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q 25

Concept: E and Z designations distinguish stereoisomers about double bonds, in which Z signifies high-priority groups on the same side of the bond and E indicates the high-priority groups on opposite sides of the bond. In disubstituted double bonds, the Z conformation is also called cis and the E conformation is called trans.

q 6

In the passage they say that this is a SN2 rxn. Sn2 only gives you the inverted product wheres SN1 gives you a mix of inversion and retention. So, when you do the R/S you get R for the compound shown and that means it started with a S configuration which was inverted to R. concept: SN2 reactions result in the inversion of configuration of the electrophile if the electrophile is a chiral center. An electrophile in the S configuration will adopt the R configuration after the reaction, and vice versa.

Reducing agent mnemonic

LiCA - laiqa from FYE Li reduces CAs meaning its a very strong reducing agent, stronger than NaBH4

q6

Qstem says which of the following can RESULT from decarboxylation so I can be eliminated cuz it still has the carbolate ion attached so it hasn't undergone the process. The other two options can be assumed to have undergone teh process bc there is no carboxylic acid group Educational objective: Decarboxylation is the removal of a carboxyl group with the release of CO2.

question 16

The peaks in a mass spec represents ionizied (charged) fragments. the higher the peak, just means that the higher the amount of that fragment in that molecule. concept: Mass spectrometry is a technique that measures the molecular weight of a molecule. Molecules in a sample are bombarded with a beam of electrons, producing positively charged ions and fragments of the molecule. The ionized fragments are detected and a mass spectrum is generated, with the y-axis representing ion abundance and the x-axis representing the mass-to-charge ratio (m/z).

q 55

D fructose has 6C so its a pyranose and all the sugars are aldoses except for fructose which is a ketose so the ans choices are I and III You can also see the ketone in fischer projction where Carbon 2 would contain a ketone Educational objective: Sugars can be classified by the number of carbon atoms in their backbone, whether they contain an aldehyde or a ketone, and by the cyclic form they adopt. The suffix ""-ose"" is used to name sugars

Boiling point comparison based on surface area

Increasing surface area (less branching) increases boiling point

q 36

1 ppm is upfield and shield and i looked for the answer choice with shielded and picked that one In general, the magnitude of shielding is dependent on the local chemical environment. For instance, protons within or near methyl groups are surrounded (shielded) by a high electron density and have a smaller chemical shift (upfield). Nearby electronegative atoms, such as oxygen and nitrogen, pull electron density away from (deshield) the proton. Therefore, an NMR peak corresponding to a carboxylic acid (-COOH) proton would have a higher chemical shift (downfield). concept: Proton nuclear magnetic resonance examines the spin properties of hydrogen nuclei as measured by chemical shifts to determine molecular structure. Protons shielded by nearby electrons have smaller chemical shifts (upfield). Deshielded protons, such as those near electronegative atoms, have larger chemical shifts (downfield).

q 7

A tertiary alcohol is more sterically hindered than a secondary alcohol because it has more bonds to large atoms that can get in the way of interactions with other molecules. A secondary alcohol is more hindered than a primary alcohol for the same reason. The alcohol on carbon 2 of glycerol is a secondary alcohol whereas carbon 1 is a primary alcohol. The added steric hindrance at carbon 2 prevents SN2 reactions from happening as readily as they do at carbon 1. Therefore, the C1 hydroxyl group reacts with an ethylhexyl halide more easily than the C2 hydroxyl group does. concept: Sterically hindered chemical groups are physically inhibited from interacting with other molecules. For this reason, they are less capable of acting as nucleophiles in SN2 reactions. Tertiary alcohols are more sterically hindered than secondary alcohols, which are more sterically hindered than primary alcohols.

q 24

Concept: A stereocenter is any atom that when two substituents switch positions, a stereoisomer is formed. A chiral center is an atom with four unique attachments. If a central atom is bonded to two or more identical atoms, the next atom in each chain should be considered to determine if the attachments are the same or different.

Boiling points of functional groups

Increasing intermolecular forces, increasing boiling point

q 27

Just had to know that enantiomers have opposite stereochemistry at every chiral carbon. So (R,R) and (S,S) shows that Educational objective: Enantiomers rotate plane-polarized light by the same magnitude in opposite directions. No correlation exists between diastereomers and the magnitude or direction of rotation. Racemic mixtures have a rotation of zero because they consist of equal amounts of each enantiomer.

question 22

Looking at the structures of ethyl acetate and chloroform in the passage, you can see that chloroform only has 1 H so there is no other H for it to be even chemically equivalent with. C and D - they are easy to see when you draw out the structure and you can see it has chemically equivalent H

q8

The carbon adjacent to the carbonyl group (carbon 2) is the α-carbon. The α-carbons in enols are susceptible to substitution because the carbocation formed is stabilized by resonance with the enol hydroxyl group. In this case, the α-carbon becomes brominated after the addition of Br2. The bromine on the carbonyl carbon (carbon 3) is eliminated during hydrolysis. Carbon 2 is the only carbon to remain brominated after hydrolysis, yielding the final product, an α-bromoacid. concept: The α-carbon of a carboxylic acid is the carbon atom adjacent to the carbonyl carbon, and it is susceptible to substitution reactions such as bromination. The carbonyl carbon of a carboxylic acid may be temporarily brominated, but this carbon is susceptible to hydrolysis when water is added.

q 13

The carbonyls of ketones and aldehydes have a partial positive charge on the carbon atom and a partial negative charge on the oxygen atom, forming a dipole moment and allowing dipole-dipole interactions to occur. Branching causes a relatively small boiling point difference in compounds with similar functional groups and MWs. Therefore, a five-carbon ketone and a branched aldehyde of the same MW have similar boiling points. less branching = higher surface area = higher boiling point concept: Boiling point depends on intermolecular forces, surface area, and molecular weight (MW). Strong intermolecular forces and greater surface area (less branching and higher MW) have higher boiling points compared to molecules with weaker intermolecular forces and a smaller surface area. Ketones and aldehydes of the same MW have similar boiling points because they both have dipole-dipole interactions.

question 11

Ultraviolet (UV) light can be used to visualize the results from TLC if the components from the reaction mixture can absorb UV light. UV light carries a large amount of energy that can excite the electrons of UV chromophores to a higher energy state. UV chromophores include double and triple bonds, carbonyls (C=O), nitroso groups, alkyl halides, and conjugated systems. concept: Absorption of ultraviolet (UV) light induces electron excitation, and a compound must contain a UV chromophore to absorb UV light.

q 47

A common protecting group used for aldehydes is known as an acetal, which is characterized by two -OR groups in place of the carbonyl. This group can be distinguished from the ketal protecting group, which is formed by two -OR groups in place of a ketone rather than an aldehyde. Nucleophilic addition of one equivalent of an alcohol to an aldehyde creates a hemiacetal, where one -OR group and one -OH are on the carbon atom that is part of the carbonyl. When another equivalent of alcohol is added, an acetal (two -OR groups) is formed. The protecting group can later be removed via acid-catalyzed hydrolysis 1,2- and 1,3-diols can also be used to form cyclic acetals. Concept: The carbonyl groups of aldehydes and ketones can be protected from nucleophiles by reaction with two equivalents of an alcohol to give an acetal and a ketal, respectively. Reaction with one alcohol equivalent (the halfway point) produces one -OR group and one -OH group in place of the carbonyl, generating a hemiacetal (for aldehydes) or hemiketal (for ketones).

q5

A conjugated system can be identified by alternating single bonds and double or triple bonds, or in other words, by alternating π bonds (p orbitals) separated by σ bonds. In general, conjugation and electron delocalization serve to stabilize molecules such as carbanions, carbocations, and radicals by creating a more favorable charge distribution, therefore reducing charge density. The quinonoid intermediate (Figure 3) delocalizes the negative charge of a carbanion over a large conjugated system of π bonds whereas the enolate (Figure 1) delocalizes the negative charge over a relatively small region via its resonance forms. Although not all of the forms are shown in the passage, the quinonoid intermediate has a greater number of resonance forms than the enolate intermediate. This causes the quinonoid intermediate to be more stable because increased delocalization of electron density is a stabilizing factor. Educational objective: A conjugated system is stabilized through electron delocalization via resonance, the distribution of electrons through alternating π bonds. Conjugated molecules can be identified by alternating single bonds and double or triple bonds

q1

A glycosidic bond is the α- or β-linkage between the hemiacetal or hemiketal group of a sugar molecule and a hydroxyl (-OH) group of another molecule. Glycosidic bonds are broken via hydrolysis reactions (addition of H2O), frequently with the aid of an enzyme, resulting in two smaller components. To complete the reaction, Glu35 must return to its protonated form and a hydroxyl group must be added to Mur2Ac to release Asp52. This can be accomplished by the addition of H2O, which is deprotonated by Glu35. This deprotonation creates a good nucleophile and facilitates a nucleophilic attack on the Mur2Ac anomeric carbon with Asp52 acting as the leaving group. Therefore, Glu35 removes a proton from H2O to create a good nucleophile for the cleavage of the glycosidic bond between Mur2Ac and Asp52. Concept: A glycosidic bond is the α- or β-linkage between a sugar and an -OH of another molecule. Hydrolysis of a glycosidic bond is cleavage of the linkage by addition of H2O, breaking the molecule into two smaller units.

q15

A retro-aldol reaction is the reverse of an aldol condensation. The retro-aldol reaction yields two products: either two ketones, two aldehydes, or one of each. The products depend on the substituents on the carbonyl carbon and the β-carbon. If the carbonyl carbon is bonded to a hydrogen atom before the reaction (ie, it was already an aldehyde), it will still be an aldehyde after the reaction. If the β-carbon is bonded to a hydrogen, it will become an aldehyde. If the carbonyl or β-carbon are not bonded to hydrogen, they will become ketones. In Compound 4 the β-carbon is bonded to hydrogen and the carbonyl carbon is not. Therefore, a ketone will form from the carbonyl and an aldehyde will form from the β-carbon when Compound 4 is heated and treated with a base. concept: The retro-aldol reaction is the reverse of the aldol condensation. When heated and treated with base, the carbon-carbon bond between the α- and β-carbons is broken, forming two ketones or two aldehydes, or one of each, depending on the substituents on the carbonyl carbon and the β-carbon. Aldehydes form if the carbonyl and/or β-carbon has hydrogen substituents, whereas ketones form if all substituents are carbon chains.

question 34

ASK LANDON For the electromagnetic spectrum remember RAV4 -- 4 is away from red so red is 700 and blue is 400 According to the given information, Compound 1 absorbs violet light (448 nm), so it appears yellow (complement of violet). After Cu2+ is added, the absorption maximum changes to 623 nm and the compound absorbs orange light, causing it to appear blue (complement of orange). Therefore, upon the addition of Cu2+, Compound 1 must undergo a change in electronic structure that causes the solution to change from yellow to blue. Concept: The color of a substance is determined by the wavelengths of light it absorbs, which is determined by the electronic structure of the molecule. In general, a substance appears to be the color that is complementary to the color of light that is maximally absorbed.

question 27

Absorption of UV light causes an electron transition, or excitation, from the ground state to a higher energy level. In the ground state, π electrons from double bonds are in the π bonding molecular orbital, and nonbonding electrons are in the n nonbonding molecular orbital. Upon interactions with UV light of sufficient energy, these electrons are excited to the Lowest Unoccupied Molecular Orbital (LUMO), called the π* anti-bonding orbital. The lone electron pairs on nitrogen in pyrimidine are nonbonding (n) electrons. Therefore, when absorbing UV light, these electrons are excited to the anti-bonding (π*) orbital and undergo a n → π* transition concept: Ultraviolet (UV) light is electromagnetic radiation that corresponds to wavelengths between 200 nm and 400 nm, and UV absorption causes an electron transition from the ground state to a higher energy level. In the ground state, π electrons from double bonds are in the π bonding molecular orbital, and nonbonding electrons are in the n nonbonding molecular orbital. These electrons can be excited to the π* antibonding orbital.

q13

According to the passage, chalcone (Compound 3) exhibits antimalarial properties. Chalcone derivatives (Compounds 4-7) were synthesized in an effort to create a more active antimalarial compound. The amino group is basic and will be fully protonated under physiological conditions. The resulting positive charge likely allows the chalcone derivative to interact with the acidic residues of PfFd, blocking the basic amino acid residues of PfFNR from interacting. Therefore, the amino group was most likely chosen to reduce the PfFd-PfFNR electrostatic interaction. Concept: Compounds with basic and acidic functional groups, such as amino acids with basic and acidic side chains, can form electrostatic interactions with each other due to positive-negative charge attraction of the respective functional groups. These interactions can be disrupted by other charged molecules that compete for binding.

q 32

According to the passage, noncanonical amino acids such as dansylalanine can be used to monitor protein structure because they are fluorescent. For a molecule to absorb UV or visible light, its electrons must be sufficiently delocalized. This delocalization can be achieved by a highly conjugated system, (several alternating single and double bonds). In general, a larger system corresponds to a higher wavelength absorbed. This is also resonance stabilization Dansylalanine contains an extended network of conjugated bonds through the naphthalene aromatic ring system and the sulfonyl group. This conjugation allows fluorescence, and therefore monitoring of protein unfolding, to occur. Educational objective: A conjugated system of alternating single and double bonds in a molecule causes electron delocalization. This delocalization allows the molecule to absorb a photon in the ultraviolet or visible region, exciting an electron to a higher energy state.

q 52

Acidic amino acids have negative charges and thats the ans choice you had to look for Educational objective: Amino acids are the fundamental components of proteins and are made of an amino group, a carboxyl group, and a side chain. The structure of amino acids differs only in the side chains, which are made of different functional groups with varying characteristics. Amino acids can therefore be grouped based on their side-chain classification, including hydrophilic, hydrophobic, acidic, or basic.

Question 17

Aldehyde carbonyls (C=O) have a characteristic strong absorption between 1740 and 1720 cm−1. The sample IR spectrum shows a strong absorption at 1730 cm−1, which corresponds to the carbonyl of butyraldehyde. Alcohols have a broad signal between 3650 and 3200 cm−1, so the absorption at 3350 cm−1 most likely corresponds to an O-H stretch. 2-propanol is the only choice that contains an OH group, and therefore it is the most likely contaminant. concept: In infrared spectroscopy, a sample is irradiated with infrared light and the percentage transmission is detected and recorded over a variety of frequencies. Absorption frequencies depend on the type of bonds present in a particular functional group. Characteristic functional group absorption ranges include 3650-3200 cm−1 (O-H stretch), 3300 cm−1 (sp C-H stretch), 3100 cm−1 (sp2 C-H stretch), 3000 cm−1 (sp3 C-H stretch), and 1810-1650 cm−1 (C=O stretch).

q 5

An acidic (pH <7) solution will change blue litmus paper to red whereas a basic (pH >7) solution will turn red litmus paper blue. Hydrolysis of an ester with the base LiOH creates a carboxylate ion and will have a pH >7 because LiOH is a strong base. Protonation of the carboxylate ion requires the addition of an acid such as HCl. The addition of acid to the basic solution causes the pH of the reaction mixture to decrease, becoming more acidic. Carboxylic acids are weak acids with pKa values near 4, so the pH of a solution of carboxylate ions must be brought below 4 to fully protonate the ions. The pH of the mixture can be monitored using litmus paper. When enough acid is added to the reaction and carboxylate ions become protonated, the blue litmus paper will turn red. Educational objective: Litmus paper is used as a pH indicator. In acidic solutions (pH <7), blue litmus paper will turn red; in basic solutions (pH >7), red litmus paper will turn blue. The addition of an acid to a solution will protonate bases such as carboxylate ions.

q 4

Anhydrides undergo nucleophilic acyl substitution and are cleaved into two molecules: A carboxylic acid and either an ester or amide, if the nucleophile is an alcohol or an amine, respectively. However, cyclic anhydrides are not cleaved into two molecules but instead the ring is broken, and the CA and the ester or amide are at opposite ends of the new molecule. The reaction described in the question forms a CA and an ester, so the starting material must be an anhydride. The prefix of the product's parent name is but- (a butanoic acid derivative), indicating a four-carbon chain. The reaction forms an ester functional group on the same molecule as the CA instead of forming two separate molecules; therefore, the original molecule must have been a cyclic anhydride with four carbon atoms. Educational objective: Carboxylic acid (CA) derivatives can undergo nucleophilic acyl substitution. Anhydrides are cleaved in this reaction into a CA and an ester or an amide. Cyclic anhydrides are converted into a single hydrocarbon chain with a CA on one end and an ester or an amide on the other.

