For this experiment, we reduced a ketone to a secondary alcohol. During the first week, everyone ran the reaction using 9-fluoreneone. The ketones for week 2 were derivatives of acetophenone, and my group elected to test the differences in reactivity between acetophenone, 4-methyl-acetophenone, and 4-bromo-acetophenone. We hypothesized that the differences in reactivity would be affected by the electronegativity of substituent. Therefore, we predicted that the 4-bromoacetophenone react the fastest since bromine is a more electronegative substituent, followed by 4-methyl-acetophenonoe, and acetophenone. For the most part, we followed the procedure from week 1. After dissolving 0.1g of 4-bromoacetophenone in 2ml of 95% ethanol, I cooled the solution in an ice bath and added 20mg of …show more content…
At the 3-minute mark, the least polar ketone was the 3-methyl-aceetophenone with a starting material (SM) Rf of 0.76 and reaction solution (RXN) Rf of 0.75. It makes sense that the ketone with the methyl substituent was less polar than the one with bromine, (SM Rf = 0.56 and RXN Rf= 0.56, 0.42). There is likely some error in TLC plates with either spotting, incorrect measurements of distance traveled by solvent, and impurities in the reaction that could have contributed to this mixture of retention time. This is supported by the fact that there were impurities, multiple spots, in the reaction lane for acetophenone and 4-bromo-acetophenone. However, for 3 of the 5 trials, the bromine substituted ketones had the lowest Rf, supporting its polar properties. Additionally, the acetophenone had the highest Rf for three of the trials, indicating that it is the least polar. However, since there were no patterns in the TLC with 4-methyl-acetopehone it was not too helpful in confidently analyzing reactivity. Lastly, we looked at IR data to see if the reactions ran to
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
A weak peak was at a position between 1600-1620 cm-1 can also be seem in the IR, which was likely to be aromatic C=C functional group that was from two benzene rings attached to alkynes. On the other hand, the IR spectrum of the experimental diphenylacetylene resulted in 4 peaks. The first peak was strong and broad at the position of 3359.26 cm-1, which was most likely to be OH bond. The OH bond appeared in the spectrum because of the residue left from ethanol that was used to clean the product at the end of recrystallization process. It might also be from the water that was trapped in the crystal since the solution was put in ice bath during the recrystallization process. The second peak was weak, but sharp. It was at the position of 3062.93 cm-1, which indicated that C-H (sp2) was presence in the compound. The group was likely from the C-H bonds in the benzene ring attached to the alkyne. The remaining peaks were weak and at positions of 1637.48 and 1599.15 cm-1, respectively. This showed that the compound had aromatic C=C function groups, which was from the benzene rings. Overall, by looking at the functional groups presented in the compound, one can assume that the compound consisted of diphenylacetelene and ethanol or
The weight of the final product was 0.979 grams. A nucleophile is an atom or molecule that wants to donate a pair of electrons. An electrophile is an atom or molecule that wants to accept a pair of electrons. In this reaction, the carboxylic acid (m-Toluic acid), is converted into an acyl chlorosulfite intermediate. The chlorosulfite intermediate reacts with a HCL. This yields an acid chloride (m-Toluyl chloride). Then diethylamine reacts with the acid chloride and this yields N,N-Diethyl-m-Toluamide.
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)
Brandon Meas Block: B Day 4 Lab day 1 5/12/17 Lab day 2 5/23/17 Lab Day 3 6/2/17 Lab 9: Acids and Bases Purpose: The purpose of the lab is to calculate the concentration of a known acid. Using the data collected from this lab, you will calculate the molarity of the acid. Introduction:
The risk of CVDs is usually predicated by a blood test that measures the level of lipoprotein because atherosclerosis is caused by high triglyceride levels, increases in low density lipoprotein (LDL) cholesterol levels, and decreases in high density lipoprotein (HDL) cholesterol levels (5). The majority of HDL consists of Apolipoprotein A-1 (ApoA-1) protein, where it exists mostly in the lipid-bound form in human plasma. ApoA-1 has anti-atherogenic properties (6), where it helps protect against the formation of abnormal fatty deposit within the walls of arteries. It has a specific role in lipid metabolism where it removes cholesterol from issues to the liver for excretion using a process called reverse cholesterol transport (7). Both ApoA-1
I believe my molecule could be a Ketone because of its very high boiling point and solubility. The boiling point is 145.5˚C. The molecule I have been given is very soluble in water. There are dispersion and Dipole-dipole forces between each Ketone molecule. Oxygen is more electronegative than carbon, so it tends to pull the electrons towards the oxygen.
...teraction of the HOMO of the diene and the LUMO of the dienophile. This reaction was done at relatively low temperatures as the dry ether has a boiling point of 34.6 °C. At low temperature the endo preference predominates unless there is extreme steric hindrance, which in this case there is not. The endo product forms almost exclusively because of the activation barrier for endo being much lower than for exo. This means that the endo form is formed faster. When reactions proceed via the endo for the reaction is under kinetic control. Under kinetic control the adduct is more sterically congested, thus thermodynamically less stable. The endo form has a lower activation energy, however, the EXO form has a more stable product. As this is a symmetrical Diels-Alder reaction there is not two possible isomers of the product.
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 this experiment, we computationally predicted the dipole moments of 5 different analyte molecules using the program Spartan. We constructed the molecules online as the program then calculated their dipole moments (polarity). We then experimentally determined the 5 analyte molecules retention factors using the TLC method in the lab. Polarity in organic chemistry refers to a separation of electric charge leading to a molecule having an electric dipole moment1. To determine whether a molecule is polar or nonpolar depends on a molecules structure. This is done by comparing the electronegativity’s of each element in the molecule that are bonded to each other. If a molecules dipole moment eliminates each other due to its symmetrical shape, it is considered to be
When performing types of chromatography, like TLC, the polarity of solvents is extremely important. This lab uses TLC plates, five solvents (hexane, toluene, ethyl acetate, dichloromethane, and acetone), and small evaporating dishes to determine which solvent would be best to use in column chromatography. The least polar solvent is hexane, which has a polarity index of 0.1. Toluene has a polarity index of 2.4, while ethyl acetate’s is almost twice that at 4.4.
The only positive reaction was with acetaldehyde, which resulted in a brown precipitate. Once, again the same procedures were followed with unknown # 4 for the tollens test and dilute permanganate, which
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.
The purpose of the lab was to be able to see the effect of Triton X – 100 on the viability of Tetrahymena thermophile, a unicellular protozoan. This organism is motile and it uses cilia, which are thick protuberances that project out from the body and beat in rhythmic waves to move around. An advantage to using T. thermophila as the test organism is that the cell can be tested directly, it is also fast, cheap and the toxicants can be added to test viability (Rowan & Goldberg, 1985). This experiment acts as a bioassay, a scientific experiment that measures the effect of toxicants or substances on living organisms (Ward & Codd, 1999). Triton X – 100 is a surfactant that is non-ionic, meaning that it contains a hydrophobic head that is uncharged and a hydrophilic chain (Dayeh et al, 2005).
1-Butanol with intermediate polarity was soluble in both highly polar water and non polar hexane as 1-butanol can be either polar or non polar compound. 1-Butanol was polar based on the general rule of thumb stated that each polar group will allow up to 4 carbons to be soluble in water. Also, 1-butanol can be non polar due to their carbon chains, which are attracted to the non polarity of the hexane.