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48 Cards in this Set

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In the ester lab, an excess of acetic acid is used in the reaction. Is this for a kinetic or a thermodynamic (equilibrium) reason? How does this help to increase the yield of product?
This is for thermodynamic reasons. Having excess acetic acid shifts the reaction toward producing more product, a manipulation of Le Chatelier's Principle (which is always a thermodynamic manipulation). By adding more reagent, the system will produce more product until equilibrium is once again reached.

In the ester lab, could you have accomplished the same increase in yield of product by using excess alcohol instead of excess acetic acid? Would it be as easy to isolate and purify the product if an excess of alcohol was used (as opposed to an excess of acetic acid)? Why or why not?
If excess alcohol were used instead of acetic acid, the same outcome would occur: an increase in product would be observed as equilibrium is re-established via Le Chatelier's Principle. If excess alcohol were used instead of acetic acid, however, it would be more difficult to isolate the ester product from the remaining alcohol, although this can still be done. Acetic acid is primarily used in excess instead of the alcohol because it can be easily washed out of the solution through the use of a neutralizing agent, such as sodium bicarbonate.

In the ester lab, sulfuric acid was added to the reaction mixture. Is this for a kinetic or a thermodynamic (equilibrium) reason? How does this help to increase the yield of product?
Kinetic reasons. H2SO4 is a catalyst in this reaction, so it doesn't change the amounts of reagent/product at equilibrium, but rather simple speeds up the reaction (less time to reach equilibrium) by lowering the activation energy needed for the reaction to occur.

In the ester lab, why was it important to vent the extraction vessel when washing the reaction mixture with sodium bicarbonate?
When you wash the organic phase of the reaction mixture with sodium bicarbonate to remove excess acetic acid, a byproduct of CO2 gas is produced. By venting the vial, you relieve the pressure that builds up under the cap as a result of this gas production. If you do not do this, you run the risk ofthe cap popping open when the pressure becomes too great, which can result in loss of product or even injury.

What was the purpose of adding the sodium bicarbonate to the reaction mixture in the ester lab?
To remove the excess acetic acid after the reaction occurred.

What was the MAIN purpose of obtaining the IR spectrum in the ester lab? How does it differ from the IR spectrum of the initial starting reagents?
The IR of the product will be that of an ester, showing a C=O stretch with a higher wavenumber (1750-1705) than a regular ketone (1750-1720). A C-O stretch will also be observed in the ester final compound between 1300-1000, which is not present in the starting reagents (no ethers used).

What is the name of the reaction used in the preparation of an ester from an unknown alcohol and acetic acid in the ester lab?
Fischer esterification (via nucleophilic acyl substitution). This is the formation of an ester from a carboxylic acid and an alcohol (the lone pairs in the -OH group attack the carbonyl in the carboxylic acid)

In the ester lab, H2SO4 was used to speed up the rate of the reaction. Why was it necessary to do this? Why couldn't we have simply let the reaction proceed without it?
Since the carbonyl in acetic acid is much less reactive than an acid chloride (R(C=O)-OH less reactive than R(C=O)-Cl) we needed a proton source to make the compound more reactive. H2SO4 also allows for proton transfers in the mechanism of this reaction, thus making one of the hydroxyl groups formed more protonated (and a better leaving group) to reform the carbonyl.

In the ester lab, what was responsible for the three peak signal around 77 ppm in the C-13 NMR?
The three signals around the 77 ppm region is due to CDCl3 splitting.

What was the purpose of the CaCl2 (calcium chloride) being added to the mixture in the ester lab after the reaction was completed?
To further dry the sample.

What physical properties of methanol suggest that it would be a good solvent to use for an esterification reaction (ester lab)?
Methanol only contains one carbon atom per -OH bond. Thus, methanol is soluble in water, and could be isolated from a ester easily since they are insoluble in one another

Odor and volatility are related. Conversion of a long-chain carboxylic acid to an ethyl ester often enhances its volatility. For example, acetic acid has a bp of 118ºC, whereas the bp of ethyl acetate is 77ºC. Why? (ester lab)
Acetic acid has a higher boiling point because it is a small compound with a hydroxyl group, and thus can participate in hydrogen bonding. In the case of the ester that is formed, the carbon chain is longer and there are no direct hydroxyl groups present, and thus no hydrogen bonding can occur. This makes acetic acid hold together more tightly at slightly higher temperatures than the ester product.

All acetate esters show a proton NMR singlet at roughly 2.0 ppm. What causes this signal?
The singlet found at 2.0 ppm is due to the isolated methyl goup in acetic acid on the opposite side of the carbonyl as the OH group. There are no adjacent hydrogens to this methyl group (thus no splitting), and the calculated chemical shift of these hydrogens yield a 2.0 ppm value.

