UW OCH

Based on the TLC plate shown in Figure 2, the best yield of Compound 4 results from which reaction conditions? A.48 h in DMF with 2.0 eq of IBX B.0.75 h in acetone with 3.0 eq IBX C.24 h in acetone with 1.5 eq IBX D.72 h in ethyl acetate with 3.0 eq IBX

0.75 h in acetone with 3.0 eq IBX The passage states that the reaction conditions for the oxidation of Compound 3 are optimized and monitored by thin-layer chromatography (TLC) to determine the best conditions for formation of Compound 4 (the product). The optimal conditions do not further oxidize Compound 4 into the side-product, Compound 5. The passage states that TLC was performed using normal-phase conditions; therefore, the compound with the weakest interaction with the stationary phase migrates the fastest up the TLC plate, and the compound with the strongest interaction with the stationary phase migrates the slowest. Compounds 3-5 have the following interaction strengths with the -OH groups on the silica stationary phase: Compound 5Two carbonyls (C=O) act as hydrogen bond acceptorsWeakest interactions with the stationary phase and fastest migration (top spot) Compound 4One hydroxyl group (-OH) acts as both a hydrogen bond donor and acceptorOne C=O acts as a hydrogen bond acceptorIntermediate interaction and migration (middle spot) Compound 3Two hydroxyl groups (-OH) act as both hydrogen bond donors and acceptorsStrongest interaction and slowest migration (bottom spot) The TLC plate shows that the reactions in acetone are the only ones in which Compound 4 is not overoxidized. Because the reactions used the same initial concentrations of Compound 3, the relative amounts of Compound 4 formed and Compound 3 remaining can be qualitatively assessed by comparing the size of the corresponding spots. For the reaction that was run for 0.75 h, the Compound 4 spot is larger and the Compound 3 spot is smaller compared to the corresponding spots for the reaction that was run for 24 h (Choice C). Therefore, the reaction that was run for 0.75 h in acetone with 3.0 eq IBX gave the greatest yield for the oxidation of Compound 3. (Choice A) Compared to 0.75 h in acetone, these conditions yield a smaller spot for Compound 4 as well as a spot for Compound 5, indicating some of the product was overoxidized. (Choice D) These conditions yield only Compound 5, indicating that all the Compound 4 that formed was overoxidized. Educational objective:The relative amounts of a product formed in different trials of the same

A mixture of 1-butanol, 1-hexanol, 1-octanol, and 1-propanol is separated by gas-liquid chromatography. What compound elutes last? A.1-butanol B.1-hexanol C.1-octanol D.1-propanol

1-octanol Gas-liquid chromatography (GC) is a technique used to separate molecules in a mixture, primarily on the basis of boiling point. The mixture to be separated is injected into a chamber that is heated to a temperature above the boiling points of all compounds in the mixture. All the compounds vaporize and enter the column, which is coated with a thin liquid layer that acts as the stationary phase. The mobile phase is an inert gas with a low boiling point (eg, N2, He, or Ar). Compounds in the mixture that have higher boiling points more readily reenter the liquid (stationary) phase, and therefore move more slowly through the column and elute later. Relative boiling points of molecules can be determined based on intermolecular forces, molecular weight (MW), and branching. The molecules in the mixture, 1-propanol, 1-butanol, 1-hexanol, and 1-octanol, are all alcohols with three-, four-, six-, and eight-carbon chains, respectively. The chain length can be determined by the compound's prefix. Because the molecules in the question all have the same functional group (alcohol) but different carbon counts, their relative boiling points depend on MW only. A higher MW indicates a higher boiling point and longer retention time because molecules with larger MWs have larger surface areas, giving a greater extent of London forces between the molecules. 1-Octanol has the greatest number of carbons (eight), the highest MW, and the highest boiling point of the four alcohols in the mixture and thus corresponds to the last peak in the GC trace (ie, has the longest retention time). (Choices A, B, and D) 1-Butanol, 1-hexanol, and 1-propanol have smaller carbon chains, lower MWs, and consequently lower boiling points than 1-octanol. These three alcohols therefore elute before 1-octanol. 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.

Which range corresponds to the expected chemical shifts of the adamantane protons in the 1H NMR spectrum of Compound 4? A.1.0-2.0 ppm B.2.5-4.5 ppm C.4.5-6.5 ppm D.7.0-8.0 ppm

1.0-2.0 ppm very upfield In proton nuclear magnetic resonance spectroscopy (1H NMR), the chemical shift (ie, the location of the signals in the spectrum relative to a reference compound) is given in parts per million (ppm) with a typical range of 0-12 ppm. Tetramethylsilane (TMS) (ie, Si(CH3)4) is commonly used as a reference compound (ie, its peak is set to 0 ppm) in 1H NMR because its methyl groups are very shielded, resulting in a signal that is upfield (ie, has smaller chemical shift) relative to most NMR signals. The chemical shift of a proton depends on its electronic environment. An upfield signal arises when a proton is surrounded by an electron cloud, shielding it from the external magnetic field B0. A downfield signal (ie, a larger chemical shift) arises when an adjacent electronegative substituent withdraws electrons from a proton, deshielding it from B0. The passage and the structure of Compound 4 in Figure 1 indicate that the adamantane substituent is a hydrocarbon (ie, made of carbon and hydrogen atoms). Aliphatic protons are surrounded by an electron cloud and shielded from B0, resulting in an upfield chemical shift typically below 2 ppm. Therefore, the signals for the protons in the adamantane substituent in Compound 4 appear between 1.0 and 2.0 ppm. The chemical shift of a proton in 1H NMR is given in ppm and depends on the electronic environment around the proton. Protons with an upfield signal (ie, small chemical shift) are more shielded from the magnetic field by their electron cloud. Protons with a downfield signal (ie, larger chemical shift) have structural features that deshield them from the magnetic field.

Compared to hands washed with no soap, washing hands with 3.0 mL of soap reduced bacteria recovered per hand by a factor of approximately: A.1/3 B.22 C.1010 D.100

100 remember log axis means 6 on the axis is 10^6= 1,000,000 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 a 100-fold change, and so on.

If Reaction 1 represents the most common process while obtaining a mass spectrum of mechlorethamine, which product mass-to-charge ratio (m/z) peak will have the highest abundance in the compound's mass spectrum? (Note: The isotopic abundances of 35Cl and 37Cl are 75% and 25%, respectively.) A.49 B.51 C.106 D.108

106 Mass spectrometry provides the ability to measure the mass of molecules and molecular fragments. An electron-impact mass spectrometer bombards a sample with electrons to remove an electron from a molecule (ionization). This creates the molecular ion, a radical cation with a mass-to-charge ratio (m/z) equal to the molecular weight of the molecule. The molecular ion can also break into a radical and a cation to produce smaller molecular fragments. Ionized material enters a magnetic field, which curves the path of the particles by their m/z. Only cations with m/z values that match the curvature of the instrument tube reach the detector. Finally, a mass spectrum (ie, a plot displaying the abundance of detected cations [y-axis] versus m/z [x-axis]) is generated. Molecules and fragments that contain isotopes are detected at different m/z values. The intensity of mass spectrum peaks containing isotopes is determined by the natural isotopic abundance. Although many elements have isotope effects that can be observed in a mass spectrum, chlorine and bromine have characteristic isotope signatures. Reaction 1 shows the fragmentation of the mechlorethamine molecular ion to a cation (m/z 106, 108) and a radical (m/z 49, 51). Each fragment contains one chlorine atom, which has two major isotopes separated by two amu (35Cl and 37Cl). Consequently, each fragment will appear at two m/z values in approximately a 3:1 ratio to reflect the natural abundance of 35Cl to 37Cl. Since a mass spectrum only detects charged particles (cations), only m/z 106 and m/z 108 will be detected. Of these, the m/z 106 peak will be the most abundant given the 75% natural abundance of the smaller 35Cl isotope.

What is the number of signals corresponding to aromatic hydrogen atoms present in the 1H NMR spectrum of Compound 3? A.1 B.2 C.3 D.4

2 Proton nuclear magnetic resonance spectroscopy (1H NMR) detects protons in a molecule when an external magnetic field and radio frequency are applied to a sample. The protons align their magnetic field either with (low-energy α spin state) or against (high-energy β spin state) the external magnetic field. Radio waves excite the protons from the α to the β spin state, and the peaks on the spectrum represent the energy difference between the spin states, known as the effective magnetic field. 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. The splitting pattern of a signal (ie, how many peaks a signal is split into) is based on the n + 1 rule. 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 (each appearing as a doublet) in the 1H NMR spectrum.

How many unique isomers are possible for a triacylglycerol molecule prepared from two molecules of stearate and one molecule of palmitate?

3 This question asks how many unique isomers are possible for a triacylglycerol molecule prepared from two molecules of stearate (C18 fatty acid) and one molecule of palmitate (C16 fatty acid). Since glycerol is symmetrical, having the palmitate attached at either end of a glycerol results in the same constitutional isomer (Compound A). However, this bonding pattern generates a stereocenter at the central glycerol carbon. As a result, both the R and S isomers of Compound A are possible outcomes. Having the palmitate attached to the middle carbon atom results in a third possible triacylglycerol isomer (Compound B). Since glycerol is symmetric, Compound B is achiral.

