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…
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…
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
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
...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.
In this experiment, four elimination reactions were compared and contrasted under acidic (H2SO4) and basic (KOC(CO3)3) conditions. The 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. The three butene products have been verified that they elute in the following order: 1-butene, trans-2-butene, and cis-2-butene.
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
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
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.
Discussion The reaction of (-)-α-phellandrene, 1, and maleic anhydride, 2, gave a Diels-Alder adduct, 4,7-ethanoisobenzofuran-1,3-dione, 3a,4,7,7a-tetrahydro-5-methyl-8-(1-methylethyl), 3, this reaction gave white crystals in a yield of 2.64 g (37.56%). Both hydrogen and carbon NMR as well as NOESY, COSY and HSQC spectrum were used to prove that 3 had formed. These spectroscopic techniques also aided in the identification of whether the process was attack via the top of bottom face, as well as if this reaction was via the endo or exo process. These possible attacks give rise to four possible products, however, in reality due to steric interactions and electronics only one product is formed.
The goal of this experiment is to determine which products are formed from elimination reactions that occur in the dehydration of an alcohol under acidic and basic conditions. The process utilized is the acid-catalyzed dehydration of a secondary and primary alcohol, 1-butanol and 2-butanol, and the base-induced dehydrobromination of a secondary and primary bromide, 1-bromobutane and 2-bromobutane. The different products formed form each of these reactions will be analyzed using gas chromatography, which helps understand stereochemistry and regioselectivity of each product formed.
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 purpose of this experiment is to determine the absolute configuration of an unknown chiral secondary alcohol using the competing enantioselective conversion (CEC) method. This method uses both R- and S- enantiomers of a chiral acyl-transfer catalyst called homobenzotetramisole (HBTM), in separate parallel reactions, and thin layer chromatography to identify the stereochemistry of the secondary alcohol, whether it be an R- or S- enantiomer. Quantitative analysis was performed using a program called ImageJ after the appropriate picture was taken of the stained TLC plate. The molecular structure of the unknown alcohol was identified using 1H NMR spectroscopy by matching the hydrogens to the corresponding peak.
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.
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.
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)