q 11

Atoms I-IV in Compound 2 are oxygen (alcohol), sp2 carbon (alkene), sp3 carbon (alkane), and methyl ketone (α-H), respectively. Upon deprotonation, the negatively charged atom in the conjugate base of these functional groups is oxygen for alcohols and carbon for the others. Because oxygen is more electronegative than carbon, a negative charge on oxygen in the conjugate base of the alcohol is more stabilized than the charge on the carbon atoms in the three other functional groups. Therefore, the alcohol oxygen atom has the most acidic proton and will be deprotonated when reacted with NaOH. concept: Acidity of a functional group is determined by the stability of the conjugate base it forms. Factors such as the size and electronegativity of the proton-donating atom as well as the ability to stabilize the negatively charged atom in the conjugate base by charge distribution or resonance contribute to acidity.

q 11

BH3 reacts most readily with carboxylic acid carbonyl groups and therefore will selectively reduce carboxylic acids. As a result, the intermediate aldehyde will be further reduced to a primary alcohol. NaBH4 reduces the reactive carbonyls—ketones and aldehydes—and will not reduce the less reactive esters and carboxylic acids. If a compound that contains a ketone, an ester, and a carboxylic acid is reacted with BH3, the carboxylic acid will be selectively reduced to a primary alcohol, leaving the ketone and ester intact. When the same compound is reacted with NaBH4, only the ketone will be reduced, generating a secondary alcohol. Therefore, the products of the two reactions will be different. concept: Carboxylic acids can be selectively reduced to primary alcohols by the reducing agent BH3. They cannot be reduced by NaBH4, which is selective for the reactive carbonyls ketones and aldehydes.

question 10

Boiling chips are made of nonreactive porous material and provide nucleation sites where small bubbles of vapor can form. This effect overcomes the surface tension and allows the liquid to boil evenly at its normal boiling temperature, thereby preventing superheating. concept: Superheating happens when a liquid is heated above its boiling point, but it does not boil. Surface tension causes the vapor pressure inside bubbles to increase as they form, causing them to explode at the surface. Addition of boiling chips gives the bubbles a surface to form on as the liquid is heated, and allows for even boiling.

q 42

Bromide and fluoride in the original compound are both primary alkyl halides, which are generally good substrates for SN2 reactions. However, halide leaving group ability increases as the size of the halide ionincreases because the increased ion surface area stabilizes the negative charge on the departing ion. The small size of fluoride makes it a poor leaving group(Choice B). Therefore, the attack by the hydroxide nucleophile will occur at the bromide site. Substitution of the bromide by the hydroxyl group forms a second -CH2OH bonded on the central carbon, causing its chirality to be lost and yielding 2-chloro-2-(fluoromethyl)propane-1,3-diol. concept: SN2 reactions are substitution reactions that occur when a nucleophile attacks the electrophile and causes the leaving group to leave in the same step. These reactions occur on the less-substituted carbon, and the halide leaving group ability increases as the size of the halide ion increases.

q 41

Bromine is bonded to the carbon that is two away from the ketone carbon in the counterclockwise direction, and chlorine is three carbons away. Both bromine and chlorine are bonded to carbon by a dashed line, indicating that they are below or behind the plane of the ring. In (3R,4S)-3-Bromo-4-chlorocyclohexanone, both bromine and chlorine are below the plane, and bromine must be two carbons away from the ketone in the counterclockwise direction. A is wrong bc its in the clockwise direction not counterclockwise concept: A molecule can be drawn in multiple ways, with different projections and orientations. For different depictions of the same molecule to be equivalent, the atoms must be drawn in the same position relative to each other.

q 32

Chiral molecules rotate plane-polarized light; the angle of rotation, known as the specific rotation, is unique for each chiral molecule and must be determined experimentally. A molecule's specific rotation contains direction (+ or −) and magnitude. Rotations that are clockwise are (+) whereas those that are counterclockwise are (−). Enantiomers have specific rotations of equal magnitude but in opposite directions. Concept: Specific rotation measures the direction (+ or −) and magnitude (angle) of rotation by which chiral molecules rotate plane-polarized light. This value is a physical characteristic of chiral molecules, is unique for each molecule, and must be determined experimentally.

q 10

Chromatography methods separate molecules based on their relative affinities for a stationary phase versus a mobile phase. For compounds to separate efficiently on any column, they need sufficient time to interact with the stationary phase. Given enough time, even subtle differences in affinity for the stationary phase can be amplified, allowing separation of compounds with similar properties. Increasing the length of the stationary phase through which the compounds must travel can provide the necessary time. Ethanol and ethyl acetate have similar, but not identical, boiling points and can be separated on a gas chromatograph by running them through a longer column. Educational objective: Chromatography techniques separate compounds based on their affinity for a stationary phase over a mobile phase. In general, adjusting experimental parameters such that molecules have more time to interact with the stationary phase improves the resolution of each compound.

q 23

Chromic acid agent is a reducing agent, anything with Cr is typically reducing so you also had to know compound C is an alcohol by drawing it out and you can tell its tertiary by the 3 methyl groups attached to the central C. Had to know tertiary alcohols cannot be oxidized bc there are no more C-H bonds for you to add a Carbon group too Educational objective: Oxidation of an organic molecule requires a decrease in the number of C-H bonds and an increase in the number of C-O bonds. Primary and secondary alcohols can be oxidized to aldehydes and ketones, respectively, by various oxidizing agents. Tertiary alcohols cannot be oxidized because the tertiary carbon does not have any C-H bonds to lose

question 29

Compound 1 has a variety of functional groups, including an aromatic ring, hydroxyl groups, an amide, and an alkyl group. The sp2 C-H stretch from an aromatic ring absorbs light at approximately 3100 cm−1 in the IR spectrum, resulting in a narrow band of weak intensity. D is wrong bc its sp3 so it gives you a strong stretch around 3000 Characteristic functional group absorptions include: 1. 3650-3200 cm−1 (O-H stretch) 2. 3300 cm−1 (sp C-H stretch) 3. 3100 cm−1 (sp2 C-H stretch) 4. 3000 cm−1 (sp3 C-H stretch) 5. 1810-1650 cm−1 (C=O stretch) concept: Functional groups absorb at different intensities and different frequencies in the infrared spectrum, depending on the type of bond present in the particular functional group. Characteristic functional group absorption ranges include 3650-3200 cm−1 (O-H stretch), 3550-3060 cm−1 (amide N-H stretch), 3300 cm−1 (sp C-H stretch), 3100 cm−1 (sp2 C-H stretch), 3000 cm−1 (sp3 C-H stretch), 1600-1475 cm−1 (C=C stretch), and 1810-1650 cm−1 (C=O stretch).

question 3

DNA is polar so it would stay in the aq layer reacting with water. The question is saying that DNA precipitates out of the aq layer using cold ethanol and sodium acetate and so they are asking how can that happen. To precipitate the DNA from aqueous solution, its charge must be neutralized through extraction with ethanol and a salt such as sodium acetate. Gentle mixing with ethanol disrupts the hydration shell around DNA molecules. Sodium cations then neutralize DNA's charge via ionic bonding with phosphate groups, making DNA less hydrophilic, decreasing its affinity for the aqueous solvent, and allowing it to precipitate more efficiently. C. Cold ethanol has a lower polarity than water and the addition of ethanol disrupts DNA h bonds but the solution is not cold enough to freeze DNA Concept: "Like dissolves like" is the driving principle for extraction procedures. During extraction, nonpolar solutes separate into the organic (nonpolar) layer whereas polar solutes separate into the aqueous (polar) layer. Modifying a molecule's charge or polarity modifies its affinity for each layer.

q11

Decarboxylation is a reaction that removes a carboxyl group from a carboxylic acid with a β-carbonyl, releasing the carboxyl group as CO2 gas. An ester with a β-carbonyl can also undergo decarboxylation if the ester is first hydrolyzed to a carboxylic acid. Once the carboxylic acid is formed, decarboxylation occurs through a cyclic transition state that incorporates both carbonyls, releases CO2, and forms an enol. The enol tautomerizes to the keto form, or the α-carbon can be alkylated and form a ketone. Compound 1, a carboxylic acid derivative, is the ester ethyl hexanoate, which does not contain a β-carbonyl. Therefore, the ester cannot undergo decarboxylation. (Choice B) The carboxylic acid derivative is an ester, but it is not conjugated. In addition, conjugation is not required for decarboxylation. (Choice C) A β-carbonyl is required for decarboxylation, but a carboxyl group is also necessary. A β-diketone, a functional group made up of two carbonyls separated by a CH2, cannot undergo decarboxylation because it does not contain a carboxyl group. (Choice D) Esters do contain an acidic proton (α-hydrogen), but acidic protons do not take part in the decarboxylation mechanism. Educational objective: Decarboxylation is a reaction that removes a carboxyl group from a carboxylic acid with a β-carbonyl, releasing the carboxyl group as CO2 gas. A β-carbonyl is necessary for decarboxylation because a cyclic transition state incorporating both carbonyls is formed. Esters with a β-carbonyl can also undergo decarboxylation if they are hydrolyzed to a carboxylic acid first.

question 6

Didn't know hexane was nonpolar. Should have known bc hexANE The use of a C18 hydrocarbon HPLC column indicates that the stationary phase is nonpolar, and therefore corresponds to RP-HPLC. A polar solvent must be used as the mobile phase in RP-HPLC. Acetone, methanol, and water are all polar solvents (Choices A, B, and D). These solvents contain oxygen (electronegative), have dipoles, and experience dipole-dipole interactions with one another. Hexanes is a nonpolar solvent and would be used as a mobile phase in NP-HPLC but would not be used in RP-HPLC. Educational objective: High-performance liquid chromatography (HPLC) is a technique that separates compounds based on polarity. The two types of HPLC are normal-phase (NP) and reverse phase (RP), in which NP-HPLC uses a polar column (stationary phase) and a nonpolar solvent (mobile phase) and RP-HPLC uses a nonpolar column and a polar solvent.

q 28

Each stereocenter can adopt two different conformations, so the expression 2^n, where n is the number of stereocenters in the molecule, provides the maximum number of stereoisomers possible for the particular compound. For example, labetalol has two stereocenters, and therefore 2^2 = 4 possible stereoisomers. A compound with three stereocenters would have 2^3 = 8 possible configurations, and so on. KNOW THAT: A molecule with n stereocenters can have a maximum of 2^n stereoisomers.

Specific rotation

Enantiomers have specific rotations of equal magnitude but opposite directions. If the molecules were enantiomers, they would either have specific rotations of +40° and −40° or +25° and −25°. Diastereomers differ in the magnitude of their specific rotations (and may differ in direction),

q6

Esters are carboxylic acid derivatives formed by a reaction between a carboxylic acid and an alcohol. Under acidic conditions, this reaction is known as a Fischer esterification. The first step of Fischer esterification is protonation of the carbonyl oxygen, followed by nucleophilic attack of the electrophilic carbonyl carbon by the alcohol. Next, a proton is transferred from the alcohol oxygen to the hydroxyl group of the carboxylic acid, resulting in the formation of water (a good leaving group). Finally, the carbonyl oxygen is deprotonated to yield an ester. For esters, the alkyl group (from the alcohol) is named first as the substituent prefix, followed by the carboxylic acid. The suffix -ic acid is then replaced by -ate. Therefore, the reaction of butanoic acid and ethanol by a Fischer esterification yields ethyl butanoate. Concept: Esters are carboxylic acid derivatives and may be formed by the condensation of a carboxylic acid and an alcohol under acidic conditions (Fischer esterification). Esters are named by using the alkyl group from the alcohol as a substituent prefix and the carboxylic acid as the root name, with the -ic acid suffix replaced by -ate.

q20

Esters are named by first stating the prefix that corresponds to the alcohol alkyl chain, followed by the name of the carboxylic acid. The carboxylic acid suffix -ic acid is replaced by -ate. The passage states that ethanol and hexanoic acid underwent an acid-catalyzed reaction to form a carboxylic acid derivative. Therefore, the ester ethyl hexanoate was formed. Extra info: Carboxylic acid derivatives = acid halide, anhydrides, ester and amide Concept: Esters are carboxylic acid derivatives formed by Fisher esterification, an acid-catalyzed reaction of a carboxylic acid and an alcohol. Esters are named by stating the alcohol chain prefix followed by the name of the carboxylic acid, with the suffix -ic acid replaced by -ate.

question 23

Ethyl acetate has three sets of nonequivalent hydrogens (two different CH3 groups and a CH2 group) that follow the n + 1 rule. The CH3 group adjacent to the carbonyl (n = 0) appears as a singlet (2.1 ppm), whereas the CH3 adjacent to the CH2 (n = 2) appears as a triplet (1.3 ppm). The CH2 is adjacent to an oxygen atom and a CH3 (n = 3); therefore, n + 1 = 4 and the CH2 corresponds to the quartet at 4.1 ppm in spectrum from the GC extraction. You can eliminate C and D bc the answer choices correspond to B&D graph and the question is asking about ethyl acetate and that wasn't used by B&D concept: Spin-spin splitting of peaks in an NMR spectrum results from interactions between nonequivalent hydrogen atoms within three bonds of each other. The splitting pattern can be determined by the n + 1 rule, where n is the number of neighboring hydrogen atoms when J is the same for all nonequivalent hydrogen atoms.

Method of extraction

Extraction is a technique used to remove compounds from complex organic mixtures according to their solubility. The principal concept of this separation method is that "like dissolves like". In other words, substances dissolve best in solvents with the same polarity. Polar compounds, which have partial positive and negative charges, dissolve in polar solvents whereas nonpolar compounds with equal charge distribution dissolve in nonpolar solvents. Therefore, immiscible solvents, which have different polarities and form layers that do not mix, are used to separate compounds in organic mixtures. Extractions are performed by adding a polar or non-polar solvent to a mixture in a separatory funnel. The mixture is shaken so that each product transfers to the solvent in which it dissolves best and is then allowed to settle into layers. The organic layer contains non-polar compounds and the aqueous (water-rich) layer contains polar compounds. The position of the layers in the funnel depends on the densities of the solvents. Denser solvents settle at the bottom of the funnel and less dense solvents float on top.

q10

Fig 1 in passage contains acidic (-ve) charged AAs so out of the ans choices only positive charged compounds will interact Educational objective: Amino acids have three components: an amino group, a carboxyl group, and a side chain. The side chains are composed of different functional groups with varying characteristics, allowing the amino acids to be categorized by these characteristics. Positively charged side chains can interact with negatively charged ones, and vice versa.

q 58

Figure 3 shows that the standard of the desired isomer has a retention time of 12 minutes. The question states that the mixture is 10:1 in favor of the desired epoxide isomer, meaning that the peak of the desired isomer will be larger than the peak of the undesired isomer. Can eliminate A and D bc the peak is not at 12 mins. Btw B and C its not C bc the elution of the undersired product should be before the desired product because it has a lower affinity for the column. concept: High-performance liquid chromatography can be used to separate mixtures into their constituents using a column that has an affinity for one of the compounds over another in the mixture. A standard shows the expected retention time of a given molecule.

q 8

Fischer esterification is the conversion of a carboxylic acid into an ester by an acid-catalyzed reaction with an alcohol. Strong acids can donate their protons to other molecules, providing the molecule with an increased positive charge, which can enhance esterification. Acid-catalyzed reactions generally do not increase the nucleophilicity of anything because nucleophiles are attracted to positive charge. Acids increase the positive charge of molecules, making them less attracted to other positive charges, and therefore making them less nucleophilic. Concept: Acids can catalyze many reactions by donating protons to a reactant, generating an increased positive charge. This creates enhanced electrophiles such as carbocations and increases the stability of leaving groups. Acids generally decrease nucleophilicity of the molecules to which they donate protons.

question 20

From the question you had to know that phosphatidylethanolamine contains an amine that can be protonated to form a water soluble ammonium salt so it acts as a weak base. Hence a strong acid is required to protonate the amino group. the dimethoxyphenol is a weak acid so it much be deprotonated with a strong base concept: Organic compounds with acidic or basic functional groups enter the aqueous layer as ionic salts when acids and bases are added to an extraction. Amines are weak bases that require strong acids to be protonated. Phenols are weak acids that are only deprotonated by strong bases.