In the 1H NMR spectrum of mechlorethamine, what is the signal area ratio for proton environments found on carbons directly bonded to the nitrogen atom?

4 to 3

Terpenes are synthesized from two or more five-carbon groups known as isoprenes. The triterpenoid lanosterol, a steroid precursor, is shown below. How many isoprene units are present in lanosterol? A.3 B.4 C.6 D.8

6 Terpenes are lipid precursors to steroids and lipid signaling molecules. 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. Terpenes are categorized according to how many isoprenes are included. For example, the most basic terpene is made up of two isoprene units (10 carbon atoms total) and is known as a monoterpene. A diterpene contains double the number of isoprene units and carbon atoms as a monoterpene, giving a diterpene four isoprene units and 20 carbon atoms. Molecules made of isoprene units that also contain nonhydrocarbon functional groups are called terpenoids. Lanosterol is a terpenoid synthesized during cholesterol synthesis in vertebrates. The question states that lanosterol is a triterpenoid, which can be thought of as three monoterpenes put together. Therefore, it is composed of six isoprene units (3 terpenes × 2 isoprenes/terpene = 6 isoprenes) and has 30 carbon atoms in total (6 isoprenes × 5 carbons/isoprene = 30 carbons). (Choice A) A triterpene can be thought of as three monoterpenes put together, which would contain six isoprenes rather than three. A terpene with three isoprenes is known as a sesquiterpene (ie, 1.5 monoterpenes). (Choices B and D) Terpenes with four and eight isoprenes are known as diterpenes and tetraterpenes, respectively. Educational objective: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 present in the molecule.

conjugated system

A conjugated system can be identified by alternating single bonds and double or triple bonds, or in other words, by alternating pi bonds (p orbitals) separated by sigma bonds. Conjugation allows for the delocalization of electron density, meaning electrons can be distributed through the alternating system's pi bonds. In general, conjugation and electron delocalization serve to stabilize molecules such as carbanions, carbocations, and radicals by creating a more favorable charge distribution. This can be represented as multiple resonance forms of the molecule. The quinonoid intermediate (Figure 2) delocalizes the negative charge of a carbanion over a large conjugated system of pi bonds, which allows for multiple resonance forms. In contrast, the enolate (Figure 3) delocalizes the negative charge over a relatively small region with fewer resonance forms. Therefore, the quinonoid intermediate is more stable.

glycosidic bond

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. The question shows that in the cleavage of the glycosidic bond between the two given sugar molecules, glutamic acid at position 35 (Glu35) donates a proton to one of the sugars in the disaccharide as aspartate at position 52 (Asp52) acts as a nucleophile to attack the anomeric carbon of the other sugar. This yields an intermediate in which a sugar is glycosidically bonded to Asp52. To complete the reaction, Glu35 must return to its protonated form and a hydroxyl group must act as a nucleophile to break the bond to Asp52. This can be accomplished if Glu35 deprotonates a water (H2O) molecule. The increased pKa of Glu35 in its local environment makes this possible. This deprotonation creates a good nucleophile (OH−) that can then attack the anomeric carbon while simultaneously restoring Glu35 to its protonated form. OH− then attacks and Asp52 acts as the leaving group. 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.

Which of the following observations would most likely indicate that the product from this reaction is pure? One absorbance peak from size-exclusion chromatography B.Two peaks in a high-performance liquid chromatogram C.Two unique −CH3 signals in the 1H NMR spectrum D.A single spot in the reaction mixture lane on a thin-layer chromatography plate

A single spot in the reaction mixture lane on a thin-layer chromatography plate Several techniques can be used to evaluate the purity of a compound, including thin-layer chromatography (TLC), which separates compounds in a mixture based on polarity. In TLC, the tested sample is dissolved in an organic solvent and a small amount is spotted onto the stationary phase (commonly a polar silica [SiO2]) coated on a TLC plate. After spotting, the TLC plate is placed in a solvent chamber, and the mobile phase (organic solvent) travels up the plate and carries the spotted compounds along at different rates, which separates the mixture components. Nonpolar compounds have less affinity for a polar stationary phase and will travel further up the plate than polar compounds. The conversion of theobromine to caffeine results in N-methylation (conversion of N-H to N-CH3), resulting in a large change in polarity. The N-H in theobromine can hydrogen bond to the polar stationary phase on the TLC plate, whereas the N-CH3 in caffeine cannot hydrogen bond. Because theobromine can hydrogen bond, it is more polar and has a smaller Rf value than caffeine, and it produces distinguishably different spots on a TLC plate. Therefore, 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 reaction mixture lane on the TLC plate, this indicates that the reaction proceeded to completion, no more reactant remained, and the isolated product is pure.

soaps

A soap is a sodium carboxylate salt of a fatty acid. Soaps can be produced through basic hydrolysis of fatty acyl-containing lipids like triacylglycerols or phospholipids. The number of unique soap molecules that can be generated is dependent on the number of unique fatty acyl groups in the source lipid.

A solution of Compound 1, shown below, absorbs light maximally at 448 nm in the absence of copper(II) ions but shifts to a 623 nm absorption maximum upon the addition of Cu2+. Which of the following best describes this process? A.Changes in electronic structure cause the solution to change from yellow to blue. B.Changes in vibrational modes cause the solution to change from green to yellow. C.Changes in the mass-to-charge ratio (m/z) cause the solution to change from violet to orange. D.Changes in nuclear spin cause the solution to change from colorless to violet.

A.Changes in electronic structure cause the solution to change from yellow to blue. In highly conjugated systems, the difference in energy between the ground and excited states of the electrons is equal to the energy of a particular wavelength of visible light. Photons of this wavelength are absorbed by the molecule, causing an electron to enter the excited state. The remaining wavelengths are either reflected or transmitted and are ultimately perceived by the eye as various colors. In general, the perceived color of a substance is complementary to the color of the wavelength that is maximally absorbed by that substance. For example, a molecule that absorbs blue light will appear orange and vice versa. When the basic colors of the visible spectrum (red, orange, yellow, green, blue, and violet, or ROYGBV) are arranged in a wheel with red and violet next to each other, a color's complement is the color directly across from it on the wheel. 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. (Choice B) Vibrational modes are excited by infrared light, which is not visible to the eye. They do not affect the color of a solution. (Choice C) The mass-to-charge ratio (m/z) is measured for ions generated from samples analyzed by mass spectrometry and pertains to the molecular mass of the ions, which does not affect color of the sample. Compound 1 absorbs violet light initially and then absorbs orange light when Cu2+ is added; the color perceived is not the color absorbed, but rather the complement of the color absorbed. (Choice D) Nuclear spin is probed by radio frequencies in nuclear magnetic resonance (NMR) spectroscopy and does not affect color. 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

Based on the passage, what can be inferred about the relative HOMO-LUMO energy difference for α-guaiene and guaiazulene? A.The HOMO-LUMO energy difference is smaller for guaiazulene B.The HOMO-LUMO energy difference is larger for guaiazulene C.The HOMO-LUMO energy difference is the same for α-guaiene and guaiazulene D.Nothing can be inferred about the relative HOMO-LUMO energy difference for these compounds

A.The HOMO-LUMO energy difference is smaller for guaiazulene When the carbon atoms in two or more alkene functional groups are directly connected to each other, the double bonds are classified as conjugated. Importantly, the pi electrons in conjugated alkenes are distributed throughout the conjugated region and are no longer isolated to one alkene functional group. Consequently, the properties and reactivity of conjugated alkenes are different than those for isolated alkenes. Understanding the properties and reactivity of conjugated alkenes requires application of molecular orbital (MO) theory. Since electron orbitals can be described by their wave function (a mathematical expression), the linear combination of atomic orbitals leads to the generation of MOs that distribute electron density throughout the nuclei in a molecule (rather than at one specific atom). Like atomic orbitals, MOs are populated by electrons starting with the lowest-energy MO. The highest occupied molecular orbital (HOMO) is the highest-energy MO containing at least one electron, and the lowest unoccupied molecular orbital (LUMO) is the lowest-energy vacant MO. Delocalization of electrons through conjugation decreases the average energy of filled orbitals. Consequently, conjugated alkenes are more stable, of lower energy, and have a smaller energy difference between the HOMO and LUMO than isolated alkenes. This question asks what can be inferred about the relative HOMO-LUMO energy difference for α-guaiene and guaiazulene. Guaiazulene is an aromatic compound containing a core of alternating single-double bonds—a type of conjugated system. In contrast, α-guaiene contains two isolated (nonconjugated) alkene groups. Because conjugation decreases the HOMO-LUMO energy difference, it can be concluded that the HOMO-LUMO energy difference is smaller for guaiazulene. Conjugated alkenes are comprised of alternating single-double bonds and are more stable than isolated (nonconjugated) alkenes. Application of molecular orbital theory demonstrates that conjugated alkenes have a smaller HOMO-LUMO gap.

Can α-D-fructose and β-D-fructose be classified as anomers? A.Yes, because they are ketoses that differ in configuration at C2 only. B.Yes, because they differ in configuration at more than one carbon. C.Yes, because they are C1 epimers. D.No, because they are enantiomers.

