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 …show more content…
and sodium borohydride. The goal of oxidation reaction is to oxidize the alcohol ,4-tert-butylcyclohexanol, to a ketone just so that it can be reduced back into the alcohol so it can be observed how the OH will act. Acetic acid and sodium hypochlorite, were added to the cyclohexanol. Aqueous sodium hypochlorite (NaOCl) is a common household bleach and it used to oxidize secondary alcohols to ketones. Acetic acid is used in the reaction because the hypochlorite oxidation reaction occurs faster in acidic conditions and it reacts with bleach. An equilibrium between hypochlorite anion and hypochlorous acid will occur when bleach is mixed in with acidic condition. The actual oxidizing agent is the hypochlorous acid because it has the Cl+ ion.The acetic acid converted the NaOCl to HOCl which serves as the oxidizing agent. When the alcohol get oxidized to form ketone, the Cl+ is reduced in reaction to a Cl-. Since the chlorine is less electronegative than the oxygen, the chlorine has a positive charge which allows the nucleophilic attack by the hydroxyl group. The tert-butyl group on one side of the molecule is very bulky so it likely to stay in the equatorial position which means changes will only occur when 4-ter-butyl-cyclohexanone is reduced to 4-tert-butylcyclohexanol. The two faces of the carbonyl group are non-equivalent because 4-tert-butylcyclohexanone is in conformation with tert-butyl group.
Depending on which face of carbonyl the hydride attacks, the ketone could result in two different diastereomers of product. Since the two ketone faces are nonequivalent, there will be stereo selectivity in reduction which means that one diastereomer will be more prevalent than the other. There are three reduction conditions can be used to reduce the 4-tert-butylcyclohexanone : NaBH4, MPV , and L-selectride. For NaBH4, the hydride attach itself to the carbonyl oxygen to become the hydroxyl group and it more likely from the top because the hydride isn’t blocked by a bulky group (Fig mech prez). Both sodium borohydride and lithium aluminum hydride are less bulky hydride reducing agents so it is expected that they will be able to attack from the top face of the molecule since the bulky tert-butyl group will not hinder the attack. For L-selectride mechanism is similar to NaBH4, the L-selectride is a source of hydride for the carbonyl oxygen but there are bulky groups that block the hydride. Since L-selectride is much larger and bulkier hydride reagent so likely not be able to attack from the top face in the presence of the bulky tert-butyl group (fig 1 and 2 like web) For MPV, the ketone is reduced with aluminum isopropoxide in isopropanol. The carbonyl oxygen attack the aluminum which causes the carbonyl oxygen to have a +1 charge, a hydrogen as …show more content…
a source of hydride, attaches itself to the carbonyl, and then the hydroxyl is formed, during a frontside attack. Correspondingly, the hypothesis is that the sodium borohydride and MPV will form more of the trans product, and the L-selectride will form the cis product. Depending on which side was attacked, the NMR will be able to determine the ratio of cis to trans. The cis and trans products in Figure 2 show the alcohol OH group in the axial (cis) and equatorial (trans) positions, respectively. It can be attack from either top or bottom face of the molecule because it has sp2 hybridization of carbonyl carbon and it is planar. The alcohol OH group is pushed to the equatorial position when attacked from the top face of the molecule which show in the trans product. Conversely, the OH group is pushed into axial position when attacked from bottom face. The trans product is the thermodynamically more stable product, hence under thermodynamic control, because both the tert-butyl and alcohol OH groups are in the equatorial positions. The equatorial position results in less steric crowding so it is more favorable. Thermodynamic control means the product that is more energetically stable will form preferential to other, less stable products. Nevertheless, the cis product has a tertbutyl group which hindered the hydride to attack at the top face of molecule but it formed faster. Thus, the cis product is regarded to be under kinetic control which means the product that forms the most rapidly is favored even though it’s not the stable. The formation of a thermodynamically favored product is reversible while the formation of a kinetically favored product is irreversible. The ratio of trans and cis product can be determined by analyzing the 1H-NMR spectra of the hydride reduction products In this experiment,TLC is used to track the progress of the reaction because purity of the product can be determined on TLC plate by observing the dot samples. KMnO4 will be used to stain the TLC plate since the starting material and product will not show up under UV light. Based off the TLC plate, it can be seen that oxidation was not a success. When spotting a reaction mixture, the disappearance of starting material and the appearance of product over time indicates that the original compound has been converted so it’s an indication that the reaction has gone to completion.Looking at the third lane of Figure 4 there are multiple dots which means that the reaction did not go to completion and impurities are still present.Since polarity of alcohol is greater polarity of ketone, starting material have a spot with lower Rf value than product. Generally speaking, the more polar a compound, the lower it's Rf value because it interacts better with the stationary phase . Looking at Figure 3, the reduction is a success because the single spot in first lane on the left represents that of the starting ketone. After time, the single spot in third lane shows that all starting material has been consumed, and the reaction reached completion. The Rf values in this case are as expected with the product, more polar compound has the lower Rf values. For both NaBH4 and MPV, there are peaks at 4.03ppm and 3.5ppm, and these peaks signify the cis and trans isomers, respectively.