q 28

Gabriel synthesis is a method used to make primary amines including alpha amino acids without overalkylation of the amine. The amine is generated from potassium phthalimide, a "protected" form of ammonia that prevents multiple alkylations due to the steric hindrance of the phthalimide group. The synthesis begins with an SN2 reaction, where potassium phthalimide is the nucleophile that attacks an alkyl halide. The amine is then "deprotected" by removal of the phthalimide group. concept: The Gabriel synthesis is a method that uses potassium phthalimide and diethyl bromomalonate as starting materials to synthesize amino acids. The starting reagents of the Gabriel and Strecker syntheses are all planar; therefore, the product is a mixture of L- and D-amino acids.

question 2

Got this question wrong because I didn't understand how to do it In the passage they say that MS is singly charged (+1) and the molecular mass is 414 amu. So, to find the m/z ratio you have to add the charge to the mass and then divide it by 1. 414 + 1 = 415 415/1 = 415 amu and thats the answer If it was doubly charge and the mass was 412 then you would do 412+2 = 414 and then 414/2 = 207 m/z ratio To find the m/z ration you have to add the charge to the mass given and then divide by the charge to get the m/z ratio concept: Mass spectrometry separates ions according to their mass-to-charge ratio. For a singly charged protonated ion, one data peak would be observed with a molecular weight one unit greater than the unprotonated form.

q 33

Graph interpretation The fluorescence of hSOD with dansylalanine at position 33 decreases significantly as the concentration of GdmCl increases, indicating the environment around dansylalanine is changing as the protein unfolds. concept: Analysis of graphical data is important for drawing conclusions about what is occurring during an experiment. A change in the response of a probe indicates a change in the environment around the probe.

q 21

HPLC chromatograms A and B in Figure 2 show the results from synthesis of bradykinin in DMF (A) and PC (B). Chromatogram A shows only one impurity peak (peak 2) whereas chromatogram B shows multiple impurities (peaks 2-6). Even though there was only one impurity when DMF was used, the impurity and product have very close retention times, causing poor resolution of the peaks and making separation difficult. There were more impurities when PC was used, but the resolution of peak 1 and the impurity peaks is good, making the separation of bradykinin and impurities easier in PC than in DMF. For this reason, PC is a better solvent. concept: The methodology of a synthesis can be optimized by an analysis of the results, including yield and ease of purification. The conditions that give high product yield and easy separation of product from impurities are considered optimal.

q 2

Had to do R and S for the given molecule Also Carbon 4 is not chiral bc its a symmetrical compound (same thing both sides of C4) concept: The R or S configuration of a chiral carbon depends on the spatial arrangement of its substituents. When the lowest-priority group is pointed into the plane (dash), the chiral carbon is R if priorities 1-2-3 are arranged in a clockwise fashion and S if they are counterclockwise. When the lowest-priority group is pointed out of the plane (wedge), the chiral carbon is S if priorities 1-2-3 are arranged in a clockwise fashion and R if they are counterclockwise.

q 35

Had to find the hybridizations for teh carbon attached to the R group and for the O2 and they both go from sp2 to sp3. The methylated carbon atom on vitamin K hydroquinone has three sigma bonds (sp2) and becomes sp3 with four sigma bonds when the epoxide is formed. The oxygen atom (from O2) has one sigma bond and two lone pairs of electrons (sp2). The epoxide oxygen atom on vitamin K 2,3-epoxide has two sigma bonds and two lone pairs of electrons (sp3). Educational objective: Hybrid orbital theory combines bound atomic orbitals into mixed hybrid orbitals of equivalent energy and shape. To determine an atom's hybridization, count the number of lone electron pairs and sigma bonds. The resulting number corresponds with the number of hybrid orbitals or electron domains.

q9

Had to know proline is the only AA that is cyclic but DOES NOT have aromaticity bc no conjugation or double bonds Educational objective: To be considered aromatic, a compound must be planar and cyclic with conjugated pi bonds. Each atom must have parallel, overlapping unhybridized p orbitals. Finally, an aromatic compound will satisfy Hückel rule and have 4n + 2 pi electrons.

a 26

Had to know that alcohols can be used as protecting groups for carbonyls as reaction with dialcohol forms an unreactive acetal. PROTECTING GROUPS of alcohols are ALWAYS UNREACTIVE under oxidizing conditions. The chromic acid test is an oxidation reaction, and the alcohol and aldehyde would both react with chromic acid to give a positive result (blue-green suspension). Primary alcohols and aldehydes will be oxidized to carboxylic acids, and secondary alcohols will be oxidized to ketones. To confirm there is an aldehyde present in the unknown compound, the alcohol can be protected. Because protecting groups are unreactive under oxidizing conditions, only the aldehyde will be oxidized by chromic acid to form a carboxylic acid. Educational objective: Chromic acid is an oxidizing agent that can oxidize primary alcohols and aldehydes to carboxylic acids and secondary alcohols to ketones. Protecting groups are organic molecules added to a functional group to prevent them from reacting under a given set of reaction conditions. Protecting groups tend to be stable against oxidation and reduction.

q 12

Had to know that tertiary alcohols are not readily oxidized while secondary and primary alcohols are. When carbon atoms are oxidized, they lose bonds to hydrogen and/or gain bonds to oxygen. Therefore, oxidizing agents convert primary alcohols to aldehydes and carboxylic acids, and secondary alcohols to ketones. However, tertiary carbon atoms have no bonds to hydrogen. Because no bonds to hydrogen can be lost, additional bonds to oxygen would result in a carbon atom with five bonds. Carbon can participate in only four bonds, so this is not possible. Therefore, tertiary alcohols cannot be oxidized by losing bonds to hydrogen or by gaining bonds to oxygen. Educational objective: Oxidation of carbon atoms involves the loss of bonds to hydrogen and is frequently accompanied by the formation of bonds to oxygen. Primary and secondary alcohols can be oxidized to carboxylic acids and ketones, respectively, but tertiary alcohols are in their highest possible oxidation state

q3

Had to know the order of increasing stability for these species is: Carbocation: methyl < primary < secondary < tertiary --> most stable Carbanion: methyl > primary > secondary > tertiary --> least stable Educational objective: Because surrounding alkyl groups are electron donors, the order of decreasing stability for carbocations is tertiary, secondary, primary, and methyl. Conversely, the order of decreasing stability for carbanions is methyl, primary, secondary, and tertiary.

q2

Had to look at ""amino acid PLP"" and recognize that an imine is a double bond N and the N has an R group or a hydrogen directly attached to it Educational objective: Imines are functional groups composed of a carbon-nitrogen double bond with hydrogen or an organic group attached to a nitrogen atom. Through tautomerization, an imine can be converted to an enamine, an amine group bonded to an alkene.

q3

Had to look for most similar sugar to the one in compound 1. Can't be a fructose even tho it has 6 Cs is bc it has an anomeric C at C-2 and compound 1 has anomeric C at C-1. Not ribose bc its 5C sugar but Compound 1 is 6C. Not glucose bc it differs in the orientation of the substituent at C-2 (OH is down when it shoudl be facing up like in compound 1). Mannose is correct bc C-2 and C-3 Carbon OH's are both pointing up and C-4 OH is pointing down just like in compound 1. Epimers are diastereomers that contain more than one chiral center but differ from each other in the absolute configuration at only one chiral center. Sugars: Usually exist in D form. D form also means that the bottom carbon in a Fischer-Diagram is facing to the right. Atoms facing right in Fischer - drawn axially downwards in cyclic form, Atoms facing left in Fischer - drawn upwards and equatorially in cyclic form. Educational objective: Sugars are classified in a number of ways including by their stereochemistry, the number of carbons in the sugar, the number of sugar units in the molecule, and the type of functional group present in the sugar. Sugars that are prevalent in nature are known by their common names (ie, ribose, glucose, fructose, mannose, galactose).

q 18

Had to understand the inductive effect which occurs with the donation of electron density through sigma bonds (single bonds). EN atoms are electron withdrawing groups (EWG) which pull electrons away from adjacent atom creating a dipole with partial negative towards the most EN atom and partial positive charge on the other atom. Carbocations are stabilized by electron donating groups (EDG), such as alkyl groups, because they donate electrons toward the positively charged carbon. EWG have the opposite effect: They tend to destabilize carbocations because the electronegative atom pulls electrons toward itself, creating a partial positive charge on the adjacent atom. With two adjacent positive charges, the carbocation is not stable. Methyl (-CH3) groups are EDG because the carbon atom is more electronegative than the adjacent hydrogen atom, creating a partial negative charge on carbon. This carbon can then donate electrons to the carbocation to stabilize it. The trifluoromethyl (-CF3) group is an EWG because the fluorine atoms are more electronegative than the adjacent carbon and draw electrons toward themselves, creating a partial positive charge on the carbon next to the positively charged carbocation. Because Compound 1 has one EWG and Compound 2 does not contain any EWG, Compound 1 is less stable than Compound 2 due to the inductive effect from the fluorine atoms on -CF3. Educational objective: The inductive effect occurs when electron density is donated through sigma bonds. Carbocations are stabilized by electron donating groups because they donate electrons to the positively charged carbon; they are destabilized by electron withdrawing groups because they pull electrons away from the carbocation, creating two adjacent positive charges.

q 39

Have to know structures of amino acids. The AA has 2 carboxylic acids and thats how you know its either Glutamate or Aspartate and it has an extra C so its glutamate or glutamic acid Educational objective: Glutamic acid is a negatively charged, polar amino acid with a three-carbon side chain containing a carboxylic acid.

q 7

Have to know that coverting an alcohol into a mesylate or a tosylate make it a better leaving group. Other choices are incorrect bc protecting the alcohol does not make it a better leaving group but rather decreases its reactivity. Deprotonation of the alcohol creates a negative charge on the alcohol oxygen atom, making it a nucleophile rather than a leaving group. Alternatively, protonation of the alcohol would create a strong electrophile and good leaving group in the form of water. Oxidation of the alcohol forms a double bond to the oxygen atom. It makes the carbon bonded to the alcohol oxygen a strong electrophile that is susceptible to nucleophilic attack, but the alcohol leaving group ability is decreased as there are now two bonds holding it in place. Alcohols are weak electrophiles and poor leaving groups; therefore, nucleophiles are unlikely to displace the hydroxyl group (-OH). The alcohol can be converted into a mesylate (-SO3CH3) to improve its leaving group ability. Mesylates are produced by the reaction of an alcohol with methanesulfonyl chloride and a base, such as triethylamine, and are good leaving groups because the mesylate anion can stabilize the negative charge through resonance. Similar to mesylates, tosylates (-SO3C6H4CH3) can also be used to make an alcohol a better leaving group. Educational objective: Alcohols are poor leaving groups and weak electrophiles. However, an alcohol's leaving group ability can be improved through the formation of a mesylate (-SO3CH3) or a tosylate (-SO3C6H4CH3). Mesylates and tosylates can be produced by the reaction of an alcohol with a base and methanesulfonyl chloride or p-toluenesulfonyl chloride, respectively. The mesylate and tosylate anions can stabilize their negative charges by the delocalization of electrons through resonance.

question 40

Have to know that the number of peaks determines the number of neighboring hydrogens that are neighboring to the C you are looking at. The number of peaks equals the number of neighbors + 1. In this case, Q stem says "doublet" which means one neighboring H so you look for that. You can also check the other locations and none of them have only one neighboring hydrogen. Ex. singlet = no neighboring hydrogens doublet = one neighboring hydrogen triplet = two neighboring hydrogens Concept: 1H NMR detects hydrogen atoms in a molecule as an external magnetic field and radio frequency pulse are applied to the sample. Hydrogen atoms in different magnetic environments within three bonds of each other cause spin-spin splitting of the peaks in the spectrum. The splitting pattern can be determined by the n + 1 rule (n = number of nonequivalent hydrogen atoms on adjacent carbons)

q22

Hexanoic acid can hydrogen bond with itself because it has a hydrogen bond donor (acidic hydrogen) and two acceptors (oxygen atoms). The reaction product ethyl hexanoate (an ester) is only a hydrogen bond acceptor, and therefore cannot hydrogen bond with itself. Esters experience dipole-dipole interactions, which are weaker than hydrogen bonds. Therefore, ethyl hexanoate has a lower boiling point and will distill before hexanoic acid. concept: Boiling points are dependent on a molecule's intermolecular forces, and strong intermolecular forces lead to higher boiling points. Carboxylic acids contain a hydrogen bond donor (hydroxyl hydrogen) and hydrogen bond acceptors (carbonyl oxygen and hydroxyl oxygen) and therefore can hydrogen bond to themselves. Because hydrogen bonds are strong intermolecular forces, carboxylic acids have higher boiling points than molecules of a similar molecular weight that cannot hydrogen bond to themselves and have weaker intermolecular forces.

question 38

High-performance liquid chromatography is a purification technique that separates compounds based on polarity and consists of a mobile phase, stationary phase, detector, and computer for data acquisition. The mobile phase is a solvent or mixture of solvents that are pumped through the system under pressure, and the stationary phase is a column made of either a nonpolar or polar material.

Fisher projection

Horizontal groups have to be facing you so thats how you know where to look from Another rule - highest carbon priority always on top

question 30

Hydrogen atoms that are in identical magnetic environments within a molecule exhibit rotational or planar symmetry. These atoms are said to be chemically equivalent because they have the same atoms surrounding them, have the same chemical shift, and appear as a single signal. Compound 3 contains four aromatic hydrogen atoms (Ha-Hd). Ha and Hd are chemically equivalent because they are both adjacent to an aromatic C-NO2 and an aromatic C-H, which is adjacent to an aromatic carbon with a side chain. Similarly, Hb and Hc are equivalent, being adjacent to identical chemical groups. Because there are two sets of equivalent aromatic hydrogen atoms and no other aromatic protons, there are two aromatic hydrogen signals in the 1H NMR spectrum. Concept: Hydrogen atoms in a molecule that have the same types of atoms surrounding them and that exhibit rotational or planar symmetry are chemically equivalent and experience identical magnetic environments. Therefore, equivalent hydrogen atoms have the same chemical shift and appear as a single signal in the 1H NMR spectrum.

Question 1

I got this question wrong because I forgot water is polar. Normal phase - Stationary is polar and mobile phase is non-polar Reverse phase - stationary is non-polar and mobile is polar Of the materials given, water (polar) would be appropriate for the mobile phase and C-18 alkyl hydrocarbon (nonpolar) would be appropriate for the stationary phase. concept: High-performance liquid chromatography (HPLC) separates molecules according to polarity. In reversed-phase HPLC, the mobile phase is a polar liquid solvent that carries the analyte through a nonpolar stationary phase.

q 12

I just rotated the molecule in my mind and looked at the structures that lined up but the actual way to do it is by finding the R or S configuration, so in this case compound 1 is R and had to look for another R in ans choices. For this one you can rotate the the groups and match it up with the correct answer choice Educational objective: Chiral compounds with the same four substituents on the central carbon are identical if they have the same stereochemical configuration. When the lowest-priority substituent is pointing away from the viewer, an R configuration is present if the substituent priorities follow a clockwise progression, whereas a counterclockwise progression indicates an S configuration. If the lowest-priority substituent is pointing toward the viewer, an R configuration is indicated by substituent priorities following a counterclockwise progression.

q12

I looked at Scheme 1 in the psg and looked to see what didn't follow with the ans choices. I used POE, A does occur in scheme 1 bc compound 1 and 2 do contain carbonyl carbons and they are substrates. C i just assumed occurred with many orgo rxns, the fact that they have resonance stabilized intermediates. Can also look at it this way: the α-carbon (carbon adjacent to the carbonyl) on Compound 1 is deprotonated when base is added. The intermediate formed is stabilized through resonance as the negative charge on the α-carbon is delocalized to the carbonyl oxygen, forming an enolate (Choice C). D also is true bc you can see conjugation in product. The elimination step requires the α-carbon to be deprotonated rather than protonated. The resulting carbanion is an enolate that can eliminate the -OH on the β-carbon and form a double bond between the α- and β-carbons, giving a conjugated product. Educational objective: Aldol condensations are carbon-carbon bond-forming reactions that require two carbonyl substrates (ketones and/or aldehydes). The reaction begins with deprotonation of the α-carbon on one of the substrates, forming a resonance-stabilized enolate intermediate. Nucleophilic addition of the enolate to the other carbonyl substrate yields the aldol product, and then deprotonation of the α-carbon, followed by -OH elimination, yields the conjugated product, an α,β-unsaturated carbonyl compound.

