Organic Chemistry

Q18

Educational objective: 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.

Q35

(Choice A) Size-exclusion chromatography separates compounds by size; therefore, compounds that are similar in size (eg, theobromine, caffeine) elute together. An absorbance peak indicates only the presence of a compound, not the number of compounds contributing to the absorbance. (Choice B) Although high-performance liquid chromatography (HPLC) can be used to determine product purity, a pure reaction product would show only one peak (not two) on the HPLC chromatogram. (Choice C) There are three unique methyl (−CH3) groups in caffeine; therefore, three unique −CH3 signals would be seen in its 1H NMR spectrum. Two unique −CH3 signals would indicate the product had not formed because theobromine has only two unique −CH3 groups. Educational objective: 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.

Q20

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), 1810-1650 cm−1 (C=O stretch). single bonds: 2700-3500 (N-H; O-H; C-H) Triple bonds: 2100-2400 (C to C; C to N) Double bonds: 1600-1800 (C=C; C=O; C=N) Fingerprint region: 600 - 1400 (C-C; C-O; C-N) Educational objective: 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).

Q17

50mL is the final volume and 20 is the DF so you do 50/20 to get the volume that was transferred. Educational objective: 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.

Q4

<---------- ----------> downfield upfield deshielded shielded higher frequency. lower frequency The chemical shift depends on the electronic environment around the proton. 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). In Compound 1, He and Hf are on a carbon adjacent to a hydroxyl group. Because oxygen is electronegative, it deshields the neighboring protons, causing a downfield shift (3.80 and 3.71 ppm, respectively) from the chemical shift of alkyl protons (0-2 ppm). Transesterification of Compound 1 transforms the hydroxyl group into an ester (Compound 3), which contains more electronegative atoms than the hydroxyl group. Therefore, the ester withdraws electrons and deshields He and Hf more so than the hydroxyl group and causes the signal to shift farther downfield (4.36 and 4.25 ppm, respectively). Educational objective: 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.

Q38 ask landon

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. Rav-4 so red is at 700 and blue is at 400 Educational objective: 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.

Q39

An atom bonded to four unique substituents is called a chiral center. If corresponding chiral centers in a pair of otherwise identical molecules have the opposite configuration (ie, R in one molecule, S in another), the molecules are known as enantiomers, or nonsuperimposable mirror images. A 50:50 mixture of enantiomers is known as a racemic mixture. Enantiomers have many of the same chemical and physical properties, including melting and boiling points, solubility, and polarity, although they differ in the way that they interact with plane-polarized (linear) light. Therefore, separation of enantiomers relies on changing the physical properties of the molecules. 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. (Choices A, B, and C) Because enantiomers have the same chemical and physical properties, they cannot be separated by physical means such as extraction, thin-layer chromatography (TLC), or fractional distillation. Enantiomers react in the same way with bases, have the same retention time on a TLC plate, and have the same boiling point. Educational objective: 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.

Q23

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. Educational objective: 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.

Q2

Characteristic functional group absorptions 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) 1810-1650 cm−1 (C=O stretch) Educational objective: 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).

Q3

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. Educational objective: 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.

Q26

Educational objective: 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).

Q34

Educational objective: 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.

Q9

Gas-liquid chromatography: mobile = inert gas (He or Nitrogen) stationary = liquid **lowest bp molecule elutes first** A small amount of a liquid mixture is injected into the gas chromatograph, and the compounds in the mixture are vaporized by heating. The vapors then travel through the column in a heated oven to the detector. 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. (Number III) Components of a mixture are separated by boiling point rather than polarity. Although polarity impacts a compound's boiling point, it is not the only determining factor. A sufficiently large nonpolar molecule can have a higher boiling point than a polar molecule. For example, a nonpolar 20-carbon alkane chain has a higher boiling point than a polar 4-carbon carboxylic acid (343 °C vs 164 °C) and would move through the column more slowly despite being non-polar. (Number IV) The column must be placed in a heated environment to allow volatile molecules to remain in the gas phase. Columns kept at room temperature would not separate the molecules. Educational objective: 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.

Q36

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 Educational objective: 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).

Q32

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.

Q21

I looked at the structure given and looked for the bond pairings given in the ans choices and realized there is no N-H bond Educational objective: 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.

Q10

In UV spectroscopy, a compound in solution is irradiated with UV light; the amount of UV light absorbed by the compound is measured at each wavelength and a spectrum is generated by plotting absorbance as a function of wavelength. 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* antibonding orbital. The lone electron pairs on nitrogen in pyrimidine are nonbonding (n) electrons. Therefore, when absorbing UV light, these electrons are excited to the antibonding (π*) orbital and undergo a n → π* transition. (Choice A) A π → π* electron transition describes excitation of a π electron from the bonding to the antibonding molecular orbital rather than a nonbonding electron to an antibonding molecular orbital. (Choices C and D) The π* → π and π* → n electron transitions describe an excited π electron and excited nonbonding electron, respectively, falling from the antibonding orbital to its ground state (π and n). Educational objective: 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.

