CHEM 238 Lab Final

There are several chemical classification tests that can discern the difference between aldehydes and ketones, and also provide partial structure determination. Identify the components of the Tollen's test reagent by dragging the reagent labels into the box. Contents of Tollen's test reagent:

- sodium hydroxide - silver nitrate - ammonium hydroxide Tollen's reagent is made by reacting silver nitrate with sodium hydroxide to make a precipitate of silver oxide, and then by redissolving the solid with ammonium hydroxide.

Which of the following is the best choice of acid catalyst in an acid-catalyzed esterification reaction? C6H5COOH + CH3OH <--> C6H5COOCH3 + H2O -A few drops of concentrated sulfuric acid. -A few drops of dilute hydrochloric acid. -A few drops of concentrated hydrochloric acid. -A few drops of dilute sulfuric acid.

-A few drops of concentrated sulfuric acid. The Fischer esterification is an equilibrium reaction in which water is a byproduct so dilute acids must be avoided; the addition of water would push the equilibrium back to the reactant side. Hydrochloric acid cannot be used because it undergoes side-reactions with alcohols to generate chloroalkanes.

After a polypeptide chain has been synthesized, certain amino acids in the peptide may become modified 1. draw a di-peptide of lysine 2. what modification could result after amino acid addition?

1. 2. addition of OH to first amino acid

After a polypeptide chain has been synthesized, certain amino acids in the peptide may become modified 1. draw a di-peptide of serine 2. what modification could result after amino acid addition?

1. 2. addition of phosphate group

ϵ units

1/[cm*M]

What is likely to be observed for a positive test for the iodoform test?

A bright yellow precipitate. For a positive test for the iodoform test, an iodoform or triiodomethane product will yield a bright yellow precipitate. An orange/red precipitate is formed in a positive 2,4-DNP test. A silver mirror will develop on the inside of the test tube in a positive Tollen's test. The solution will become red for a positive Schiff's test.

For recrystallization, rapid cooling gives the best crystals t/f

.false For recrystallization, slow cooling gives the best crystals.

write a reaction scheme that would lead to the formation of toluene [methylbenzene] from 4-bromotoluene by using a grignard reagent

4-bromotoluene + Mg --> + water --> toluene + OH- + Mg+Br

Sodium borohydride react slowly with water and even more slowly with ethanol to generate hydrogen gas. what observation would allow you to deduce that hydrogen is being generated either during your reaction or when you pour the reaction mixture into water?

Bubbles of hydrogen gas are formed in the reaction flask.

Rank the following molecules in terms of their expected λmax in the UV-visible spectrum. 2,4-cyclohexadiene 2-Cyclohexenone Cyclohexanone

longest wavelength 2,4-cyclohexadiene 2-Cyclohexenone Cyclohexanone Shortest Wavelength of Absorbance All three molecules have π electrons that absorb UV energy. 2-cyclohexenone and 2,4-cyclohexadienone have conjugated π systems that act as chromophores. The chromophore in 2,4-cyclohexadienone displays the greatest amount of conjugation. The electron delocalization due to resonance in the extensively conjugated system lowers the energy of the whole molecule and increases the observed wavelength of absorbance of ultraviolet energy. This shift to longer wavelength by increased conjugation is known as a bathochromic shift.

Ammonia is corrosive. t/f

true Ammonia is corrosive and should be used with caution.

Recrystallization is a common laboratory method to purify solids by dissolving them in hot solvent, filtering while hot, then allowing the filtrate to cool so that crystals form. Use clamps to secure the filter flask to avoid tipping and unnecessary exposure to hot liquids or the solute.

true Clamps must be used in this experiment in order to secure the filter flask in order to avoid tipping and unnecessary exposure to hot liquids or the acetanilide.

Enzymes hydrolyze only double bonds. t/f

true Enzymes do not just hydrolyze double bonds exclusively. For this experiment in particular, the enzymes hydrolyze amide bonds.

t/f Recrystallization is the process of dissolving a solid material into a suitable solvent or mixture of solvents and allowing the material to crystallize from this solution

true Recrystallization is the process of dissolving a solid material into a suitable solvent or mixture of solvents and allowing the material to crystallize from this solution.

Benzoyl chloride undergoes hydrolysis when heated with water to make benzoic acid. Calculate the molar mass of the reactant and product. Report molar masses to 1 decimal place. Molar mass of benzoyl chloride g/mol Molar mass of benzoic acid g/mol

benzoyl chloride has chemical formula C7H5OCl. (12.011g/mol×7)+(1.008g/mol×5)+(15.999g/mol×1)+(35.45gmol×1)= 140.6g/mol benzoic acid has chemical formula C7H6O2. (12.011g/mol×7)+(1.008g/mol×6)+(15.999g/mol×2)= 122.1g/mol

circle the types of reactivity that carbon bonded to Mg in the above reagent is expected to exhibit

basic and nucelophilic

Methanol is flammable. t/f

true Methanol is flammable and toxic.

Match the procedural step to its purpose by dragging each step in the written procedure for the aldol condensation of dibenzyl ketone and benzil in ethanol solution in the presence of potassium hydroxide base, followed by the UV analysis of the product in hexane solution. Purpose / Motivation---Procedure? To avoid potassium hydroxide from seizing the glass pieces together.

Grease all ground glass joints prior to heating under reflux.

Match the procedural step to its purpose by dragging each step in the written procedure for the Fischer esterification of benzoic acid with methanol in the presence of mineral acid into the appropriate box Purpose: Methanol, which is used as a reagent and solvent, is flammable.

Prodedure: The solution of benzoic acid and methanol was heated on a steam bath. The reaction was heated on a steam bath rather than by electric heating mantle or open flame because the methanol in the flask is flammable.

Unripe fruit with little fragrance begins to give off the characteristic ester smell of that fruit as it ripens. What type of chemistry must be occurring as the fruit ripens?

esterification

The zwitterionic form of the amide bond is the most favored resonance form. t/f

f Amide bonds can be hydrolyzed under either acidic or basic conditions. However, in most cases, strong heating is required for the reaction to proceed at an acceptable rate.

Only the methyl ester of aspartame is sweet. t/f

f Soft drinks contain an reasonable amount of phosphoric acid. The expiration date on the bottom of soft drink cans gives a rough estimate of the expected survival time of aspartame in the can. After the expiration date, it would be expected that the level of sweetness of the drink had dropped below acceptable levels. It can be inferred that the phosphoric acid in soda hydrolyses the aspartame ester making it unsweet.

10% sodium carbonate solution must be used in the fume hood. t/f

false Sodium carbonate solution may be irritating to the skin, but is not otherwise hazardous

Enzymes can selectively cleave specific peptide linkages between specific amino acids within a protein, thereby creating larger peptide fragments. t/f

false Some enzymes remove single amino acids from proteins by beginning at the amine or N-terminus; others hydrolyze proteins into individual amino acids beginning at the free carboxylate end or C-terminus.

what is the final oxidation product of methanol when subjected to an excess of aq. KMnO4?

formic acid

Dimethyformamide, DMF, is a simple amide. Draw the other contributing resonance structure for DMF.

http://www.softschools.com/formulas/images/dimethyl_formamide_resonance_formula_1.png

Azo dyes are hypoallergenic. t/f

t Azo dyes do not fade when exposed to air or light and are technically too small to provoke an allergic response. Most azo dyes are not readily used as food colorants because some of the aromatic amines used to prepare azo dyes are considered carcinogenic. A few azo dyes are an exception, including Yellow #5, Yellow #6, Red #40, Citrus Red #2, and Orange B. Vat dyes must be reduced before being used on fabric. In this case, indigo is reduced to form the colorless, water‑soluble derivative, leucoindigo. After application the fabric is air dried, which oxidizes leucoindigo and turns it back into indigo, producing an intense color.

My Attempt Identify the safety concerns that are associated with each of the given compounds. Aromatic amine Sodium hydroxide Phenols

Aromatic amine: irritant mutagen Sodium hydroxide: caustic Phenols: irritant Aromatic nitrogen compounds can be irritants and mutagens. Sodium hydroxide is a caustic agent. Phenolic compounds are skin irritants. Wear gloves when handling any of these chemicals and avoid skin contact. Keep in mind that indigo has an additional concern that it can be absorbed by the skin and will stain skin and clothing.

Rank the following carbocations in terms of their energy. Carbocation A Carbocation B Carbocation C

Highest Energy Carbocation B Carbocation C Carbocation A Lowest Energy Solution Carbocation A is the most stable carbocation because it is a secondary carbocation that can be conjugated into the allylic π system. It benefits from inductive stabilization and resonance stabilization. Carbocation B is an isolated primary carbocation. It is the least stable because it is only bonded to one carbon atom. Carbocation C is an isolated secondary carbocation. It is more stable than the primary carbocation because it is able to spread its positive charge across two carbons, due to the weak positive inductive effect of carbon

1,3-butadiene undergoes electrophilic addition with HBr to make the two products shown. Label the steps in the pathway in terms of their energy.

In the first part of this reaction, two carbocations are formed (A). A secondary allylic carbocation and a primary allylic carbocation. The secondary allylic carbocation is lower in energy and more stable than the primary allylic carbocation due to the inductive effect of the carbon. Transition state #1 is the lower energy transition state which will lead to product 1, the higher energy product. The higher energy product (kinetic) is a result of the lower transition state, which makes the fastest product from first available carbocation. Transition state #2 is the higher energy transition state which will lead to product 2, the lower energy product. The lower energy product (thermodynamic) is due to the higher transition state that occurs from the rearranged carbocation.

Carboxylic acid derivatives undergo hydrolysis to make carboxylic acids. The rate of hydrolysis depends on the leaving group, L. Rank the carboxylic acid derivatives according to their reactivity, from the most reactive to the least reactive.

Most Reactive: acid chlorides acid anhydrides esters amides Least Reactive: Solution The rate of reaction of carboxylic acid derivatives depends on the basicity of the leaving group. The weaker the basicity of the leaving group, the faster the reaction proceeds.

Polymers are classified by the identity of their repeating unit. Match each of the given polymers to the correct type 1. Polymeric acrylonitrile 2. Polysaccharide 3. Polyamide Nylon 66 Creslan Orlon wool cotton silk

Polymeric acrylonitrile Creslan Orlon Polysaccharide cotton Polyamide Nylon 66 wool silk Creslan and Orlon are both acrylics, or polymeric acrylonitriles. Acrylics contain one CNCN per repeating unit. Although chemically identical, Creslan and Orlon differ slightly in their molecular weight and fiber characteristics due to manufacturing differences. Cotton is a fibrous form of a polysaccharide made up of glucose units connected to each other by covalent bonds. Wool, silk, and Nylon are polyamides. Wool and silk are naturally occurring polymers made up of repeating amino acid units. Nylon is a polymer made with a diamine and a dicarboxylic acid, resulting in a polymeric amide of repeating units from the amine and carboxylic acid.

What is sublimation?

Sublimation is a phase change in which a solid transitions directly to the gas phase.