q 8

I looked at scheme 1 and looked at megestrol and megestrol acetate and saw that the ester from ethyl acetate got transferred to megestrol acetate so its transesterification Transesterification is the reaction between an ester and an alcohol that results in an exchange of alkoxy and alcohol R groups. The reaction can be either acid-catalyzed or base-catalyzed, and involves nucleophilic attack by the alcohol on the carbonyl carbon of the ester to form a new alkoxy group. The original alkoxy group leaves and becomes protonated to form a new alcohol. The conversion of megestrol to megestrol acetate described in Scheme 1 is a reaction between an alcohol (megestrol) and an ester (ethyl acetate). It results in the formation of a bond between megestrol and acetate, and the ethoxy group leaves as ethanol. This is an example of transesterification. Educational objective: Transesterification is the exchange of alcohol and alkoxy R groups between an ester and an alcohol

q 59

IR spectroscopy gives you info about the amount of energy needed to stretch a bond by giving you wavenumber (which is inverse of the wavelength) and bc its related to wavelength it is also directly related to freq by the equation c = (f)(lambda) so you can find E through the energy of photon equation. The signals also do correspond to stretching vibrations and rotations and it tells you the amt of light absorbed at a certain freq. It doesn't tell you about the spin-spin splitting of atoms. Which is a phenomenon that occurs in nuclear magnetic resonance (NMR) spectroscopy. It revolves around how protons act in a magnetic field and the effect they have on the signals of neighboring protons. Educational objective: Infrared spectroscopy is a technique used to determine the functional groups present in a sample; the data collected are plotted as percent transmittance vs. wavenumber. The signals in the spectrum correspond to bond-stretching vibrations and rotations at a certain frequency, and the signal intensity is dependent on the amount of energy absorbed.

question 24

If thin-layer chromatography shows that the distillate is not pure, the distillation was not successful at separating the compounds in a mixture. Because a mixture should be heated at a slow rate for separation, a possible error in the distillation is heating the mixture too quickly, causing the molecules to evaporate together. B. using a tall fractionating column in fractional distillation would improve the the separation of two molecules concept: Distillation is a technique that separates molecules based on their boiling points, and there are three main types: simple, fractional, and vacuum. The molecule with the lowest boiling point distills before the molecules with higher boiling points. The mixture must be heated slowly to ensure that a lower boiling-point compound evaporates before a higher boiling-point compound and that the molecules are separated.

q 4

If you can rotate plan polarized light then you had to be chiral so had to look for a chiral molecule and it was only B bc A has two substituents that are the same Educational objective: Chiral molecules contain at least one atom that has four different substituents, known as a stereocenter, or chiral center. If the atom is bonded to two or more of the same atom, the subsequent atoms are compared to determine if the substituents are different. Chiral molecules can interact with and rotate plane-polarized light.

q 43

In Compound 1, four functional groups are present, with an aldehyde group as the highest-priority group. Therefore, the compound is named as an aldehyde using the -al suffix. The longest continuous carbon chain containing the aldehyde is six carbon atoms long, and the chain is numbered to give the lowest position to the aldehyde carbon. The amine, ketone, and alcohol groups are subsequently located along the position reference numbers and are named as substituents using their respective prefixes. The configurations of the three stereogenic carbon atoms are then specified by position number at the beginning of the name to give (2R, 3S, 5S)-2-amino-5-hydroxy-3-methyl-4-oxohexanal. concept: Organic compounds with multiple functional groups are named using the highest-priority functional group to determine the suffix. All other functional groups are named as substituents using prefixes. Numbering of the longest carbon chain provides location references for substituents and stereocenter configurations and must assign the lowest possible number to the highest-priority group.

question 31

In H-NMR, upfield is shield (right, 0 ppm ) and downfield is deshielded (left, 12 ppm). An electron cloud around the protons shields them from the external magnetic field, causing an upfield signal (small chemical shift). An electronegative substituent withdraws electrons and deshields neighboring protons from the magnetic field, causing a downfield signal (larger chemical shift) concept: The location of signals in the 1H NMR spectrum is known as the chemical shift, measured in ppm. The chemical shift depends on the electronic environment around protons. Protons with an electron cloud surrounding them are shielded from the magnetic field and exhibit an upfield signal, whereas protons adjacent to electronegative substituents are deshielded from the magnetic field and have a downfield signal.

q 37

In HPLC, stationary phase = polar and mobile phase = non-polar so that means higher retention time = more polar bc more interaction with the stationary phase Retention time is correlated to affinity for the mobile phase versus the stationary phase. The contaminant has a longer retention time than trans-epoxy vitamin K1, insinuating a comparatively lower mobile phase affinity (ie, higher stationary phase affinity) during HPLC. concept: High-performance liquid chromatography is a column-based analytical technique used to separate, quantify, and identify components of a mixture. Compounds are separated based on their affinity for the mobile or stationary phase. Compounds with a larger absorption value and area under the curve are greater in concentration.

q2

In both the linear and cyclic forms, the anomeric carbon is characterized as the carbon in a carbohydrate with two bonds to oxygen. The carbonyl carbon (C-1 in aldehydes and labeled IV in the question stem) is bonded to the C-5 hydroxyl oxygen and can be identified as the anomeric carbon because it has two bonds to oxygen. (Choices A, B, and C) These carbon atoms only have one bond to oxygen and do not correspond to the carbonyl carbon in the linear form. Therefore, they cannot be the anomeric carbon and do not show a distinct C-H coupling in the 13C NMR spectrum. Educational objective: The carbonyl carbon (C-1 in aldoses and C-2 in ketoses) of a sugar in linear form and the carbon bonded to two different oxygen atoms in the cyclic form of a sugar is known as the anomeric carbon. This carbon is characterized by two bonds to oxygen and can form glycosidic bonds with biological molecules.

question 26

In gas liquid chromatography, mobile phase is an inert gas like helium or nitrogen and the stationary phase is a liquid that coats a solid support on the inside of the column. The most volatile molecules (low boiling points) spend more time in the gas phase than they spend interacting with the stationary phase of the column, so they migrate to the detector rapidly. However, molecules with higher boiling points condense more readily and spend more time interacting with the liquid stationary phase. These molecules make their way through the column slowly as the temperature in the oven increases. Polarity impacts a compound's boiling point but its not the only determining factor. concept: Gas-liquid chromatography is a technique that separates compounds in a mixture based on boiling point. The mobile phase is an inert gas, and the stationary phase is a liquid that coats the column, which is in a heated oven. The molecule with the lowest boiling point will reach the detector before molecules with higher boiling points.

q 38

In the psg, they say that unsaturated unbranched 22 carbon FA was formed. Basically needed to look for a straight hydrocarbon chain with 22C's in the answer choices. concept: Fatty acids (FAs) are made up of a nonpolar hydrocarbon chain with a polar carboxyl head group, and can be classified as either saturated or unsaturated. Free FAs are usually not found in blood plasma but rather exist as derivatives, such as a triacylglyceride or phospholipid. Hydrolysis of triacylglycerides and phospholipids releases free FAs.

q 20

In the solid-phase peptide synthesis, the variable was the solvent used. Synthesis of bradykinin was carried out in two different solvents, DMF and PC, to determine which gave better results. If a different solvent was used in the rinse step than in the reaction step, the synthesis results could not be solely attributed to the solvent in which the reaction occurred. C is wrong bc DMF and PC are both polar aprotic solvents, and are unlikely to produce different side reactions. concept: Variables are components of an experiment that change while other components of the experiment are held constant. These parameters must be isolated to determine the effect a variable has on the results of an experiment.

q 14

In the wedge-dash projection, the wedges indicate that the substituent is above or in front of the plane of the ring, and the dashes indicate that the substituent is below or behind the plane. In a chair conformation, the wedges and dashes can be shown as axial or equatorial positions. The axial positions are perpendicular to the plane of the ring and alternate pointing above or below the ring plane. The equatorial positions are oriented radially with the ring plane and alternate positions slightly above or below the plane. The chair conformations that depict −OH above the plane of the ring ("up") and −Br and −CH3 below the plane ("down") and on carbons 3 and 5 clockwise from −OH, respectively, are the same molecule as Compound 1. concept: Compounds can be illustrated in different orientations and forms, including wedge-dash projections and chair conformations. For two depictions of a molecule to be considered the same compound, the atoms must have the same relative position and stereochemical orientation.

q10

In this experiment, positive control would show the results where no bacteria are transferred. Counting the bacteria recovered from untouched spheres would serve as a good positive control and also demonstrate that the spheres have been thoroughly sterilized. A negative control would show the number of bacteria transferred if the soap used is completely ineffective. Washing hands with water only or not at all prior to handling spheres would serve as good negative controls, showing the effect of water without soap and no soap and water Recovering bacteria directly from unwashed hands instead of from spheres that were handled by unwashed hands would not demonstrate the number of bacteria transferred. concept: Experimental controls, including positive and negative controls, are required to demonstrate the validity of a test result. Positive controls show what the maximum effect would look like, and negative controls show what no effect would look like. Both must relate to the dependent variable being measured without influence from the independent variable.

q17

Inductive effect = dipole (partial neg charge on the electronegative atom and partial positive charge ont he adjacent atom) Acetic anhydride and N,N-diisopropylisobutyramide both have two electronegative atoms bonded to each carbonyl carbon. The anhydride carbonyl carbons are bound to two oxygen atoms whereas the amide carbonyl carbon is bound to an oxygen and a nitrogen. Oxygen more electroneg than nitrogen, it draws more electron density away from the carbonyl carbon than nitrogen does, creating a larger dipole on the anhydride than on the amide, as well as a more electrophilic carbonyl carbon on the anhydride. Therefore, an anhydride, such as acetic anhydride, has a greater inductive effect than an amide, such as N,N-diisopropylisobutyramide.

q 42

Isomers are molecules with the same molecular formula but different structural arrangements. They can be generally classified as constitutional isomers and stereoisomers. Constitutional isomers differ in atom connectivity whereas stereoisomers have the same atom connectivity but differ in spatial arrangement. Cis and trans are known as geometric isomers and they can be separated by gas chromatography because they have slightly different boiling points. In the psg, docosenoic acid is an unsaturated fatty acid, meaning it contains a disubstituted double bond, and can be in the cis or trans form. Both isomers have the same molecular weight and m/z ratio, but will come off the GC column at different times, yielding two distinct peaks. Therefore, it can be inferred that Peaks A and B are cis/trans isomers of docosenoic acid. Concept: Isomers are molecules with the same molecular formula but different structural arrangements, and are either constitutional or stereoisomers. Disubstituted double bonds give rise to cis or trans isomers, depending on the relative positions of nonhydrogen substituents. These isomers have slightly different boiling points and can be separated by gas chromatography.

question 33

It was stated in the passage that one reason for coming up with an analogue was to make it taste better. In some cases, the analogue is converted back to the parent drug in the body. One way is thru transesterification, which converts a hydroxyl group to an ester. Esters are readily hydrolyzed back to the alcohol (along with a carboxylic acid byproduct) in the acidic environment of the stomach. The passage states that chloramphenicol tastes bitter, making the drug unpleasant to take. To reduce the bitterness, an analogue of chloramphenicol was synthesized via transesterification. Because chloramphenicol analogues with a substituent on the primary hydroxyl group, such as Compound 3, are hydrolyzed back to the parent drug in vivo, taking an analogue has the same effect as taking chloramphenicol. Therefore, the biological activity of chloramphenicol is preserved. concept: Transesterification produces an ester from a carboxylic acid and an alcohol. The ester can be hydrolyzed back to its alcohol and carboxylic acid constituents in acidic or basic conditions.

question 37

Its a gas liquid chromatography so you have to look at boiling point. the higher the number of carbons = the higher the bp = more intermolecular forces. That is why the answer is 1-octanol concept: Gas-liquid chromatography separates molecules based on boiling point. A number of factors contribute to a molecule's boiling point: Intermolecular forces, molecular weight, and branching. For compounds with the same functional group but different carbon counts, a higher molecular weight indicates a higher boiling point and a longer retention time.

Question 28

Just had to do the Rf value for this question concept: Thin-layer chromatography is a technique used to separate compounds based on polarity. The Rf value is expressed as the ratio of the distance the compound of interest traveled to the distance the mobile phase (solvent front) traveled and is always less than 1. A larger Rf value corresponds to a less polar compound and vice versa.

q 33

Just have to know the equation and substitute 0.2g/mL for c and 2dm for l and +10 for aobs and you get +25 Educational objective: Specific rotation is the angle by which chiral molecules rotate plane-polarized light and contains direction (+ or −) and magnitude (degrees). The observed rotation αobs is dependent on the concentration (c in g/mL) and sample cell length (l in dm). Therefore, the specific rotation is standardized to account for these experimental parameters: (look at the equation on the image)

q 45

Ketones and aldehydes contain carbonyls, in which the electronegative oxygen has a partial negative charge and the carbon has a partial positive charge. Like alcohols, ketones and aldehydes have dipoles and therefore have stronger intermolecular interactions than alkanes. Unlike alcohols, aldehydes and ketones have hydrogen bond acceptors but not donors, so they cannot form hydrogen bonds with themselves. Therefore, they have lower boiling points than their corresponding alcohols. Concept: Ketones and aldehydes contain carbonyls, which create dipoles and increased intermolecular forces. Because the oxygen of the carbonyl can accept but not donate hydrogen bonds, ketones and aldehydes cannot form hydrogen bonds with themselves. Alcohols have dipoles and can form hydrogen bonds with themselves, resulting in a higher boiling point than ketones and aldehydes.

q 40

Know nucleotide structures but guanine is a purine (bigger molecule with two rings) so eliminate C and D and also guanine has one carbonyl group so its B but also tautomer means that the double bond in the crabonyl becomes a single bond and an H bonds to the O Tautomerization is a type of isomerization that involves the transfer of hydrogen from one position to another within a molecule and the movement of a double bond to an adjacent atom. These two forms, called tautomers, are in equilibrium. Tautomerization frequently involves a keto form (compound with C=O) as the major tautomer whereas the enol form is the minor tautomer. Tautomers are classified as constitutional isomers rather than resonance structures because a bond is broken during the hydrogen transfer to yield a different compound. On the other hand, resonance structures are forms of the same compound in which electrons move from one atom to another without breaking any sigma bonds. Educational objective: Tautomerization is a type of isomerization involving the transfer of a hydrogen atom from one position to another in a molecule and the movement of a double bond. The nucleotides can tautomerize, with guanine and thymine shifting between keto (major) and enol (minor) forms whereas adenine and cytosine shift between amines and imines."

q 25

LiAlH4 is a reducing agent that oxidizes 2ndary ketones to secondary alcohols. Compound E is a secondary ketone so had to pick an ans with secondary alcohol which is choice C concept: Reduction of an aldehyde or ketone with a reducing agent, such as LiAlH4, reduces the number of C-O bonds and increases the number of C-H bonds to form a 1° and 2° alcohol, respectively. Ketones are named using the suffix -one; alcohols are named using the suffix -ol.

q 57

LiAlH4 will react with a carbonyl by transferring a hydride to the carbonyl carbon, thereby increasing the number of C-H bonds, decreasing the number C-O bonds, and reducing the oxidation state of the carbonyl carbon. The ester in Compound 1 contains a carbonyl, and therefore will react with LiAlH4 and form two alcohols, splitting the molecule in two. The amide also contains a carbonyl and will react with LiAlH4 to yield an amine. concept: Hydride reagents, such as LiAlH4, act as nucleophiles and attack electrophilic centers, most commonly carbonyl carbons (C=O). These reagents cause a reduction of the electrophile, decreasing the number of bonds to electronegative atoms and increasing the number of bonds to less electronegative atoms.

question 12

Look at acetaminophen in reaction 3 to see that it is more polar than 4-aminophenol so it will be closer to the origin and elute much more slowly and then do the Rf value. concept: Thin layer chromatography is used to separate compounds based on polarity. The Rf value of a compound is a ratio of the distance up the plate a compound travels to the distance the solvent travels. A polar compound will have a smaller Rf value than a nonpolar compound

q1

Look at compound 1 and realize its a 6 carbon sugar but don't forget the CH2 outside of the chain as part of the substituent. Pyranose = 6C (i want more pie, so more Cs) Furanose = 5C So eliminate B and C. If the carbonyl is an aldehyde (at the end of the chain), the sugar is an aldose; if the carbonyl is a ketone (within the chain), the sugar is a ketose. A linear sugar can cyclize to adopt either a hemiacetal (aldose) or a hemiketal (ketose) form. A hemiacetal originates from an aldehyde and consists of a carbon bonded to a hydroxyl group (-OH), an alkyl group, an -OR group (where R is any carbon chain), and a hydrogen atom. The structure from which Compound 1 is derived from is a hemiacetal because C-1 is bonded to two oxygen atoms (a hydroxyl group and an -OR group), a hydrogen atom, and an alkyl group that formed from an aldose. Educational objective: The cyclic structure of a sugar is classified by the size of the ring as well as whether the linear form contains an aldehyde (aldose) or a ketone (ketose). Furanoses are sugars with five-membered rings (four carbons and one oxygen), and pyranoses are sugars with six-membered rings (five carbons and one oxygen). Aldoses cyclize to form hemiacetals, and ketoses cyclize to form hemiketals.

