OCHEM Final PostLab Questions

Exp. #8: Propose a synthesis of the following compound. Hint: Use a SN2 reaction. (CH3-double bonded ring-O-CH2CH2CH3)

The answer should include the phenol and propyl halide (or another leaving group). They need a base, if students was to propose using NaOH as a base they should have that in the 1st step and then react the resulting phenoxide with the propyl halide.

Exp. #2: What solvent (diethyl ether or water) would glucose (shown below) be most soluble in? Briefly explain why

Water. The molecule is covered with polar hydroxyl groups that hydrogen bond well with water.

Exp. #5: Where did you observe the O-H and/or C=O peaks (depending on which compound you used) in the IR spectrum of the compound for which you acquired a spectrum?

We tested benzaldehyde in the IR spectrum. A C=O bond was observed at 1702.06 cm-1. Another peak was observed at 3394.82 cm-1, indicating the Csp2-H bond.

Exp. #9: Which of the following will have the fastest rate of reaction in a SN1 reaction? Why?

tertiary alkyl halide will react fastest in SN1 because it has the greatest carbocation stability

Exp. #10: List all of the of the elimination products of the following reaction. Indicate which one is the most stable. (5C chain with methyl at C2 and Br at C3 reacting with NaOCH3 and CH3OH)

trans and cis products, trans is most stable

Exp. #12: Resveratrol undergoes photoisomerization when exposed to UV light. Answer the following: Draw the isomer produced in the following reaction? Identify which isomer (starting material or product) is trans and which is cis? Which isomer is most stable?

trans is most stable

Exp. #12: The following compound undergoes photoisomerization upon exposure to light. What is the product of the following reaction?

turns into a cis isomer (rings on the same side)

Exp.#6: What are the products (if any) of a reaction between each of the following compounds and TEMPO/TCCA in dichloromethane?

(ring with a C=O and H) Primary alcohols would not react. (4C chain with methyl and OH at C2) The second compound contains a tertiary carbon, or carbon with 4 bonds already. This compound could not react further because if the carbon double-bonded to the oxygen during a reaction, it would then have 5 bonds, which is not allowed. (6C chain with OH at C3) The third compound is a secondary alcohol that would react to form a ketone.

Exp. #10: What is the SN2 product of the following reaction? (7C chain with Br at C2 with NaOCH3 and CH3OH)

7C chain with OCH3 at C2

Exp. #6: If the IR spectrum of a reaction product contains a broad peak near 3300 cm-1, what else could be present in the sample other than an alcohol?

A broad peak near 3300 cm-1 could indicate the presence of not only an O-H bond (alcohol) but also the presence of a N-H bond.

Exp. #4: Why does the IR spectrum of caffeine contain two carbonyl peaks?

A carbonyl peak, or peak indicating the presence of a C=O, is present twice in the IR Spectrum of caffeine because caffeine has 2 of these bonds in its structure.

Exp. #8: What key feature should be present in the IR spectrum of the product if unreacted acetaminophen is present present?

A peak at 3400 cm-1, indicating an O-H bond, would be seen on an IR spectrum of a sample containing unreacted acetaminophen.

Exp. #7: If a student submitted a sample of water for GC analysis because they were confused about what was their organic layer and what was their aqueous layer, would a solvent peak be observed in the gas chromatogram of the sample assuming it was analyzed? Remember our gas chromatograph uses a FID.

A solvent peak for the GC analysis of water would not be observed because in FID, the detector is very sensitive towards organic molecules, but relatively insensitive to a few small molecules, such as carbon dioxide or water. The more carbon atoms that are in the molecule being analyzed, the more fragments are formed and the more sensitive the detector is. Water contains no carbon atoms; therefore, it would be relatively insensitive to the FID.

Exp. #1: As the distillation continues eventually the observed temperature of the vapor will approach but never exceed 100oC (assuming there is nothing wrong with the thermometer, no bubbles in the alcohol column, etc.). Why is 100oC the upper limit observable in the experiment?

Because water has the higher boiling point of the 2 components in the mixture (methanol's only being 64.7°C) and the presence of a boiling chip prevents the mixture from boiling over, the temperature can reach water's boiling point of 100°C, but will not exceed it.

Exp. #5: Which of the three compounds has a C=O group and how do the C=O peak locations in the IR spectra compare?

Benzoic acid and benzaldehyde both contain a C=O bond. Both peak locations are found at about 1700 cm-1. These peaks are also relatively similar in appearance/shape as well. Benzoic acid has a long, narrow, prominent peak at the mentioned location, while benzaldehyde also has a long, narrow peak here. One difference between the two is the peaks located to the immediate right of the prominent peaks. In benzoic acid, the peaks are shorter and very narrow. In benzaldehyde, the peaks are also narrow, but are significantly longer.

Exp. #5: Which of the three compounds has a O-H group and how do the O-H peak locations in the IR spectra compare?

Benzyl alcohol and benzoic acid contain an O-H group. Both peak locations are found on or around 3000 cm-1; however, in benzyl alcohol, the peak is long and broad/wide. Contrasting this, the peak in benzoic acid is significantly shorter and instead of showing one broad peak, many smaller narrower peaks are observed.