A.Yes, because they are ketoses that differ in configuration at C2 only. Ketoses differ at anomeric C2 and Aldoses differ at anomeric C1 Stereoisomers are molecules that have the same atom connectivity but differ in the spatial orientation of the atoms. Stereoisomers contain one or more stereocenters, or chiral centers, which consist of an atom (typically carbon) with four different substituents. Diastereomers, a type of stereoisomer, have more than one stereocenter and differ in orientation at some, but not all, stereocenters. 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. Carbon atoms in sugars are numbered beginning with the carbon that is closest to the anomeric carbon when the sugar is in linear form. 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. (Choice B) Anomers differ in configuration at only one position (the anomeric carbon), not multiple positions. (Choice C) Epimers are diastereomers that differ in configuration at one position, and anomers are a specific type of epimer that differs at the anomeric carbon. However, for fructose, the anomeric carbon is C2, not C1. (Choice D) α-D-fructose and β-D-fructose are not enantiomers because they differ in configuration at only one stereocenter, whereas enantiomers differ in configuration at every stereocenter. The enantiomer of α-D-fructose is α-L-fructose, and the enantiomer of β-D-fructose is β-L-fructose. Educational objective: 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.

Why was the optimization for the oxidation of Compound 3 monitored by thin-layer chromatography? The resulting TLC plate given in the passage is used to determine: A.which reaction conditions provide the greatest amount of Compound 4 and the least amount of Compound 5. B.the percent yield of the reaction. C.whether Compound 4 or Compound 5 has a greater Rf value. D.which solvent and reaction time provide the greatest amount of Compound 5 and the least amount of Compound 4.

A.which reaction conditions provide the greatest amount of Compound 4 and the least amount of Compound 5. Reaction optimization is necessary to determine under what conditions the desired product readily forms while preventing or minimizing side-product formation. Thin-layer chromatography (TLC) is a separation technique commonly used to visually monitor reaction progress when the starting material and products have different relative polarities. As such, TLC can be used to determine whether starting material is still present and whether desired products and/or side-products have formed. The rate at which each compound travels up the TLC plate depends on the strength of the competing intermolecular interactions between the compound and the stationary and mobile phases. The passage states that the oxidation of Compound 3 yielded a mixture of Compound 4 (the desired product) and Compound 5 (the overoxidized side-product). TLC was then performed using a nonpolar mobile phase and a polar silica stationary phase. Compounds 3-5 have different strengths and numbers of interactions with the stationary phase: Compound 5 has the fewest and weakest interactions and migrates the fastest and farthest up the TLC plate whereas Compound 3 migrates the slowest. The given TLC plate shows that different solvents and reaction times yield different relative amounts of Compounds 3-5. This information can be used to determine the optimal conditions—those that yield the greatest amount of Compound 4 (the product) and the least amount of Compound 5 (the side-product). (Choice B) Although TLC can show which conditions produce more or less of a compound, it is not a quantitative technique and does not allow determination of exact amounts of a compound present in a mixture. Therefore, the percent yield of the reaction cannot be calculated from the TLC plate. (Choice C) An Rf value is the ratio of the distance a compound migrates up the TLC plate to the distance traveled by the mobile phase. Relative Rf values of two compounds can be compared without running TLC because the relative strengths and numbers of interactions two compounds have with silica in the stationary phase are known previously. A compound that has a stronger intera

The (R)-enantiomer of the antiasthma drug albuterol is the active isomer. If a researcher wants to separate a racemic mixture of albuterol, which of the following methods will most likely separate the enantiomers? A.Extraction with dilute base B.Thin-layer chromatography C.Fractional distillation D.Addition of a resolving agent

Addition of a resolving agent 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.

aldol condensation

Aldol condensations are carbon-carbon bond-forming reactions between two carbonyl compounds (ketones and/or aldehydes). The first step (aldol addition) involves enolate formation and nucleophilic attack of a carbonyl, forming the aldol product, a β-hydroxy ketone (or aldehyde). The second step of an aldol condensation is dehydration of the β-hydroxy ketone (or aldehyde) to yield an α,β-unsaturated ketone (or aldehyde).

how to make alcohol a good leaving group

Although a hydroxyl group is a poor leaving group, conversion to a mesylate or tosylate makes it an improved leaving group. Conversion of chiral alcohols to mesylates or tosylates does not change the stereochemistry of the carbon bonded to the hydroxyl group.

amino acid configurations

Amino acids contain a central carbon bonded to an amino group, a carboxyl group, a hydrogen atom, and a side chain. The central carbon is known as the α-carbon and is a stereocenter in all amino acids except glycine, which has only three unique substituents bonded to the α-carbon (ie, two groups are hydrogen). All other amino acids contain a stereocenter because the side chain is different than the other three substituents (ie, four unique substituents are attached to the α-carbon). These amino acids can be classified in two ways: the ʟ/ᴅ system and the R/S system. The ʟ/ᴅ system compares chiral molecules to glyceraldehyde whereas the R/S system assesses the arrangement of substituents around the central carbon. To determine the ʟ/ᴅ configuration of an amino acid, it is easiest to first determine the R/S configuration by considering the priority of the α-carbon substituents. For most amino acids, the order of priority is -NH3+ -COO− Side chain -H Cysteine is an exception because its side chain (-CH2SH) has a higher priority than -COO−. All natural chiral amino acids have an S configuration except for cysteine, which has an R configuration. However, both R-cysteine and the S form of other amino acids are in the ʟ configuration; therefore, the S form of serine is the ʟ form. From highest to lowest priority, the α-carbon substituents shown in Choice B are arranged in a clockwise fashion. Because the lowest priority group is in front of the plane (wedge) in the given depiction, the clockwise arrangement indicates the molecule is the S form, and is therefore the ʟ configuration of serine. Amino acids are chiral molecules (except glycine) that contain an α-carbon atom with four unique substituents: -NH3+, -COO−, a side chain, and -H. The natural chiral amino acids all have an S configuration except cysteine, which has an R configuration. All the natural chiral amino acids have an ʟ configuration, and the enantiomers of these amino acids have a ᴅ configuration.

f the compound shown below is the product of a Fischer esterification, which of the following compounds is a starting reagent for the reaction and must contain oxygen-18?

An atom in a compound can be substituted with an isotope, which can be tracked through a reaction. This procedure is called labeling. The final position of the isotopic label in the reaction product gives information about how the reaction occurs. Esters are carboxylic acid derivatives that can be formed under acidic conditions through a Fischer esterification using a carboxylic acid and an alcohol. The reaction begins with protonation of the carbonyl oxygen of the carboxylic acid, followed by nucleophilic attack of the carbonyl carbon by the alcohol oxygen. A proton is then transferred from the alcohol oxygen to the hydroxyl group of the carboxylic acid, resulting in water (a good leaving group). Lastly, the carbonyl oxygen is deprotonated to yield an ester. 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). (Choice B) Although it is correct that the 18O in the ester comes from an alcohol, this structure has too many carbon atoms. The four-carbon chain in the ester comes from a carboxylic acid (butanoic acid) rather than an alcohol. (Choice C) It is true that the four-carbon chain originates from a carboxylic acid, but the 18O in the ester must come from the alcohol rather than the carboxylic acid. If 18O were in the hydroxyl group of the carboxylic acid prior to the reaction, then that 18O atom would be found in a water molecule at the end of the reaction, not in the ester. (Choice D) A carboxylic acid is involved in a Fischer esterification, but this structure has the incorrect number of carbon atoms, and the 18O in the ester comes from the alcohol starting reagent. Educational objective: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

anhydride formation

Anhydrides are carboxylic acid derivatives produced by condensation of two carboxylic acids or nucleophilic acyl substitution of an acid chloride with a carboxylate. The anhydride product is characterized by two carbonyl groups joined by an oxygen atom. An organic R group (typically, an alkyl group) is also bonded to each carbonyl carbon; these alkyl groups may be identical to each other or they may be distinct. Anhydride formation results in the loss of a water molecule from the carboxylic acid starting material. 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. 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.

heterocycles

Aromatic heterocycles are commonly found in biological systems. A heterocycle is any cyclic compound containing at least one heteroatom (ie, an atom other than carbon or hydrogen). To be classified as aromatic, a compound must have: Conjugated pi bonds in a cyclic structure Unhybridized p orbitals present in each atom in the ring Planar geometry, forming a continuous ring of parallel, overlapping unhybridized p orbitals 4n + 2 pi electrons (Hückel rule), where n is a non-negative integer Tryptophan, thymine, and adenine are all considered aromatic heterocycles because they have nitrogen-containing rings (indole, pyrimidine, and purine, respectively) with conjugated pi bonds and overlapping unhybridized p orbitals (Choices A, C, and D). Although thymine may not appear to have conjugated pi bonds, a resonance structure can be drawn to show a conjugated system similar to pyrimidine. Basic nitrogen atoms, such as in pyrimidine and purine, have a double bond placing the lone pair electrons in an sp2 orbital perpendicular to the conjugated p orbitals containing the pi electrons. Electrons in orbitals perpendicular to the conjugated plane cannot undergo side-to-side overlap with the p orbitals and therefore do not participate in aromaticity. Nonbasic nitrogen atoms, such as in purine and indole, have three bonds, putting the lone pair of electrons into a p orbital parallel to other pi electron-containing p orbitals. Parallel orbitals permit side-to-side overlap; therefore, these electrons do participate in aromaticity. As a result, thymine has 6 pi electrons, and tryptophan and adenine each have 10 pi electrons; all satisfy Hückel rule. (Choice B) Proline can be classified as a heterocycle, but it is not considered aromatic because it is not conjugated and does not contain any pi electrons. 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.

physiological pH

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. Acidic amino acids have a net negative charge Basic amino acids have a net positive charge

When a student uses mass spectrometry to verify that they made Compound 2, what is the expected m/z of the molecular ion peak (M+) for Compound 2? (Note: Acetic acid and propanol each have a molecular weight of 60 g/mol.) A.60 B.102 C.120 D.138

B.102 whole thing is 120 but you need to account for H2O leaving which is 120-18

Which of the following statements most accurately describes the component that remains in the reaction flask during the steam distillation? A.2-nitrophenol remains in the reaction flask because it has more intermolecular hydrogen bonding. B.4-nitrophenol remains in the reaction flask because it has more intermolecular hydrogen bonding. C.2-nitrophenol remains in the reaction flask because it has less intermolecular hydrogen bonding. D.4-nitrophenol remains in the reaction flask because it has less intermolecular hydrogen bonding.

