer. Whereas the L-selectride has bulky groups that block from the carbonyl oxygen which means that it must perform bottom attack and because of this, the isomer that gets made is the cis at 91%. In MPV, the proton is free to attack the carbonyl oxygen in a frontside attack to give more of the trans isomer The MPV reaction using aluminum isopropoxide gives reversible reduction of ketones and aldehydes and the cis or trans can revert back to starting ketone. Each step in the mechanism is reversible so the reaction is driven by the formation of the more stable product which favored thermodynamic. Overall, the stereoselectivity of reaction is affected how the hydride is opened was when it was attacking the carbonyl
...e 3. Both letters A and B within the structure of trans-9-(2-phenylethenyl) anthracene, that make up the alkene, have a chemical shift between 5-6 ppm and both produce doublets because it has 1 adjacent hydrogen and according to the N + 1 rule that states the number of hydrogens in the adjacent carbon plus 1 provides the splitting pattern and the number of peaks in the split signal, which in this case is a doublet.1 Letters C and D that consist of the aromatic rings, both are multiplets, and have a chemical shift between 7-8 ppm. 1H NMR could be used to differentiate between cis and trans isomers of the product due to J-coupling. When this occurs, trans coupling will be between 11 and 19 Hz and cis coupling will be between 5 and 14 Hz, showing that cis has a slightly lowered coupling constant than trans, and therefore have their respective positions in a product. 2
Discussion and Conclusions: Interpreting these results have concluded that relative reactivity of these three anilines in order of most reactive to least reactive go; Aniline > Anisole > Acetanilide. Aniline, has an NH2 , the most active substituent , and adds to any ortho/para position available on the ring. This data is confirmed with the product obtained, (2,4,6 tribromoaniline, mp of 108-110 C). As for anisole, it has a strongly activating group attached, OMe an alkoxy group, and it added in two of the three available spots, both ortho. The results conclude: (2,4-Dibromoanisol mp 55-58 C ). Acetanilide has a strong activating group attached, acylamino group, but this group is large and the ortho positions are somewhat hindered so the majority of the product obtained added at the para position, results conclude: (p-bromoacetanilide mp 160-165 C). Since all the substituents attached to the aromatic rings were activators the only products able to be obtained were ortho/para products.
People have often pondered the reasons for the greatness of Edith Wharton's novel, Ethan Frome. What is it that causes this story to be considered an all-time American classic? One journalist quotes a humanities professor at MIT who states that, "We turn to Wharton because the truths she tells are a bracing tonic in a culture steeped in saccharine sentimentality." The journalist goes on to describe the typical, "popular" story and how they often have endings where "romantic ideals are magically fulfilled..." There is much more to Ethan Frome than simply an unhappy ending to contrast with the many other stories that have sugar-coated and sanguine endings. At first glance, the ending of Ethan Frome may appear to be only depressing. In truth, Wharton offers the reader a complex ending through the careful incorporation of poetic justice and irony.
Tymoczko, J. L., Berg, J. M., & Stryer, L. (2013). Biochemistry: A Short Course, 2nd Edition. New York, NY: W. H. Freeman and Co.
...teraction of the HOMO of the diene and the LUMO of the dienophile. This reaction was done at relatively low temperatures as the dry ether has a boiling point of 34.6 °C. At low temperature the endo preference predominates unless there is extreme steric hindrance, which in this case there is not. The endo product forms almost exclusively because of the activation barrier for endo being much lower than for exo. This means that the endo form is formed faster. When reactions proceed via the endo for the reaction is under kinetic control. Under kinetic control the adduct is more sterically congested, thus thermodynamically less stable. The endo form has a lower activation energy, however, the EXO form has a more stable product. As this is a symmetrical Diels-Alder reaction there is not two possible isomers of the product.
As a final point, the unknown secondary alcohol α-methyl-2-naphthalenemethanol had the R-configuration since it reacted the fastest with S-HBTM and much slower with R-HBTM. TLC was a qualitative method and ImageJ served as a quantitative method for determining which reaction was the faster esterification. Finally, 1H NMR assisted in identifying the unknown from a finite list of possible alcohols by labeling the hydrogens to the corresponding peaks.