q 7

Looking at scheme 1, you have to realize that the substitution is taking place with a tertiary carbon. Tertiary carbons readily readily undergo SN1 reactions and SN1 reactions form carbocations. Thats why the first step will be formation of a carbocation once the iodide leaves. concept: Nucleophilic substitution reactions can proceed by SN1 or SN2 mechanisms. Tertiary carbons are more likely to participate in SN1 reactions due to steric hindrance and the ability to form stable carbocations.

q6

Mass spectrum can be used to identify the mass of a molecule's fragments by taking the m/z difference between peaks. The m/z difference between each major peak in the mass spectra is 162, which corresponds to a single hexose unit. The mass spectrum for the molecular ion of product A has two major m/z peaks (464 and 302), indicating that one hexose unit was cleaved. The mass spectrum for the molecular ion of product B has three major m/z peaks (626, 464, and 302), indicating that two hexose units were cleaved. concept: Mass spectrometry is a technique that ionizes molecules in a sample, and the ions can fragment. The ions are accelerated toward a magnet, deflected according to mass, and detected. A plot of ion mass abundance vs. m/z ratio is generated; fragments of the sample can be identified by the m/z difference between two peaks in the mass spectrum.

question 5

More polar molecules interact preferentially with the stationary phase (typically a silica gel that forms hydrogen bonds with polar solutes) and move more slowly through the HPLC column. Of the adducts shown, Adduct C is the most polar due to the presence of four -OH groups and two -NH groups. These groups allow Adduct C to form hydrogen bonds to the polar stationary phase with greater affinity than the other adducts. Therefore, Adduct C moves through the column most slowly and has the longest retention time and the latest elution. Concept: In normal-phase high-performance liquid chromatography, the stationary phase is polar whereas the mobile phase is nonpolar. Molecules of greater polarity move through the column more slowly (due to longer retention times) and elute later.

q 29

Most naturally occuring amino acids are L amino acids and they are S configuration so if you R/S stereochemistry fo Dansylalanine you get S configuration and you don't need to flip it bc the lowest priority group is a dash (behind the plane) so since its an S configuration it does not differ from most naturally occurring amino acids and bc cysteine has an R config Educational objective: All naturally occurring amino acids have an L configuration with the exception of glycine, which is not chiral, and therefore neither L nor D. Of the chiral amino acids, all are in the S configuration except cysteine.

q4

Numbering the carbons in Compounds 5 and 6 starting with the anomeric carbon, C-5 is the chiral center with the highest number. C-5 has an R configuration in both compounds, making these compounds D-sugars. Because the hydroxyl substituents of the anomeric carbons are on the opposite side of the ring plane as the C-5 substituents, these compounds are the α-anomers. Therefore, the configuration of Compounds 5 and 6 is α-D. MNEMONIC: DOMINICAN REPUBLIC (R CONFIGURATION = D SUGAR) concept: Sugars are classified as L or D based on the configuration of the highest numbered chiral carbon (the anomeric carbon is assigned the lowest number possible). An R configuration at the highest numbered stereocenter designates a D-sugar, and an S configuration designates an L-sugar. Sugars are also classified as α or β based on the configuration of the anomeric carbon. In α-anomers, the anomeric carbon substituent is on the opposite side of the ring as the highest chiral center and the β-anomer has the substituent on the same side.

q 48

PCC can only oxidizes primary and secondary alcohols to aldehydes and ketones, respectively. PCC cannot oxidize aldehydes further bc it does not contain the water (anhydrous reagent) necessary to convert the aldehyde to a hydrate, which is a necessary intermediate to be oxidized to the carboxylic acid Educational objective: Aldehydes are readily oxidized to carboxylic acids in aqueous environments by a number of oxidizing agents, including chromium compounds. Aldehydes are believed to go through a hydrate intermediate before being oxidized to a carboxylic acid. The anhydrous oxidizing reagent pyridinium chlorochromate (PCC) cannot oxidize aldehydes because it lacks the water needed to form the hydrate intermediate.

q 16

Peptides contain two or more amino acids bonded together. The amino group of one amino acid reacts with the carboxyl group of another amino acid to form a peptide bond. This is a dehydration reaction

q 30

Protecting groups are added to organic molecules to modify the functional group by masking it. Protecting groups prevent unwanted reactions under given reaction conditions. They tend to be stable to oxidation and reduction conditions and prevent nucleophilic reactions from occurring After a reaction is complete, the protecting group can then be removed to reveal the original functional group. In this question, the Boc group provides steric hindrance to prevent the amino group from acting as a nucleophile. concept: Protecting groups are organic molecules added to functional groups to mask them and keep them from reacting under a given set of reaction conditions. There are different protecting groups for different functional groups.

q 44

Protons on the carbon atom that is adjacent (α) to a carbonyl (C=O) group are more acidic than other protons bonded to a carbon atom for the following reasons: - the oxygen on the carbonyl group is electron withdrawing which gives the carbonyl carbon a partial positive charge - charge delocalization over multiple atoms creates a more stable resonance structure In acetyl-CoA, the methyl (CH3) group is α to a type of carbonyl called a thioester, which makes it susceptible to deprotonation by a base. The resulting negative charge can be transferred from the α-carbon to the oxygen of the thioester, producing a stable enolate. Educational objective: Protons α to a carbonyl are more acidic than other protons bonded to a carbon atom because they are electron-deficient and can be abstracted by a base to yield an anion. The charge is delocalized to form a more stable resonance structure, known as an enolate.

q 43

Psg states all of the other ans choices except for D. The omega-6 FA concentrations observed in the patient plasma and serum samples were lower than normal ranges, and an increased TT ratio was observed. Arachidonic acid is an omega-6 FA, and an increased concentration of arachidonic acid would decrease the TT ratio rather than increase it because arachidonic acid is in the denominator of the ratio. Therefore, the concentration of arachidonic acid is not expected to be higher than normal. Educational objective: Drawing conclusions based on experimental results requires analysis of given information. The effect that results have on another variable can be applied to make conclusions about what changes should occur to that variable. When given information about how a ratio changes, assess what increases or decreases to cause the ratio to change in the specified way.

q 16

Q asking for the most nucleophilic compound out of the ans choices. You had to know that nucleophilicity increases with decreasing electronegativity. The conjugate base (deprotonated form) of a compound is always the better nucleophile when compared to its corresponding acid because the conjugate base has a greater electron density due to its negative charge. so you had to look at the conj base of the ans choices. CH3(CH2)2CH- would be the least electronegative and therefore the most nucleophilic. Weak acids (high pKa) produce strong conjugate bases (and vice versa), and strong bases (low pKb) tend to be strong nucleophiles. In general, as the electronegativity of a negatively charged atom decreases from right to left across a period on the periodic table, its nucleophilicity increases because less electronegative atoms stabilize a negative charge less effectively, have a weaker hold on electrons, and more readily donate electrons to an electrophile. Educational objective: When comparing the nucleophilicity of atoms of equal negative charge, nucleophilicity tends to increase from right to left across a row of the periodic table as electronegativity decreases. Atoms of higher electronegativity more effectively stabilize negative charge and less readily donate electron density to electrophiles whereas those of lower electronegativity stabilize a negative charge less effectively and more readily donate electrons to electrophiles.

q 15

Q asks which compound can do INTRAmolecular hydrogen bonding, so bonding within itself so had to look for ans where substituents are closer together so both point down in B would do that. When both functional groups are in the axial position, they align and can hydrogen bond with one another (intramolecular), stabilizing the molecule. Educational objective: Hydrogen bonding is an intermolecular or intramolecular force that occurs between a hydrogen bond donor and acceptor; this type of force helps stabilize molecules. Hydrogen bond acceptors are electronegative atoms with a lone pair of electrons (eg, oxygen and nitrogen), and hydrogen bond donors are hydrogen atoms bonded to an electronegative atom. Alcohols and amines contain groups that can act as both hydrogen bond donors and acceptors.

q 1

Q basically asking which bond forms when compounds 1 + 2 combine. I looked at Reaction 1 in passage and saw the bond formed between a CA and an amine and that bond is a peptide bond Educational objective: Amides are carboxylic acid (CA) derivatives made up of a carbonyl bonded to a nitrogen atom. These compounds form from the condensation of any CA derivative with a primary or secondary amine.

q 9

Q is asking by what factor did bacteria recover when no soap was used and when 3mL of soap was used. At 0 soap the bacteria on hands is 10^6 and at 3mL its 10^4.1 so 6-4.1 = ~2 and 10^2 = 100 so 80 is the closest to that answer Educational objective: Graphs may display data on a linear scale, a logarithmic (log) scale, or a combination with one scale on each axis, depending on the size of the data being displayed. Log scales display changes in order of magnitude: one step corresponds to a 10-fold change, two steps is a 100-fold change, and so on.

q 34

Q is asking what charcaterizes the C=O bond which is a double bond so just had to know that a pi bond is a covalent bond created by side-to-side overlap of nonhybridized p orbitals. Although a pi bond alone has less overlap of electrons and therefore less bond energy (ie, the energy required to break a bond) than a sigma bond, the overall bond energy of triple and double bonds is greater due to the combined contributions of the sigma and pi bonds. a sigma bond is a covalent bond formed by direct end-to-end overlap of hybridized atomic orbitals such as sp, sp2, or sp3. (Note that the exception is hydrogen, whose atomic orbital is s.) Educational objective: A pi bond is created by side-by-side overlap of p orbitals whereas a sigma bond is formed by direct end-to-end overlap of atomic orbitals. A single bond contains one sigma bond; a double bond contains one pi bond and one sigma bond. A triple bond contains two pi bonds and one sigma bond

q21

Q stem stated 3-methylbutan-1-ol so i looked for an answer choice with that compound attached to the O. Transesterification is a reaction between an ester and an alcohol in which the ester -OR group is replaced with the alcohol -OR group. Transesterifications can be either acid-catalyzed or base-catalyzed. The alcohol acts as a nucleophile and attacks the electrophilic carbonyl carbon of the ester, adding a new -OR group to the ester. The original -OR group leaves and is protonated, becoming an alcohol and yielding a new ester. The carboxylic acid derivative formed from the reaction described in the passage is the ester ethyl hexanoate, which contains a two-carbon chain derived from ethanol and a six-carbon chain derived from hexanoic acid. The alcohol 3-methylbutan-1-ol contains a four-carbon chain (butyl group) with a hydroxyl group (alcohol) on C-1 and a methyl group (-CH3) on C-3. When ethyl hexanoate reacts with 3-methylbutan-1-ol, the ethyl group on the ester is replaced with the alcohol alkyl chain (3-methylbutyl), forming a new ester, 3-methylbutyl hexanoate, shown in Choice B. Educational objective: Transesterification is a reaction between an ester and an alcohol where the ester -OR group is replaced with the alcohol -OR group, forming a new ester and a new alcohol. Alcohols are named based on the longest carbon chain containing the hydroxyl group. The prefix corresponds to the number of carbons in the alkyl chain, and the suffix -ol corresponds to the hydroxyl functional group.

q9

Qstem says that L-arginine is oxidized via NOS enzyme to make L-citrulline so had to look for a similar structure to arginine amino acid and D is the only answer Educational objective: Amino acids play an important role in biological processes. They have three substituents bonded to the same carbon (α-carbon): a carboxyl group, an amino group, and a side chain. Amino acids are distinguished by their side chains, which have functional groups that can participate in a reaction

q 22

Qstem states that the tertiary halide will have more products than the primary alkyl halide after undergoing substitution rxns so you had to know that tertiary halides form more stable carbocations than less sterically hindered alkyl halides and they readily undergo SN1. Primary alkyl halides readily undergo SN2 rxns only. Had to know that SN1 rxns are not stereospecific and will form 2 products. Educational objective: Stereospecific reactions occur when stereoisomers undergo the same reaction and give different stereoisomers of the product. Tertiary alkyl halides readily undergo SN1 reactions, which are not stereospecific because the nucleophile can attack the planar carbocation intermediate from the top or the bottom of the carbocation, forming two stereoisomers. Primary alkyl halides readily undergo SN2 reactions, which are stereospecific because the nucleophile attacks from the opposite side, where the leaving group is eliminated (backside attack)

q 19

R = D and S = L for amino acids (except cysteine) and sugars. Here H is not in the place (not in the front or back) so you can't switch it. You have to approach this differently. You have to draw it to make the H in the back and then you find the R/S. Here after drawing the groups come out to be S which is L in sugars. The question also wanted you to know what the structure of glyceraldehyde (an aldehyde and 2 hydropxyl groups) Amino acids have a stereocenter at the α-carbon because there are four different substituents attached (an amino group, a carboxyl group, a hydrogen atom, and an R group). The exception is glycine because there are only three unique substituents on the α-carbon rather than four (two hydrogen atoms, one of them being the side chain). There are two systems for labeling the absolute configuration of amino acids: the D/L system and the R/S system. The D/L system is based on the absolute configuration of D- and L-glyceraldehyde. Amino acids that resemble the configuration of L-glyceraldehyde are designated as L-amino acids, and those that resemble D-glyceraldehyde are D-amino acids. L-amino acids predominate in nature. Glyceraldehyde is an oxidized form of glycerol. It is a three-carbonaldose with one aldehyde group and two hydroxyl groups. In Figure 1, Compound 1 is an L-amino acid with the R group at the top of the molecule shown as a wedge (coming out of the plane). L-glyceraldehyde has a similar configuration; therefore, it should also have a functional group (the hydroxyl group) at the top of the molecule shown as a wedge. (Choices B and D) Both of these three-carbon structures contain an aldehyde and one hydroxyl group. However, an aldehyde and two hydroxyl groups are present in glyceraldehyde. Educational objective: Glyceraldehyde is an aldose that can be derived from the oxidation of glycerol. It contains an aldehyde and two hydroxyl groups. D/L designations for all other molecules are based on D- and L-glyceraldehyde.

q1

Reactivity of the alkyl halide substrate is dependent on its level of substitution. Tertiary alkyl halides form the most stable carbocations and undergo SN1 reactions more readily than less-substituted alkyl halides. As alkyl halide substitution decreases, the carbocations become less stable and the reaction slows. The tertiary alkyl halide 2-bromo-2-methylpropane will react the fastest because it forms the most stable carbocation in the first step. The secondary alkyl halide will react second fastest, followed by the primary and then the methyl alkyl halides. Therefore, the order of reactivity from fastest to slowest in an SN1 reaction is 2-bromo-2-methylpropane → 2-bromobutane → 1-bromobutane → bromomethane. Concept: SN1 reactions are nucleophilic substitutions that occur in two steps: Formation of a carbocation and nucleophilic attack of the carbocation. Reactivity of alkyl halide substrates depends on the alkyl halides' substitution. Tertiary alkyl halides are the fastest to react because they form a stable carbocation; as substitution decreases, the alkyl halide will react more slowly because it forms a less stable carbocation.

q24

Reducing agents decrease an atom's oxidation state. LiAlH4 and NaBH4, are common reducing agents and are made of a metal surrounded by one or more hydride ions. The hydride ion acts as a nucleophile and attacks an electrophilic atom. LiAlH4 is a strong reducing agent and will reduce carbonyl compounds to alcohols. Aldehydes, esters, and carboxylic acids will be reduced to primary alcohols, and ketones will be reduced to secondary alcohols. A: A base-catalyzed aldol condensation does require carbonyl compounds, but the required compounds are ketones and aldehydes rather than carboxylic acids and esters. B: Decarboxylation occurs with β-dicarbonyl compounds. Neither ethyl hexanoate nor hexanoic acid contains a β-carbonyl, so they will not undergo decarboxylation. C: NaBH4 is a reducing agent rather than an oxidizing agent. Furthermore, NaBH4 is a mild reducing agent and cannot reduce carboxylic acids or esters; therefore, hexanoic acid and ethyl hexanoate would not react with NaBH4. Concept: Reducing agents decrease an atom's oxidation state and reduce the number of bonds carbon has to electronegative atoms while increasing the number of bonds to less electronegative atoms (commonly hydrogen). Hydride reagents, such as LiAlH4 and NaBH4, contain a metal atom surrounded by hydride ions. LiAlH4 is a strong reducing agent and will reduce carbonyl compounds, including esters and carboxylic acids, to alcohols.