Exp. #2: Why is biphenyl more soluble in diethyl ether than water?

Biphenyl is a non-polar hydrocarbon. Polar solvents will not solvate it as well as non-polar or less polar solvents like diethyl ether.

Exp. #10: What is the major elimination product of the following reaction? (CH3CBr with a tert-butyl reacting with NaOCH3 and CH3OH)

CH2=CH2 with a tert-butyl

Exp. #7: Why is CH2Cl2 the first peak observed in the gas chromatograms for this experiment?

CH2Cl2 is the first peak observed in the gas chromatographs for this experiment because it is a low boiling point-solvent, and in most chromatographs, low boiling point-solvents, such as dichloromethane, are commonly used as solvents in order to dissolve the sample.

Exp. #11: What is the E1 product of the following compound? (tert-butyl with a Cl)

CH3-double bond going up-CH3

Exp. #8: What additional reaction product would result if NaOH was used as a base in this reaction instead of K2CO3?

CH3CH2OH would result assuming NaOH is in the reaction mixture with iodoethane present.

Exp. #4: Tea contains various carboxylic acids. Given that, why is it helpful to us Na2CO3 in this experiment?

Deprotonating the carboxylic acid makes them more water-soluble which prevents them from being extracted into the organic solvent with the caffeine.

Exp. #9: How would doubling the concentration of NaOH(aq) used in this experiment affect the rate of the reaction? Why?

Doubling the concentration of NaOH used in the experiment would not affect the rate because it is not the substrate in this SN1 reaction. First-order kinetics is used in these types of reaction, meaning only k and the concentration of the rate-determining step are considered. The NaOH concentration being doubled would not change the overall rate.

Exp. #9: How would doubling the concentration of t-BuCl affect the rate of the reaction? Why?

Doubling the concentration of t-BuCl would double the rate because SN1 kinetics depend on the concentration of the substrate only (rate= k [substrate])

Exp. #7: What affect is expected for retention times in a gas chromatography experiment if the temperature of the column is reduced?

If the temperature of the column is reduced, retention times would increase. This is because a lower column temperature causes a compound to spend most of its time condensed in the stationary phase. This would mean only a small amount would be able to evaporate and transfer down the column, resulting in longer retention times.

Exp. #1: Why is it important to have a boiling chip present in the boiling flask when distilling liquid mixtures?

It is important to have a boiling chip present when distilling liquid mixtures because it ensures the liquid heats and continues to boil evenly, as well as prevents it from boiling over.

Exp. #9: What would happen to the rate of the reaction if 9 mL of 0.0333 M t-BuCl in acetone and 1 mL of water were combined and mixed instead of the 3 mL of 0.1 M t-BuCl in acetone and 7 mL of water mixture used in this experiment? Why? (Hint: In addition to being a nucleophile, what other role does water have in this reaction?)

Less water present in the reaction mixture would cause the rate to be slower. Water is polar protic which helps stabilize charged intermediates of the reaction. With less water available, there is less stabilization, causing the rate to decrease.

Exp. #6: If a small amount of carboxylic acid is produced in the oxidation reaction of a primary alcohol with the TEMPO/TCCA, why would you not likely observe it in the IR spectrum of the product mixture? Hint: What aqueous solutions are used to extract the organic layer during the product isolation portion of the experiment?

One of the aqueous solutions used to extract the organic layer, Na2CO3, would neutralize any acids present during the oxidation reaction, so carboxylic acid would not show up in the IR spectrum.

Exp. #7: If a significant clog in a 30 m capillary column is so severe that it has to be trimmed to 20 m, how will this affect its ability to effectively separate mixtures of compounds?

Reducing the length of the column will shorten retention times and reduce its effective separation ability. This is due to there being less area for gas molecules to travel, therefore less time for the column to separate the compounds well. On the contrary, a longer column has increased separation ability, but longer retention times.

Exp. #11: What is the product of the following reaction? Hint: This is a SN1 reaction. (ring with OH at the top with HCl)

Ring with Cl replacing OH

Exp. #11: What are the elimination products of the following reaction? Which of these is the major elimination product? (Ring with OH at the top and then methyl to the left with H3PO4 and H2SO4)

Ring with double bond b/w 1 and 2, ring with double bond b/w 1 and 6, or ring with double bond on the methyl at 6

Exp. #3: Why should the vial of hot solution be allowed to cool to room temperature slowly during crystallization rather than putting it in an ice bath to speed the process?

Slow cooling means slower crystal growth which makes for purer crystals

Exp. #2: Outline a procedure for separating anthracene and 4-chloroaniline (shown below) using extraction and possibly acid/base chemistry.

Step 1-dissolve the mixture in a relatively to non-polar solvent that is volatile. Diethyl ether and dichloromethane are good candidates. Step 2-extract the organic layer with an acidic solution like hydrochloric acid Step 3-seperate the layers. maybe do a couple of more extracts for good measure Step 4-dry and evaporate the organic solvent to recover the anthracite Step 5-treat the aqueous acidic layer that contains the protonated indoline with enough base like NaOH to neutralize the acid and deprotonate the conjugate of indoline. Indoline should precipitate out.

