Tanraj Bains Elimination Lab Introduction: 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. Theory: There are two types of reactions that can take place in this lab, E1, unimolecular elimination, and E2, bimolecular elimination. An E1 …show more content…
This is because 1-butene is the most highly substituted alkene. This reaction follows Hoffman elimination because when forming the least substituted alkene there is steric hindrance, which is not preferred product. 2-bromobutene also undergoes a similar mechanism as 2-butanol, Hoffman E2 reaction, producing 1-butene as the major product. However since 2-bromobutene is less satirically hindered than 2-butanol more 2-trans-butene will be formed. When 1-bromobutane is reacted with potassium t-butoxide there is only one product formed, 1-butene. This is because the halide is on a primary carbon thus producing only one product. 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 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 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
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
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
The goal of this lab is to exemplify a standard method for making alkyne groups in two main steps: adding bromine to alkene groups, and followed by heating the product with a strong base to eliminate H and Br from C. Then, in order to purify the product obtained, recrystallization method is used with ethanol and water. Lastly, the melting point and IR spectrum are used to determine the purity of diphenylacetylene.
This is called the rate determining step. The rate determining step must be equal to the experimental determined step because the rate of reaction will be controlled. For the majority of the experiment, a catalyst (Im) will be utilized to help speed up the reaction. A catalyst will generate an alternative path when driving the reaction forward by lowering the energy of activation, resulting in a faster reaction and isn’t consumed during the reaction but can be included in the law due to it being a reactant in the determining step. Because the catalyst stabilizes in the high energy transition state structure, it doesn’t affect the free energy reaction.
The purpose of this lab was to perform an electro-philic aromatic substitution and determine the identity of the major product. TLC was used to detect unre-acted starting material or isomeric products present in the reaction mixture.
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.
In a small reaction tube, the tetraphenylcyclopentadienone (0.110 g, 0.28 mmol) was added into the dimethyl acetylene dicarboxylate (0.1 mL) and nitrobenzene (1 mL) along with a boiling stick. The color of the mixed solution was purple. The solution was then heated to reflux until it turned into a tan color. After the color change has occurred, ethanol (3 mL) was stirred into the small reaction tube. After that, the small reaction tube was placed in an ice bath until the solid was formed at the bottom of the tube. Then, the solution with the precipitate was filtered through vacuum filtration and washed with ethanol. The precipitate then was dried and weighed. The final product was dimethyl tertraphenylpthalate (0.086 g, 0.172mmol, 61.42%).
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
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.
Figure 1: Initial anti attack approach of bromine to the bottom side of the trans-cinnamic acid:
Although not shown in the fermentation reaction, numerous other end products are formed during the course of fermentation Simple Sugar → Ethyl Alcohol + Carbon Dioxide C6 H12 O6 → 2C H3 CH2 OH + 2CO2 The basic respiration reaction is shown below. The differences between an-aerobic fermentation and aerobic respiration can be seen in the end products. Under aerobic conditions, yeasts convert sugars to
In addition, the additional water produced from each reaction may allow production of H2CO3 when react with CO2 as in Equation 2.4 and thus a continuation carbonation process might occur.