Identify which of the following structures is carboxylic acid. formaldehyde acetic acid methyl acetate acetic anhydride

The carboxylic acid shown is acetic acid. This compound is a carboxylic acid, because it contains the RCOOH functional group.

When you take your IR Spectra next week, what evidence would indicate that some unreacted camphor was still present in your isolated product ?

The original camper had a carbonyl group instead of a hydroxyl group, so there should be a peek at 1715 cm-1 on the IR spectrum for the ccarbpnyl group

What is likely to be observed for a positive test for the Schiff's test?

The solution becomes red For a positive test for the iodoform test, an iodoform or triiodomethane product will yield a bright yellow precipitate. An orange/red precipitate is formed in a positive 2,4-DNP test. A silver mirror will develop on the inside of the test tube in a positive Tollen's test. The solution will become red for a positive Schiff's test.

Using the relationship between energy and wavelength (E = hc/lambda) and the information given in figure B-2 in the lab manual about the relative and energies of the pi-pi* transitions for butene versus butadiene, explain which direction longer wavelength or shorter wavelength the lamda_max will be found for butadiene compared with that found for a butadiene (see figure B1).

The wavelength would be longer, since butadiene does not have as much energy in its state as butene.

Which molecule would give a positive result for the Tollen's, Schiff's, and 2,4-DNP tests.

Tollen's Test This test is positive for aldehydes, reducing sugars, and both aldoses and ketoses, but negative for ketones. Schiff's Test Schiff's test is primarily used to distinguish between aliphatic and aromatic aldehydes. With the exception of aromatic ketones, all aldehydes and ketones will produce a color change when treated with Schiff's reaction. 2,4-dinitrophenylhydrazine Test The 2,4-DNP test gives positive results for the presence of an aldehyde or ketones

draw the structure of the simplest possible organic ketone and indicate the polarization of the carbon and the oxygen atom of the carbonyl group using delta +/- symbols at these atoms

c = delta + o = delta -

My Attempt When is recrystallization not a suitable method to purify an organic material? a. when the organic material is insoluble in water b. when the organic material is impure c. when the organic material is a liquid instead of a solid d. when the organic material is thermally unstable near its melting point

c. when the organic material is a liquid instead of a solid

which polymer is more receptive to dye: cotton or cellulose triacetate? Why?

cotton The degree to which a dye will bind to a fabric and resist fading, or fastness, depends on the number of available polar groups in the fabric polymer. Cotton is more polar than cellulose triacetate because cellulose triacetate replaces three OH groups with less polar acetates. As a result, cotton is more receptive to dye than cellulose triacetate.

Four major methods of purification in the chemistry laboratory.

distillation chromatography recrystallization extraction

structure of silk

http://wwwchem.uwimona.edu.jm/courses/CHEM2402/Textiles/silk_fibroin.gif

what is the structure of these compounds: ester lactone macrolide

https://d2vlcm61l7u1fs.cloudfront.net/media%2F584%2F58484a2f-f7b1-4fab-826a-58a8e1434d7c%2FphpkQRqqF.png

give the starting amine and the phenolic compound that would be used to synthesize the following commercial azo dye called Sudan Red G?

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write a complete mechanism showing intermediates and electron flow with curvy flow with curvy arrows for the electrophilic aromatuc substitution reaction shown below: https://i.groupme.com/1125x1500.jpeg.1eb9d0ace0b64b4eb407171ba0d12d47.large

https://i.groupme.com/1125x1500.jpeg.1eb9d0ace0b64b4eb407171ba0d12d47.large

perfumes lose their fragrance over time. What type of chemistry must be occurring?

hydrolysis of the esters

Wichterman from the tables in parts 3 and 4 of the data sheet if you can identify a sunscreem component that has significantly greater absorption in the long wavelength portion of the UV a band solutions with greatest long wavelength UVA absorption: name of the strongest you be a absorbing ingredient:

solutions with greatest long wavelength UVA absorption: -solution 7 and 3; neutrogena ultrasheer dry touch sunscreen w/ helioplex -coppertone sport sunblock cotton name of the strongest you be a absorbing ingredient: -avobenzone

The phosphoric acid in soda hydrolyses the aspartame ester making it sweet. t/f

true Amide bonds can exist in two resonance structures. The contribution of the zwitterionic form to the structure lowers the energy of the amide bond, thus making it slower to react (increased stability). Zwitterionic form of amide bond

4 requirements of recrystallization solvent

-The solvent should not chemically react with the compound. -The solvent should dissolve the compound while hot. -The solvent should not dissolve the compound when cold. -The solvent should either dissolve the impurities at all temperatures or not dissolve the impurities at all.

a unknown in this experiment has a measured boiling point of 98 C. Classification tests were positive for iodine and negative for tollen's and schiff's test. The spectra data is givwn on the following page. Give the structure of the unknown: https://i.groupme.com/1125x1500.jpeg.501e846b318c4c74846bc45b1a9dea98.large

3-methylbutanone

A Grignard reagent and a ketone are reacted in ether solution and, followed by an acid workup, form a tertiary alcohol. Recall that Grignard reactions must be scrupulously dry in order to work effectively. A common method of drying glassware is to rinse with acetone prior to use. Why is rinsing with acetone not suitable for the reaction described? -Magnesium dissolves in acetone. Adding acetone will remove a vital reactant from the flask. -Acetone is a ketone. Grignard reagents will react with acetone to make an unwanted byproduct. -Magnesium does not dissolve in acetone. -Water dissolves in acetone. Adding acetone will add water to the reaction flask.

-Acetone is a ketone. Grignard reagents will react with acetone to make an unwanted byproduct. Grignard reagents are strong nucleophiles that react with ketones to form tertiary alcohols. Although acetone can be used as a rinse to dry glassware, its use in this reaction is unwise unless the glassware is then put into an oven to drive off all the acetone. Acetone is a ketone and it will react with the Grignard reagent to form an unwanted byproduct.

Select the statements that are TRUE. -Benzene and its derivatives tend to undergo electrophilic aromatic addition reactions. -Benzene and its derivatives tend to undergo electrophilic aromatic substitution reactions. -Substitution reactions have energies of activation that are very low. -Benzene and its derivatives undergo a type of substitution reaction in which a hydrogen atom is replaced by a substituent, but the stable aromatic benzene ring is regenerated at the end of the mechanism. -Addition reactions with benzenes lead to the loss of aromaticity -Dienes and alkenes are much more stable than benzene rings.

-Benzene and its derivatives tend to undergo electrophilic aromatic substitution reactions. -Benzene and its derivatives undergo a type of substitution reaction in which a hydrogen atom is replaced by a substituent, but the stable aromatic benzene ring is regenerated at the end of the mechanism. -Addition reactions with benzenes lead to the loss of aromaticity Benzene and its derivatives tend to undergo electrophilic aromatic substitution reactions rather than addition reactions. Both addition and substitution reactions require very high activation energies, but the substitution reaction is far more favorable due to the restoration of aromaticity that is lost during the reaction. Addition reactions with benzenes lead to the loss of aromaticity as they form dienes or alkenes, which are significantly less stable than aromatic benzenes. In a substitution reaction, aromaticity is sacrificed as a substituent replaces a hydrogen atom. However, substitution reactions give the restored aromatic benzene product at the end of the reaction

identify two vacuum filtration set up and name the pieces of the set up

-separation of a solution and a precipitate -drying a filtered solid

Why are ethers commonly used as a solvent for Grignard reagent preparations? -Ethers have very low vapor pressure which allows them to be used cold. -Ether excludes moisture from the reaction, because water is not very soluble in it. -Ethers are more polar than alcohols and more likely to dissolve the magnesium. -Ether's high vapor pressure helps to prevent moisture from entering the reaction vessel. Solution

-Ether excludes moisture from the reaction, because water is not very soluble in it. -Ether's high vapor pressure helps to prevent moisture from entering the reaction vessel. Ethers, such as THF or diethyl ether, are commonly used as solvents in the preparation of Grignard reagents for the following reasons: Water must be excluded from the reaction vessel because the Grignard reagent reacts vigorously with water. Water is not particularly soluble in ethers. Protic solvents such as alcohols would destroy the Grignard reagent. Ethers typically have high vapor pressures which prevent atmospheric moisture from getting into the reaction vessel during the reaction.

My Attempt The safety hazard warning for acetyl chloride labels it as a lachrymator. Which of the given statements best defines what a lachrymator is? -Exposure to the chemical causes cancer. -Exposure to the chemical vapors make you cry -.Exposure to the chemical causes birth defects. -The chemical is flammable on contact with water

-Exposure to the chemical vapors make you cry Lachrymators are substances that cause eye irritation, causing watering or tearing of the eyes.

What to do in the event of a spill of concentrated sulfuric acid. [2]

-First, rinse the affected area with copious amount of water. -Second, treat the area with aqueous sodium bicarbonate solution. A spill of concentrated sulfuric acid is best treated by first washing with copious amounts of water followed by a mild base like sodium bicarbonate to neutralize the remaining acid. Combating the spill with a strong base such as sodium hydroxide may result is a highly exothermic neutralization. Sodium thiosulfate is used to combat skin burns from exposure to bromine.

Sodium hypochlorite (household bleach) solution is a strong oxidizing agent. Bleach solutions can also emit chlorine gas, which is a respiratory and eye irritant. Which of the following actions should you follow if you spill bleach on your arm? -There's no need for immediate action, your jeans will absorb most of the acid. -Rub some sodium hydroxide in it. Base neutralizes acid. -Immediately flush the affected area with running water and notify your TA. -Wait until the lab period is over before washing the affected area with lots of water.

-Immediately flush the affected area with running water and notify your TA. You must immediately flush the affected area with running water and notify your TA. Any delay may lead to severe chemical burns. Lab mates will be instructed to hold a modesty blanket for you, and other members of the lab will be instructed to leave. The clothing should be removed and the bare skin rinsed with copious amounts of water. The sooner the intervention, the less likely any significant burn will occur. Do not wash with soap or apply any lotions until the affected area has been inspected by a healthcare professional. Bases also cause severe tissue damage, especially to eyes, and no attempt to neutralize the acid on your skin should ever be attempted

There are several chemical classification tests that can discern the difference between aldehydes and ketones, and also provide partial structure determination. Identify the components of the iodoform test reagent by dragging the reagent labels into the box. Contents of iodoform test reagent: .

-sodium hydroxide -iodine The iodoform test reagent contains a mixture of iodine and potassium iodide in sodium hydroxide base. Ammonium hydroxide is not a powerful enough base to be a viable reagent choice in this case

Other interactions can destabilize the structure.

-Interactions between two charged side chains can destabilize the structure. For example, a segment of Glu residues, which are negatively charged at physiological pH, repel each other. Lys and Arg have positively charged side chains whose charges may interfere with formation of a stable helix. -Gly, with the side chain −H,−H, has great conformational flexibility. Segments with glycine residues do not favor α helix formation, and glycine is not a major component of α helices. -Pro is structurally constrained, and there is no rotation about the N-CαN-Cα bond. Proline also does not take part in hydrogen bonding. Therefore, Pro is not common in α helices. -Bulky or large residues (e.g., Tyr, Ile, or Trp) can also strain the structure if they are close together.