q 19

Results from experiments can be compared to draw conclusions about aspects of the reactions performed. Observations for the SN2 reaction conditions (Table 1) show that 1-bromobutane (a primary halide) formed a precipitate whereas tert-butyl chloride (a tertiary halide) did not react, even after heating. Conversely, results for the SN1 reaction conditions (Table 2) record that tert-butyl chloride formed a heavy precipitate, but 1-bromobutane only formed a trace amount of precipitate after heating. This data demonstrates that SN1 and SN2 reactions are dependent on substrate substitution because primary and tertiary alkyl halides react only in certain conditions. Concept: Results from experiments can be compared to draw conclusions about aspects of the reactions performed, including the substrate reactivity, leaving group ability, and plausible intermediates formed. In particular, experimental results show that nucleophilic substitution reactions (SN1 and SN2) are dependent on substrate substitution.

q 40

Saponification is the hydrolysis of ester bonds with a strong base. Triacylglycerols contain three fatty acids bonded to a molecule of glycerol through ester linkages. Saponification of a triacylglycerol with a strong base, such as NaOH, releases free fatty acids (as sodium salts) and a molecule of glycerol. One equivalent of base is defined as the amount of base needed to hydrolyze one ester linkage in a triacylglycerol. If three equivalents of NaOH are used to hydrolyze a triacylglycerol, the molecule will be completely hydrolyzed because each equivalent hydrolyzes one ester, and there are three esters in a triacylglycerol molecule. Concept: Saponification is the hydrolysis of an ester with a strong base. Triacylglycerols contain three fatty acids bonded to a molecule of glycerol through ester linkages, and saponification of a triacylglycerol with a strong base releases free fatty acids (as sodium salts) and a molecule of glycerol. One equivalent of base is needed to hydrolyze one ester linkage; therefore, three equivalents of base are needed to completely hydrolyze a triglyceride.

q 41

Separation of glycerol from fatty acid sodium salts can be accomplished by protonation of the negatively charged carboxyl group, which can be achieved by the addition of a strong acid, such as HCl. Hexanes are non-polar organic molecules that can readily interact with the protonated fatty acids. Therefore, protonated fatty acids will enter the organic layer after extraction with the organic solvent hexanes. concept: Fatty acids sodium salts are soluble in water due to ion-dipole interactions of the charged carboxyl group and sodium ion with water. Protonation of fatty acid sodium salts renders the molecule insoluble in water because the fatty acid is no longer a charged species, and therefore cannot interact as readily with water. The long-chain hydrocarbon makes fatty acids hydrophobic, allowing for solubility in organic solvents.

q 15

Sn1: In the first step of an SN1 reaction, the bond between the leaving group and carbon is broken, transferring the bonding electrons to the leaving group and forming a carbocation intermediate. This is the rate-determining (slowest) step. The second step involves formation of a bond between the nucleophile and the carbocation to form the final product of the reaction. Tertiary alkyl halides undergo SN1 reactions more readily than less substituted alkyl halides because tertiary alkyl halides form a more stable carbocation. Conversely, tertiary alkyl halides are unreactive in an SN2 reaction because they are more sterically hindered, preventing nucleophilic attack and the breaking of the leaving group-carbon bond in one step. The substrate tert-butyl chloride is a tertiary alkyl halide, meaning the chlorine leaving group is bonded to a carbon with three alkyl substituents therefore, tert-butyl chloride is sterically hindered and cannot undergo an SN2 reaction with NaI, as described for Experiment 1. Breaking the bond between the chlorine and the tertiary carbon (the first step of an SN1 reaction) causes a stable, tertiary carbocation to form, permitting tert-butyl chloride to go through an SN1 reaction with AgNO3 in Experiment 2. concept: SN1 reactions occur in two steps: carbocation formation and nucleophilic addition. These reactions readily occur with tertiary alkyl halides because they are sterically hindered and form stable carbocations.

q 21

Sn2 gives you inversion product. You know this is Sn2 bc tertiary will go thru sn1 and second and primary will go thru sn2 In this Sn2, −OH (nucleophile) attacks carbon 2 in (2R,3R)-2-bromo-3-methylpentane, while bromide (leaving group) leaves. Because the electrophile (carbon 2) is chiral, its configuration is inverted from R to S. Carbon 3 is also chiral but retains the R configuration because it does not participate in the reaction. concept: An SN2 reaction is a concerted substitution reaction where a nucleophile forms a bond with an electrophile while a leaving group is displaced. If the electrophile is a chiral center, its stereochemistry will be inverted in the product. Therefore, if the electrophilic carbon has an R configuration, it will adopt an S configuration after the reaction and vice versa, provided the priority ranking of the nucleophile is the same as the leaving group.

q 23

Some sugars can be classified as a special kind of diastereomer, known as an anomer, that differs in configuration at only the anomeric carbon. The α and β designations in a sugar's name refer to different orientations of the hydroxyl group at the anomeric carbon. For aldoses such as glucose, the anomeric carbon is C1, whereas for ketoses such as fructose, the anomeric carbon is C2. α-D-fructose and β-D-fructose are diastereomers that differ in orientation at C2 (the anomeric carbon) only. Therefore, α-D-fructose and β-D-fructose are anomers. Fructose is a ketose and all the other sugar are aldoses Concept: Stereoisomers are molecules containing the same atom connectivity but different spatial orientation of the atoms. Sugar stereoisomers that differ in orientation at only one stereocenter are known as epimers, a specific kind of diastereomer. If the sugars differ in configuration at only the anomeric carbon (denoted by α and β in the sugar's name), they are classified as anomers, a special kind of epimer.

q 38

Sp3-, sp2-, and sp-hybridized atoms have bond angles of 109.5°, 120°, and 180°, respectively. Lone pairs and electrons involved in pi bonding, especially in molecules with larger dipoles, may further distort predicted bond angles. For example, the water molecule (H2O) displays a bond angle of 104.5° due to increased electron repulsion of its lone pairs despite its sp3-hybridized orbitals, which would normally suggest a bond angle of 109.5°. The bond angles of water, the methylated carbon atom on vitamin K 2,3-epoxide, and the newly formed alkene in vitamin K quinone are 104.5° (sp3), 109.5° (sp3), and 120° (sp2), respectively. concept: Valence shell electron pair repulsion theorizes that atoms within a molecule strive to achieve a geometry that minimizes electron repulsion. Hybridization of sp, sp2, and sp3 correlates with 180°, 120°, and 109.5° bond angles, respectively. Despite the water molecule's sp3 hybridization, it displays a bond angle of 104.5° due to lone electron pair repulsion.

q 3

Specific rotation, the degree to which chiral molecules rotate plane-polarized light, is unique to each chiral molecule. The value of specific rotation consists of direction (+ or −) and magnitude (number of degrees). Clockwise rotations are designated as (+), whereas counterclockwise rotations are (−). Isomers are molecules with the same molecular formula but have either different connectivity (constitutional isomers) or spatial orientation (stereoisomers). Diastereomers differ at one or more stereocenters but have the orientation of at least one stereocenter in common. The specific rotations of diastereomers differ in magnitude and may differ in direction. The same is true for constitutional isomers. Because the question states that the two molecules have the same connectivity, they cannot be constitutional isomers. Because they have different specific rotations (direction and magnitude), it can be concluded that the two molecules are diastereomers.

q 26

Stereoisomers are molecules that contain the same atom connectivity but differ in the spatial arrangements of the atoms. Diastereomers are stereoisomers that are not mirror images. These molecules contain at least two stereocenters in which one or more, but not all, of the corresponding stereocenters are in opposite configurations. Number I: The two compounds have different configurations at the stereocenter in position 1 (one S and one R) and have the same configuration at the stereocenter in position 2 (both R). Therefore, (S,R)- and (R,R)-labetalol are diastereomers. Number II: Enantiomers are stereoisomers that are nonsuperimposable mirror images. These molecules contain one or more stereocenters, and each stereocenter is the opposite configuration from the corresponding stereocenter in the other molecule. For example, the enantiomer of (S,R)-labetalol is (R,S)-labetalol. Number III: Conformational isomers are different forms of the same molecule that are generated as atoms rotate about their bonds. Unlike other stereoisomers, conformational isomers can rapidly interconvert by rotation without the need to break any bonds. concept: Diastereomers are stereoisomers that are not mirror images. They contain at least two stereocenters in which one or more, but not all, are in opposite configurations.

q 29

Structural isomers have the same molecular formula and the same molecular weight but differ in the location of atom connectivity in at least one position. Because there is a methylene group (-CH2-) missing from the carbonyl (C=O)-containing alkyl chain in Choice C, this structure does not have the same molecular formula as the original compound, and therefore is not a structural isomer of valsartan. Educational objective: A structural isomer, or constitutional isomer, is a molecule that contains the same molecular formula as another molecule. The atoms in each molecule differ in their connectivity.

q 30

Structural isomers same as constitutional isomers The wedge-dash projection of Compound 1 shows the -OH and the -CH3 both on a dash (down). Likewise, the chair conformation of Compound 2 shows the -OH and -CH3 as equatorial down. To represent this wedge-dash projection as a chair conformation, both substituents can be shown as axial down. A chair flip puts the substituents in an equatorial down position (still on the same side of the ring), yielding an identical structure to Compound 2. Because these compounds have the same connectivity and can be interconverted from one to the other through bond rotations (via chair flip), they are conformational isomers. concept: Conformational isomers are structures that have the same formula and connectivity, and can be interconverted by the rotation of σ bonds. Because conformational isomers are identical except for structural bond rotations, they are the same compound.

What is Tautomerization

Tautomers are isomers that readily convert through the migration of a hydrogen atom and alternative placement of the double bond. Because tautomers are separate compounds with unique structures and their conversion requires a chemical reaction (ie, rearrangement of bonded atoms), tautomers are not resonance structures of the same compound

q5

Terpenes are lipid precursors to steroids and lipid signaling molecules, and are generally isolated from plants. These compounds are made up of branched five-carbon groups, called isoprene units, in which the branched end is referred to as the head and the unbranched end is the tail. Isoprene units can be joined to form terpenes in one of three ways: Head-to-tail, tail-to-tail, or head-to-head. Lanosterol is a triterpenoid, which can be thought of as three monoterpenes together. Therefore, it is composed of six isoprene units (3 terpenes × 2 isoprenes/terpene = 6 isoprenes) and has 30 carbon atoms in total because 6 isoprenes × 5 carbons/isoprene = 30 carbon atoms. Concept: Terpenes are a type of lipid made up of branched five-carbon units known as isoprenes. Terpenes are classified according to the number of isoprene units in the molecule.

q7

The Strecker synthesis is used to generate α-amino acids from an aldehyde using ammonium chloride (NH4Cl) and potassium cyanide (KCN). Because the imine intermediate formed during the reaction is planar (no stereocenters), nucleophilic addition can occur from either above or below the plane. Therefore, the Strecker synthesis is not a stereospecific reaction and produces a mixture of L- and D-amino acids. Gabriel synthesis also gives you a mix of L and D because the final step of decarboxylation is not stereospecific

q13

The Strecker synthesis is used to make α-amino acids from aldehydes using NH3 and potassium cyanide (KCN). The first step of the reaction proceeds with protonation of the carbonyl oxygen by H3O+, followed by nucleophilic attack of the carbonyl carbon by NH3, resulting in dehydration and imine formation. Therefore, an aldehyde and NH3 are used to form the imine intermediate in the Strecker synthesis. concept: An imine is an analogue of ketones and aldehydes that contains a carbon-nitrogen double bond. Imines are formed from a ketone or aldehyde and NH3 or a primary amine via an acid-catalyzed addition of the amine followed by an acid-catalyzed dehydration.

q 46

The aldol condensation begins with enolate formation by deprotonation of a carbon atom adjacent to a carbonyl group (C=O), known as the α-carbon. A reaction at a low temperature (−78 °C) with a bulky base, such as lithium diisopropylamide (LDA), favors formation of the kinetic over the thermodynamic enolate because the kinetic enolate requires a low activation energy to form. The kinetic product is a result of deprotonation of the less substituted α-carbon. A reaction at a higher temperature with a smaller base such as sodium hydride (NaH) will lead to deprotonation of the more substituted α-carbon and formation of the thermodynamic enolate and therefore of the thermodynamic product. A shows the kinetic enolate B shows the aldol product after the kinetic enolate has formed but not undergone dehydration. C and D These structures show the dehydration and aldol products after the thermodynamic enolate, but not the kinetic enolate, has formed. concept: Kinetic enolates require low activation energy and form rapidly under low temperature and bulky base reaction conditions, whereas the less-substituted α-carbon is deprotonated. The base-catalyzed aldol condensation has four steps: formation of an enolate, nucleophilic addition to a carbonyl, protonation, and acid- or base-catalyzed dehydration to form an α,β-unsaturated carbonyl.

q 37

The anhydride product is characterized by two carbonyl groups joined by an oxygen atom. When a carboxylic acid and NaOH react, hydroxide (−OH) deprotonates the carboxylic acid, forming a carboxylate (carboxylic acid salt) and H2O. When an acid chloride is added to the reaction, the negatively charged carboxylate acts as a nucleophile to attack the acid chloride carbonyl, yielding a tetrahedral intermediate. Chloride is then eliminated from the tetrahedral intermediate, and an anhydride is formed. Concept: Anhydrides are carboxylic acid derivatives characterized by two carbonyl groups joined by an oxygen atom. Anhydrides can be prepared by condensation of two carboxylic acid molecules or nucleophilic acyl substitution of an acid chloride by a carboxylate. Both reactions result in the loss of a water molecule from the carboxylic acid starting material.

question 1

The aromatic rings of the indole group on the tryptophan side chain and the Fmoc group are made up of conjugated systems of double bonds, and therefore can absorb UV light. Because tryptophan, Fmoc-Cl, and compound 2 are visible on a TLC plate under UV light and differ in relative polarity, the addition of the Fmoc protecting group to tryptophan can be monitored by TLC. concept: Thin-layer chromatography is a technique that separates mixture components based on polarity, and can be used to monitor reactions. After separation, the mixture components are visualized, usually by UV light. Molecules with UV chromophores (double and triple bonds, carbonyls, conjugated systems) can be visualized with UV light.

q5

The assay analyzed in lane 4 contains the acceptor substrate (Compound 5 or 6) and the enzyme; the donor substrate was not included. The results from the TLC show that the fluorescent acceptor substrate is still present, but no other fluorescent compounds are present. Therefore, glycosidases did not cleave Compound 5 or Compound 6 to form the new product observed in lane 5. The assay in lane 4 functions to ensure that glycosidases do not degrade acceptor substrate. concept: To draw conclusions from the results of a study on enzyme activity, control experiments should be done to ensure that no confounding factors, such as side reactions, occur.

q 2

The compound has 1 ester group (carbon 4) and 3 amides (1-3). Oxygen is electron withdrawing which makes the carbonyl more electrophilic and the nitrogen of the amides makes the carbon less reactive and less electrophilic. Ester = more reactive so milder conditions like weak acid and room temp are enough for it to undergo hydrolysis concept: Carbonyls are electrophiles whose reactivity depends on whether an electron donating or electron withdrawing atom is bonded to the carbonyl carbon. Electron donating atoms decrease reactivity and increase the electron density around carbonyl carbons whereas electron withdrawing atoms increase reactivity and decrease the electron density around carbonyl carbons. More reactive carboxylic acid derivatives can hydrolyze under mild conditions whereas less reactive derivatives require harsh conditions.

q 54

The disaccharide sucrose is made up of fructose and glucose joined together by a glycosidic bond. To obtain fructose from sucrose, the glycosidic bond must be hydrolyzed. concept: A glycosidic bond is the linkage between monosaccharides and is made up of a hemiacetal or hemiketal from one sugar and a hydroxyl group of another molecule. The glycosidic bond is broken through a hydrolysis reaction in which water cleaves the bond, breaking up the sugar molecule.

q8

The ester given in the question is formed from butanoic acid and methanol. The isotopically labeled oxygen (18O) in the ester is bonded to the carbonyl carbon and a methyl (−CH3) group. Because the ester oxygen atom bonded to the carbonyl carbon and the alkyl group (methyl, in this instance) always originates from the alcohol reagent, the labeled starting material must be the alcohol. Therefore, the 18O originates from 18O-labeled methanol (CH3−18OH). concept: Esters are carboxylic acid derivatives that can be formed from a carboxylic acid and an alcohol through a Fischer esterification. The ester oxygen bonded to both the carbonyl carbon and the alkyl group originates from the alcohol, a fact that can be confirmed by isotopically labeling the hydroxyl oxygen of the alcohol starting reagent.