Q7

In distillation if a compound turns out impure it is due to heating the mixture too quickly and that can cause contaminated or impure products. Otherwise the lower bp and higher bp substances won't be able to separate properly. And because they say that greater than 150 degrees then a vacuum distillation must be used and not fractionated distillation. And using a tall fractionated column in fractional distillation would help improve the separation of compounds by increasing the length of the path the vapor must travel before condensing and dripping into the receiving flask. Educational objective: 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.

Q22

Intramolecular H bonding = decreased BP and polarity Intermolecular H bonding = increased BP and polarity Hydrogen bonding can be intermolecular (between two molecules) or intramolecular (within the same molecule), and therefore can have a significant impact on a molecule's boiling point. The difference in connectivity of the constitutional isomers 2-nitrophenol and 4-nitrophenol causes the molecules to experience different intermolecular forces, which contributes to the difference in their boiling points. Because the hydroxyl and nitro groups in 2-nitrophenol are ortho and therefore close in proximity to each other, the hydrogen from the hydroxyl group can hydrogen bond intramolecularly with the lone pair of electrons on the nitro group. This intramolecular bonding decreases the number of intermolecular bonds that can form, thereby decreasing the boiling point of the compound. Because the hydroxyl and nitro groups on 4-nitrophenol are para, 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. Educational objective: 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.

Q1

Just had to find the Rf of the product in figure 2 in the passage and Rf equation is in picture and also Rf is always <1 Educational objective: 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.

Q5

Just needed to look at table 1 and see which answer choice/variable wasn't a part of what helped it get to 99% conversion Table 1 shows the results of the reaction condition optimization for the transesterification of chloramphenicol. The reaction was catalyzed by the enzyme LipBA, and its concentration was 4.5 g/L in each trial (Choice D). The solvent 1,4-dioxane (Choice A) gave a considerably higher conversion efficiency (91% or higher) and purity than dichloromethane (72% conversion). Of the temperatures tested, 55 °C yielded the highest conversion efficiency (Choice B), and a reaction time of 10 hours, rather than 20, gave the highest conversion efficiency at 55 °C (99%). A longer reaction time likely led to product degradation or side reactions. Therefore, a 20-hour reaction time is not part of the optimal set of reaction conditions. Educational objective: 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.

Q25

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. Educational objective: 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.

Q15

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 H for it to be chemically equivalent with. (Choices A, C, and D) Methanol, ethanol, and ethyl acetate have one, two, and three sets of equivalent protons, respectively. Educational objective: Hydrogen atoms or groups on a molecule that display rotational or planar symmetry are in identical magnetic environments. These atoms, called chemically equivalent hydrogens, have the same types of atoms surrounding them. Equivalent hydrogens have identical chemical shifts and are represented as a single signal on the NMR spectrum.

Q4

Mass spec depends on m/z ratios and Qstem is asking what would increase intensity of peaks for a smaller m/z ratio and the only way to get a smaller ratio is to increase the denominator which is charge so you would have to increase the number of ions or protons in solution and you would do that by decreasing the pH. (Choice A) An increase in the solvent's pH would cause a decrease in the concentration of protons and therefore would not increase the number of multiply charged molecules or the signal intensity of the smaller m/z values. (Choice C) Decreasing the concentration of the sample would decrease the amount of the sample present and cause a decrease in the overall signal intensity. (Choice D) Increasing the concentration of the sample would increase the amount of the sample present, but this would not affect the relative heights of the peaks because all the peaks would increase. The concentration of protons would affect how often a molecule gets protonated. Educational objective: 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.

Q2

Mass spectrometry (MS) separates ions according to their mass-to-charge ratio m/z. Adduct 1 in passage is given as 414 g/mol and in Qstem they say "singly charged adduct 1 ion" meaning you have to add the mass of that one proton to adduct 1 and divide by 1 (charge bc of m/z ratio) to get 415. (414+1)/1 = 415 **if it was double charged then you: (414+2)/2 = 208 **if they give you m/z ratio you can find charge or the mass by going backwards** 207 is the m/z ratio charge has to be given (+2) so you would multiply 207 by 2 and subtract 2. (207*2) - 2 = 412 Educational objective: 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.