The use of rinse acetone to remove traces of water from glassware is a very common laboratory technique. When performing qualitative tests in glass test tubes, such as the iodoform test or dinitrophenylhydrazine test, why should you avoid rinsing the glassware with acetone prior this experiment? -It may initiate an unwanted precipitate in your product. -It will give a false negative for some carbonyl classification tests. -None of the classification tests in this experiment are harmed by traces of acetone. -It will give a false positive for some carbonyl classification tests.

-It will give a false positive for some carbonyl classification tests. Acetone is a solvent that is commonly used to clean glassware in the laboratory. However, because it is a ketone, it gives a strong positive outcome for some carbonyl classification tests. In this experiment, this is an easy problem to fix because none of the classification tests used are hurt by traces of water. If you use acetone to clean your glassware for this experiment, be sure to rinse the glassware with water in order to remove the acetone.

What might happen if a low molecular weight carboxylic acid is exposed to a slightly acidic aqueous environment such as skin? -It will produce burns. -A "fruity" odor will be produced. -There will be no effect. Low molecular weight carboxylic acids are completely safe for handling. -It will have a cooling sensation on the skin.

-It will produce burns. Lower molecular weight carboxylic acids are volatile irritants and are corrosive. They will produce burns if they come into contact with your skin, which is a slightly acidic aqueous environment. If you or your partner spills a low molecular weight carboxylic acid on your skin, wipe up any spills with a paper towel, wash the area with copious amounts of water, and treat the area with aqueous sodium bicarbonate. Dispose of the paper towel immediately in the container provided for solid paper waste in the hood.

Potassium permanaganate and chromic acid can both be used as classification tests for alcohols. methanol and two amino acids are the hydrolysis products of aspartame. Why is potassium permanganate a better choice as a classification test reagent than chromic acid to detect the methanol produced by aspartame hydrolysis? Select all that apply. -Potassium permanganate oxidizes amino acids slower than it oxidizes alcohols. -Chromic acid also oxidizes amino acids, and will give a false positive color change. -Chromic acid is too weak of an oxidizing agent to oxidize alcohols. -Potassium permanganate loses its purple color as it oxidizes alcohols, so the reaction can be followed visually. -Chromic acid doesn't display a color change when it oxidizes alcohols and so cannot be followed visually.

-Potassium permanganate oxidizes amino acids slower than it oxidizes alcohols. -Chromic acid also oxidizes amino acids, and will give a false positive color change -Potassium permanganate loses its purple color as it oxidizes alcohols, so the reaction can be followed visually. The hydrolysis of aspartame will produce methanol. The traditional method for detecting alcohol is using a chromic acid solution, with a red to green color change being positive. Consequently, the solution that you are using in this experiment will contain amino acids. Chromic acid will oxidize these amino acids and produce a similar color change. Therefore, chromic acid is not a reliable test for the detection of methanol for this experiment because it will give a false positive result. A potassium permanganate solution, aqueous KMnO4, is traditionally used for the detection of double bonds. It oxidizes amino acids slower than it oxidizes alcohols. Therefore, it is a good oxidizing agent and can be used as an appropriate test for the identification of an alcohol in this laboratory experiment.

During the preparation of a tertiary alcohol by a Grignard reaction, the reaction mixture was shaken with saturated sodium chloride solution. What is the purpose of the saturated sodium chloride solution? -Saturated sodium chloride raises the density of the aqueous layer so that it will be the bottom layer. -Saturated sodium chloride causes the water to be immiscible with the THF. -Saturated sodium chloride lowers the boiling point of the water. -Saturated sodium chloride decreases the solubility of the organic product in water.

-Saturated sodium chloride decreases the solubility of the organic product in water. Adding a strong ionic electrolyte, like sodium chloride, to the aqueous solution removes traces of water from the organic layer and decreases the solubility of organic molecules in the aqueous layer.

A secondary alcohol was oxidized to a ketone using hypochlorous acid. Assign the characteristic functional group stretch in each of the spectra 1. diphenyl methanol, alcohol 2. benzophenone, ketone

1. OH stretch 3350 cm-1 2. carbonyl bond at 1680 cm-1

During the recrystallization of a solid, the sample is dissolved in hot solvent then filtered while hot. The filtrate is allowed to cool and the mixture is filtered a second time. What effect will result if the filtrate solution is cooled too rapidly? -The maximum amount of the compound as a purified solid will be obtained. -Small pure crystals will form. -Small impure crystals will form. -Large impure crystals will form.

-Small impure crystals will form. The purpose of using hot vacuum filtration is to remove insoluble solid impurities such as dust, inorganic salts, polymeric material, or other unidentifiable insoluble material. Solid impurities need to be removed before crystallization occurs because they can interfere with the crystallization process and contaminate the final product.

Grignard reactions are highly exothermic and are performed in ether solvent. There is a real risk of fire during this reaction. In the case that a fire should occur in your distillation apparatus, what is the best course of action? -Use a fire extinguisher. -Pour water on the fire to put it out. -Smother the flames with a watch glass or beaker. -Run outside for help.

-Smother the flames with a watch glass or beaker. First, do not panic. Get the attention of the people around you, then smother the flames with a watch glass or beaker. The fire is already contained within the reaction vessel, so do not disrupt the container.

Which of the following statements about semi-empirical methods are true: [2] -Some experimentally measured parameters are substituted into the calculation. -It can determine precisely both the location and the energy of bonding electrons. -It is slower to calculate than ab initio methods. -It is good at predicting structural properties that arise from subtle interactions of orbitals.

-Some experimentally measured parameters are substituted into the calculation. -It is good at predicting structural properties that arise from subtle interactions of orbitals. Solution A semi-empirical method is a way to simplify a quantum mechanical calculation so that it can be readily solved. It is done by substituting some experimentally measured parameters for parts of the equations that are difficult or impossible to calculate exactly. The parameters are determined for a reference set of molecules and are then assumed to be acceptable for all other molecules as well. For the question above, true statements about semi-empirical methods include: Some experimentally measured parameters are substituted into the calculation. Combines theoretically sound model of bonding with some experimentally determined parameters. Faster to calculate than ab initio methods. Good at predicting structural properties that arise from subtle interactions of orbitals. For a more in depth explanation of semi-empirical methods, please reference p. 275 a

Benzophenone reacts with phenylmagnesium bromide in ether solution to make triphenylmethanol after acid workup. Which of the following experimental mistakes will decrease the effectiveness of a Grignard reaction? [2] -The magnesium ribbon was not wiped clean before use. -The magnesium ribbon is insoluble in ether. -The glassware was not dried before use. -The fume hood was too cold.

-The magnesium ribbon was not wiped clean before use. -The glassware was not dried before use. Grignard reagents are very powerful bases and can react violently with water or acid. If the glassware or reagents are not dry prior to the reaction then Grignard reagent will be lost to side-reactions. Magnesium metal develops a coating of magnesium oxide on its surface if left in the air for too long a period of time. The oxide should be removed by wiping with a dry abrasive cloth prior to reaction. Magnesium metal is itself largely insoluble in ether but it manages to react with the bromoalkane and form a soluble Grignard reagent. Grignard reactions are exothermic, and are therefore aided by cold conditions to which they can lose their heat.

Under what conditions does the hydrolysis of an amide bond occur? [3] -The reactants must be heated for hydrolysis to occur. -The reactants must be cooled for hydrolysis to occur. -Amide hydrolysis occurs under basic conditions. -Amide hydrolysis occurs under acidic conditions.

-The reactants must be heated for hydrolysis to occur. -Amide hydrolysis occurs under basic conditions. -Amide hydrolysis occurs under acidic conditions. Amide bonds can be hydrolyzed under either acidic or basic conditions. However, in nearly all cases, strong heating is required for this reaction to proceed at an acceptable rate.

2‑Butanone is converted into 3‑methyl‑3‑hexanol using a Grignard reagent prepared from 1‑bromopropane and magnesium metal in THF solution. List the procedural steps required to collect the alcohol product by microdistillation.

1. Decant dried THF solution into 25 mL round‑bottom flask. 2. Add boiling chips to the solution in round‑bottom flask. 3. Grease all joints and attach the flask to a microdistillation apparatus. 4. Set apparatus in heating mantle and use heat shield tiles. 5. Adjust variac to distill THF at 60 ∘C.60 ∘C. 6. Increase variac setting to cause distillation of higher‑boiling product. 7. Stop distillation before the flask becomes dry.

Which statements about the operation of an automated micropipettor are true? [2] -The volume dial has not been set, adjust it to the desired setting. -The volume dial has already been set, do not alter it. -Rapid release of the plunger may introduce bubbles into the pipette tip. -After using the pipette, one should lay it down in the hood.

-The volume dial has already been set, do not alter it. -Rapid release of the plunger may introduce bubbles into the pipette tip. The given statements are true about the use of the micropipettor and should be followed in this laboratory experiment. The micropipette is already set to the correct settings. Please do not change these settings. The plunger will stop at two different positions when it is depressed. The first of these stopping points is the point of initial resistance and is the level of depression that will result in the desired volume of solution being loaded into the pipette tip. The second stopping point is when the plunger is depressed beyond the initial resistance until it is in contact with the body of a pipettor. The second stopping point is only used for the complete discharging of solutions from the tip. Do not eject the plastic tip from the micropipette. Everyone is using the same solution for each micropipette, so one tip is sufficient for an entire class. The micropipettes should be stored upright. Do not lay them down in the hood. Do not release the plunger rapidly after loading and discharging your sample. The rapid release of the plunger may introduce bubbles into the pipette tip.

Select the true statements about protein secondary structure and explain why they're correct. -The β‑pleated sheet is held together by hydrogen bonds between adjacent segments. -In an α‑helix, the side chains are located inside the helix. -In a β‑pleated sheet, the side chains extend above and below the sheet. -Disulfide bonds stabilize secondary structure. -The secondary level of protein structure refers to the spatial arrangements of short segments of the protein. Solution

-The β‑pleated sheet is held together by hydrogen bonds between adjacent segments. -In a β‑pleated sheet, the side chains extend above and below the sheet. -The secondary level of protein structure refers to the spatial arrangements of short segments of the protein. Solution A protein, or polypeptide, is a chain of amino acids folded up into a stable 3‑D structure. Protein structures are described by a series of four levels that increase in complexity from the primary level (amino acid sequence) to the more complex levels: secondary (short segments of helices and sheets), tertiary (overall 3‑D structure), and quaternary (arrangement of subunits). The secondary level of structure describes the spatial orientation of short segments of amino acids. The two most common secondary structures are the α‑helix and the β‑pleated sheet (often shortened to β-sheet). The α‑helix is a compact coiled structure. The β‑sheet is made up of short strands that are lined up side by side to form an open surface with a pleated appearance. Both the α‑helix and the β‑sheet are stabilized by hydrogen bonding. Both structures form hydrogen bonds between the amide backbone N−HN−H of one amino acid and the C=OC=O of another amino acid further down the chain. The α‑helix is a compactly wound coil with the side chains protruding from the backbone of the helix. In the β‑sheet, side-by-side segments are closely aligned, and the side chains extend away from both sides of the sheet's surface. There are two types of bonds that are not directly involved in the stabilization of secondary structure: peptide bonds that join the amino acids, and disulfide bonds that form between some cysteine residues. The disulfide bonds stabilize the third (tertiary) level of protein structure.