B.4-nitrophenol remains in the reaction flask because it has more intermolecular hydrogen bonding. Distillation is a purification technique used to separate molecules based on boiling point. A liquid mixture is heated to a temperature that overcomes the intermolecular forces keeping the compound in the liquid phase. The vapors then condense in the collection flask as a liquid. A molecule that participates in strong intermolecular forces will have a higher boiling point than molecules with weak intermolecular forces. 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 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.

A student needs to separate a mixture of chloroform (bp 61°C) and benzene (bp 80°C). What type(s) of distillation would be expected to give the best separation of the two compounds? Simple distillation Fractional distillation Vacuum distillation A.I only B.II only C.I and III only D.II and III only

B.II only Distillation is a technique used to separate liquid molecules based on their boiling points. The compound with the lowest boiling point evaporates first. The mixture is heated slowly so that the boiling temperatures of compounds with higher boiling points are not reached before the compound with a lower boiling point (distillate) evaporates and is collected. The vapors of the distillate travel up the distillation column connected to a thermometer and a condenser and then condense into the receiving flask, which is in an ice bath to prevent the distillate from evaporating. This type of distillation is known as a simple distillation. It is best suited for compounds with boiling points <150°C and >25°C apart from each other. Chloroform and benzene have boiling points separated by <25°C (Number I). There are two variations of simple distillation: Fractional distillation works best for compounds that have boiling points that are <25°C apart, such as chloroform and benzene (19°C difference in boiling points). The setup is the same as a simple distillation, with the addition of a fractionating column between the flask and the condenser to provide a larger surface area on which the vapors of the distillate condense before distilling into the receiving flask. The increased area allows for better separation of compounds with similar boiling points (Number II). Vacuum distillation is appropriate for compounds that have boiling points >150°C, which may decompose when heated beyond this temperature. To avoid decomposition, the distillation apparatus is connected to a vacuum, decreasing the pressure of the system and consequently lowering the boiling points of the mixture components. Neither chloroform nor benzene has a boiling point >150°C (Number III). 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).

Which of the following statements does NOT correctly describe a keto-enol tautomerism? A.The reaction results in a pi bond migration, forming a C=C double bond. B.The reaction can be acid-catalyzed by protonating an enol oxygen atom. C.The reaction involves removal of a proton from an α-carbon. D.The reaction results in an equilibrium mixture of keto and enol tautomers, which are constitutional isomers.

B.The reaction can be acid-catalyzed by protonating an enol oxygen atom. Keto-enol tautomerization is a type of isomerization that involves a proton removal from the α-carbon of a ketone or aldehyde (Choice C) and addition of a proton to the carbonyl (C=O) oxygen atom. Additionally, a pi bond migrates from the C=O double bond to form a C=C double bond between the carbonyl carbon and the α-carbon (Choice A). The effective proton transfer and pi bond migration result in the formation of an enol. The ketone (or aldehyde) and enol forms of the molecule are called tautomers and exist in equilibrium. Because the tautomers have the same molecular formula and their structures only differ in the atom connectivity, they are classified as constitutional isomers (Choice D). Keto-enol tautomerization can be acid catalyzed when an acid protonates the carbonyl oxygen in the ketone (or aldehyde), resulting in a cation intermediate. Then, the α-carbon is deprotonated, and the pi bond migrates, forming an enol. Additionally, keto-enol tautomerizations can be base catalyzed. The question asks which statement does NOT describe keto-enol tautomerization. Acid-catalyzed tautomerism does not involve protonation of the enol oxygen atom; instead, it involves protonation of the carbonyl oxygen atom. Educational objective:Keto-enol tautomerization is a base- or acid-catalyzed isomerization that converts a ketone or aldehyde into an enol. Tautomerism involves a proton transfer from the α-carbon of a ketone or aldehyde to the carbonyl oxygen atom along with pi bond migration. The keto and enol tautomers are in equilibrium and are classified as constitutional isomers.

The pKa values of the functional groups of lysine are shown in the following table. Functional group pKa α-carboxyl 2.18 α-amine 8.95 Side chain amine 10.53 Which of the following statements best explains why the pKa value of the α-amine is lower than the pKa value of the side chain amine? A.The α-carboxylate stabilizes the α-amine through charge-charge interactions. B.The α-carboxylate withdraws electrons from the α-amine due to the inductive effect. C.The protonated α-amine is stabilized through resonance. D.Different resonance forms stabilize the deprotonated amine.

B.The α-carboxylate withdraws electrons from the α-amine due to the inductive effect. The pKa value of an ionizable functional group is the pH value at which the protonated and deprotonated forms of the functional group are present in equal concentrations. As such, the pKa value is influenced by local environmental and structural factors that affect the stabilities of the protonated vs. deprotonated forms. If the protonated form is stabilized, the functional group resists deprotonation until higher pH values, raising the pKa. Similarly, destabilization of the deprotonated form allows the group to pick up a proton more easily, raising the pH at which significant protonation can occur, and thus raising the pKa. The inductive effect occurs when one or more functional groups donate or withdraw electron density from nearby ionizable groups. This effect works through covalent bonds. By withdrawing electrons, the electronegative oxygens of a carboxylate ion stabilize negative charges and destabilize positive charges of nearby functional groups. Thus, the α-carboxylate destabilizes the positively charged α-amine. In contrast, the side chain amine is farther away from the α-carboxylate and less affected by the inductive effect. Destabilization of the protonated α-amine means the α-amine gives up its proton more easily at a lower pH, and therefore the α-amine has a lower pKa due to the inductive effect. (Choice A) Charge-charge stabilization of the positive, protonated α-amine would tend to increase its pKa by making the group resistant to deprotonation. (Choices C and D) The α-amine of an isolated amino acid like lysine does not participate in resonance. The α-amine is connected only to the sp3 hybridized α-carbon, which does not have any π-bonds to allow for delocalized electrons. Educational objective:The α-connected functional groups of amino acids have lower pKa values than similar side-chain functional groups. This variation in pKa value is due to effects such as the inductive effect, through which nearby electronegative atoms destabilize protonated functional groups and stabilize deprotonated groups.

A wax is made up of a long-chain fatty acid and a long-chain alcohol joined by: A.an amide linkage. B.an ester linkage. C.an anhydride linkage. D.a glycosidic linkage.

B.an ester linkage. Lipids are hydrophobic molecules that are classified based on characteristics such as length, linkage type, and solubility. These molecules are broadly classified as hydrolyzable (lipid contains hydrolyzable ester or amide linkages) or nonhydrolyzable (lipid does not contain hydrolyzable ester or amide linkages). The hydrolyzable lipids, such as phospholipids, glycolipids, and waxes, are further classified based on the structure of their backbone (glycerophospholipids, glycosphingolipids, etc.). 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. (Choice A) Amide linkages are present in sphingolipids but not in waxes. For an amide to form, the carboxyl group of the fatty acid would need to combine with an amine rather than an alcohol. (Choice C) Like esters, anhydrides are carboxylic acid derivatives. However, for an anhydride to form, the carboxyl group from the fatty acid would need to combine with an acid chloride rather than an alcohol. (Choice D) A glycosidic bond joins a sugar to another molecule. Because waxes are composed of a fatty acid and an alcohol and do not contain sugars, a glycosidic bond cannot be present. Educational objective: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.

Two separate reactions are conducted in which a compound containing a ketone, an ester, and a carboxylic acid is reacted with borane (BH3) in THF in one reaction and with NaBH4 in methanol in the other. Which of the following explains why different products are observed? A.BH3 will selectively reduce ketones, and NaBH4 will only reduce carboxylic acids. B.BH3 does not reduce ketones, and NaBH4 will selectively reduce esters. C.BH3 reduces carboxylic acids and esters, and NaBH4 will only reduce esters. D.BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce ketones.

BH3 will selectively reduce carboxylic acids, and NaBH4 will selectively reduce ketones. Reduction of a functional group results in a gain of electrons from the transfer of a hydride, a decreased oxidation state of carbon, and a decreased number of C-O bonds. Several reagents can be used to reduce a functional group, including borane (BH3) and NaBH4. These reducing agents transfer one or more hydrides to the carbon atom being reduced. 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. (Choice A) BH3 and NaBH4 are both reducing agents, but BH3 selectively reduces carboxylic acids rather than ketones, and NaBH4 is not strong enough to reduce carboxylic acids. NaBH4 selectively reduces ketones and aldehydes.

