2. The gas chromatography calculations offer the area values for under the peak curves. Those area values are directly correlated with the relative concentrations of each alkene product in the mixture. When the ratio of the two values is calculated, 3-methylcyclohexene being the 2nd peak area value and 1-methylcyclohexene being the 3rd peak area (represented on table 1), (11152:1283) = 8.7 1-methylcyclohexene molecules: 1 3-methylcyclohexene molecule. 3. 0.02 mol observed alkene yield/ 0.048 mol theoretical alkene yield = 0.41(100%) = 41% yield 4. The 1-methylcyclohexene product is the major product because of its increased stability due to hyperconjugation along the pi bond. The alkene of this product is trisubstituted on the pi bond which is more stable than the bisubstitution on the pi bond of 1-methylcyclohexene which is the minor product. 5. The likely products of that reaction would be 2-methylcyclohexene, 3-methylcyclohexene and 4-methylcyclohexene. The major product would be the 2-methylcyclohexene because it is the most highly substituted alkene of the group, giving it the highest stability both in this form and its transition state during the reaction. 6. The starting material IR expressed IR spectral data representative of 2-methylcyclohexanol with …show more content…
The results of the gas chromatography test expressed that the alkenes within the final product were produced at a ratio of 8.7:1 or 9 molecules of 1-methylcyclohexene to every 1 molecule of 3-methylcyclohexene. Additionally, the IR spectroscopy test yielded spikes that aligned with the desired wavenumber for alkene components. The graph also did not have spikes within the OH bond region as the starting material graph did, which indicates that OH was not in the final product and the purification of the sample went
The experiment was not a success, there was percent yield of 1,423%. With a percent yield that is relatively high at 1,423% did not conclude a successful experiment, because impurities added to the mass of the actual product. There were many errors in this lab due to the product being transferred on numerous occasions as well, as spillage and splattering of the solution. Overall, learning how to take one product and chemically create something else as well as how working with others effectively turned out to be a
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.
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
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
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
...form 〖PbCrO〗_4 and then process it through a filter. After filtering the 〖PbCrO〗_4 I had to dry the 〖PbCrO〗_4 residue in the drying oven for 30 minutes at 80℃. Then let it cool for 5 minutes and weigh it and finally make a few calculations to obtain the theoretical, actual, and percent yields of 〖PbCrO〗_4. I was able to fulfill the experiment because I obtained all the answers to the equations in an accurate amount. I believe this experiment was a success due to my hypothesis of, If physical methods are used to separate 〖 PbCrO〗_4 precipitate from the reaction mixture then I can successfully calculate the theoretical, actual, and percent yields, being correct.
...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
...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.
A weak peak was at a position between 1600-1620 cm-1 can also be seem in the IR, which was likely to be aromatic C=C functional group that was from two benzene rings attached to alkynes. On the other hand, the IR spectrum of the experimental diphenylacetylene resulted in 4 peaks. The first peak was strong and broad at the position of 3359.26 cm-1, which was most likely to be OH bond. The OH bond appeared in the spectrum because of the residue left from ethanol that was used to clean the product at the end of recrystallization process. It might also be from the water that was trapped in the crystal since the solution was put in ice bath during the recrystallization process. The second peak was weak, but sharp. It was at the position of 3062.93 cm-1, which indicated that C-H (sp2) was presence in the compound. The group was likely from the C-H bonds in the benzene ring attached to the alkyne. The remaining peaks were weak and at positions of 1637.48 and 1599.15 cm-1, respectively. This showed that the compound had aromatic C=C function groups, which was from the benzene rings. Overall, by looking at the functional groups presented in the compound, one can assume that the compound consisted of diphenylacetelene and ethanol or
The boiling point of the product was conducted with the silicone oil. Lastly, for each chemical test, three test tubes were prepared with 2-methylcyclohexanol, the product, and 1-decene in each test tube, and a drop of the reagent were added to test tubes. The percent yield was calculated to be 74.8% with 12.6g of the product obtained. This result showed that most of 2-methylcyclohexanol was successfully dehydrated and produced the product. The loss of the product could be due to the incomplete reaction or distillation and through washing and extraction of the product. The boiling point range resulted as 112oC to 118oC. This boiling point range revealed that it is acceptable because the literature boiling point range included possible products, which are 1-methylcyclohexene, 3-methylcyclohexene, and methylenecyclohexane, are 110 to 111oC, 104oC, and 102 to 103 oC. For the results of IR spectroscopy, 2-methylcyclocahnol showed peaks at 3300 cm-1 and 2930 cm-1, which indicated the presence of alcohol and alkane functional group. Then, the peak from the product showed the same peak at 2930 cm-1 but the absence of the other peak, which indicated the absence of the alcohol
The product was recrystallized to purify it and the unknown filtrate and nucleophile was determined by taking the melting points and performing TLC. Nucleophilic substitution reactions have a nucleophile (electron pair donor) and an sp3 electrophile (electron pair acceptor) with an attached leaving group. This experiment was a Williamson ether synthesis usually SN2, with an alkoxide and an alkyl halide. Conditions are favored with a strong nucleophile, good leaving group, and a polar aprotic solvent.
0.091 moles, giving a yield of 76% or based on 10.5 cm3, a yield of
If the elimination happens with either protons in the terminal methyl group, the resultant product is 1-butene, a Hoffman product (monosubstituted alkene). In contrast, elimination of either -hydrogens by the conjugate base of sulfuric acid (HSO4) on the methylene group leads to an alkene that is dissubstituted. Either the cis- or trans-2-butene can form depending on which hydrogen is deprotonated. These are the Saytzeff products since they are the most popular ones. In this case, the trans-alkene is the most stable, thus one should see the greatest area for the second peak when analyzing the gaseous products under gas chromatography.
Km is the michaelis constant in moles in the solution which is measured as half the maximal velocity (Km=1/2 Vmax). This varies between the different enzymes substrates if a small Km is recorded this suggests that the substrates are bonded tightly to the enzyme. A larger recording shows that the binds are much weaker. Vmax and Km is important to represent the amount of enzyme added and the time required to become synthesised from the amount of subtract used (Alberts et al., 2013).