Microwave-Assisted Suzuki-Miyaura Coupling Beginning Question: How effectively and efficiently can biphenyl be synthesized by a Suzuki coupling reaction? Claims: Biphenyl synthesized by a Suzuki coupling reaction was effectively isolated and then characterized using TLC, melting point, IR and 1H NMR tests. However, this process was not efficient because it had a poor pure yield of 22%. Evidence: The goal of the experiment was to synthesize biphenyl product by a Microwave-assisted Suzuki reaction and this was carried out using 1 equivalent bromobenzene, 1 equivalent phenylboronic acid, 1.3 x 10-2 equivalent Pd(OAc)2, 3.0 equivalent Na2CO3, and 0.25 equivalent TBAB (Figure 1). All starting reagents were combined in a microwave tube and the …show more content…
We obtained a purified yield of 22% for a white, crystalline solid, which weighed 0.033 g. Then, we used deuterated-chloroform to dissolve a small amount of the reaction product for 1H NMR testing. Because we had a small amount of purified product, this required us to rotovap the 1H NMR solution so that we can reuse the remaining for IR testing as well. We added an extra IR test for isopropanol to help us analyze the IR results for the product. Melting point was measured with the plateau temperature set at 64°C, which is 5°C below biphenyl’s expected boiling point of 69°C (U.S. National Library of Medicine). The melting point for the greyish solid was determined to be 64°C. Reasoning: In the TLC, the more organic compounds, such as biphenyl and the unknown reaction product, travel further down the plate due to its lower affinity for the polar silica gel and thus had greater Rf values. On the other hand, phenylboronic acid had a very low Rf value of 0, but this was expected due to the polar nature of its two hydroxyl groups. Since the standard biphenyl and the reaction product had the most similar Rf values, this suggested that the identity of the product is most likely …show more content…
In addition, there was also an interesting peak at 3465.8 cm-1. This peak suggested the presence of an OH group, specifically maybe that of the isopropanol to clean the IR instrument. Because of this, we decided to run an extra IR for that isopropanol solution. However, the OH peaks did not seem to correlate or resemble each other. Isopropanol’s OH peak at 3326.6 cm-1 was broader and much more prominent that that of the product (Figure 4). Thus, the 3465.8 cm-1 could represent more carbon-carbon interactions, which then further supports the possibility that the product is
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
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
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
In a separate beaker, acetone (0.587 mL, 8 mmol) and benzaldehyde (1.63 mL, 16 mmol) were charged with a stir bar and stirred on a magnetic stirrer. The beaker mixture was slowly added to the Erlenmeyer flask and stirred at room temperature for 30 minutes. Every 10 minutes, a small amount of the reaction mixture was spotted on a TLC plate, with an eluent mixture of ethyl acetate (2 mL) and hexanes (8 mL), to monitor the decrease in benzaldehyde via a UV light. When the reaction was complete, it was chilled in an ice bath until the product precipitated, which was then vacuum filtrated. The filter cake was washed with ice-cold 95% ethanol (2 x 10 mL) and 4% acetic acid in 95% ethanol (10 mL). The solid was fluffed and vacuum filtrated for about 15 minutes. The 0.688 g (2.9 mmol, 36.8%, 111.3-112.8 °C) product was analyzed via FTIR and 1H NMR spectroscopies, and the melting point was obtained via
Triphenylmethyl Bromide. A 400 mL beaker was filled with hot water from the tap. Acetic acid (4 mL) and solid triphenylmethanol (0.199 g, 0.764 mmol) were added to a reaction tube, with 33% hydrobromic acid solution (0.6 mL) being added dropwise via syringe. The compound in the tube then took on a light yellow color. The tube was then placed in the beaker and heated for 5 minutes. After the allotted time, the tube was removed from the hot water bath and allowed to cool to room temperature. In the meantime, an ice bath was made utilizing the 600 mL plastic beaker, which the tube was then placed in for 10 minutes. The compound was then vacuum filtered with the crystals rinsed with water and a small amount of hexane. The crude product was then weighed and recrystallized with hexane to form fine white crystals, which was triphenylmethyl bromide (0.105 g, 0.325 mmol, 42.5%). A Beilstein test was conducted, and the crystals produced a green to greenish-blue flame.
Craig, D. Q. (2002). Pharmaceutical Applications of Micro-Thermal Analysis. Journal of Pharmaceutical Science, 91(5), 1201-1213.
The sole purpose of performing the lab was to utilize aldol condensation reactions to synthesize a cyclopenta-dienone, while using UV spectrophotometry and computer visualization to further understand the dienone. In the beginning of the lab, the tetraphenylcyclopentadienone (TPCP) was synthesized using dibenzyl ketone and benzyl under extremely basic conditions. The synthesis process could be further understood by observing the mechanism portrayed in Figure 1. According to the figure, the dibenzyl ketone will first loose an alpha hydrogen to form the enolate intermediate.
Biphenyl was fully soluble in dichloromethane at room temperature and partially soluble in hexanes. Biphenyl was insoluble in methanol, acetone, toluene, and hexanes at room temperature but soluble near their respective boiling points. At room temperature and at 100°C, biphenyl was insoluble in water. Biphenyl recrystallized out from hexanes and methanol at room temperature but remained dissolved in acetone and toluene.
The IR spectrum that was obtained of the white crystals showed several functional groups present in the molecule. The spectrum shows weak sharp peak at 2865 to 2964 cm-1, which is often associated with C-H, sp3 hybridised, stretching in the molecule, peaks in this region often represent a methyl group or CH2 groups. There are also peaks at 1369 cm-1, which is associated with CH3 stretching. There is also C=O stretching at 1767 cm-1, which is a strong peak due to the large dipole created via the large difference in electronegativity of the carbon and the oxygen atom. An anhydride C-O resonates between 1000 and 1300 cm-1 it is a at least two bands. The peak is present in the 13C NMR at 1269 and 1299 cm-1 it is of medium intensity.
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 theoretical yield of the m-nitrobenzoate was de-termined to be 4.59 grams. The actual amount of crude product was determined to be 3.11 grams. The percent yield of the crude product was determined to be 67.75 %. The actual amount of pure product formed was found to be 4.38 grams. The percent yield of the pure product was determined to be 95.42%. Regarding the thin layer chromatography, the line from the solvent front was 8 centimeters.
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.
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%).
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.