List at least three factors that enhanced the yield of the mixed aldol condensation reaction, as opposed to either of the self-aldol condensation reactions (Aldol Lab).
1) Excess aldehyde (>2 eq) so likelier for ketone enolate to reach with aldehyde as opposed to ketone carbonyl. 2) The ketone carbonyl is less reactive, so there is less self-condensation of ketone. 3) No alpha hydrogens are present on the aldehyde so it can't self condense or act as an enolate.

(Aldol lab) Consider the double mixed aldol condensation reaction between excess 3-methylbenzaldehyde and 3-hexanone. Why does this non-symmetrical ketone only give one, as opposed to two, different mixed double aldol condensation products?
Since both alpha hydrogens of the ketone react (double aldol) in this sample reaction, it does not matter that the two alkyl groups on the ketone carbonyl are different.

(Aldol lab) How can the formation of oils, instead of crystals, be avoided during a recrystallization attempt? (Note: This was a risk during the aldol lab, but certainly applies to multiple labs and organic chemistry recrystallization techniques in general)
The formation of oils can be avoided by carefully choosing a recrystallization solvent that has a lower boiling point, especially if the compound being recrystallized has a relatively low melting point. If oil does form rather than crystals during recrystallization, more solvent can be added so the compound does not come out of the solution at high temperature.

(Aldol lab) Explain how you could quickly distinguish between the following double aldol condensation products by using proton NMR: benzaldehyde + 4-ethylcyclohexanone, benzaldehyde + 4-ethoxycyclohexanone, and 4-ethoxybenzaldehyde + cyclohexanone. Quickly means that you only need to provide one spectral feature that would distinguish each compound from the other two, provided that one is sufficient.
The quickest way to distinguish between the products of these different compounds revolves around the presence of the ethoxy group in the different compounds (in one case it is attached to the benzene ring in benzaldehyde, in another it is found in the ketone, and in the last case there is no ethoxy group present). The oxygen and its location will affect the shifting of nearby aromatic or cyclohexane hydrogens.

According to Dr. Kline, what is the driving force behind the single aldol condensation reactions in the aldol lab? (Note: This is also why "double condensation" readily occurs and why the rate of dehydration (forming of the product) is more favorable)
The formation of conjugated π bonds in the final compound and its stability.

What type of mechanism (specifically) was the aldol condensation reactions observed in the aldol lab?
They proceeded via an E1Cb mechanism.

Although it did vary depending on which reagents you were assigned at the beginning of the lab, what was one common outcome of most of the products obtained from the aldol lab (appearance)?
Most of the products formed yellow solids.

Let's say you are given a proton NMR for a product from an aldol condensation reaction (aldol lab). Because you know the starting reagents, you account for chemical shifts of the aromatic protons present in one of the starting compounds, however you notice in the proton NMR there are a few more peaks in this region that do not come from the aromatic hydrogens. What is type of hydrogens are most likely causing these signals and why are they so downfield?
The reactions performed in the aldol lab generate compounds with alkene groups (C=C), which often have vinyl hydrogens attached to them. Vinyl hydrogens (H's attached to a C=C) are known to show up far downfield in H NMR spectra, in the aromatic region.

(General Unknown - Solubility) What types of compounds are soluble in H2O (think of functional groups)? Why are they soluble in H2O?
The functional groups that are soluble in H2O are the low molecular weight carboxylic acids, low molecular weight neutral compounds, and low molecular weight amines. They are soluble in water because they have a small enough carbon chain and (in the case of amines and carboxylic acids) can participate in hydrogen bonding with water.

(General Unknown - Solubility) Let's say you have a compound that dissolves in water. How can you tell if it is a neutral compound, carboxylic acid, or an amine?
By testing the pH with pH paper. Carboxylic acids are acidic, amines are basic, and neutral compounds are neutral.

(General Unknown - Solubility) Let's say you have a compound that is soluble in NaOH but insoluble in NaHCO3. What do you likely have? Why? What if it were soluble in NaOH as well as NaHCO3?
A phenol. Phenols are weakly acidic, so they dissolve in a strong base (NaOH), but not NaHCO3. If they also dissolved in NaHCO3 you know you're dealing with a strong acid, which is some phenols and higher molecular weight carboxylic acids.

(General Unknown - Solubility) Let's say your compound was insoluble in both H2O and NaOH, but soluble in HCl. What do you likely have?
Amines of higher molecular weight. Since they are basic and of larger chains, they don't dissolve in H2O or NaOH, but readily dissolve in a strong acid such as HCl (the HCl makes the amine soluble in H2O by protonating it).

(General Unknown - Solubility) Let's say you have a compound that is insoluble in H2O, NaOH, and HCl, but soluble in H2SO4. What functional groups do you possibly have?
An assortment: Esters, aldehydes, alkenes, alcohols, and ketones. These are insoluble in the previous tests because they are not strong acids or bases, and have no available hydrogen bonding to dissolve in H2O. Since they are weakly basic, however, they will dissolve in H2SO4.