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
oxygen IR spectroscopy can be used to determine the functional group in compound which can be helpful to determine whether the reduction product formed. The IR spectrum of starting material contains a peak at 1720.59 cm-1, representing a carbonyl C=O .In the IR spectrum of the hydride reduction product, the peak is not as intense. This aligned with TLC plate indicated starting material was present in the reaction mixture, and thus, the reaction did not reach completion . The IR spectrum of the hydride reduction product exhibits a broad peak at 3300 cm-1, indicating the presence of an alcohol O-H. The appearance of this stretch shows that the hydride reduction product was formed. Nevertheless, the presence of the carbonyl C=O stretch in this spectrum shows the impurity of product as it contains a significant amount of starting material. The percent yield for oxidation was calculated to be 53% but it is not certain of how much of the ketone has be formed since there are impurities according to TLC. For first part of the lab, the solution needed to be fully converted from hexanol to hexanone so more bleach was used initially. Some particles from other elements might have still been in the beaker which might caused the impurity of product. In the reduction part of the lab, it’s hard to clearly distinguish two layers when performing the extraction so some product might have been dissolved in aqueous layer. Thus, not all product was completely formed since .103g was used at the beginning while only .064g was recovered. To improve this issue, the reaction can rinse again with a solvent that can be added to the seperatory funnel so that any of the product that might have been missed could be added to the final mass. For the future, the experiment should extend reaction time to make sure there are no two spots on the TLC plate in order to confirm that the reaction go to full completion. The purpose of this experiment was to examine the stereoselectivity of various hydride reducing agents. The reduction of 4-tert-butylcyclohexanone using sodium borohydride, lithium aluminum hydride, and L-selectride was studied in detail. After analysis of 1H-NMR spectra with cis and trans ratio, it’s clear that sodium borohydride favored the trans product over cis product in a ratio 24.2:75.8, lithium aluminum hydride favored the trans product over cis product in a ratio of 43.6:56.3, and the L-selectride favored the cis product over trans product in a ratio of 91.3:.86. Because the hydride of sodium borohydride and lithium aluminum hydride less bulky, it attacked the carbonyl carbon from the top which form trans product under thermodynamic control. Conversely, the more bulky L-selectride attacked the bottom of the molecule which form the cis product under kinetic control. Thus, the hypothesis was correct.
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...
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.
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
Alcohol, which is the nucleophile, attacks the acid, H2SO4, which is the catalyst, forming oxonium. However, the oxonium leaves due to the positive charge on oxygen, which makes it unstable. A stable secondary carbocation is formed. The electrons from the conjugate base attack the proton, henceforth, forming an alkene. Through this attack, the regeneration of the catalyst is formed with the product, 4-methylcyclohexene, before it oxidizes with KMnO4. In simpler terms, protonation of oxygen and the elimination of H+ with formation of alkene occurs.
...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
The 1H NMR spectrum shows that there are 18 protons in 11 different proton environments. This fits with the Diels-Alder reaction taking place a...
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.
