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
This week’s lab was the third and final step in a multi-step synthesis reaction. The starting material of this week was benzil and 1,3- diphenylacetone was added along with a strong base, KOH, to form the product tetraphenylcyclopentadienone. The product was confirmed to be tetraphenylcyclopentadienone based of the color of the product, the IR spectrum, and the mechanism of the reaction. The product of the reaction was a dark purple/black color, which corresponds to literature colors of tetraphenylcyclopentadienone. The tetraphenylcyclopentadienone product was a deep purple/black because of its absorption of all light wavelengths. The conjugated aromatic rings in the product create a delocalized pi electron system and the electrons are excited
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.
Many reactants want to form the more stable product, whether that be in terms of sub-stituents (Markovnikov), or stability in terms of reduc-ing charges on the molecules. The more stable a product is, the quicker the reaction will take place, and the more stable product will also be formed in more quantity. This stability of charges comes into play while discussing ortho, para, and meta addition. 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 electrophile is positively charged, so it will not go to the ortho and para positions, but to the meta positions in greater abundance. Therefore, the majority of EWGs (with the exception of halogens), are meta directors. In this experiment a meta director is used. If the product added to the ortho or para positions would produce a carbocation intermediate that has a positive charge on a carbon that is directly touching the EWG. This carbocation intermediate has more energy, and is therefore less stable. Therefore, it is expected that the methyl meta-nitrobenzoate would be the product formed faster and in greater quantities because it has the more stable intermedi-ate. Thin layer chromatography uses a solvent (in this case 85% hexane–15% ethyl acetate) to separate dif-ferent products based on differences in polarity of the molecules. Typically more polar compounds will have more interaction with the stationary phase, and will not move as from the solvent front. This means that the less polar a substance is, the farther it will move. Using the mechanism o electrophilic benzylic substitution, it can be determined at where each step of the mechanism is occurring, and at
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
Wittig reactions allow the generation of an alkene from the reaction between an aldehyde/ketone and a ylide (derived from phosphonium salt).The mechanism for the synthesis of trans-9-(2-phenylethenyl) anthracene first requires the formation of the phosphonium salt by the addition of triphenylphosphine and alkyl halide. The phosphonium halide is produced through the nucleophilic substitution of 1° and 2° alkyl halides and triphenylphosphine (the nucleophile and weak base) 4 An example is benzyltriphenylphosphonium chloride which was used in this experiment. The second step in the formation of the of the Wittig reagent which is primarily called a ylide and derived from a phosphonium halide. In the formation of the ylide, the phosphonium ion in benzyltriphenylphosphonium chloride is deprotonated by the base, sodium hydroxide to produce the ylide as shown in equation 1. The positive charge on the phosphorus atom is a strong EWG (electron-withdrawing group), which will trigger the adjacent carbon as a weak acid 5 Very strong bases are required for deprotonation such as an alkyl lithium however in this experiment 50% sodium hydroxide was used as reiterated. Lastly, the reaction between ylide and aldehyde/ketone produces an alkene.3
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.
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.
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.
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 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...
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.
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.
Oxidation-reduction reactions can be used to stereochemically control and produce many different organic molecules. The oxidation step in this process increases the number of carbon oxygen bonds by losing a hydrogen and breaking that bond. Through the reduction step, carbon-oxygen bond is broken and the hydrogen is returned.
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.