question 2

The formation of Compound 2 was monitored by TLC, and the conjugated double bonds in tryptophan, Fmoc-Cl, and Compound 2 each absorb UV light on the TLC plate. The chromophore electrons jump to a higher energy excited state as the light is absorbed, and the compounds are visible as black spots on the TLC plate. concept: When UV chromophores (conjugated systems, double and triple bonds, carbonyls) absorb UV light, energy from the UV light excites electrons to a higher energy excited state.

question 21

The formula to use to find dilution factors is Vfinal volume/Vtransferred for each individual dilutions and then to find the overall you multiply the DF of each dilutions In this question, 1st they transferred .5mL for a total volume of 10 mL volume. So, you would do 10/.5 = 20. Thats the first DF. And then they dilute again: .5mL standard solution with .5 mL solvent. That means the final volume is 1 mL and amount transferred is .5mL. 1/.2 = 2 To find the overall DF: 20x2 = 40 concept: Dilutions reduce the concentration of a solute by transferring a small volume of solute VT into a larger volume of solvent VS. The dilution factor is calculated by dividing VT from the stock solution or the previous diluted solution by the final volume of the solution VF.

q 9

The four peaks given correspond to ethanol, acetic acid, acetaldehyde and ethane so just had to figure out which peaks correspond to what. You have to figure out the boiling point bc we are talking about gas chromatography. Alkanes have the weakest intermolecular forces and the lowest boiling points, followed by aldehydes and ketones, then alcohols, and carboxylic acids, which have the highest boiling points. Acetic acid is a carboxylic acid and has the highest boiling point of the compounds produced; therefore, it has the longest retention time and must elute at 15 min. concept: Gas chromatography separates compounds primarily by boiling point. Molecules with low boiling points elute before those with high boiling points. For compounds with the same number of carbon atoms, alkanes have the lowest boiling point, followed by aldehydes and ketones, alcohols, and carboxylic acids.

q 3

The fused rings in b lactam makes it a trigonal pyramidal. This decreases resonance bc the orbitals don't overlap very well. This lack of double bond character makes it sp3 hybridized and the decreased resonance donation of electron density to the carbonyl carbon makes it more partially positive = more reactive The small ring size makes the bond angles to be less than 109.5 which is the normal angle for sp3 hybridized atoms. This causes ring strain and this strain makes the carbonyl carbon more reactive than it would be in a noncyclic amide concept: β-lactams are a class of cyclic amides that exhibit significant ring strain due to their small ring size, making the carbonyl carbon more reactive than noncyclic amides. The nitrogen atom in a β-lactam is trigonal pyramidal rather than planar and has decreased resonance. Unlike noncyclic amides, this nitrogen is sp3 hybridized and substantially more reactive.

q51

The general structure of a phospholipid includes a polar head group and nonpolar tail joined together by a molecule referred to as the backbone. The hydrophilic head contains a substituted phosphate, and the hydrophobic tail is made of two fatty acid chains The hydrophilic head and the hydrophobic tail are linked together by a glycerol backbone Two adjacent carbons of glycerol are bonded to the fatty acid chains through an ester bond, and the third glycerol carbon is bonded to phosphate through a phosphodiester bond Phenols are present in biological molecules, such as the amino acid tyrosine; however, they are not a component of the phospholipid structure. Concept: Phospholipids are the main component in the cell membrane and provide structure to the cell membrane. These molecules are made of a hydrophilic head containing a substituted phosphate group and a fatty acid hydrophobic tail. The head and tail are linked together by a glycerol backbone.

q 5

The halogens are generally good leaving groups because they can form stable anions. Larger halogens are better leaving groups than small halogens because they can spread the negative charge over a larger surface area. Of the options shown, iodide is the best leaving group because it is the largest halide and can best stabilize the negative charge upon anion formation. Therefore, ethylhexyl glycerin would be most efficiently synthesized from glycerol and 2-ethylhexyl iodide concept: Bimolecular nucleophilic substitution (SN2) reactions occur when a nucleophile donates electrons to an electrophile and the electrophile loses a leaving group. Good leaving groups must be able to stably carry the electrons they receive. Halogens are generally good leaving groups, and large halogens are better leaving groups than small halogens.

q11

The labeled protons on the carbon adjacent to the carbonyl (the α-carbon) are acidic and have a lower pKa compared to other protons in the compound. Therefore, these α-protons can be removed by a base to form an enolate by delocalization of the negative charge. Concept: Protons on an α-carbon (adjacent to a carbonyl) are more acidic than other protons bonded to a carbon atom because the carbonyl oxygen is electron withdrawing, resulting in less electron density around the α-protons. Therefore, α-protons have lower pKa values and can be more easily removed by a base to form an enolate, which can be stabilized by charge delocalization.

q 35

The leaving group gains a pair of electrons when its bond with the electrophile is broken, and the leaving group is more likely to depart if the electrons gained can form a stable arrangement. Weak bases make better leaving groups than strong bases because they are often neutral molecules or have delocalized charge. Alcohols must be modified to form a good leaving group for substitution reactions because hydroxide (−OH) is a strong base. The addition of acid to an alcohol causes the hydroxyl group to become protonated, forming bound water with a positive charge on oxygen. Water is a weak base, and becomes neutral once it leaves. Therefore, protonation of an alcohol forms a good leaving group. Concept: Hydroxyl groups in alcohols are poor leaving groups because hydroxide is a strong base. Hydroxide must be converted to a good leaving group to participate in substitution reactions. The addition of acid to an alcohol protonates the hydroxyl group, forming a bound, positively charged water molecule. Water is a weak base and forms a neutral molecule upon leaving, which makes it a good leaving group.

question 15

The organic and aqueous layers are immiscible (meaning they do not mix) so after shaking, the layers separate based on their densities. Hydrophilic ("water-loving") molecules are polar and will remain in the aqueous layer. Hydrophobic ("water-fearing") molecules are nonpolar and will dissolve better in the organic layer. Therefore, a nonpolar organic solvent is used in an extraction to remove the hydrophobic molecules from the polar aqueous layer while the hydrophilic molecules remain in the aqueous layer. concept: Extraction is a technique used to separate molecules based on their solubility in aqueous (polar) or organic (nonpolar) phases. The mixture components will dissolve in a solvent of the same polarity ("like dissolves like"). Hydrophobic (nonpolar) molecules will dissolve in a nonpolar organic solvent, and hydrophilic (polar) molecules will dissolve in an aqueous solution.

q 24

The passage demonstrates that when a secondary alcohol gives a positive result in the chromic acid test, it is converted to a ketone in the process. The increase in the number of C-O bonds as a result of reaction indicates that the chromic acid test is an oxidation. Compound D is a four-carbon aldehyde (butanal), and the question states that it gives a positive result for the chromic acid test (meaning it is oxidized). Aldehydes already have two C-O bonds (as a C=O group), but when aldehydes are oxidized, they form a compound with an additional C-O bond (a carboxylic acid). Therefore, a four-carbon carboxylic acid will be produced when Compound D reacts with the chromic acid reagent. concept: Oxidation of organic molecules often increases the number of C-O bonds and decreases the number of C-H bonds in the molecule. Oxidizing agents, such as chromic acid, will convert primary alcohols to aldehydes, secondary alcohols to ketones, and aldehydes to carboxylic acids.

question 4

The passage states that piperidine, a heterocyclic secondary amine, is used to remove Fmoc, a protecting group. The two variables in the amine's structure are the substitution of the nitrogen atom (secondary) and whether the amine is cyclic. To conclude that the amine must be cyclic for Fmoc group removal, a noncyclic amine should be used for comparison. Therefore, researchers should react Compound 2 with a noncyclic secondary amine and determine whether the Fmoc group was removed. If the Fmoc group is removed by a noncyclic amine, the amine does not need to be cyclic for deprotection whereas if the Fmoc group is not removed, the cyclic nature of the amine is important for deprotection. Educational objective: Conclusions about a variable can be drawn from experimental data. To determine whether a result is dependent on a specific variable, the experiment is repeated and the variable in question is changed while all other variables are held constant so that only one parameter of the experiment changes.

question 18

The position of the layers in the funnel depends on the densities of the solvents. Denser solvents settle at the bottom of the funnel and less dense solvents float on top. In the B&D and GC extractions (Figure 2), chloroform and ethyl acetate serve as the nonpolar solvents, respectively. However, the organic layers are in different positions in each method because chloroform is denser than the aqueous phase (ρ = 1.5) whereas ethyl acetate is less dense than the aqueous phase (ρ = 0.9). Consequently, in the B&D method sterols settle in L3, but in the GC method sterols are in L1. Also, you could cross out choice A and choice C bc L2 is the cellular debris so it has nothing to do with extraction concept: The principal concept of the extraction technique is that "like dissolves like." In other words, substances dissolve best in solvents with the same polarity. Extraction requires immiscible solvents of different polarities.

question 32

The question is asking which of the answer choices will not contribute to the highest conversion efficiency. You had to look at the table and look at the one that did give you the highest conversion efficiency and then see what was used in that. Whatever was not used is the correct ans which is C concept: When experimental conditions are optimized, one variable should be evaluated in each trial while other variables are kept constant. The conditions that give the highest yield are optimal.

question 41

The question states that theres one spot near the origin using hexane (aka polar) and another spot near the solvent front using ethyl acetate ( aka nonpolar). So, what would happen if you mix hexane and ethyl acetate? The 1:1 hexanes/ethyl acetate mixture is more polar than hexanes and less polar than ethyl acetate and achieved the compound separation essential for column chromatography. This solvent mixture decreased the compounds' affinity for the mobile phase relative to ethyl acetate. concept: Thin-layer chromatography is a technique used to separate compounds based on polarity. The rate at which a compound travels up the plate is a function of the compound and solvent polarities. Nonpolar solvents decrease a compound's affinity for the mobile phase, whereas polar solvents increase a compound's affinity for the mobile phase.

q4

The racemase-catalyzed conversion of L-glutamate to D-glutamate is stereospecific. The planar intermediate (imine) is formed in step 3 of the PLP-dependent reactions (Figure 3) after hydrogen is removed from the chiral carbon atom in L-glutamate. In step 4, hydrogen is added to the opposite side, which allows for the conversion to D-glutamate. If you look at the picture you can see it started with hydrogen away from you and changed pointing towards you. Extra info: Stereospecific: the configuration of the substrate completely determines the configuration of the product. concept: Conversion of one chiral molecule to its enantiomer often requires the formation of a planar intermediate. The reaction is stereospecific if a side group is preferentially added to one side of this intermediate.

q16

The reactant 4-tert-butyl-2-octanone has an eight-carbon parent chain denoted by the prefix oct-, and the suffix -one indicates the compound is a ketone with the carbonyl at carbon 2. The substituent is a tert-butyl on carbon 4. The reaction of a ketone or an aldehyde with a cyanide ion (from NaCN) acting as a nucleophile and attacking the electrophilic carbonyl (C=O) carbon produces a cyanohydrin. The cyanohydrin functional group contains a hydroxyl group (-OH) and a cyano group (C≡N) attached to the same carbon. Concept: The reaction of a ketone or aldehyde with sodium cyanide (NaCN) results in the nucleophilic attack of the carbonyl by the cyanide ion, and produces a cyanohydrin. Organic molecules are named after the number of carbon atoms in the longest carbon chain, adding the suffix of the highest priority functional group. Substituent names precede the chain name, and the location of the high-priority group is denoted by placing its location between the substituent and chain names.

q1

The reaction they are talking about forms D-glutamate from L-glutamate. Meaning L-glutamate is the substrate and thats what B is showing. All amino acids are L All amino acids are S except for cysteine which is R and glycine which is achiral R = clockwise S = counterclockwise Extra info: Enantiomers differ by their absolute configuration around a chiral center, a carbon atom bound by four nonidentical side groups concept: Of the 19 common L-amino acids, all have an S configuration except for cysteine (glycine is achiral). Absolute configuration of R or S enantiomers is determined by the priority of groups attached to a chiral carbon.

q 49

The retro-aldol reaction is the reverse of the aldol reaction or aldol condensation and is catalyzed by a base and heat. Water is lost, and the carbon-carbon bond between the α- and β-carbon atoms is broken, yielding two aldehydes, two ketones, or one of each. In this 2 identical aldehydes are produced. Look at the picture concept: The retro-aldol reaction is the reverse of the aldol reaction, in which the bond between an α- and a β-carbon is broken into two fragments. Aldehydes are named after the number of carbon atoms in the longest carbon chain, replacing the -e in the parent alkane name with the suffix -al. Substituent names precede the chain name and are numbered from the carbonyl carbon.

question 35

The separation of enantiomers, such as those in the racemic mixture of albuterol, requires the addition of a resolving agent (a chiral molecule). When a resolving agent is added to a racemic mixture, it reacts with each enantiomer, forming a covalent bond or an ionic salt. Because the resolving agent is chiral, it incorporates a new chiral center into each enantiomer, creating a pair of diastereomers. Diastereomers can be separated from each other because, unlike enantiomers, they have different physical properties. Once the diastereomers are separated, the resolving agent is removed, yielding the original molecules as single enantiomers. content: A racemic mixture is a 50:50 mixture of enantiomers, which have the same chemical and physical properties, and therefore cannot be directly separated. Consequently, the separation of enantiomers requires the addition of a resolving agent to change their physical properties by creating a pair of diastereomers. The resolving agent is removed once the diastereomers are separated to yield the original molecules as single enantiomers.

q 6

The table shows minimal inhibitory concentration (MIC) which means that the higher the number the more concentration is needed for it to be inhibited whereas if the number is small it means a smaller concentration is needed for it to be inhibited. Choice C says that compound 3 demonstrates the greatest antimicrobial activity against S. aureus and thats true because looking at the table, compound 3 has the lowest MIC concentration so it has the greatest activity against S. aureus. concept: The minimal inhibitory concentration (MIC) gives the lowest concentration necessary to inhibit the growth of a bacterium. A compound with the lowest MIC value in a set of tested compounds indicates that the compound has the greatest antimicrobial activity of the set.

question 8

There is no N-H bond in the molecule The compound 4-nitrophenol contains the following functional groups: 1. Hydroxyl group (-OH) 2. Aromatic ring 3. Nitro group (-NO2) However, 4-nitrophenol does not contain any N-H bonds, and therefore would NOT contain an N-H stretch. Concept: An infrared (IR) spectrum displays absorption signals characteristic of the functional groups in a particular molecule based on the IR frequencies absorbed by those functional groups. Functional group bands appear in the same region of the IR spectrum regardless of the overall structure of the molecule.

q 50

Think electrophiles and nucleophiles For this Q- the carbonyl carbons are the electrophiles; Here the oxygens are not the nucleophiles, rather all the alpha carbons are because they are the ones losing the H and they end up becoming carbanion (it has to be alpha carbon bc of resonance stabilization, if you lose an alpha hydrogen, the electrons are pushed up leading to resonance stabilization) Here in the Q they say that the asterisk carbonyl is the carbonyl in the final product. And the carbonyl thats attacked is gone in the end product

question 19

To separate long-chain amides from carboxylic acids in the organic layer, a base must be added. Bases deprotonate the carboxylic acid and generate carboxylate anions that are more soluble in the aqueous layer than in the organic layer. Dilute LiOH is strong enough to deprotonate a carboxylic acid but not to deprotonate an amide N-H. Therefore, addition of the base LiOH will produce carboxylate anions that enter the aqueous layer and allow long-chain amides to remain in the organic layer. concept: Acylation reactions between anhydrides and amines generate amides and carboxylic acids. The solubility of carboxylic acids in water increases when they are converted into carboxylate anions by a base.

q 56

Tollens test is used to identify the presence of aldehydes and hydroxy ketones, including reducing sugars, which have a free anomeric carbon. It utilizes the oxidizing agent Ag(NH3)2+ to oxidize aldehydes to carboxylic acids. A positive Tollens test results in the formation of metallic silver that looks like a silver mirror deposited on the glassware. Ketoses such as fructose are hydroxy ketones. Ketones cannot be directly oxidized to carboxylic acids; however, hydroxy ketones can undergo tautomerization via an enediol intermediate to produce an aldehyde under Tollens test basic conditions. Because of this tautomerization, ketoses give a positive result to Tollens test. (tautomerization - think of enol and keto) Concept: Tollens test is used to identify the presence of aldehydes and hydroxy ketones, and uses the oxidizing agent [Ag(NH3)2]+ to oxidize aldehydes to carboxylic acids. Ketoses can undergo tautomerization via an enediol intermediate to their aldose form, resulting in a positive Tollens test and the formation of a silver mirror.