My Attempt Methyl benzoate was prepared by the reaction of benzoic acid with methanol in the presence of a mineral acid. The ester product was extracted into dichloromethane and then shaken with saturated sodium chloride solution. What is the purpose of the sodium chloride? -To lower the boiling point of the water. -To raise the density of the aqueous layer so that it will be the bottom layer. -To cause the water to be immiscible with the methylene chloride. -To decrease the solubility of the organic product in water.

-To decrease the solubility of the organic product in water. Adding a strong ionic electrolyte, like sodium chloride, to the aqueous solution decreases the solubility of organic molecules in the aqueous layer and promotes the transfer of the organic molecule to the organic layer.

Sodium borohydride and lithium aluminum hydride are two of the most commonly used reducing agents. sodium borohydride

-is the least reactive of the two hydrides -is the safest to use -can be used in reactions with methanol solvent -reduce aldehyde to primary alcohols -reduce ketones to 2nd alcohols

What two amino acids contribute to the structure of aspartame? -aspartic acid and tyrosine -alanine and tyrosine -phenylalanine and aspartic acid -phenylalanine and glutamic acid

-phenylalanine and aspartic acid Aspartame is L-aspartyl-L-phenylalanine methyl ester. It is the dipeptide of aspartic acid and phenylalanine.

Sodium borohydride and lithium aluminum hydride are two of the most commonly used reducing agents what do they have in common?

-reduce aldehyde to primary alcohols -reduce ketones to 2nd alcohols

Sodium borohydride and lithium aluminum hydride are two of the most commonly used reducing agents. what does lithium aluminum hydride reduce?

-reduce esters -reduce carboxylic acids to primary alcohols -reduce aldehyde to primary alcohols -reduce ketones to 2nd alcohols

During the recrystallization of a solid, the sample is dissolved in hot solvent then filtered while hot. The filtrate is allowed to cool and the mixture is filtered a second time. What is the purpose of the initial hot vacuum filtration in the recrystallization process? -to obtain the maximum amount of the compound as a purified solid -to remove soluble solid impuritiesiesto form larger crystals Solution -to remove insoluble solid impurities -to form larger crystals

-to remove insoluble solid impurities The purpose of using hot vacuum filtration is to remove insoluble solid impurities such as dust, inorganic salts, polymeric material, or other unidentifiable insoluble material. Solid impurities need to be removed before crystallization occurs because they can interfere with the crystallization process and contaminate the final product.

Could the structures below undergo a Fischer esterification reaction? [benzoic acid] and [thionyl chloride] -Yes, if the carboxylic acid is chilled in the SOCl2 reagent and a small amount of H2SO4 is used as a catalyst. -No, the structures above cannot undergo a Fischer esterification reaction to form an ester. -Yes, if the carboxylic acid is heated in the SOCl2 reagent and a small amount of H2SO4 is used as a catalyst. -Yes, if the carboxylic acid is heated in the SOCl2 reagent and a small amount of NaOH is used as a catalyst

.-No, the structures above cannot undergo a Fischer esterification reaction to form an ester. A Fischer esterification yields esters by heating a carboxylic acid in an alcohol solution containing a small amount of a strong acid catalyst. Carboxylic acids are not reactive enough to be attacked by neutral alcohols, but they are made much more reactive in the presence of a mineral acid, such as H2SO4. Heat is also needed in order for the Fischer esterification reaction to proceed efficiently. In this question, the two structures given are benzoic acid, a carboxylic acid, and thionyl chloride, a reagent. A Fischer esterification will not occur in this reaction because there is no alcohol present. Instead of an ester, an acid chloride would be formed with the combination of a carboxylic acid and thionyl chloride.

Avobenzone is used in sunscreen because it offers protection in the UVA and UVB regions. Use Beer's Law and data provided in the table to determine how many milligrams of avobenzone are present in 2.2 mL of sunscreen. Molar mass of avobenzone = 310.4 g/mol Absorbance: 0.75 Wavelength: 300 nm Molar Extinction Coefficient: 9120 M-1cm-1 Pathlength of UV cell: 1.0 cm Volume of Sunscreen Sample: 2.2 mL How many milligrams of avobenzone are present in 2.2 mL of sunscreen? mg

0.056 g

Oxybenzone is used in sunscreen because it offers protection in the UVA and UVB regions. Use Beer's Law and data provided in the table to determine how many milligrams of oxybenzone are present in 2.6 mL of sunscreen. Molar mass of oxybenzone = 228.2 g/mol Absorbance: 1.00 Wavelength: 330 nm Molar Extinction Coefficient: 9130 M-1*cm-1 Pathlength of UV cell: 1.0 cm Volume of Sunscreen Sample: 2.6 mL How many milligrams of oxybenzone are present in 2.6 mL of sunscreen? mg

0.065 mg Beer's law can be used to determine the concentration of oxybenzone in the sunscreen sample. c= A/ϵl = 1.00/???e???⋅1.0 = 0.000110 M The concentration can be converted to mass: Mass (g) = Concentration (mol/L) × Volume (L) × Molar Mass (g/mol) Mass= 0.000110× 2.6/1000 × 228.2 =6.5×10−5 g = 0.065 mg

List the procedural steps, from start to finish, that are required to convert 2-butanone into 3-methyl-3-hexanol using a Grignard reagent prepared from 1-bromopropane and magnesium metal.

1- Dry all glassware and reagents prior to starting the experiment. 2- Add 1-bromopropane to clean magnesium. 3- Wait for reaction to turn cloudy. Sonicate if necessary. 4- Add 2-butanone dropwise to the Grignard reagent, allowing a gentle reflux. 5- add aqueous ammonium chloride to quench the reaction. 6- Wash the reaction mixture with sodium bicarbonate and sodium chloride solutions in a separatory funnel. 7- Dry over magnesium sulfate and distill product from THF. Since Grignard reagents are very powerful nucleophiles, reagents and glassware must be dried prior to use. The Grignard reagent is generated by reacting solid magnesium with a haloalkane in ether solution. Sonication or warming may help initiate the Grignard reaction. Once the Grignard reagent has performed as a nucleophile, the reaction will require acid workup with dilute acid. Reaction byproducts are removed by extraction with sodium bicarbonate and saturated sodium chloride. Magnesium sulfate is used to remove traces of water and the product is distilled from THF solution.

Consider the nucleophilic addition reaction of 2‑butanone with excess propyl magnesiumbromide, made in situ by reacting 1‑bromopropane with metallic magnesium, to make 3‑methyl‑3‑hexanol. 2-butanone: d=0.81 g/mL 1-bromopropane: d=1.35 g/mL magnesium: - 3-methyl-3-hexanol: d=0.82 g/mL A reaction was performed in which 0.50 mL0.50 mL of 2‑butanone was reacted with an excess of propyl magnesiumbromide to make 0.47 g0.47 g of 3‑methyl‑3‑hexanol. Calculate the theoretical yield and percent yield for this reaction. theoretical yield = g percent yield = %

1. 0.653 2. 72.0 Step 1. The mass of 2‑butanone is 0.50 mL×0.81 g/mL=0.41 g Step 2. The number of moles in 0.41 g0.41 g of 2‑butanone is (0.41 g)(1 mol/72.1 g)=0.0056 mol Step 3. The number of moles of 3‑methyl‑3‑hexanone formed is 0.0056 mol 2-butanone =0.0056 mol 3-methyl-3-hexanone Step 4. The molar mass of 3‑methyl‑3‑hexanone is 116.2 g/mol. Step 5. The theoretical yield of 3‑methyl‑3‑hexanone is (0.0056 mol)(116.2 g 1 mol)=0.65 g Step 6. The percent yield is actual yield of product / theoretical yield of product ×100% = [0.47 g / 0.65 g] ×100% = 71%

Consider the reaction of a Grignard reagent prepared from 1‑bromobutane and magnesium in THF solution, with hexanone. Calculate the molar masses of the reactants and product. molar mass of 1‑bromobutane: g/mol molar mass of 2‑hexanone: g/mol molar mass of 5‑methyl‑5‑nonanol: g/mol

1. 137 2. 100.1 3. 158.2 Solution 1‑bromobutane has chemical formula C4H9Br.: (12.011×4)+(1.008×9)+(79.904×1)=137.0 2‑hexanone has chemical formula C6H12O: (12.011×6)+(1.008×12)+(15.999×1)=100.2 5‑methyl‑5‑nonanol has chemical formula C10H22O: (12.011×10)+(1.008×22)+(15.999×1)=158.3

My Attempt Consider the iodination of salicylamide by sodium iodide and sodium hypochlorite via an electrophilic aromatic substitution to form iodo-salicylamide. Calculate the molar masses of the reactants and product. Report molar masses to 1 decimal place. Molar mass of iodosalicylamide: g/mol Molar mass of salicylamide: g/mol Solution

1. 263 2. 137.1 Salicylamide has chemical formula C7H7NO2 (12.011g/mol×7)+(1.008g/mol×7)+(14.0067g/mol×1)+(15.999g/mol×2)=137.1gmol Iodosalicylamide has chemical formula C7H6INO2 (12.011 g/mol×7)+(1.008 g/mol×6)+(126.904 g/mol×1)+(14.0067 g/mol×1)+(15.999 g/mol×2)=263.0 g/mol

Write the structure of the organic products for the following reactions. Name the class of compounds to which the product(s) belong 1. Cyclohexanecarboxaldehyde and silver nitrate hydroxide 2. 2-pentanone and I2 and NaOH

1. Cyclohexane carboxylic acid [carboxylic acid class] 2. butyric acid [carboxylic acid class]

Salicylamide can undergo an iodination by electrophilic aromatic substitution. Arrange the procedural steps in order to iodinate salicylamide.

1. Start of Lab 2. Dissolve salicylamide in ethanol. 3. Add sodium iodide. 4. Add sodium hypochlorite to the ice cold solution of salicylamide and sodium iodide. 5. Add sodium thiosulfate. 6. Acidify by adding 10% HCl. 7. Collect the crude product by vacuum filtration. 8. Recrystallize from hot ethanol.End of Lab The full procedure is listed in the lab manual.