(General Unknown - Solubility) Let's say you have a compound that is insoluble in H2O, NaOH, HCl, and H2SO4. What do you likely have?
An alkyl halide, aromatic hydrocarbon, or an aryl halide. These are all the neutral compounds, that are insoluble in water (no hydrogen bonding) and all the acids or bases tried (since they are neutral).

In the General Unknown lab, how did you purify your sample if you had a liquid? A solid?
All liquid samples were purified via hickman distillation, and all solid samples were purified via recrystallization.

What was the main purpose of the initial solubility tests for the General Unknown lab?
To give you a basic idea of the dominating functional group present in your unknown sample.

For the compounds soluble in H2O, they are caracterized as haveing "low molecular weight". What kinds of compounds are soluble in water (how many carbons)?
Compounds with 5-6 carbons or less (and possibily hydrogen bonding) are soluble in water. Above this, even compounds with hydrogen bonding are insoluble in water.

What tests were there for amine groups in the general unknown lab?
None, we had no reliable tests to test for amines. Determination of the amine group came solely from the IR, solubility, and pH data.

What test was performed if you thought you had a carboxylic acid (general unknown)?
Fizz test, in which you react your carboxylic acid with NaHCO3, which will produce CO2 gas in the form of bubbling if a carboxylic acid is present.

What reaction was performed if you thought you had a phenol (general unknown)?
The reaction between phenol and bromine (Br2) a SEAr reaction. Because the -OH is strongly activating, if the bromine is the limiting reagent, the bromine color in solution will disappear in the presence of a phenol.

What two tests could you have used to confirm the presence of an alcohol (General Unknown)?
The Jones oxidation (which reacts the alcohol with a cromium VI reagent, which will change color from yellow to green as electrons are transferred, or the Lucas test, in which the alcohol is reacted with HCl and ZnCl2 (catalyst for 1º case).

Which test does not work for tertiary alcohols, the jones oxidation or the lucas test (general unknown)?
Jones oxidation does not work for tertiary alcohols.

The Jones oxidation can be used to determine the presence of a 1º or 2º alcohol. What functional group gives a false positive for this test?

What test could you have performed if you believed you had a methyl alcohol (general unknown)? What functional group does this test also give a positive for?
A reaction with Iodine (I2) in base to form a ketone, and eventually a carboxylic acid (with Iodoform as a by product CH3I). This test also gives a positive for a methyl ketone, so make sure you have an alcohol first.

What did the 2,4-DNF reaction test for (General Unknown)? How could you tell a positive test result?
It tested for a carbonyl (C=O). It was a positive result if a yellow precipitate formed.

What test could positively ID an aldehyde (general unknown)?
A Tollen's Test, which reacts the aldehyde with silver (Ag) in ammonia (NH3). The positive test is identified when silver is reformed in the sample after the reaction proceeds.

What did the hydroxylamine test, test for? What functional groups gave a false positive for this test?
The hydroxylamine test (H2N-OH) tests for esters, which produces a colored hydroxamic acid with Fe3+. These tests also give a false positive for phenols and amides.

How do primary and secondary amides show up in IR spectra compared to regular ketones? (General unknown)
The C=O stretch in amides is at lower wavenumbers in IR spectra when compared to regular ketones.

How could you test for aromatic compounds (general unknown)? There are two.
One, burn it! And two, react it with CHCl3 in a friedel-crafts reaction (only if no meta director is on the ring)

How could you test for an alkene in the general unknown lab (2)?
reacting it with bromine (Br2) that adds to both carbons (Br2 color will disappear if it is the limiting reagent) OR react it with MnO4^- in base, which will add two hydroxyl groups to the compound and a MnO2 byproduct (which will appear brownish in color).

What if you thought you had an ether? What two tests could you do? (General unknown)
You could react it with Fe(NH4)2SO4/KSCN, which works for an oxygen containing compound, or I2 over CH2Cl2.

When a sugar interconverts between the alpha and beta forms, what is this called?

Why is it imperative that you work quickly once you mix your sugar in water?
The initial value for specific rotation is different for alpha and beta anomers of a sugar, but the same for both at equilibrium. Since they start to mutarotate after the addition of water, you must work quickly to get an initial value to see which one you have.

What are reducing sugars? Why do they give a positive tollen's test?
Reducing sugars are those that have an available OH group on the anomeric carbon (one that is attached to two carbons). This hemiacetal is thus available to react with a tollen's reagent. Non-reducing sugars often have an alkoxy group instead of a hydroxyl group on the anomeric carbon.

Why is it that the addition of a base (NH3) causes a reducing sugar to mutarotate faster?
Carbohydrates that are reducing sugars are also hemiacetals, which are not stable in base (and thus interconvert). Acetals on the other hand (non-reducing sugars) are stable in base, and thus do not interconvert.