Based on the passage, which solvent will most likely promote the desired reaction outcome? A.(CH3CH2)2O B.CH3(CH2)4CH3 C.CH3SOCH3 D.CH3CO2H

C.CH3SOCH3 DMSO The solvent that is used for a nucleophilic substitution reaction can have a profound impact on the reaction outcome. Since first-order (SN1) substitution reactions proceed through a carbocation intermediate, polar solvents are required to promote the formation and solvate the essential ionic intermediates. Many of the best solvents for SN1 reactions are also protic and contain functional groups that are hydrogen bond donors. Polar protic solvents (eg, water, alcohols) tend to have higher relative boiling points to accommodate heated SN1 reactions and, in many cases, double as the nucleophile for the substitution reaction. Since second-order (SN2) substitution reactions proceed through a polar transition state, polar solvents are also required. However, because SN2 reactions require the use of strong nucleophiles, solvents that lack acidic hydrogen atoms (ie, aprotic solvents) are commonly used to ensure that the nucleophile is not surrounded by a rigid solvent shell. Common polar aprotic solvents include acetone (CH3COCH3), acetonitrile (CH3CN), dimethylformamide (DMF, HCON(CH3)2), and dimethylsulfoxide (DMSO, CH3SOCH3). Because the passage states that the transformation is intended to be an SN2 reaction, the correct solution should be a polar aprotic solvent. CH3SOCH3 contains a polar sulfoxide functional group (S=O) and lacks the ability to hydrogen bond, making it the best solvent of the given choices to promote an SN2 reaction. (Choice A) The molecule (CH3CH2)2O (diethyl ether) contains an ether functional group. Ethers are relatively nonpolar and are not the best candidates to promote an SN2 reaction. (Choice B) The molecule CH3(CH2)4CH3 (hexane) is a nonpolar alkane that will be unable to solvate the polar reactants and transition state in an SN2 reaction. (Choice D) The molecule CH3CO2H (acetic acid) contains a carboxylic acid functional group. Although carboxylic acids are polar, the acidic proton will rapidly neutralize the basic alkoxide nucleophile, greatly decreasing the efficiency of the SN2 displacement. Educational objective:SN2 reactions are promoted by polar aprotic solvents, which facilitate formation of the transition state without diminishing the strength of the nucleo

Researchers purify two samples of the same mixture, one by simple distillation and the other by vacuum distillation. Which of the following expressions correctly compares the relationship between the mixture components' boiling points (bp) during each distillation? A.Simple distillation bp = vacuum distillation bp B.Simple distillation bp < vacuum distillation bp C.Simple distillation bp > vacuum distillation bp D.Simple distillation bp = 2(vacuum distillation bp)

C.Simple distillation bp > vacuum distillation bp Distillation is a technique used to separate liquid molecules based on their boiling point, the temperature at which the vapor pressure equals the pressure of the system. There are three common types of distillation: simple, fractional, and vacuum. The type of distillation suitable for a mixture is dependent on the boiling points of the mixture components. 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). (Choices A, B, and D) Reducing the pressure of a system reduces the boiling point of a compound. Therefore, the boiling points in simple and vacuum distillations cannot be equal and the vacuum distillation boiling point would not be greater than the simple distillation boiling point. 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.

A linear five-carbon ketone has a boiling point most similar to: A.a five-carbon alkane. B.a five-carbon carboxylic acid. C.a branched aldehyde with the same molecular weight. D.an alcohol with a higher molecular weight.

C.a branched aldehyde with the same molecular weight. The boiling point of a substance depends on the intermolecular forces holding the molecules together. Strong intermolecular forces require more energy to break, causing higher boiling points. Molecules with functional groups that participate in strong bonds experience stronger intermolecular forces and have higher boiling points than molecules with similar molecular weights (MWs) that participate only in weak bonds. Molecules with a greater surface area (higher MW or less branching) experience more intermolecular forces and have higher boiling points than similar molecules with smaller surface areas. 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.

If thin-layer chromatography shows that a distillation of molecules with boiling points above 150 °C results in an impure compound, one can conclude that a possible error made during the distillation was: A.reducing the pressure of the distillation system. B.using a tall fractionating column. C.heating the mixture too quickly. D.placing boiling chips in the mixture.

C.heating the mixture too quickly. Distillation is a technique used to separate liquid molecules based on their boiling points, and there are three common types: simple, fractional, and vacuum. In each, a liquid mixture is heated, and the molecule with the lowest boiling point distills first. The mixture must be heated slowly so that the molecule with the lowest boiling point can evaporate before the molecules with the higher boiling points. The vapors of the lower boiling-point compound travel up the distillation column (which is fitted with a thermometer), condense and travel through the condenser, and then drip into the receiving flask. The flask is placed in an ice bath to prevent the distillate from evaporating. 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. (Choice A) Compounds with boiling points above 150 °C must be distilled under reduced pressure using vacuum distillation. This decreases the boiling points, preventing decomposition of the compounds. (Choice B) Fractional distillation uses a column to improve separation of compounds by increasing the length of the path the vapor must travel before condensing and dripping into the receiving flask. (Choice D) Boiling chips are used to evenly heat a liquid and prevent superheating. Their use in the distillation would prevent the mixture from abruptly forming large bubbles that could spill over 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.

In the oxidation of vitamin K hydroquinone to vitamin K 2,3-epoxide, hydroxyl groups are converted to carbonyls. Compared to the carbon-oxygen bonds of the hydroxyl groups, the carbonyl carbon-oxygen bonds are characterized by: A.decreased overall bond strength. B.longer total bond length. C.side-to-side overlap of p orbitals. D.increased bond rotation.

C.side-to-side overlap of p orbitals. A sigma bond is a covalent bond formed by direct end-to-end overlap of hybridized atomic orbitals such as sp, sp2, or sp3 orbitals. (Note that the exception is hydrogen, whose atomic orbital is s.) In contrast, a pi bond is a covalent bond created by side-to-side overlap of nonhybridized p orbitals. A double bond contains one sigma bond and one pi bond, and a triple bond consists of one sigma bond and two pi bonds. During Reaction 1, two hydroxyl groups (C-OH) are converted to carbonyl groups (C=O). Each of the original single C-O bonds consisted of one sigma bond between two atoms whereas the newly created C=O double bonds contain a sigma bond and a pi bond. The pi bond is made from side-to-side overlap of p orbitals and is found in the carbonyl but not the hydroxyl group. (Choices A and B) A pi bond alone is weaker than a sigma bond, but a double bond such as that found in a carbonyl consists of both a sigma bond and a pi bond. As such, the double bond of the carbonyl is stronger than the single bond of the hydroxyl group. Bond length is inversely proportional to bond strength and therefore the double bond of the carbonyl has a shorter total bond length than the C-O bond of the hydroxyl group. (Choice D) Atoms bonded by only a sigma bond can freely rotate around the bond. The addition of pi bonds inhibits rotation and introduces molecular rigidity. 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.

Suppose that NaI was replaced with another nucleophile in Experiment 1. Which of the following anions would be the most nucleophilic? A.CH3(CH2)2CH2− B.(CH3)2N− C.CH3CH2O− D.F−

CH3(CH2)2CH2− Nucleophiles donate electrons to electrophiles in nucleophilic substitutions. There are several factors that contribute to nucleophilicity, including charge and electronegativity. 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. To compare the relative nucleophilicity of given anion, the electronegativity of the negatively charged atoms must be considered. The charged carbon atom of CH3(CH2)2CH2− is less electronegative than the nitrogen atom of (CH3)2N−, the oxygen atom of CH3CH2O−, and the fluorine atom of F−. Therefore, electrons on a negatively charged carbon atom are less stabilized and more easily shared with an electrophile, making CH3(CH2)2CH2− the best nucleophile. (Choices B, C, and D) Nucleophiles consisting of a negatively charged nitrogen, oxygen, or fluorine contain charged atoms that are more electronegative and have a stronger hold on their electrons, making these species less nucleophilic than carbon nucleophiles. 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.

Based on the NMR results described in the passage, what effect does the addition of calcium ions have on the protons that corresponded to the 1 ppm peak? A.Calcium ions cause additional splitting of the 1 ppm peak. B.Calcium ions reduce the splitting of the 1 ppm peak. C.Calcium ions result in shielding of the protons that corresponded to the 1 ppm peak. D.Calcium ions result in deshielding of the protons that corresponded to the 1 ppm peak.