List the procedural steps, from start to finish, that are required to convert benzoic acid into methyl benzoate via a Fischer esterification. [7]

1. The reaction involves heating an alcohol so a round bottom flask is required. Boiling chips are advised because alcohols "bump" when they boil. 2. The carboxylic acid, methanol, and acid are mixed and heated on a steam bath because methanol is flammable. 3. The reaction mixture is poured over ice, and the ester product is extracted into dichloromethane. 4. Washing with sodium carbonate solution neutralizes the acidic conditions, and washing with saturated sodium chloride removes byproducts from the organic layer. 5. Magnesium sulfate removes water from the organic layer. 6. The dichloromethane solvent is removed by distillation on a steam bath because dichloromethane has a low boiling point. 7. The pure ester is collected by distillation on a heating mantle because it has a high boiling point. "Dissolve the benzoic acid in methanol in a round‑bottom flask.Add concentrated sulfuric acid and heat under reflux on a steam bath.Pour the reaction mix into ice, and rinse the reaction flask with dichloromethane. Collect the combined layers in a test tube.Wash the dichloromethane layer with sodium carbonate solution, followed by a wash with sodium chloride solution.Dry the dichloromethane solution over magnesium sulfate.Remove the dichloromethane by distillation on a steam bath.Distill the remaining liquid ester in a vial on a heating mantle"

what is the structure of lysine bonded to serine [with serine at the C-terminal]? what is the full name?

1. https://pubchem.ncbi.nlm.nih.gov/image/imgsrv.fcgi?cid=151428&t=l 2. lysylserine

Indicate which compounds present in [a] the reactants and [b] the products from this lab experiment could be expected to give positive tests with [1] permangate and [2] ceric ammonium nitrate tests.

1. permangante a. reactants: phenylalanine, aspartic acid, base-hydrolyzed aspartame b. Products: methanol 2. ceric ammonium nitrate a. reactants: base-hydrolyzed aspartame b. Products: methanol

A reaction was set up to convert an alcohol to a carbonyl. Samples of the reaction mixture were checked by infrared spectrometry every couple of minutes. Use the infrared spectra provided to determine if the reaction mixture contains: pure alcohol, a mixture of alcohol and carbonyl, or pure carbonyl. The reaction mixture contains.... -pure alcohol. -pure carbonyl. -mixture of alcohol and carbonyl. [choose one for each of the 2 spectra]

1. pure carbonyl 2. -mixture of alcohol and carbonyl.

Two solutions of the same UV-absorbing molecule were analyzed by UV-vis spectroscopy on the same instrument using 1 cm pathlength cells. Use data from the table to calculate by what percentage the concentration of Solution B is compared to that of Solution A. Sample- Absorbance at 340 nm A- 0.40 B- 0.85 In regards to concentration, what percentage is sample B compared to sample A? %

210% Because the two samples contained the same UV-absorbing molecule, and because the pathlength was the same in both measurements, the comparison of B to A can be simplified as shown: AB/AA = ϵcBl/ϵcAl = cB/cA The value of ε is the same for both samples. The value of l is the same for both samples. The percentage that sample B is more concentrated than sample A is: cB/cA = [0.85/0.40]⋅100 = 210%

none of the given molecules display much solubility in water at room temperature. However, two of the molecules display appreciable solubility in boiling water. Based on its structure, which molecule will be the least soluble in hot water? 2‑naphthol salicylic acid acetaminophen

2‑naphthol

draw a product for the intramolecular aldol reaction [with dehydration] of 2,5-hexandione

3-Methyl-2-cyclopenten-1-one

The aldol condensation involves nucleophilic attack of an enolate to a carbonyl. Use the pop-up menus to identify the total number of aldol products that each of the following reactions can produce. 1. 3-dimethyl-butan-2-one and cyclopentanone 2. Cyclohexanecarbaldehyde 3. benzaldehyde and Cyclohexanol

4 aldol products 1 aldol product No aldol products. 1. 2,2-dimethyl-3-butanone can generate an enolate nucleophile. Because cyclopentanone is symmetrical, it can generate the same enolate nucleophile on either side of the carbonyl. 2,2-dimethyl-3-butanone can act as a nucleophile to another molecule of 2,2-dimethyl-3-butanone or a molecule of cyclopentanone. Cyclopentanone can act as a nucleophile to another molecule of cyclopentanone or a molecule of 2,2-dimethyl-3-butanone. This combination of reactants generates four possible aldol products. 3. Cyclohexanecarbaldehyde is asymmetrical, but only has α hydrogens on one side of the carbonyl, so it can only generate one type of enolate nucleophile. Cyclohexanecarbaldehyde can act as a nucleophile to another molecule of Cyclohexanecarbaldehyde. There is only one possible aldol product for this reaction. 3. Benzaldehyde is a carbonyl that does not possess any α hydrogens, and so cannot generate an enolate nucleophile. Cyclohexanol is not a carbonyl and does not participate in an aldol condensation. This combination of reactants does not generate any aldol products.

My Attempt Use the values provided below to calculate the predicted isoelectric point, pI, value for aspartame. aspartic acid: pka 1: 1.88 pka 2: 9.60 pka 3: 3.65

6.63 To solve this problem, use the pKa values for aspartic acid's amine group and its carboxylic side chain. In aspartame. all other amines and carboxylic acids have been converted into amides or esters. Isoelectricpointforaspartame=3.65+9.602=6.63

My Attempt The plot shows the absorbance spectra for solutions of caffeine, benzoic acid, and Mountain Dew® soda, each in 0.010 M HCl. What is the approximate absorbance of benzoic acid at 228 nm? A= What is the molar absorptivity (ϵ)(ϵ) of benzoic acid at 228 nm? Assume that the path length is 1.00 cm. ε = M−1*cm−1

A = 0.8 ε = 11178 M−1*cm−1 Beer's Law describes the relationship between solution concentration (C), absorbance (A), path length (b), and molar absorptivity (ϵ.)(ϵ.) A=ϵbC Before using this equation to solve for the molar absorptivity, the concentration given on the graph must be converted to molarity. The concentration of the benzoic acid solution is 8.74 mg/L. Convert this concentration to molarity. The formula for benzoic acid is C7H6O2. 8.74 mg benzoic acid / L × 1 g/1000 mg × 1 mol benzoic acid / 122.12 g= 7.16×10−5 M benzoic acid Use the absorbance at 228 nm (0.80), the path length given (1.00 cm), and the concentration in molarity to solve for the molar absorptivity of benzoic acid. ϵ=0.80(1.00 cm)(7.16×10−5 M)=11000 cm−1⋅ M−1

consider two different solutions of the same UV-absorbing compound. Upon UV-analysis, solution B gave an absorbance of 0.8 and solution A gave an absorbance of 0.4 at the same wavelength. Solution B is higher in concentration then solution a by what ratio or percentage?

A = e*c*l Absorbance A / Absorbance B = 0.8/0.4 = 2

My Attempt Which of the following statements relates the value and magnitude of ΔHf to stability of a product? A value of zero of ΔHf indicates a stable product. The value and magnitude of ΔHf cannot be in relation to the stability of a product. A large negative value of ΔHf indicates a stable product. A large positive value of ΔHf indicates a stable product. Solution

A large negative value of ΔHf indicates a stable product. The more stable the species, the lower its energy. Because exothermic reactions have negative values for enthalpy change, a large negative value of ΔHf would indicate a more stable product.

why do we use an ammonium chloride solution rather than only water in the work-up of the grignard reaction?

Alcohols [ROH] have a pKa of about 16 and water has a pKa of 15. Ammonium ion is more acidic then these with a pKa of 9.24 Since the pKa's are similar for the water and the alcohol product, it would require a much greater amount of water to be able to protonate the grignard product. However, ammonium chloride donates its protons much faster and readily [bc lower pKa] and would allow the product to form w/o having to add a much greater amount of ammonium chloride that the reactants.

Classify the safety concerns that are associated with the given molecules. Aspartame Methanol Ninhydrin Ceric ammonium nitrate Potassium permanganate

Aspartame is an irritant. Methanol is toxic. Ninhydrin is toxic, an irritant, and an oxidizer. Ceric ammonium nitrate is toxic, an irritant, and an oxidizer. Potassium permanganate is toxic, an irritant, and an oxidizer. Keep in mind that if you come in contact with potassium permanganate and ninhydrin, they will stain your skin temporarily.

Indicate whether each of the given amino acids is acidic, basic, neutral polar, nonpolar aliphatic, or aromatic Aspartate Arginine Histidine Asparagine Threonine Cysteine Valine Isoleucine Phenylalanine Tryptophan`

Aspartate is the salt form of aspartic acid. It has an ionized carboxyl group in the side chain at neutral pH, so it carries a negative charge and is acidic. Arginine has a protonated guanidinium group in the side chaing at neutral pH, which carries a positive charge, so it is basic. Histidine has a partially protonated imidazole group in the side chain at neutral pH (the pKpK is about 6), so it carries a partial positive charge and is considered basic. Asparagine has an amide group in the side chain which is not charged, so it is neutral polar. Threonine has a polar hydroxyl group in the side chain which is not charged, so it is neutral polar. Cysteine has a thiol group in the side chain which is not charged and which is polar enough to be considered neutral polar rather than hydrophobic. Valine has an isopropyl group in the side chain which is not charged, so it is nonpolar aliphatic. Isoleucine has a secondary butyl group in the side chain which is not charged, so it is nonpolar aliphatic. Phenylalanine has an aromatic benzene ring in the side chain which is not charged, so it is aromatic. Tryptophan has an indole aromatic ring in the side chain which is not charged, so it is aromatic.

Right the reaction of the borrow hydride ion with water using curvy arrow's to show electron movement 2 form one molecule of hydrogen gas. Draw a complete line bond structure for the bow hydride ion.

BH4 + H2O --> BH3 + OH- + H2

why would it be difficult to observe a positive ceric nitrate test on an old bottle of a diet cola drink [ex. diet coke it if contained a hydrolyzed aspartame]? hint: what indicates a positive test and what color is the starting solution from cola drinks?

Because a positive result for ceric nitrate test would have a color change to a brownish color, which is a color similar to cola, a + result would be difficult to detect.

True or False. Some enzymes remove single amino acids from proteins by beginning at the amine or N-terminus.

Enzymes can selectively cleave specific peptide linkages between specific amino acids, within a protein, thereby creating smaller peptide fragments. true

what are key structural differences between the chemical structures of the fibers which gave intense color and the fibers that showed much less color intensity? (see discussion section of the on-line manual for structures of fabrics or figure 1 of lab manual)

For nylon 66, wool, and silk, the highly polar amide groups provide sites for hydrogen bonding to dye molecules For disperse polyester and poly-acrylic, they have weakly polar groups that do not bind well to polar dyes

My Attempt Select the group that is ortho‑, para‑ directing, but also deactivating in electrophilic aromatic substitution

HALOGEN Like all groups that contain lone pair electrons on the atom that is directly attached to the aromatic ring, halogens are ortho‑, para‑directing in electrophilic aromatic substitution reactions. However, unlike most ortho‑, para‑directing groups, halogens are deactivating in these reactions. This is because the destabilizing inductive effect of the electronegative halogen is stronger than the halogen's ability to stabilize the carbocation intermediate on the ring via resonance. These two competing effects are illustrated.

