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 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...
Then the reaction tube was capped but not tightly. The tube then was placed in a sand bath reflux to heat it until a brown color was formed. Then the tube was taken out of the sand bath and allowed to cool to room temperature. Then the tube was shaken until a formation of a white solid at the bottom of the tube. After formation of the white solid, diphenyl ether (2 mL) was added to the solution and heated until the white solid was completely dissolved in the solution. After heating, the tube was cooled to room temperature. Then toluene (2 mL) was added to the solution. The tube was then placed in an ice bath. Then the solution was filtered via vacuum filtration, and there was a formation of a white solid. Then the product was dried and weighed. The Final product was hexaphenylbenzene (0.094 g, 0.176 mmol,
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.
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
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
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
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
As this is a symmetrical Diels-Alder reaction there is not two possible isomers of the product. Figure 5. Mechanism of the formation of 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.
In the lab we added bromine (Br2) to trans cinnamic acid, which formed the product 2,3-dibromo-3-phenylpropanoic acid. This product could be in either Erythro form or Threo form. We know what form was produced based on the melting point. A high melting point, between 200 and 202 °C, or 115-160 °C for an impure form, means it is an erythro type. If the melting point of the product was between 80-91 °C it was a threo form compound. There are 2 places or centers on the molecule where the bromine could bond, which makes 4 stereoisomers possible in the reaction. The product could attack either anti or syn, and could attack to either the top or bottom side. Anti means the bromines would attack to the opposite sides, so they aren’t in the way of each other. Syn means on the same side, so both bromine’s
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.
The purpose of the experiment is to study the rate of reaction through varying of concentrations of a catalyst or temperatures with a constant pH, and through the data obtained the rate law, constants, and activation energies can be experimentally determined. The rate law determines how the speed of a reaction occurs thus allowing the study of the overall mechanism formation in reactions. In the general form of the rate law it is A + B C or r=k[A]x[B]y. The rate of reaction can be affected by the concentration such as A and B in the previous equation, order of reactions, and the rate constant with each species in an overall chemical reaction. As a result, the rate law must be determined experimentally. In general, in a multi-step reac...
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.