Reacting 1-butanol produced 2-trans-butene as the major product. 1-butanol produces three different products instead of the predicted one because of carbocation rearrangement. Because of the presence of a strong acid this reaction will undergo E1 Saytzeff, which produces the more substituted
The spots moved 3.8cm, 2.3cm, 2.1cm, 1.8cm, and 2.5 cm, for the methyl benzoate, crude product, mother liquor, recrystallized product, and isomeric mixture, respectively. The Rf values were determined to be.475,.2875,.2625,.225, and.3125, for the methyl benzoate, crude product, mother liquor, recrystallized product, and isomeric mixture, respectively. Electron releasing groups (ERG) activate electrophilic substitution, and make the ortho and para positions negative, and are called ortho para directors. In these reactions, the ortho and para products will be created in a much greater abundance. Electron Withdrawing groups (EWG) make the ortho and para positions positive.
Enantiomers, a type of isomer, are non-superimposable, mirror images of each other. Diasteriomers, another type of isomer, are non-superimposable, non-mirror images of each other. Dimethyl maleate and dimethyl fumarate are diasteriomers, as they are not mirror images but instead vary in the orientation of the carbomethoxy groups around the double bond. Dimethyl maleate is the cis-isomer because both groups are on the same side and dimethyl fumarate is the trans-isomer because the two groups are on opposite sides. A bromine free radical mechanism was required for this conversion. First, energy from light is required to create two bromine free radicals from Br2. Then one of the free radicals attacks the double bond in dimethyl maleate, breaking it and creating a carbon radical on the other carbon. The bond then rotates and reforms, freeing the bromine radical and creating the trans-isomer, dimethyl fumarate. Bromine in this reaction is acting as a catalyst in this reaction and then cyclohexane is added at the end to neutralize the bromine free radicals. The activation reaction of the radical reaction is lower than the activation energy of the addition reaction, which is why it occurred more quickly. This reaction was successful because the percent yield was 67.1%, which is greater that 65%. It also demonstrated the expected principles, as the reaction did not occur without the presence of both light and bromine. The dimethyl fumarate had a measured boiling point of 100C to 103C, which is extremely close to the expected boiling point of 102C to
In order to separate the mixture of fluorene, o-toluic acid, and 1, 4-dibromobenzene, the previously learned techniques of extraction and crystallization are needed to perform the experiment. First, 10.0 mL of diethyl ether would be added to the mixture in a centrifuge tube (1) and shaken until the mixture completely dissolved (2). Diethyl ether is the best solvent for dissolving the mixture, because though it is a polar molecule, its ethyl groups make it a nonpolar solvent. The compounds, fluorene and 1, 4-dibromobenzene, are also nonpolar; therefore, it would be easier for it to be dissolved in this organic solvent.
In set 1 the bond angle for ethane, propane, 2-methylprpane, and 2,2-dimethylpropane were found. One that that was noticed was that all of the values were fairly close ranging from 108-111 degrees.this may be due to the fact that the structures are similar to one another. On the other hand, because the numbers are not identical due to it being different structures it makes sense for the values to be different. In data set 2 the bond angle was observed again but instead for methyl fluoride, methyl chloride, methyl bromide, and methyl iodide. A difference that was noticed was unlike the first set the angles are all identical that being 109.5. However, both of the sets are searching for the bond angle of similar structures however this set is
Identifying Five Unknown Chemicals I was given 5 unknown samples, which are sodium chloride, sodium thiosulfate, calcium carbonate, sodium bicarbonate and sodium nitrate and these samples are all white and solid. However, I don’t know which substance is which sample and I was only given some information about the physical and chemical properties of these substances. In order to identify the 5 unknown samples, knowing which sample is which substance, I have to carry out this experiment, finding out the chemical and physical properties of these samples. How can we identify five unknown chemical samples that seem to look the same?
Structural theory, which developed in the 1860’s, started the second major period of growth in the organic chemistry field. The de...