Aspartame, phenylalanine, and aspartic acid were examined by thin‑layer chromatography (TLC) using ninhydrin as the visualization indicator. The presence of methanol was detected using permanganate test and ceric ammonium nitrate test. What methods should be used to promote a safe and efficient laboratory environment for this experiment? 5

Lay the TLC plate on a paper towel because it makes picking it up from the bench top easier. If you have the slightest doubt about contamination, rinse the pipet, glassware, or both again with distilled water. Clean up any spills and wash all contaminated surfaces with water. Wash everything, including pipets, with distilled water. Place the used TLC spotters and the used TLC plates in the appropriate waste containers provided for this purpose in the hood. Laboratory methods that should be implemented for this experiment include: Place a paper towel under the TLC plate to make it easier to pick up from the bench top. Clean up any spills and wash all contaminated surfaces with water. Potassium permanganate, ninhydrin, and ceric ammonium nitrate solutions are irritants and oxidizers. Potassium permanganate will react with your skin and produce a brown stain. Ninhydrin reacts with your skin and produces a purple stain. These two stains are not harmful and will eventually wear off. To prevent contamination, you should wash everything, including your pipets, with distilled water. Trace contaminants found naturally in tap water can destroy the accuracy of your analysis. If you have the slightest doubt about contamination, rinse the pipet, glassware, or both with distilled water again. For disposal in this experiment: Place the used TLC spotters and the used TLC plates in the appropriate waste containers that are provided in the hood. Dispose of all solutions in the appropriate bottles that are labeled in the hood.

Rank the following molecules in terms of their expected λmax in the U.V.-visible spectrum. 1,5-diphenyl-1,4-pentadien-3-one 1-phenyl-1-buten-3-one Benzaldehyde

My Attempt Rank the following molecules in terms of their expected λmax in the U.V.-visible spectrum. Longest Wavelength of Absorbance 1,5-diphenyl-1,4-pentadien-3-one 1-phenyl-1-buten-3-one Benzaldehyde Shortest Wavelength of Absorbance All three molecules have conjugated π systems that act as chromophores, absorbing U.V. energy. The chromophore in 1,5-diphenyl-1,4-pentadien-3-one displays the greatest amount of conjugation. The electron delocalization due to resonance in the extensively conjugated system lowers the energy of the whole molecule, and increases the observed wavelength of absorbance of ultraviolet energy. This shift to longer wavelength by increased conjugation is known as a bathochromic shift.

Please indicate which of the compounds will yield postive results for the permanganate, ninhydrin, and ceric ammonium nitrate tests: Permanganate test Ninhydrin test Ceric ammoniumnitrate test

Permanganate test: methanol ethene Ninhydrin test: aspartic acid phenylalanine Ceric ammoniumnitrate test: methanol The potassium permanganate solution, aqueous KMnO4,KMnO4, is traditionally used for the detection of double bonds. Potassium permanganate is a good oxidizing agent and can be used as a reasonable test for the presence of an alcohol (ethene and methanol). Aspartame, phenylalanine, and aspartic acid will be examined in this experiment by thin‑layer chromatography (TLC) using ninhydrin as the visualization indicator. Ninhydrin is the best known of all tests for the presence of amino acids (aspartic acid and phenylalanine). Ceric ammonium nitrate is used to detect the presence of alcohols (methanol).

what are the three levels of polypeptide assembly and what types of connections/structure go in each level?

Primary structure: order of amino acids Secondary structure: coil ββ‑pleated sheet αα‑helix Tertiary structure: overall shape of a single polypeptide unit The primary structure of a protein refers to a unique sequence of amino acids joined by peptide bonds. Most proteins have regions in their polypeptide chains that are repeatedly coiled or folded. These coils and folds, which mainly result from hydrogen bonding, contribute to the protein's overall shape and are collectively referred to as secondary structure. Common secondary structures are the αα‑helix and ββ‑pleated sheet. The regions of secondary structure and the side chains interact to form the protein's tertiary structure, which is its overall shape. Hydrogen bonds, ionic attractions, disulfide bridges, and hydrophobic interactions contribute to this level of structure.

Match the procedural step to its purpose by dragging each step in the written procedure for the Fischer esterification of benzoic acid with methanol in the presence of mineral acid into the appropriate box Purpose: Water needs to be removed before the liquid ester product is collected

Procedure: Magnesium sulfate is added to the dichloromethane layer before the solvent is distilled off. Magnesium sulfate is a drying agent. Water must be removed from the dichloromethane solution before the solvent is distilled off because it will be much harder to remove from the liquid ester product.

Match the procedural step to its purpose by dragging each step in the written procedure for the Fischer esterification of benzoic acid with methanol in the presence of mineral acid into the appropriate box Purpose: The unreacted carboxylic acid was neutralized under mild conditions.

Procedure: The reaction mix was washed with sodium carbonate solution. Because the Fischer esterification is an equilibrium process, there is likely to be unreacted carboxylic acids in the reaction mix. The carboxylic acids are weak acids and can be neutralized and extracted into aqueous solution using mildly basic conditions.

Match the procedural step to its purpose by dragging each step in the written procedure for the Fischer esterification of benzoic acid with methanol in the presence of mineral acid into the appropriate box .Purpose: Flat-bottomed flasks transfer heat in an uneven fashion, and may fracture if heated strongly.

Procedure: The reagents were added to a round bottom flask prior to the reaction being heated under reflux. Reactions should be heated in round-bottomed flasks because the round shape disperses the heat more evenly than flat-bottomed flasks do.

Ultraviolet (UV) radiation can be subdivided into three regions: UVA, UVB, and UVC, based on their energy. What are the minimum and maximum values of wavelength in the UVA, UVB, UVC region? in nm

Region-Min-Max Wavelength UVA 315 nm; 400 nm UVB 290 nm; 315 nm UVC 100 nm; 290 nm

difference between ortho/para and meta directing substituents

Ring substituents that have lone pairs at the point of ring attachment are ortho/para activators because they have the ability to donate electrons into the ring by resonance. Electronegative halogens withdraw electrons inductively, and donate electrons by resonance conjugation of their lone‑pair electrons with the π electron system of the ring. The inductive withdrawal is stronger, resulting in ring deactivation, but the lone-pair conjugation leads to ortho/para substitution. Groups with oxygen or nitrogen (and with lone pairs) attached to the ring inductively withdraw electrons, and they donate electrons by resonance conjugation of their lone‑pair electrons with the π electron system of the ring. In this case, though, the resonance conjugation is favored and the ring is activated. Substitution is favored at the ortho/para positions. Groups containing π bonds that can conjugate with the π electron system of the ring, such as −CO2H−CO2H and −C≡N−C≡N, remove electrons from the ring through resonance conjugation and thereby deactivate the ring. (Note that the point of attachment for each group, a carbon atom, does not have lone pairs of electrons to donate into the ring.) The deactivation is stronger at the ortho and para positions, so substitution at the meta position is favored. In the case of phenyltrimethylammonium ion (C6H5N(CH3)+3)C6H5N(CH3)3+), the nitrogen atom lacks lone pairs, so there can be no donation by resonance. Similarly, there are no π bonds in the substituent for electron withdrawal by resonance conjugation. The positive charge strongly withdraws ring electrons by induction, deactivating the ring's reactivity and directing substitution to the meta position.

My Attempt Match the hazard label(s) to the appropriate reagents. Salicylamide Sodium Iodide Sodium Hypochlorite Hydrochloric Acid These hazards are outlined in the laboratory manual under the highlighted SAFETY! portion of the text.

Salicylamide: toxic, irritant Sodium Iodide: irritant Sodium Hypochlorite: Hydrochloric Acid: corrosive, irritant

My Attempt Arrange the procedural steps, from start to finish, that are required to generate a low molecular mass ester, and to characterize it by smell. Drag and drop the answers into the box in the correct order. All answers must be used.

Start of Lab 1. Add 10 drops of liquid carboxylic acid to a clean dry test tube. 2. Smell the acid by wafting your hand over the open tube. 3. Add 10 drops of alcohol to the carboxylic acid. 4. Add 5 drops of concentrated sulfuric acid to the mixture. 5. Shake the tube then place it in a warm water bath for 20 minutes. 6. Allow to cool, and pipet out the organic layer onto a watch glass. 7. Allow any excess alcohol time to evaporate, then smell the ester by wafting your hand over the watch glass. End of Lab A liquid carboxylic acid is placed in a dry test tube. The acid is smelled, then converted to an ester by adding alcohol and concentrated sulfuric acid. The reaction is warmed in a water bath for 20 minutes, and the organic layer is removed. Any excess alcohol is allowed to evaporate, and then the ester product is smelled by wafting a hand over the watch glass. The full procedure is listed in your lab manual.

Arrange the procedural steps, from start to finish, that are required for the correct use of an automatic micropipettor.

Start of Lab 1. Check that the pipette is set to dispense the correct volume. 2. Depress the knob until the point of initial resistance is reached. 3. Immerse the tip of the pipette into the solution. 4. Slowly release the plunger to draw solution into the tip. 5. Withdraw the tip from the solution, then place the tip on the internal wall of the container the solution should be dispensed into. 6. Depress the knob to the initial point of resistance for a second time. 7. Pause, then depress the knob as far as it will travel.End of Lab Answer Bank

Arrange the procedural steps, from start to finish, that are required to prepare indigo from nitrobenzaldehyde and acetone in base, and then test its ability as a dye. 7

Start of Lab 1. Mix 2-nitrobenzaldehyde and acetone in a test tube, then slowly add sodium hydroxide. 2. Collect the purple precipitate by filtration and wash with water followed by ethanol. 3. Fit a slit pipette bulb over the side arm of a filter flask. 4. Place solid indigo into the pre-prepared filter flask and heat vigorously with sodium hydroxide solution in the stoppered flask. 5. Add sodium dithionate until the hot solution is a clear yellow. 6. Place a fabric strip into the hot solution and heat for 10 minutes. 7. Remove the strip, rinse with water, and record the color when dry. End of Lab

Match the procedural step to its purpose by dragging each step in the written procedure for the aldol condensation of dibenzyl ketone and benzil in ethanol solution in the presence of potassium hydroxide base, followed by the UV analysis of the product in hexane solution. Purpose / Motivation---Procedure? The product has a very high molar extinction coefficient, and is sparingly soluble in hexane.

The UV spectrum was obtained within 30 seconds of preparing the solution. Ethanol is flammable; it should be heated on a steam bath for safety.Bases can cause ground glass joints to fuse together. A barrier of grease prevents this.The product has a very high molar extinction coefficient but is not very soluble in hexane. Leaving the sample in the cuvette too long before the spectrum is recorded will lead to a solution that is too concentrated to reveal any detail.The aldol condensation produces water as a product, and all the reaction steps are equilibria. Excluding water by using absolute ethanol as a solvent will help push the equilibrium forward.For a detailed safety protocol and specific directions, please consult your laboratory manual.