Exp. #3: Why does ethanol have to be added to the hot water/carboxylic acid mixture to get all of the solid carboxylic acid to dissolve?

The carboxylic acid is not soluble enough in hot water to dissolve by itself so a co-solvent

Exp. #1: During the beginning of the distillation experiment (either simple or distillation), what compound in the starting water/methanol mixture predominates in the distillate? Briefly explain why.

The compound that predominates the distillate in the beginning would be methanol. This is due to the fact that more volatile compounds have lower boiling points and come off first, therefore the distillate coming out into the graduated cylinder would be methanol first. Then, as the apparatus got hotter and reached closer to 100°C, water would be present in the distillate.

Exp. #3:Why is it important to use hot solutions during recrystallization of your compounds?

The goal is to get the compound being recrystallized just barely soluble at a higher temperature so it will not be soluble at lower temperatures. By using hot solutions, you maximize the temperature difference allowing for a bigger difference in solubility.

Exp. #7: What is typically the mobile phase in gas chromatography?

The mobile phase, or carrier gas, is typically a chemically inert gas, like helium, argon, or nitrogen, that carries the molecules of the analyte through a heated column in gas chromatography

Exp. #5: If benzyl alcohol, benzaldehyde, and benzoic acid were chromatographed using silica gel TLC plates, which should have the largest Rf value? Explain why.

The one with the highest Rf value will be a non-polar compound that has to travel the distance of the TLC plate through the mobile phase. Benzaldehyde will have the largest Rf value because it is non-polar and will therefore be in the mobile phase when introduced to silica gel. Benzaldehyde is the most non-polar out of the three because it is the only one that contains the non-polar C-H bond.

Exp. #5: Are these consistent with those seen in the reference spectra (located in Experiment 5 folder on Blackboard) for the same compound?

The reference spectra on Blackboard had a peak on or around 1700 cm-1 and another significant peak a little after 3000 cm-1. The reference spectra is consistent with our experimental IR spectrum in this regard.

Exp. #8: What role does the 5% NaOH(aq) extraction play in product isolation?

The role the 5% NaOH (aq) extraction plays in the product's isolation is that it reacts with the acetaminophen that is still left in the solution. Specifically, the NaOH reacts with the phenol to move the acetaminophen into the aqueous layer, removing it from the organic layer and therefore the final product.

Exp. #1: What role does the steel wool serve in the fractional distillation setup used during the experiment?

The steel wool inside the reflux condenser serves as a way to increase surface area in the tube and can improve separation of the water and methanol as they distill. This packing material, among others like glass or plastic beads, allow the vapor to condense, re-evaporate, and condense again which is more beneficial in the distillation process.

Exp. #3: Why does the addition of water to the hot ethanol solution of hydrocarbon help increase crystal growth? Hint: Which solvent is the hydrocarbon more soluble in?

The water makes the solution more polar, thereby making the hydrocarbon less soluble.

Exp. #6: What compound (ketone, carboxylic acid, or alcohol) best matches the following IR spectrum? Explain your reasoning.

This IR Spectrum does not match an alcohol or a carboxylic acid because the peak at 3000 cm-1 is not a broad, intense one which would indicate and O-H bond. It is also not an alcohol IR because there is a narrow, prominent peak at 1700 cm-1, indicating a C=O bond which is not seen in alcohols. This spectrum most closely resembles a ketone IR spectrum because the peak at 3000 cm-1 is narrow and there are 2 long, narrow peaks seen in between 1000 cm-1 and 1500 cm-1.

Exp. #2: Why is sodium benzoate more soluble in water than benzoic acid?

Water is a polar poetic solvent that solvates ions much better than diethyl ether can.

Exp. #3: What affect on the melting point might be observed if the crystals of carboxylic acid are still wet when measuring the melting point of the sample.

Water would be an impurity that could broaden melting points and possibly lower them.

Exp. #12: The pi electrons of an alkene can act as a base. Explain why the cis-dibenzoylethylene produced in this experiment can be isomerized back into trans-dibenzoylethylene by heating an ethanol solution of cis-dibenzoylethylene with a few drops of H2SO4.

When the pi bond is protonated, there is free rotation around the remaining single sigma bond allowing the more stable trans isomer to form when deprotonation occurs reforming the pi bond between the central carbon atoms.

Exp. #11: Which of the following alcohols will undergo dehydration most readily when treated with sulfuric acid? Explain why.

a tert-butyl with an OH would undergo dehydration most readily because it is carbocation is the most stable

Exp. #4: Why might the melting point of your isolated caffeine be lower than the literature value? What might be done to get your isolated caffeine to exhibit a melting point closer to the literature value?

impurities in crude caffeine may depress the melting point, caffeine may be purified possibly by recrystallization.

Exp. #9: Put the following carbocations in order of increasing stability starting with the least stable.

least stable is methyl (no Carbon groups), then primary, then secondary, and most stable is tertiary (3 carbon groups attached)