Q19

Mass spectrometry is a technique that measures the molecular weight of a molecule. Molecules in a sample are injected into a mass spectrometer, where they are bombarded with a beam of electrons. This beam removes electrons from the molecule and forms a positively charged ion known as the molecular ion. The molecular ion can also fragment during bombardment to form smaller ions. An electric field accelerates the ions toward a magnet, which deflects them according to mass. The strength of the magnetic field is gradually changed during the experiment, and each field strength causes ions of a specific mass to reach the detector while all others are deflected into the walls of the tube. The ions are detected, and a mass spectrum is generated. The y-axis of the mass spectrum represents the ion mass abundance, and the x-axis represents the mass-to-charge ratio (m/z). The mass spectrum can be used to identify the mass of a molecule's fragments by taking the m/z difference between peaks. Samples analyzed by mass spectrometry are ionized and fragmented before detection. Therefore, the peaks observed in the mass spectrum represent ionized fragments of the sample. (Choice A) The peak that corresponds to the highest m/z ratio is typically the molecular ion. This peak may, or may not, be the tallest peak. (Choice B) A sample analyzed by mass spectrometry is ionized when injected, and only ions are detected. Therefore, the peaks represent charged species rather than uncharged molecules. (Choice C) Because the y-axis of a mass spectrum reflects abundance and the x-axis reflects molecular weight, the tallest peak represents the most abundant fragment. Therefore, peak height is not dependent on molecular weight but rather on abundance. Educational objective: 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).

Q27

Needed to know the polarities of the FGs: alkane < ester < alcohol < CA 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.

Q30

Passage stated that piperidine was a heterocyclic secondary amine. 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. (Choice A) This is a secondary amine but, it is not heterocyclic because the nitrogen atom is not part of the ring. (Choice B) This is an aromatic amine. It contains three carbon-nitrogen bonds, so it is not a secondary amine. (Choice C) This amine is heterocyclic, but it is a 3° amine rather than a 2° amine. Educational objective: Amines can be classified as primary, secondary, tertiary, or quaternary (ie, one, two, three, or four non-hydrogen substituents, respectively).

Q1

Q asking for RP-HPLC what would be the mobile and stationary phase? Normal phase: mobile = nonpolar stationary = polar RP: mobile = polar stationary = nonpolar Water is polar and hydrocarbons are nonpolar Educational objective: 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.

Q12

Q is asking how to separate the long-chain amide and CA in an extraction. So both the long chain amide and CA are in the organic layer and you need one to go into the aqueous layer. Acylation reactions between anhydrides and amines generate amides and carboxylic acids. Amides and carboxylic acids are polar, water-soluble molecules because they form hydrogen bonds in an aqueous solution. However, long alkyl side chains (R groups) of amides and carboxylic acids increase the hydrophobic (nonpolar) nature of the molecules and make them insoluble in aqueous solvents. Long-chain amides and carboxylic acids are the products of the reaction between long-chain anhydrides and the amine group of phenylethylamine. 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. (Choice A) NaHSO4 (pKb ∼12) is not basic enough to deprotonate the carboxylic acid. NaHSO4 is the salt of the conjugate base of sulfuric acid (a strong acid), and strong acids produce weak conjugate bases. (Choices B and D) HNO3 and HClO4 are both acids. Acids will not deprotonate carboxylic acids, and therefore cannot be used to separate long-chain amides and carboxylic acids. Educational objective: 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.

Q13

Q is asking how would phosphatidylethanolamine and 2,6-dimethoxyphenol enter the aqueous layer consecutively. Had to recognize that converting the acidic or basic FGs to ionic salts would help them enter the aqueous layer. phosphatidylethanolamine - weak base (amine) 2,6-dimethoxyphenol - weak acid (alcohol) so a weak base needs to be mixed with strong acid to turn it into ionic salt and a weak acid needs to be mixed with strong base to turn it into ionic salt The acid-base properties of organic compounds can be used to increase their separation during extraction. Compounds with acidic or basic functional groups can enter the aqueous layer as ionic salts, which are formed by deprotonation or protonation by a base or acid of varying strengths, respectively. Strong organic bases form ionic salts easily when weak acids are added to a mixture. On the other hand, weaker bases such as amines are only protonated by strong acids. Likewise, carboxylic acids, which are considered relatively strong organic acids, can be deprotonated by strong and weak bases. However, phenols are much weaker acids that require strong bases to be deprotonated. For phosphatidylethanolamine and 2,6-dimethoxyphenol to enter the aqueous layer consecutively, their ionic salts must be formed in sequence. Phosphatidylethanolamine contains an amine group that can be protonated to form a water-soluble ammonium salt; therefore, a strong acid such as H2SO4 or HCl should be added to protonate the amino group. In contrast, 2,6-dimethoxyphenol has a weakly acidic hydroxyl group on the benzene ring that must be deprotonated to form an ionic salt. Only strong bases such as NaOH or KOH are able to remove protons from phenols. Consequently, the correct sequence of extraction steps would involve the addition of a strong acid followed by a strong base. (Choice A) The order in which phosphatidylethanolamine and 2,6-dimethoxyphenol enter the aqueous layer is reversed. (Choice B) The aqueous layer will contain only phosphatidylethanolamine. NaHCO3(aq) is a weak base that cannot deprotonate 2,6-dimethoxyphenol. (Choice C) Neither phosphatidylethanolamine nor 2,6-dimethoxyphenol will enter the aqueous layer. Amines cannot be protonated by weak acids such as H2CO3(aq), and NaHCO3 is a weak base that will not deprotonate phenols. Educational objective: 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.