Complete the generic mechanism for an electrophilic aromatic substitution (EAS) reaction using E as the electrophile, and show how the sigma complex is resonance stabilized. Use curved arrows to show the mechanism and the conversion between resonance structures. Make sure to add any missing charges. Note the use of a generic base (BB) in the last step. Then, label the reaction coordinate diagram for a typical EAS reaction by correctly placing the structures on the diagram. Step 1: add a curved arrow.SelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseCH slow−−→→slow Step 2: add the missing charge, then and add a curved arrow.SelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseCH sigma complex←−−−−−−−→↔sigma complex Step 3: add the missing charge, then and add a curved arrow.SelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseCH ⟷⟷ Step 4: add the missing charge, then add curved arrows.SelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseSelectDrawRingsGroupsMoreEraseCHB fast−−→→fast Complete the reaction coordinate diagram by placing the starting benzene, transition states, intermediate and product.

The benzene ring has π electrons that will attack a very strong electrophile to form the cationic sigma complex shown in the second step. Curved arrows show how the π electrons of the sigma complex stabilize the positive charge by resonance, delocalizing the positive charge by distributing it to the carbons that are para and ortho to the electrophile as shown in the third and fourth steps, respectively. The reaction coordinate diagram starts with the starting material, benzene. The energy rises to the first transition state between the starting aromatic ring and the intermediate. Here the carbon-electrophile bond is forming, as demonstrated by the dotted bond. Once the bond is fully formed, the intermediate is the σ complex. The energy rises again as the carbon-hydrogen bond begins weakening, giving the other transition state, where the hydrogen has a partial positive charge. Once the hydrogen is lost, the ring is once again a stable aromatic benzene ring and has the lowest energy level. The product is the substituted benzene.

Match the procedural step to its purpose by dragging each step in the written procedure for the aldol condensation of dibenzyl ketone and benzil in ethanol solution in the presence of potassium hydroxide base, followed by the UV analysis of the product in hexane solution. Purpose / Motivation---Procedure? To exclude water from the reaction.

The hydroxide pellets were dissolved in absolute ethanol. Ethanol is flammable; it should be heated on a steam bath for safety.Bases can cause ground glass joints to fuse together. A barrier of grease prevents this.The product has a very high molar extinction coefficient but is not very soluble in hexane. Leaving the sample in the cuvette too long before the spectrum is recorded will lead to a solution that is too concentrated to reveal any detail.The aldol condensation produces water as a product, and all the reaction steps are equilibria. Excluding water by using absolute ethanol as a solvent will help push the equilibrium forward.For a detailed safety protocol and specific directions, please consult your laboratory manual.

In a Fischer esterification, a carboxylic acid and an alcohol combine in the presence of acid to make an ester and a molecule of water. The infrared spectrum shown below represents the substance isolated at the end of the reaction. Use the infrared spectrum to identify the substance, and infer the success of the reaction. What would indicate if infrared spectrum shows that isolated substance is: a. pure ester. b. unreacted alcohol. c. wet ester. d. unreacted carboxylic acid.

The infrared spectrum shows that the substance isolated from the reaction is unreacted carboxylic acid. d. Carboxylic acids display a very broad O−HO−H stretch between 3000 cm−1 and 2500 cm−1, along with a sharp C=O stretch at 1700 cm−1. b. Alcohols display a broad O−H stretch between 3500 cm−1 and 3000 cm−1 a. Esters display a sharp C=O stretch at 1700 cm−1. c. Wet esters display a sharp C=O stretch at 1700 cm−1cm−1, and a broad O−H stretch between 3600 cm−1 and 3000 cm−1 due to the water. Note that the broad O−H stretch in a carboxylic acid occurs at significantly lower frequency than the O−H stretch in free water.

My Attempt Which of the following statements are true of the ratio of products formed by a reaction under thermodynamic control?

The ratio of products is determined by energies of products. The ratio of products is not determined by relative energies of transition states. A reaction is said to be under thermodynamic control when sufficient energy is available to make the reaction reversible. When a reaction is under thermodynamic control, the relative amounts of the products depend on their stabilities.

Which of the following statements about the ratio of products formed by a reaction under kinetic control are true?

The ratio of products is determined by the relative energies of transition states. The ratio of products is not determined by the energies of products. A reaction that is irreversible under the conditions employed in the experiment, is said to be under kinetic control. When a reaction is under kinetic control, the relative amounts of the products depend on the rates at which they are formed.

Match the procedural step to its purpose by dragging each step in the written procedure for the aldol condensation of dibenzyl ketone and benzil in ethanol solution in the presence of potassium hydroxide base, followed by the UV analysis of the product in hexane solution. Purpose / Motivation---Procedure? To minimize the flammability risk associated with ethanol.

The reaction flask was heated under reflux on a steam bath. Ethanol is flammable; it should be heated on a steam bath for safety.Bases can cause ground glass joints to fuse together. A barrier of grease prevents this.The product has a very high molar extinction coefficient but is not very soluble in hexane. Leaving the sample in the cuvette too long before the spectrum is recorded will lead to a solution that is too concentrated to reveal any detail.The aldol condensation produces water as a product, and all the reaction steps are equilibria. Excluding water by using absolute ethanol as a solvent will help push the equilibrium forward.For a detailed safety protocol and specific directions, please consult your laboratory manual.

what is the gas in the bubbles observed when the crude methyl benzoate product is washed with sodium bicarbonate? write a balanced chemical equation that produces this gas hint: consider the acidity of the solution [some HCl will be present]

The reaction produces carbon dioxide [and H3O+] (1) sodium carbonate + (1) water --> CO2 + H3O+

Camphor is reduced to isoborneol by sodium borohydride in ethanol. The instructions for safe disposal of the chemical waste generated by this reaction state that the lids should not be screwed back onto the waste containers. Why should the lids not be put back on the waste containers?

The reaction waste is still generating hydrogen gas that would get trapped in the container. Borohydride reduces aldehydes and ketones to alcohols and produces hydrogen gas as a byproduct. Sealing the waste bottle after the reduction mixture has been deposited would build up pressure of explosive hydrogen gas within the container.

reactivity of substituents on an aromatic ring

The substituents on an aromatic ring influence the ring's reactivity towards electrophilic aromatic substitution. Electron donation and withdrawal can occur by either induction or resonance. Induction is limited to one bond's length of influence and is a consequence of the position of that atom on the periodic table. For example, chlorine has a strong inductive effect because it is electronegative; it pulls electrons towards itself within a covalent bond. Resonance occurs when a substituent's lone pair or π bond delocalizes to the ring's conjugated π system. Substituents that withdraw electrons from the ring deactivate the ring towards electrophilic aromatic substitution. In a deactivating group, the atom directly attached to the ring is also multiply bonded to an electronegative atom. Substituents that donate electrons into the ring activate the ring towards electrophilic aromatic substitution. Activating groups possess a lone pair of electrons on the atom that is directly attached to the ring. Note that halogens are an exception; their strong electronegativity outweighs their donation by resonance. Aniline is the strongest activator. Its lone pair of electrons delocalizes into the ring by resonance, making the ring more negative and more reactive toward electrophiles. Toluene experiences mild electron donation into the ring by the inductive effect of the methyl carbon substituent, so toluene is mildly activated. Benzene has only hydrogen substituents that neither donate nor withdraw electrons. Fluorobenzene is capable of donating electrons into the ring by resonance, but it has powerful electronegative induction that deactivates the ring. Both acetophenone and nitrobenzene deactivate the ring by withdrawing electrons by resonance, but nitrobenzene is most heavily deactivated because it more strongly withdraws by induction and has more resonance contributors.

Consider the sodium borohydride reduction of camphor to isoborneol. A reaction was performed in which 1.200 g of camphor was reduced by an excess of sodium borohydride to make 1.024 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction. Theoretical yield: Percent yield:

Theoretical yield: 1.217 g Percent yield: 84.17 %

Which functional groups are common to all amino acids?

There are 20 amino acids found in the human body. The side chain is the only variable group and is usually symbolized by the letter R. Every amino acid has an amine group and a carboxyl group.

A Fischer esterification is performed in which acetic acid is placed in a test tube along with ethanol and concentrated sulfuric acid. After the test tube was warmed for twenty minutes, it was noticed that the reaction mixture contained two layers. Top or Bottom Layer? Organic Layer Ester Aqueous Layer Sulfuric Acid Carboxylic Acid Alcohol .

Top Layer: Organic Layer Ester Bottom Layer: Aqueous Layer Sulfuric Acid Carboxylic Acid Alcohol The ethyl acetate ester produced by the reaction forms an upper layer on the aqueous reaction mix. Ethyl acetate has a density of 0.89 g/mL while water has a density of 1.00 g/mL. The sulfuric acid, carboxylic acid and alcohol are all soluble in water

My Attempt Match the statements to the appropriate ultraviolet region. -lowest energy wave -causes skin cancer -highest energy wave -causes mild skin aging -totally absorbed by earth's ozone layer -potentially lethal

UVA: -lowest energy wave -causes mild skin aging UVB: -causes skin cancer UVC: -highest energy wave -totally absorbed by earth's ozone layer -potentially lethal Ultraviolet A contains the lowest energy waves in the ultraviolet region and is associated with skin damage and aging. Ultraviolet B is most commonly associated with skin cancer. Ultraviolet C contains the highest energy waves in the ultraviolet region and would be lethal were it not for the fact that it is blocked by the Earth's ozone layer.

Only one aldehyde gives a positive test with the iodoform test reagent. Give the name and draw the structure of that aldehyde in the boxes below.

acetaldehyde The iodoform reaction converts methyl carbonyls into carboxylic acids. Aldehydes have a hydrogen directly attached to the carbonyl group, so the only possible aldehyde that can give a positive test for the iodoform test is ethanal, also known as acetaldehyde.

A small generic section of the primary structure of an α helix is given. -amino acid1-amino acid2-amino acid3-amino acid4-amino acid5-amino acid6-amino acid7- Which amino acid residue's backbone forms a hydrogen bond with the backbone of the second (2nd)(2nd) residue? EXPLAIN

amino acid 6 In α helices, the hydrogen bonds that form the secondary structure form between atoms of the peptide backbone. The HH atom attached to the NN atom of the peptide linkage forms a hydrogen bond with a carbonyl oxygen atom of the fourth amino acid on the N-N-terminal side of the peptide bond. Therefore, for the peptide given, hydrogen bonds form between the backbones of amino acids 2 and 6, and between amino acids 1 and 5. -amino acid1-amino acid2-amino acid3-amino acid4-amino acid5-amino acid6-amino acid7- The backbone of residue 66 forms a hydrogen bond with the backbone of the second residue. Hydrogen bonds, ion pairs, and hydrophobic interactions between amino acid side chains can stabilize the α helix. These interactions commonly occur between amino acids spaced three, or sometimes four, residues apart.