Calcium ions result in deshielding of the protons that corresponded to the 1 ppm peak. Proton nuclear magnetic resonance spectroscopy (1H NMR) detects hydrogen atoms (protons) in a molecule by applying an external magnetic field and radio frequencies to a sample. NMR measures the relative resonance frequencies of protons, which are displayed as chemical shifts on an NMR spectrum. The chemical shift of a proton depends on its magnetic environment. Protons surrounded by high electron density are said to be shielded from magnetic effects and have small chemical shifts (ie, they are upfield on the spectrum), whereas protons with low electron density are deshielded and have large chemical shifts (ie, they are downfield). In general, the magnitude of shielding is dependent on the local chemical environment. For instance, protons within or near methyl groups are highly shielded and have a smaller (upfield) chemical shift. Nearby electronegative atoms, such as oxygen and nitrogen, pull electron density away from (deshield) the proton and have a downfield chemical shift. For example, double bonds can cause an anisotropic effect, which deshields protons, resulting in a downfield chemical shift relative to alkyl groups. The passage states that a 1H NMR peak at approximately 1 ppm is diminished when calcium ions are added, with a new peak appearing farther downfield. The 1 ppm peak has a comparatively small chemical shift (ie, it is found upfield), which suggests that protons corresponding to this peak are more shielded by the surrounding electron density. The reduction of this peak and the appearance of a downfield peak upon addition of calcium ions suggests that calcium ions cause deshielding of the corresponding protons. (Choices A and B) Peak splitting is related to the number of hydrogen atoms covalently bonded to carbons that are adjacent to the C-H bonds of interest. Calcium ions are unlikely to change the covalent bonds in FIX, and the passage does not mention any changes in splitting. (Choice C) The passage states that the new peak is downfield of the original peak. If calcium ions caused shielding, the new peak would have appeared upfield of 1 ppm. Educational objective:Proton nuclear magnetic resonance

If Compound A is NOT subjected to Step 1 before proceeding to Step 2, what type of chemical transformation will occur in Step 2? reaction between carboxylic acid and amine

Carboxylic acid derivatives differ in the group bonded to the carboxylic acid carbonyl and are interconverted through nucleophilic acyl substitution reactions. The relative reactivity of carboxylic acid derivatives results from a combination of steric effects and electronic effects. The general order of carboxylic acid derivative reactivity is, from most reactive to least: acid halide, acid anhydride, ester, amide, and carboxylic acid. Viable carboxylic acid derivative interconversions are thermodynamically favorable when they proceed from higher-reactivity derivatives to lower-reactivity derivatives. For this reason, the conversion of a carboxylic acid to an acid halide (through reaction with SOCl2) has a special role in synthesis design, as it transforms the lowest-reactivity derivative into the highest-reactivity derivative and facilitates most other acid interconversions. This question asks what type of chemical reaction occurs if Compound A skips Step 1 and proceeds directly to Step 2. The condensation of a carboxylic acid and an amine into an amide (Compound C) is not spontaneous, as the carboxylic acid lacks a good leaving group and is less electrophilic than the acid halide (Compound B) that is formed during Step 1. Instead, Step 2 would promote a neutralization reaction between the carboxylic acid in Compound A and the basic reagents from Step 2: CH3NHOCH3 (an amine) or pyridine. Interconversions of carboxylic acid derivatives generally proceed from derivatives of higher reactivity to lower reactivity. The direct reaction between a carboxylic acid and a basic amine will result in a neutralization reaction, rather than a condensation reaction.

Which of the following statements provides the most plausible explanation for why UV light was used to visualize the thin-layer chromatography plate?

Conjugated double bonds of the aromatic rings on the indole and Fmoc groups absorb UV light. Thin-layer chromatography (TLC), a technique that separates components of a mixture based on polarity, can be used to monitor reactions. The stationary phase (TLC plate) is made of a polar material, usually silica (SiO2). Therefore, polar molecules will have a greater affinity for the TLC plate and travel a smaller distance up the plate than nonpolar molecules. After separation of the reaction mixture, the components are visualized. Molecules that have UV chromophores, which include double and triple bonds, carbonyls, and conjugated systems, can be visualized with UV light. Thin-layer chromatography, a technique that separates mixture components based on polarity, 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.

Which of the following statements do NOT correctly describe a contributing factor for the increased reactivity of cyclopropane versus cyclohexane? A.The carbon-carbon sigma bonds have poor orbital overlap. B.The bond angle in cyclopropane is compressed to 60°. C.Cyclopropane has a lower boiling point. D.The C-H bonds in cyclopropane are permanently eclipsed.

Cyclopropane has a lower boiling point. Alkanes are organic molecules that are composed solely of carbon and hydrogen and lack the presence of pi bonds. Alkanes are among the least reactive functional groups due to the nonpolar nature of their C-H and C-C bonds. Cycloalkanes are alkanes that contain a cyclic region of at least three carbon atoms. The reactivity of cycloalkanes varies with ring size due to differences in ring strain. Ring strain encompasses two structural components: angle strain and torsional strain. Angle strain is present when cycloalkanes have bond angles that deviate significantly from the preferred 109.5° bond angle of a tetrahedral sp3 carbon atom. Cyclopropane has 60° bond angles and is the cycloalkane with the largest angle strain (Choice B). In contrast, the 120° bond angles of planar cyclohexane become 109.5° when it adopts a chair conformation, fully eliminating the angle strain in cyclohexane. A secondary effect of cyclopropane's compressed bond angles is poor head-to-head overlap of sp3 orbitals in its carbon-carbon sigma bonds (Choice A). Unlike a normal sigma bond, where the electron density lies on an axis between the nuclei, cyclopropane carbon-carbon sigma bonds have electron density extended outward from this axis—a feature called banana bonds. Banana bonds are weaker than a traditional carbon-carbon sigma bond due to decreased overlap between sp3 orbitals. Cyclohexane does not contain banana bonds. The second component of ring strain is torsional (conformational) strain (which results from eclipsed bonds). The Newman projection of cyclopropane is unable to adopt alternate conformations to relieve torsional strain of the eclipsed C-H bonds (Choice D). In contrast, planar cyclohexane adopts one of two chair conformations, which allow a lower-energy, staggered conformation in the Newman projection. The question asks which statement does NOT describe a contributing factor for the increased reactivity of cyclopropane versus cyclohexane. Although cyclopropane has a lower boiling point than cyclohexane due to its smaller molecular weight and smaller amount of London dispersion forces, a compound's boiling point does not directly impact its inherent reactivity. Educational objective: Cyclopropane has a higher relative reactivity than most other alkanes due to its increased ring strain. Ring strain is due to a combination of angle strain and torsional (conformational) strain.

In Experiment 1, which aqueous solutions could be included in the extraction protocol to cause phosphatidylethanolamine, then 2,6-dimethoxyphenol to sequentially enter the aqueous layer? A.0.05 M NaOH(aq), then 0.01 M H2SO4(aq). B.0.05 M HCl(aq), then 0.01 M NaHCO3(aq). C.0.01 M H2CO3(aq), then 0.05 M NaHCO3(aq). D.0.05 M H2SO4(aq), then 0.01 M NaOH(aq).

D.0.05 M H2SO4(aq), then 0.01 M NaOH(aq). A mixture of organic compounds with acidic and basic functional groups that are initially in the organic layer can be induced to enter the aqueous layer by converting them into ionic salts. Ionic salts are formed by either deprotonation or protonation of the compounds through treatment with an aqueous base or acid, respectively. Strong bases form ionic salts easily when weak acids are added to a mixture. Conversely, weaker bases such as amines are only protonated by strong acids. Carboxylic acids can be deprotonated by both strong and weak bases; however, phenols are much weaker acids and require strong bases to be deprotonated. Because the question states that phosphatidylethanolamine should be extracted to the aqueous layer first, followed by 2,6-dimethoxyphenol, 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 should be first used to protonate the amino group. In contrast, 2,6-dimethoxyphenol has a weakly acidic phenol group (pKa ≈ 10) that must be deprotonated to form an ionic salt. Only strong bases such as NaOH can deprotonate phenols. A mixture of organic compounds with acidic or basic functional groups can be separated by acid-base extraction. The basic or acidic molecules enter the aqueous layer as ionic salts by extraction with an acid or base, respectively. Amines are weak bases that require strong acids to be protonated. Phenols are weak acids that are only deprotonated by strong bases.

Water is added to the concentrated residue from Reaction B. Which liquid is the most likely to be effective in extracting the desired amine product? A.Methanol B.Ethanol C.n-Hexane D.Ethyl acetate

D.Ethyl acetate A liquid-liquid extraction is a laboratory technique that facilitates the separation of materials based on differences in solubility. Liquid-liquid extraction requires liquids that are not soluble in one another (immiscible) and form two distinct layers when placed in a container. Extraction liquids typically have different polarities, and dissolved solute(s) partition between the two layers based on the principle of "like dissolves like." A separatory funnel is used to perform liquid-liquid extractions. The liquid with the higher density will be the bottom layer in the funnel and can be drained into a different container. The liquid with the lower density remaining in the separatory funnel is then poured out the top of the separatory funnel. Liquid-liquid extractions are most effective when performed iteratively, where the layer requiring extraction is returned to the funnel and the opposing immiscible liquid is added to repeat the extraction process. In this question, water is added to the residue of Reaction B. Water is polar and tends to dissolve polar solutes and ions. Consequently, the second liquid needs to be an organic liquid that does not dissolve in water. To effectively extract the polar organic amine (Compound 6N), the organic layer must also be moderately polar. Ethyl acetate is the only liquid listed that meets the criteria of being both moderately polar and immiscible in water.