Transesterification vs fisher esterification

Transesterification: ester + alcohol = new ester where ester -OR group is replaced with the alcohol -OR group. Can be acid or base catalyzed Fischer esterification: CA + alcohol = ester catalyzed by an acid and the alkyl group bonded to the carbonyl carbon comes from the carboxylic acid, and the alkyl chain bonded to the ester oxygen atom comes from the alcohol.

q 39

Triacylglycerols are made up of 3 FAs and attached to a glycerol molecule through ester linkages. There are three fatty acids released during the hydrolysis; however, the number of distinct fatty acids released corresponds to the number of distinct hydrocarbon chains in the triacylglyceride. The triacylglycerol has two distinct hydrocarbon chains: one with 16 carbons and two with 18 carbons. The two 18-carbon chains are identical to each other and will collectively form only one product. Because there are only two distinct hydrocarbon chains, hydrolysis of this triacylglycerol will yield two distinct fatty acids in addition to glycerol. Educational objective: Triacylglycerols are made up of three fatty acids and a glycerol molecule connected through ester linkages. Hydrolysis of a triacylglycerol releases free fatty acids and glycerol. The number of distinct fatty acids released corresponds to the number of distinct hydrocarbon chains in the triacylglyceride.

q 31

Under neutral conditions (pH ~7), the N- and C-termini are charged. Neutral pH is above the carboxyl group's pKa of 2, causing it to become deprotonated and form its conjugate base. The carboxyl group in this form contains a −1 charge. A protonated amine contains a +1 charge. This doubly charged species is known as a zwitterion, or a dipolar ion, in which the charges at both ends cancel each other out. The passage states the unfolding experiment is done at pH 7.2. Under these neutral conditions, the N-terminus will be protonated and exist in the conjugate acid form, giving it a +1 charge. pH < pKa = protonated concept: In a neutral environment (pH ~7), amino acids are zwitterions, also known as dipolar ions, because the N- and C-termini are charged (+1 and −1, respectively). The carboxyl group is deprotonated and in its conjugate base form. The amine group is protonated and in its conjugate acid form.

q 36

Waxes are an example of a hydrolyzable lipid that serves as a form of protection in plants and animals. Waxes are made up of a long-chain fatty acid and a long-chain alcohol. The wax forms when the acid and the alcohol combine through dehydration to form an ester bond. concept: Lipids are hydrophobic molecules broadly classified as hydrolyzable or nonhydrolyzable, and more specifically classified based on their backbone structure. Waxes are hydrolyzable lipids that contain an ester bond formed by the linkage between a long-chain fatty acid and a long-chain alcohol.

q 4

Z = zame zide E = opposite side concept: Alkenes are geometric isomers that differ in the substituent arrangement about a double bond and can be classified as either E or Z. When the higher-priority substituents are on the same side of the double bond (top or bottom), the alkene is Z, and it is E when the higher-priority substituents are on opposite sides of the double bond.

q 14

aspartic acid is not a zwitterion at physiological pH bc it has a -1 charge concept: At physiological pH (7.4), amino acid backbones have a positively charged amino group (+1 charge) and a negatively charged carboxyl group (−1 charge), and the charges cancel one another out, yielding a net charge of 0. These compounds are known as zwitterions. Acidic and basic amino acid side chains are charged at pH 7.4, so the overall charge of the molecule is not zero and these amino acids are not zwitterions at physiological pH.

q 31

be careful Qstem says stereoISOMERS not stereocenters so first count number of stereocenters and you find that there are only 2 bc the middle C is not chiral and then you do 2^2 to find the total number of stereoisomers so that woudl be 4 Educational objective: The maximum number of stereoisomers possible for a compound is determined by the expression 2n, where n is the number of stereocenters. However, the actual number of stereoisomers for a particular compound could be less than 2n if two of the possible stereochemical configurations have an internal mirror plane of symmetry (meso compound), making them conformations of the same compound.

q 17

compound 3 contains a peptide bond so a dehydration rxn took place concept:Peptide bonds are formed through a dehydration reaction, in which the carboxyl group of one amino acid loses a hydroxyl group and the amino group of another amino acid loses a hydrogen atom. H2O is released as a byproduct in a dehydration reaction.

question 13

concept: Functional groups show absorption in the infrared spectrum at different frequencies depending on the bond type present in the particular functional group. Characteristic functional group absorptions include 3650-3200 cm−1 (alcohol and phenol O−H stretch); 3550-3060 cm−1 (amide N-H stretch); 3100 cm−1 (sp2 C-H stretch); 3000-2875 cm−1 (sp3 C-H stretch); 2260-2,100 cm−1 (triple bonds); and 1850-1650 cm−1 (C=O stretch).

q 17

concept: Steric hindrance is a characteristic bulkiness in compound structures that either blocks a portion of a molecule so that it cannot react or keeps a reaction from occurring altogether. Decreasing the substitution on (or near) an atom decreases the steric hindrance of the atom.

question 5

even tho tryptophan is nonpolar you have to look at this in comparison to what they are comparing to I looked at Scheme III structures and noticed that tryptophan is more polar than compound 2 meaning that it would elute slower in normal phase HPLC. But Qstem says it has a shorter retention time = elutes faster meaning it is RP-HPLC so nonpolar stationary phase and polar mobile phase Educational objective: In high-performance liquid chromatography (HPLC), two types of columns—normal-phase (NP) or reverse-phase (RP)—can be used, depending on the polarity of the compounds being separated. NP-HPLC is used to separate nonpolar compounds and consists of a polar stationary phase and a nonpolar mobile phase. RP-HPLC is used to separate polar compounds and consists of a nonpolar stationary phase and a polar mobile phase.

q23

first count the longest parent carbon chain which comes out to be 5. So the ans has to contain pentane and this eliminates A and C. The carboxylic acid will take the priority so the ending will be with oic acid and C1 will be from that end which makes the ethyl group on C2 and methyl group on C3 and thats ans choice B concept: Carboxylic acids are named by identifying the parent chain (longest continuous carbon chain containing the carboxyl group) and numbering the carbons beginning with the carbonyl carbon. The substituents are named beginning with the parent chain carbon number it is bonded to and followed by the substituent prefix. The parent chain is named last, replacing the -e from the parent alkane with -oic acid.

question 3

i got it wrong bc the key part was the heterocyclic part. Heterocyclic compounds are defined as cyclic compounds which consist of at least two different elements in the ring system, one of them mostly nitrogen, oxygen, or sulfur. so based on that definition and the secondary part, D is the correct answer concept: Amines can be classified as primary, secondary, tertiary, or quaternary (ie, one, two, three, or four non-hydrogen substituents, respectively).

q12

know that L configuration is an S stereochemistry so did that for all of them but make sure you switch the stereochemistry for all of them bc the H is a wedge and it should be in the back Educational objective: Amino acids are chiral molecules (except glycine) that contain an α-carbon atom with four unique substituents: -NH3+, COO-, side chain, and H. The natural chiral amino acids have an S configuration (except cysteine, which is R). All the natural chiral amino acids have an L configuration, and the enantiomers of these amino acids have a D configuration.

q 14

looking at the table you can see when both are present the percent inhibition is high and thats why C is the ans concept: Structural modifications to biologically active molecules are performed to determine whether the biological activity of the molecule can be increased. To determine the effect of the structural modifications, the activity of each modified molecule should be compared to that of the original molecule.

q 53

needed to kno val lue and IIe are hydrophobic, uncharged and branched AA concept: Amino acids are made up of an amino group, a carboxyl group, and a side chain, all bonded to the same carbon atom. The side chains contain different functional groups that give amino acids specific characteristics, which allow the classification of amino acids based on their properties. Four main classifications of amino acids include acidic, basic, hydrophilic, and hydrophobic.

q2

needed to know what tosylate contains To improve its leaving group ability, an alcohol can be converted into a tosylate (-SO3C6H4CH3) or a mesylate (-SO3CH3). Tosylates are produced by the reaction of an alcohol with TsCl and a base, such as pyridine. The hydroxyl O acts as the nucleophile to attack the S of TsCl, and Cl acts as the leaving group. Then pyridine deprotonates the O atom, giving the tosylate product. concept: Alcohols are poor leaving groups and weak electrophiles; therefore, nucleophiles are unlikely to displace the hydroxyl (-OH) groups of alcohols in substitution reactions. To improve its leaving group ability, the -OH can be converted into a tosylate (-SO3C6H4CH3) or a mesylate (-SO3CH3) through a reaction with a base and p-toluenesulfonyl chloride (TsCl) or methanesulfonyl chloride (MsCl), respectively. The hydroxyl oxygen attacks the sulfur of TsCl or MsCl, forming a bond between S and O. Cl acts as the leaving group, breaking the bond between S and Cl.

q19

needed to look at the compound and see which amino acid given is in the compound. The only one present is trp concept: Each amino acid has a unique side chain that confers specific chemical and biological properties on that amino acid. Side chains may have relatively simple functional groups, such as alkyl chains, alcohols, and carboxylic acids, or they may have more complex groups, such as indole or imidazole.

q 27

table 2 shows you that iodoform reacts in compound A (secondary alcohol) and E (methyl ketone). So, had to look at the choices given for those. choice III is a methyl ketone and IV is a secondary alcohol. concept: Analysis of the results from an organic reaction with respect to structural features of the reacting compounds is important for drawing conclusions about what types of functional groups will react with the particular reagents. Such an analysis helps predict what compounds will or will not react with the same reagent under similar conditions.

q 18

the AA chain for bradykinin was given in the psg and only arg and ser are hydrophobic and arg shows up twice so the ans is 3 concept: Amino acids are made up of an amino group, a carboxyl group, and a side chain, all bonded to the α-carbon. Side chains are the point of difference in amino acid structure and contain different functional groups, which allow for amino acid classification based on the functional group characteristics. The main classifications include hydrophilic (including charged and polar uncharged) and hydrophobic.

question 39

the answer i picked is wrong because even tho HPLC can be used, for a pure product you wouldn't get two peaks TLC can be used to assess product purity by comparing a standard of pure theobromine to the isolated product. If a single spot with a larger Rf than that of theobromine is visible on the product sample track on the TLC plate, then the isolated product is pure. concept: Thin-layer chromatography (TLC) is a technique used to evaluate a compound's purity. The components of a mixture are separated based on polarity, and a single spot on a TLC plate is indicative of a compound's purity.

question 36

the diff btw the boiling point is 19 so that means you need to use fractional distillation which separates molecules that are less than 25 degrees bp diff concept: Distillation is a purification technique that allows for separation of liquid mixtures based on the boiling points of the mixture's components. The three main types of distillation are simple (bp <150°C and >25°C apart), fractional (bp <25°C apart), and vacuum (bp >150°C).

question 14

the final volume of the solution is 50mL and you got that by doing a 1:20 dilution factor. So to get the volume that was diluted just divide 50 by 20 and that will give you 2.5mL concept: Dilutions are done to decrease the concentration of a solution when a small amount is needed and cannot be accurately measured. A small volume of concentrated solution (VT) is transferred to a certain volume of solvent (VS), and the final volume (VF) is the sum of VT and VS. The dilution factor can be determined by the ratio of VT to VF.

q10

the main acid that will NOT form would be tyrosine bc of the mutation in phenylalanine hydroxylase which would convert phenylalanine to tyrosine Have to know amino acid structures, tyrosine has an OH group and a 6C ring that aromatic and phenylalanine is the same but without OH group Educational objective: Amino acids are important biological molecules that make up proteins. Amino acids contain an amino group, a carboxyl group, and a side chain all bonded to an α-carbon. The amino acids can be grouped together based on the classification of the side chains and functional group. Classifications include aromatic, hydrophilic or hydrophobic, acidic or basic, and charged (positive or negative) or uncharged.

question 4

the question is asking about how to increase the signal intensity for smaller m/z values The way to do that would be to increase the charge which corresponds to z. And increasing z would give you a small ratio. In mass spec, smaller m/z gives you a greater signal intensity. For this q, lowering the pH would increase H+ and that would results in a greater # of multiply charged molecules (z>1). An increase in multiply charged molecules would therefore result in greater signal intensity for smaller m/z values. Concept: Signal intensity of a mass spectrometry peak corresponds to the relative quantity of ions at a given mass-to-charge ratio m/z. Increased charge yields a decreased m/z ratio; therefore, an increase in the number of multiply charged particles increases signal intensity at lower m/z ratios.

question 25

the question is saying how would the boiling point of one compound compare if its run under vacuum and simple distillation. So, simple distillation is open to the atmospheric pressure so the boiling point would not change. Vacuum distillation on the other hand is closed by a vacuum pump so it's not equal to atmospheric pressure. Since the pressure in this is reduced, the boiling point is also reduced. Thats why simple distillation bp of a molecule is greater than the bp of that molecule under vacuum concept: Distillation is a technique used to separate molecules based on their boiling points. The common types of distillation are simple, fractional, and vacuum. Simple distillations are done at atmospheric pressure, whereas vacuum distillations are done at a reduced pressure, decreasing the compound's boiling point relative to the boiling point at atmospheric pressure.

question 9

these 2 structures are constitutional isomers. In 2 nitrophenol the OH and NO2 are closer together (ortho) so the Hydrogen from the OH group can hydrogen bond intramolecularly with the lone pair of electrons on the nitro group. This is INTRAmolecular bonding and it decreases the number of INTERmolecular bonds that can form = decrease in boiling point of the compound in 4 nitophenol the 2 groups are on opposite side(para position) so they are able to hydrogen bond intermolecularly but not intramolecularly. Intermolecular bonds hold the molecules of 4-nitrophenol together, thereby increasing the boiling point and causing it to stay in the flask while 2-nitrophenol distills. concept: Distillation separates compounds based on their boiling point. Constitutional isomers can experience different intermolecular forces, contributing to the difference in their boiling points. The isomer that experiences increased intermolecular hydrogen bonding has a higher boiling point compared to the isomer that experiences increased intramolecular hydrogen bonding.

q 22

they say in the psg that lucas reagent is used for secondary and tertiary alcohols and occurs thru sn1 so you needed to see which one is 2ndary or tertiary. Compound A is secondary but since its secondary its less stable than the tertiary carbocation and will react more slowly in the lucas test than compound C. Compound C is tertiary= will react faster. concept: An SN1 reaction occurs in two steps. First, a carbocation intermediate is formed when the bond between the leaving group and adjoining carbon atom is broken; next, the nucleophile adds to the carbocation to form the reaction product. Substrates that form a more stable carbocation react faster than substrates that form a less stable carbocation.

q18

ubiquinone receives 2 electrons from NADH so the 2 carbonyls on the ring structure will become OH. The electrons will distribute to the to the 6 membered ring and so the ring structure will become aromatic and thats ans choice D concept: Oxidation-reduction reactions result in the transfer of electrons from one atom to another. Oxidation is a loss of electrons and an increase in the number of carbon-heteroatom bonds; reduction is a gain of electrons and a decrease in the number of carbon-heteroatom bonds. In the electron transport chain, ubiquinone receives two electrons from NADH, which reduces the carbonyl carbons to hydroxyl groups and forms an aromatic ring, yielding ubiquinol.

q3

you can get the answer by using the name α-bromoacid. This tells you that the Br group will be at the alpha position to the carboxylic acid and the only ans that has a carboxylic acid and br at alpha position is C A is wrong bc that the intermediate formed after the addition of PBr3 and Br2 to the carboxylic acid. B is wrong bc This carboxylic acid contains bromine atoms on both the α- and β-carbons rather than the α-carbon only. d is wrong bc This carboxylic acid is a β-bromoacid, meaning bromine is on the β-carbon (two carbon atoms away from the carbonyl carbon) rather than the α-carbon. concept: The α-carbon of a carboxylic acid is the carbon adjacent to the carbonyl carbon and can undergo substitution reactions to form α-substituted carboxylic acids. An α-bromoacid is a carboxylic acid with bromine on the α-carbon and is formed when a carboxylic acid is reacted with PBr3, Br2, and then H2O.

question 7

you needed to know the polarity of the functional groups Educational objective: In high-performance liquid chromatography (HPLC), two types of columns—normal-phase (NP) and reverse-phase (RP)—can be used, depending on the polarity of the compounds being separated. NP-HPLC consists of a polar stationary phase and a nonpolar mobile phase, whereas RP-HPLC consists of a nonpolar stationary phase and a polar mobile phase. Molecules with similar polarity to the stationary phase interact with it more and have longer retention times.