The hydrogen bonding responsible for the secondary structure of a protein generally takes place between backbone carbonyl oxygen atoms and amide hydrogen atoms. In contrast, the hydrogen bonding that contributes to the tertiary structure generally occurs between amino acid side chains. Which bonds depict the hydrogen bonding that occurs between serine residues and contributes to the tertiary structure?

between the two hydroxyl groups Solution Hydrogen bonds can form between the hydrogen atom bonded to an atom of a very electronegative element (F, N, or O) and another atom of one of these electronegative elements. An example of a hydrogen bond is O-H⋅⋅⋅O where the dots represent a hydrogen bond. The side chain of serine is -CH2OH-CH2OH. Oxygen has the highest electronegativity value. Therefore, the hydrogen bonds that form between the O atom of one hydroxyl group and the H atom of the other hydroxyl group (b and c) contribute to the tertiary structure of the protein. The amide group is not part of the side chain, so the hydrogen bond formed between the O atom of one amide group and the H atom of the other amide group (d) contributes to the secondary structure of the protein.

My Attempt A series of six solutions was prepared at the given concentrations. SampleABCDEFConcentration (mM)50100150200300400 The absorbance of each solution was then measured using a colorimeter. Sample: ABCDEF Absorbance: 0.25, 0.50, 0.75, 1.00, 1.50, 2.00 What is the concentration of the solution with an absorbance of 0.5 ? c= mM What is the absorbance of a 50 mM solution? A= As the concentration of the solution increases, the absorbance of the solution.... a. increases b. decreases. c. stays the same.

c = 100 A = 0.25 increases The colorimetry data allows you to observe the effect of light passing through the solution at each concentration. First, identify the solution with an absorbance of 0.500.50 . At that absorbance reading, the concentration is 100 mM.100 mM. Next, identify the solution with a concentration of 50 mM.50 mM. The absorbance reading at that concentration is 0.250.25 . Finally, notice that the concentration, or amount of solute in a solution, does impact the amount of light that the solution absorbs, as demonstrated by the changing absorbance readings as the concentration data changes. The data shows that the absorbance of the solution increases as the concentration of the solution increases. Therefore, the absorbance and concentration of a solution are directly related. The specific relationship is an equation known as Beer's Law, A=abcA=abc , where AA is absorbance, aa is the absorptivity of solute, bb is the path length of the light, and cc is the concentration of the solution.

Rank the following carbocations in terms of their energy. least carbocation A Carbocation B Carbocation C most

carbocation A Carbocation c Carbocation b Carbocation A is an isolated primary carbocation. It is the least stable because it is only bonded to one carbon atom. Carbocation B is the most stable carbocation because it is a tertiary carbocation that can be conjugated into the allylic π system. It benefits from inductive stabilization and resonance stabilization. Carbocation C is an isolated secondary carbocation. It is more stable than the primary carbocation because it is able to spread its positive charge across two carbons, due to the weak positive inductive effect of carbon.

structure and class of asparagine

class: neutral polar

which fabric(s) give the most intense colors with each of the dyes synthesized in this lab? indigo? azo?

indigo: the fabrics that gave the most intense colors was filament acetate, viscose rayan, amd worsted wool azo dye: the fabrics were nylon 66, silk, and worsted wool

Leucoindigo is deep purple in color. t/f

f Azo dyes do not fade when exposed to air or light and are technically too small to provoke an allergic response. Most azo dyes are not readily used as food colorants because some of the aromatic amines used to prepare azo dyes are considered carcinogenic. A few azo dyes are an exception, including Yellow #5, Yellow #6, Red #40, Citrus Red #2, and Orange B. Vat dyes must be reduced before being used on fabric. In this case, indigo is reduced to form the colorless, water‑soluble derivative, leucoindigo. After application the fabric is air dried, which oxidizes leucoindigo and turns it back into indigo, producing an intense color.

Benzoic acid is a lachrymator. t/f .

false Benzoic acid is an irritant but is not a lachrymator.

Dichloromethane is flammable. t/f

false Dichloromethane, also known as methylene chloride, is not flammable but is listed as a possible human carcinogen by the Environmental Protection Agency (EPA).

t/f For hot vacuum filtration, the filter paper should be completely dry when pouring the hot solution into the Buchner funnel to filter.

false For hot vacuum filtration, the filter paper will adhere to the bed of the funnel when the water aspirator is turned on full blast. The filter paper should then be carefully moistened with hot, clean solvent. This will help the paper to stick to the plate and gives an immediate sign if something is incorrect.

Select the structure of the intermediate carbocation in the reaction. E is an abbreviation for electrophile. C6H6+E+⟶Intermediate⟶C6H5X+H+C6H6+E+⟶Intermediate⟶C6H5X+H+ draw structure of the intermediate is:

https://docs.google.com/document/d/1lmwHSuTs1rCIFNqGpg_u1V1CSEs_X-YQQLAJu2PSRH0/edit?usp=sharing During electrophilic aromatic substitution, one of the C=CC=C bonds breaks to form the new C-EC-E bond. The carbocation forms on the carbon atom adjacent to the C-EC-E bond, so compound C can be eliminated. Only one E atom is transferred to the product, so compound D can be eliminated. The carbon atom of the C-EC-E bond in compound A is vinylic, but in an electrophilic aromatic substitution the C=CC=C bond breaks to form a new C-EC-E bond and a carbocation, so compound A can be eliminated. Compound B contains a single new C-EC-E bond attached to a carbon that is not vinylic. The structure shown is the major contributing resonance structure, but there exists another equivalent structure in which the carbocation is adjacent to the new C-EC-E bond. Therefore, compound B is the intermediate carbocation.

Consider the structures of salicyclic acid, aspirin, and oil of wintergreen. Aspirin and oil of wintergreen are both esters of salicylic acid. Write equations for the conversion of salicylic acid into each of these esters using Fischer esterification.

https://i.groupme.com/1125x1500.jpeg.2312d3f7f59d4e94ae1f335939d7fd27.large

labeled ir spectrum of salicylamide

https://i.groupme.com/1125x1500.jpeg.e3b442468f314e14860489dec0b1d89b.large

draw the tripeptide of Glu-Leu-Ser how many peptide bonds are in this structure?

https://i.groupme.com/320x157.png.1e8ee78c1a1b464fb7b9de878591ae04 2 peptide bonds

Draw aspartic acid (aspartate) at pH 1, pH 7, and pH 13. Include all hydrogen atoms.

https://us-static.z-dn.net/files/d07/24803f1ba76ac738b9389185f348d97a.png Aspartic acid (aspartate) is one of the 20 standard amino acids. The structure of a generic amino acid with side chain R at physiological pH (7.4) is The side chain (R group) for aspartic acid is −CH2COOH.−CH2COOH. At pH 1, the amine is present as the conjugate acid. The carboxylic acid is present in its protonated form (COOH).(COOH). There is an overall +1 charge. At pH 7 (and at physiological pH, 7.4), the amine is present as the conjugate acid and the carboxylic acid is present as the conjugate base. There is an overall −1 charge. This form of aspartic acid is called aspartate. At pH 13, the amine group is present as NH2,NH2, and the carboxylic acid is present as the conjugate base. There is an overall −2 charge.

Structure of.. m‑bromonitrobenze .

https://www.sigmaaldrich.com/content/dam/sigma-aldrich/structure1/092/mfcd00024298.eps/_jcr_content/renditions/mfcd00024298-medium.png When two functional groups are on a benzene ring, their position on the benzene ring can be described as ortho, meta, or para. One functional group is assigned to be bonded to carbon number one and if the second functional group is bonded to an adjacent carbon atom, carbon atom number two, then it is ortho to the first functional group. If the second functional group is on carbon atom number three, then it is meta to the first functional group. If the second functional group is on carbon atom number four, then it is para to the first functional group. The prefixes ortho, meta, and para are sometimes abbreviated o‑, m‑, and p‑, respectively. General structures for otho, meta, and para substituted benzene rings are depicted in the image, where X and Y represent any group

what is the structure and color of the oxidized and unoxidized version of indigo

indigo: color: purple leucoindigo: color: colorless The molecule on the left is indigo. Indigo is an insoluble purple dye that can be converted into colorless leucoindigo by reduction. From left to right, the reaction above shows the oxygen double bond becoming a single bond, which indicates reduction.

what is the identity of the acid present in all carbonated beverages that hastens the hydrolysis of aspartame?

phosphoric acid

Azo dyes do not fade in light t/f

t Azo dyes do not fade when exposed to air or light and are technically too small to provoke an allergic response. Most azo dyes are not readily used as food colorants because some of the aromatic amines used to prepare azo dyes are considered carcinogenic. A few azo dyes are an exception, including Yellow #5, Yellow #6, Red #40, Citrus Red #2, and Orange B. Vat dyes must be reduced before being used on fabric. In this case, indigo is reduced to form the colorless, water‑soluble derivative, leucoindigo. After application the fabric is air dried, which oxidizes leucoindigo and turns it back into indigo, producing an intense color.

Most azo dyes are not used as food coloring. t/f

t Azo dyes do not fade when exposed to air or light and are technically too small to provoke an allergic response. Most azo dyes are not readily used as food colorants because some of the aromatic amines used to prepare azo dyes are considered carcinogenic. A few azo dyes are an exception, including Yellow #5, Yellow #6, Red #40, Citrus Red #2, and Orange B. Vat dyes must be reduced before being used on fabric. In this case, indigo is reduced to form the colorless, water‑soluble derivative, leucoindigo. After application the fabric is air dried, which oxidizes leucoindigo and turns it back into indigo, producing an intense color.

Amide bonds can be hydrolyzed under only acidic conditions. t/f

t Aspartame is L-aspartyl-L-phenylalanine methyl ester. Only the methyl ester of this dipeptide tastes sweet.

why does acetyl chloride [2 carbons with 1 polar functional group] react with water almost violently, but you had to warm and shake the mixture of water and benzoyl chloride [7 carbons]? hint: what did you observe when you first added benzoyl chloride to water? why did you need to shake the mixture while heating it?

the reaction of acetyl chloride and water is significantly more exothermic and reacts more readily than benzoyl chloride and water. Also acetyl chloride is more soluble in water than benzoyl chloride. [benzoyl chloride requires additional mixing and heating to increase its solubility]

My Attempt Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.8 g3.8 g of benzoic acid was reacted with excess methanol to make 3.4 g3.4 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction. theoretical yield: g percent yield % Solution

theoretical yield: 4.2 g percent yield 81 % Step 1. The number of moles in 3.8 g3.8 g of benzoic acid is (3.8 g)(1 mol/122.1 g)=0.031 mol Step 2. The number of moles of methyl benzoate formed is 0.031 mol benzoic acid=0.031 mol methyl benzoate Step 3. The molar mass of methyl benzoate is 136.2 g/mol Step 4. The theoretical yield of methyl benzoate is (0.031 mol)(136.2 g/ 1 mol)=4.2 g Step 5. The percent yield is actual yield of product theoretical yield of product × 100% = 3.4 g/4.2 g × 100%= 81%

the IR spectrum of an unknown sample idicates a C-H stretch at 2700 cm-1. Which classification test should give a "positive" result based on this IR data?

tollen's test

For recrystallization, the material that is to be purified must be a solid. t/f

true Recrystallization requires the material that is to be purified be a solid.