Which of the following characteristics is(are) shared by constitutional isomers? Same molecular formula Nonsuperimposable mirror images Different connectivity of atoms Usually have different physical properties A.I only B.I and III only C.II and IV only D.I, III, and IV only

D.I, III, and IV only Same molecular formula, different connectivity of atoms, different physical properties Isomers are molecules with the same molecular formula, and they are classified based on the connectivity (ie, constitutional isomers) and spatial orientation (eg, stereoisomers) of the atoms in their molecular structures. Constitutional isomers, also known as structural isomers, are molecules that contain the same molecular formula (Number I) and differ in the order of atom connectivity at one or more points within the structure (Number III). Because the atoms in constitutional isomers are connected differently, constitutional isomers are different molecules. Therefore, constitutional isomers usually have different physical properties (eg, boiling point, melting point) (Number IV). Stereoisomers are molecules that have the same molecular formula and atom connectivity but differ in spatial orientation. Types of stereoisomers include diastereomers, geometric isomers, and enantiomers, which are nonsuperimposable mirror images (Number II). Educational objective:Isomers are molecules with the same molecular formula, and they are classified based on atom connectivity and spatial orientation. Constitutional isomers (structural isomers) are molecules with the same molecular formula that differ in the order of atom connectivity and have different physical properties.

Which of the following stretching vibrations is expected to occur at the highest wavenumber in the infrared spectrum of mechlorethamine? A.The N-C stretch B.The C-C stretch C.The C-Cl stretch D.The C-H stretch

D.The C-H stretch Infrared (IR) spectroscopy is an analytical technique in which a sample is irradiated with IR light and a spectrometer detects and records the percentage of radiation that passes through the sample over a range of frequencies. Absorption of IR radiation causes vibrational and rotational motion in a molecule. Different bonds and functional groups absorb IR radiation at different intensities and frequencies, and therefore the signal from each bonding environment appears in a characteristic region of the IR spectrum. The peaks in an IR spectrum are measured in wavenumbers, with units of cm−1. The wavenumber of an IR absorption is proportional to the frequency of the motion and can be approximated by Hooke's law. In this model, atoms act as weights on either side of a spring (bond), and heavier atoms lead to slower frequencies of oscillation. This question asks which of the bond choices will have the highest wavenumber stretching vibration in the IR spectrum of mechlorethamine. As wavenumbers are proportional to frequency, and frequency is inversely proportional to the mass of atoms connected by the bond, the highest wavenumber stretching vibration occurs for the bond with the lightest combined mass of atoms. Of the given choices, C-H is the only bond between a heavy atom and a light hydrogen. Therefore, the C-H stretching absorption will have the highest wavenumber.

Which of the following statements correctly describes the peaks displayed in a mass spectrum? A.The tallest peak is the molecular ion. B.The peaks are uncharged molecules. C.The height of each peak is dependent on molecular weight. D.The peaks represent ionized fragments of the sample.

D.The peaks represent ionized fragments of the sample. 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).

decarboxylation

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.

Which neutrally charged functional group forms in an acid-catalyzed reaction between a secondary amine and a ketone? A.Enamine B.Nitrile C.Imine D.Alkene

Enamine Ketones and aldehydes react with primary (1°) and secondary (2°) amines to form a carbon-nitrogen bond. The reaction begins with an acid-catalyzed addition of the amine to the carbonyl, which involves protonation of the carbonyl, nucleophilic attack of the carbonyl by the amine, and deprotonation of the amine. This forms an intermediate known as an α-aminoalcohol. The second part of the reaction is an acid-catalyzed dehydration involving protonation of the -OH group, loss of H2O, and deprotonation. The product formed is dependent on the type of amine used (1° or 2°). If a 1° amine is used, an imine (nitrogen analogue to ketones and aldehydes) is formed because the nitrogen has a proton available for removal in the last step, permitting C=N bond formation. If a 2° amine is used, an enamine (nitrogen analogue to an enol) is formed because the nitrogen does not have a proton available to be removed in the last step. Therefore, a proton is removed from the α-carbon, forming a C=C bond. (Choice B) A carbon-nitrogen bond is formed in the reaction, but the functional group is not a nitrile because the bond between carbon and nitrogen is a single bond rather than a triple bond. (Choice C) Although a positively charged iminium forms as a reaction intermediate, this cation is not neutrally charged. Neutrally charged imines will form only if a ketone reacts with a 1° amine rather than a 2° amine. (Choice D) Although a carbon-carbon double bond is formed in the reaction, an alkene indicates that the functional group contains only a double bond. Because a nitrogen atom is bonded to the sp2 carbon, the functional group would be classified as an enamine rather than an alkene. Educational objective:Ketones and aldehydes undergo acid-catalyzed addition of 1° and 2° amines to form imines or enamines, respectively. Like ketones and aldehydes, imines contain a carbon-heteroatom double bond (nitrogen in imines, oxygen in ketones and aldehydes). Enamines are the nitrogen analogue to enols, containing a carbon-carbon double bond.

Which extraction procedure would be most useful to isolate mechlorethamine from a mixture? A.Extract mechlorethamine into an organic layer from an acidic aqueous layer. B.Extract mechlorethamine into an acidic aqueous layer from an organic layer. C.Extract mechlorethamine into a basic aqueous layer from an organic layer. D.Extract mechlorethamine into an organic layer from a basic aqueous layer.

Extract mechlorethamine into an acidic aqueous layer from an organic layer. Liquid-liquid extraction uses immiscible solvents (ie, an organic solvent and an aqueous solution) to separate molecules in a mixture based on differences in solubility. The principle "like dissolves like" indicates that molecules dissolve in a solvent with a similar polarity. Hydrophilic molecules containing polar or ionic bonds have a greater affinity for the polar aqueous layer. Hydrophobic or neutrally charged molecules have fewer polar functional groups, and therefore these molecules have a greater affinity for the nonpolar organic layer. A molecule's charge can be modified to change its affinity for the organic or aqueous layer. This question asks which extraction procedure would be most useful to isolate mechlorethamine from a mixture. Mechlorethamine contains a basic tertiary amine functional group. Under acidic conditions, the amine lone pair is protonated to generate a hydrophilic ammonium ion. Under basic conditions, the amine exists in its hydrophobic neutral form. By these criteria, mechlorethamine will have the greatest solubility in either: An acidic aqueous solution as an ammonium ion (acidic pH) An organic solvent as a neutral molecule (basic pH) Liquid-liquid extraction can be used to separate molecules based on differences in solubility between solutions. Basic functional groups (eg, amines) can be extracted into an acidic aqueous solution through reversible conversion to their ionic, hydrophilic form.

branching and bp

For compounds with the same functional group and same number of carbons, branching (smaller surface area) decreases the boiling point. As branching increases, imf forces decrease, bp decreases

Which separation technique is optimal for purification of small sample sizes and employs a stationary phase, a liquid mobile phase (under pressure), and a detector to separate compound mixtures based on polarity? A.Gas chromatography B.Extraction C.High-performance liquid chromatography D.Thin-layer chromatography

High-performance liquid chromatography Organic reactions generally do not produce a single product; therefore, separation and purification methods are imperative to obtain a compound of interest. Several techniques exist to separate compounds based on their physical properties, including boiling point, solubility, and polarity. High-performance liquid chromatography (HPLC) is a purification technique comprised of a mobile phase, stationary phase, detector, and computer for data acquisition. The mobile phase is a liquid solvent or mixture of solvents pumped through the system under pressure. The stationary phase is a column made of either a polar (normal-phase) or nonpolar (reversed-phase) material, depending on the nature of the compounds to be separated. HPLC is optimal for small sample sizes as it separates compounds based on their polarity, giving each compound a different affinity for the mobile and stationary phases and causing the compounds to pass through the column at different rates. Compounds are detected as they come off the column and go into the waste or are collected separately. Data is transmitted to the computer, creating a chromatograph. (Choice A) Gas chromatography requires small sample sizes but separates compounds based on boiling point. This method contains a stationary and mobile phase; however, the mobile phase during separation is a gas rather than a liquid solvent. (Choice B) Extraction separates compounds based on their relative solubility in different solvents. This method uses two immiscible solvents that make up the aqueous and organic layers; no stationary phase, mobile phase, or detector is required for extraction. (Choice D) Thin-layer chromatography (TLC) separates small sample sizes and includes a stationary phase, solvent mobile phase, and detector (UV). However, TLC is not performed under pressure. 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

Which of the following characterize or are included in gas-liquid chromatography? Gas mobile phase Liquid stationary phase Separation based only on polarity Room-temperature column conditions

I and II Gas-liquid chromatography is a technique used to separate molecules in a mixture based on their boiling points. A gas chromatograph consists of an injection port, mobile and stationary phases, a column in a heated oven, a detector, and a computer for data analysis. The mobile phase is an inert gas such as helium or nitrogen (Number I), and the stationary phase is a liquid that coats a solid support on the inside of the column (Number II). 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 (ie, low boiling points) spend more time in the gas phase than they spend interacting with the stationary phase of the column, so they rapidly migrate to the detector. 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) In gas-liquid chromatography, mixture components are separated primarily based on boiling point. Although the boiling point is determined by several factors (eg, polarity, intermolecular forces, molecular weight), each of these factors is already accounted for by the boiling point itself. For molecules with the same number of carbons, the boiling point increases with the increasing strength of intermolecular forces and relative polarity. However, a sufficiently large nonpolar molecule can have a higher boiling point than a small 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 relatively nonpolar. (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 l

key points of techniques

In mass spectrometry, a sample is ionized by high-energy electrons, but the electrons do not change energy levels. An external magnetic field is applied to a sample in NMR spectroscopy. Radio waves, which are low in energy, are used to detect hydrogen atoms and excite them from the α spin to the β spin state. 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.

carboxylic acid IR spectrum notables

Infrared (IR) spectroscopy measures the amount of light transmitted through a sample over a range of frequencies. Absorption frequencies and peak intensities depend on the type of bond present in a particular functional group. Carboxylic acids have two characteristic absorptions: 3300-2400 cm−1 (O-H stretch, broad and irregular band) and 1710 cm−1 (C=O stretch, strong peak).