Purpose/Introduction: In this experiment, four elimination reactions were compared and contrasted under acidic (H2SO4) and basic (KOC(CO3)3) conditions. Acid-catalyzed dehydration was done on 2-butanol and 1-butanol; a 2o and 1o alcohol, respectively. The base-induced dehydrobromination was performed on 2-bromobutane and 1-bromobutane isomeric halides. The stereochemistry and regiochemistry of the four reactions were analyzed by gas chromatography (GC) to determine product distribution (assuming that the amount of each product in the gas mixture is proportional to the area under its complementary GC peak).
After performing the second TLC analysis (Figure 4), it was apparent that the product had purified because of the separation from the starting spot, unlike Figure 3. In addition, there was only spot that could be seen on the final TLC, indicating that only one isomer formed. Since (E,E) is the more stable isomer due to a less steric hindrance relative to the (E,Z) isomer, it can be inferred that (E,E) 1,4-Diphenyl-1,3-butadiene was the sole product. The proton NMR also confirmed that only (E,E) 1,4-Diphenyl-1,3-butadiene formed; based on literature values, the (E,E) isomer has peaks between 6.6-7.0 ppm for vinyl protons and 7.2-7.5 ppm for the phenyl protons. Likewise, the (E,Z) isomer has vinyl proton peaks at 6.2-6.5 ppm and 6.7-6.9 ppm in addition to the phenyl protons. The H NMR in Figure 5 shows multiplets only after 6.5 ppm, again confirming that only (E,E) 1,4-Diphenyl-1,3-butadiene formed. In addition, the coupling constant J of the (E,E) isomer is around 14-15 Hz, while for the (E,Z) isomer it is 11-12 Hz. Based on the NMR in Figure 5, the coupling constant is 15.15 Hz, complementing the production of (E,E)
In this lab experiment, three milliliters of pure cyclohexane was placed within a test tube and lowered into an ice-water bath. The test tube had a temperature probe within it, which measured the cyclohexane lowering in temperature. Once the cyclohexane solution started to solidify, the cooling curve could be observed and the freezing temperature could be determined. The pure cyclohexane was then thawed, with 0.60 grams of biphenyl being added to the cyclohexane. The experiment was then run again. The result was a freezing point of around 8.6℃ for the pure cyclohexane and 7.0℃ for the cyclohexane-biphenyl solution. To confirm the results that the cyclohexane-biphenyl solution had a lower freezing point, the experiment was ran again. The results
Simple metal hydrides contain hydrogen in their crystal structure. These simple metal hydrides include binary and intermetallic hydrides. Binary hydrides contain only one metal and generally represented as MHx, were M stands for metal. The intermetallic hydrides contain at least two metals in addition to the hydrogen and generally represented by the formula AmBnHx, where A, B are metals. These are further classified in to AB (CsCl structure), AB2 (Laves phase), A2B (AlB2 structure type), AB5 (CuC5 structure type). In these hydrides, metal A has strong affinity for hydrogen and B does not interact with hydrogen.
Although some of the results obtained were not totally satisfactory, the general trend expected was persistent. Fluorenone as the limiting reagent was rapidly consumed by the Sodium Borohydride (NaHB4), this is confirmed by noticing the rapidly diminishing activity of Fluorenone, to the point that on the 3rd TLC plate Fluorenone is already absent (consumed by the reaction). One rationale Fluorenone was so rapidly consumed during the reaction, it is because, a hydrogen from the borohydride first attacks the carbonyl group on the Fluorenone molecule, leaving the oxygen with a nucleophilic site. Such a nucleophilic site can then speed up the reaction to the point that after 90 seconds all Fluorenone was consumed; methanol (CH3OH) is attracted
3. Using a reaction from the citric acid cycle, please explain oxidation and reduction. Be specific in your description. Include products and reactants, which molecules are oxidized and reduced, and what the oxidizing and reducing agents are in reaction. Why must we speak of oxidation and reduction together? (10 points)
For this Module, I chose to explain and elucidate the topic, “Oxidation-Reduction Reaction” also known as Red-Ox Reaction not just because it is required for us to elaborate the most complicated matter for us, but also for the reason that I wanted to improve my understanding and comprehension regarding this topic. I really had a difficult time especially during assessments because I wasn’t a hundred percent sure that my answers were correct—I even had questions left out unanswered. I reflected on what I must do. Gladly, our teacher thought of this so that we can help ourselves by looking back at the topic and, slowly and surely understand it again step by step.