Q41

Q is asking which of the following will elute last in gas-liquid chromatography. In gas-liquid chromatography the compound with the highest bp elutes last so you have to look at the compound that has the most number of carbon bonds which means that its gonna have a higher BP Educational objective: 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.

Q8

Q stem is testing one substance under both simple and vacuum distillation and asking how their boiling points compare? Simple distillation is used for compounds with boiling points less than 150 °C, and that are more than 25 °C apart from each other. Vacuum distillation is used for compounds that decompose at their boiling point (typically boiling points greater than 150 °C), causing the boiling point to decrease and thereby preventing degradation. Simple and vacuum distillations are set up in the same way, except that simple distillations are done at atmospheric pressure, whereas vacuum distillations are connected to a vacuum pump and performed at reduced pressure. Because vacuum distillations are performed under reduced pressure, the boiling point of a compound under vacuum will decrease relative to its boiling point at atmospheric pressure. Therefore, the simple distillation boiling point of a molecule is greater than the boiling point of that molecule under vacuum (ie, simple distillation bp > vacuum distillation bp). Educational objective: 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.

Q3

Q stem says that DNA precipitated from aqueous layer using cold ethanol and sodium acetate so modifying DNAs charge or polarity modifies its affinity for each layer. If sodium acetate binds to DNA it neutralizes its charge and makes it go from a polar molecule to a nonpolar one. 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. Ethanol has a lower polarity than water. The addition of ethanol disrupts the solvation shell surrounding DNA, allowing for sodium cations to bond with DNA. (Choice A) Sodium acetate, the sodium salt of acetic acid, is a weak base. The addition of sodium acetate to water increases the solution's pH instead of reducing it. (Choice B) Ethanol has a lower polarity than water. The addition of ethanol disrupts the solvation shell surrounding DNA, allowing for sodium cations to bond with DNA. (Choice C) Cold ethanol is used because it is less likely than warm ethanol to disrupt DNA hydrogen bonds. However, the solution is not cold enough to freeze DNA. Educational objective: "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.

Q6

Stated in the passage: 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. Educational objective: 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.

Q11

Sterols are hydrophobic so you had to find the hydrophobic layer in both BD and BC. The densities of ethyl acetate (0.9006 g/mL) and choloroform (1.489 g/mL) was given to you and the density of water is 1. 0.9 < 1 < 1.5 densest layer sinks so in the BD diagram water would be in L1 and L3 would be organic layer. In the GC one the organic layer is L1 and the aqueous layer in L3. we know its not L2 because the passage states "The layers obtained in each extraction method, separated by films of cellular debris (L2)" Educational objective: 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.

Q28

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. Educational objective: 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.

Q14 (ask landon)

The caption under figure 2 tells you the conc of the lipids in the organic layer as 160 mg/mL Educational objective: 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.

Q29

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. Educational objective: 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.

Q31

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.

Q37

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. (Choice A) When using only hexanes, the compounds did not separate and remained near the origin. Therefore, hexanes are too nonpolar, and this results in the compounds' having a higher affinity for the stationary phase than the mobile phase. (Choice C) When ethyl acetate was used alone, both compounds moved with the solvent front. Therefore, ethyl acetate is too polar of a solvent, and this results in the compounds' having a higher affinity for the mobile phase than for the stationary phase. (Choice D) The 1:1 hexanes/ethyl acetate mixture increased the compounds' affinity for the mobile phase, rather than the stationary phase, relative to hexanes. Educational objective: 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.

Q33

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.

Q24

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. Educational objective: Absorption of ultraviolet (UV) light induces electron excitation, and a compound must contain a UV chromophore to absorb UV light.

Q16

ethyl acetate was only used in GC and not in BD so right from the beginning you can eliminate choices C and D. Look at figure 3 and knew you only had to look at (1) cuz GC. 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. (Choice A) The singlet at 0 ppm from the GC extraction (spectrum 1) is from the four -CH3 groups on tetramethylsilane (TMS), an internal standard used to calibrate the chemical shifts of the NMR sample components dissolved in an organic solvent. Educational objective: 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.

Q5

normal phase HPLC means polar molecules elute last so adduct c is the most polar and will elute last. 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. Educational objective: 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.

Q40

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 Educational objective: 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).