Thermodynamic vs Kinetic

Isomers are compounds with the same molecular formula that have a distinguishing structural difference that makes them different compounds with different properties. There are several classifications of isomers, including constitutional isomers, diastereomers, enantiomers, and cis/trans isomers. When a reaction is performed, a mixture of isomers can occur with one isomer predominating over the other. In such a case, the reaction conditions dictate which isomer is the major product (such as in the formation of kinetic and thermodynamic enolates). The kinetic product in a mixture requires a lower activation energy to form; therefore, it forms faster than the thermodynamic product but is also higher in energy (less stable). When a reaction is performed at low temperatures (kinetic conditions), insufficient energy is present to supply the larger activation energy required to form the thermodynamic product in high yield. Although the kinetic product still forms faster at higher temperatures (thermodynamic conditions), the increased thermal energy enables the kinetic product reaction to occur in reverse (back to the intermediate) and supplies the additional activation energy needed to form the lower energy (more stable) thermodynamic product, which then predominates. Educational objective: Isomers are different compounds with the same molecular formula, and some reactions produce a mixture of isomers. The major product of a reaction is dependent on the reaction conditions: The more stable product will predominate under thermodynamic conditions, and the less stable product will predominate under kinetic conditions.

ketal vs acetal

Ketals and acetals are formed by acid-catalyzed nucleophilic addition of two alcohols to a ketone or aldehyde carbonyl, respectively. An acid catalyst must be used to protonate the hemiketal (or hemiacetal) -OH group to create a good leaving group so the second alcohol molecule can be added. cannot be base catalyzed

grignard reagent and alcohols

Ketones and aldehydes can be converted to alcohols using different reagents, including reducing agents and organometallic reagents. Grignard reagents are used in carbon-carbon bond-forming reactions between its alkyl group and an electrophilic molecule. When the electrophile is the carbonyl carbon of a ketone or aldehyde, the Grignard reagent causes a nucleophilic addition to the carbonyl, generating an alkoxide. Protonation of the alkoxide during an acidic workup results in an alcohol, and the Grignard reagent adds one alkyl group to the electrophilic carbon. Therefore, the type of alcohol formed (ie, primary, secondary, or tertiary) contains one more alkyl group than the number of alkyl groups bonded to the electrophilic carbonyl used in the reaction: Reaction with formaldehyde (CH2O) (zero alkyl groups) leads to formation of a primary alcohol Reaction with aldehydes (one alkyl group) leads to formation of a secondary alcohol Reaction with ketones (two alkyl groups) leads to formation of a tertiary alcohol The question asks which statement describes an approach to form a secondary alcohol. All the choices describe a reaction with methylmagnesium bromide, a Grignard reagent. Because secondary alcohols have two alkyl groups bonded to the hydroxyl group, an aldehyde must be the electrophile in the Grignard reaction. Aldehydes are named using the suffix -al. Therefore, the reaction between pentanal and methylmagnesium bromide, followed by an acidic workup, results in the formation of a secondary alcohol. Ketones and aldehydes can be transformed into alcohols by nucleophilic addition of an alkyl group to the carbonyl by an organometallic reagent (eg, Grignard reagent), followed by acidic workup. The type of alcohol formed (ie, primary, secondary, or tertiary) contains one more alkyl group than the number of alkyl groups bonded to the carbonyl reactant.

Which of the following compounds can be used to synthesize a compound containing a five-carbon alkyl chain with a cyanohydrin at carbon 2?

Ketones and aldehydes can be transformed into other functional groups via nucleophilic addition to their carbonyl. In a nucleophilic addition, a nucleophile attacks the electrophilic carbonyl carbon and forms a new sigma bond, and the pi bond of the carbonyl breaks to form an alkoxide. A cyanohydrin is a functional group combination where a hydroxyl group (-OH) and a cyano group (-C≡N) are attached to the same carbon. Cyanohydrins are formed by nucleophilic addition of the cyanide anion (CN−) to the carbonyl of a ketone or aldehyde. The resulting alkoxide is protonated by an acidic workup. Because cyanohydrins are formed through the nucleophilic addition of cyanide to a carbonyl, the compound used to form the cyanohydrin described in this question must contain a carbonyl. The question states a five-carbon alkyl chain contains a cyanohydrin group at carbon 2. The starting material must be a ketone because the cyanohydrin is not located at the end of the chain. Therefore, the carbonyl of the ketone should also be at carbon 2. The structure shown in Choice D is a five-carbon ketone with a carbonyl at carbon 2. A cyanohydrin is a functional group combination where a hydroxyl group and a cyano group are attached to the same carbon. Cyanohydrins are generated by the nucleophilic addition of a cyanide ion to the carbonyl of a ketone or aldehyde, followed by an acidic workup.

lactones

Lactones (cyclic esters) are carboxylic acid derivatives named for the number of carbons in the ring system (excluding the carbonyl). Lactones are formed by an acid catalyzed reaction between a carboxylic acid and an alcohol in the same molecule (ie, intramolecular Fischer esterification).

mass spec identification

Mass spectrometry provides the ability to measure the mass of molecules and molecular fragments. An electron-impact mass spectrometer bombards a sample with electrons to remove an electrron from a molecule (ionization). This creates the molecular ion (M+), a radical cation with a mass-to-charge (m/z) ratio equal to the molecular weight of the molecule. The molecular ion can also break into a radical and a cation to produce smaller molecular fragments. Finally, a mass spectrum plot displaying the abundance of detected cations (y-axis) versus m/z (x-axis) is generated. The presence of functional groups leads to characteristic signatures in a mass spectrum: Amides, amines, and compounds that contain an odd number of nitrogen atoms have a molecular ion with an odd m/z and fragments with an even m/z. Some alcohols are prone to dehydration, where the molecular ion is apparent 18 m/z below the expected value. Chlorinated and brominated compounds have peaks determined by their natural isotopic abundance.

Specific rotation of diastereomers

May/may not be opposite in sign, but they definitely won't be equal in magnitude The same is true for constitutional isomers.

What base and reaction temperature are needed to add a methyl group to Compound 1 as shown in the reaction below? A.LDA; −78 °C B.LDA; 25 °C C.NaH; −78 °C D.NaH; 25 °C

NaH; 25 °C An enolate is formed when a base deprotonates an α-carbon (ie, removes an acidic proton from the carbon adjacent to a carbonyl C=O). For unsymmetrical carbonyl compounds, the type of base used and the reaction temperature affect which α-carbon is deprotonated. The deprotonated α-carbon can then be methylated by reacting with a methyl halide, such as CH3I. The kinetic enolate results from deprotonation of the least substituted α-carbon. This enolate forms relatively quickly because the base can access the less substituted carbon more easily. Consequently, a large, bulky base, such as lithium diisopropylamide (LDA), and a lower reaction temperature (eg, −78 °C) may be used to yield this product. Alternatively, the thermodynamic enolate forms by deprotonation of the more substituted α-carbon. A small base such as sodium hydride (NaH) and a higher reaction temperature (eg, 25 °C) are needed to facilitate this reaction. The thermodynamic enolate is more stable but forms more slowly than the kinetic enolate. Compound 1 in this question is an unsymmetrical ketone in which one α-carbon has one substituent (the 6-membered ring) and the other has two substituents (the 6-membered ring and an ethyl group). The reaction shows Compound 1 becomes methylated at the more substituted α-carbon. For this to occur, the thermodynamic enolate must be created using a small base and a higher reaction temperature. Of the choices given, NaH and 25 °C are ideal for thermodynamic enolate formation. (Choices A and B) Because of steric hinderance, LDA is too bulky to deprotonate the more substituted α-carbon and is used to form the kinetic rather than the thermodynamic enolate. (Choice C) Thermodynamic enolates are formed at higher reaction temperatures. −78 °C is too low for the formation of the thermodynamic product. Educational objective:Kinetic enolates are formed by deprotonation of the least substituted α-carbon of a carbonyl compound using a bulky base, such as LDA at low temperatures. Thermodynamic enolates are formed by deprotonation of the more substituted α-carbon using a small base, such as NaH, at higher temperatures.

If (2R,3R)-2-bromo-3-methylpentane undergoes an SN2 reaction with OH−, which of the following compounds is the reaction product?

Only the electrophilic carbon undergoes change of configuration so the other carbons need to stay the same The stereochemistry of a chiral center is designated as R or S based on the priority and circular arrangement of its substituents: The highest priority substituent (ie, priority 1) is the substituent with the greatest atomic number. If two substituent atoms are the same, the atomic numbers of the next attached atoms are considered. If the lowest priority group is positioned to point into the plane (dashed), a clockwise arrangement of groups 1, 2, and 3 gives an R configuration whereas a counterclockwise arrangement gives an S configuration. If the lowest priority group is pointing out of the plane (wedge), a clockwise arrangement of groups 1, 2, and 3 gives an S configuration whereas a counterclockwise arrangement gives an R configuration. An SN2 reaction is a substitution reaction in which a nucleophile attacks an electrophile and displaces a leaving group in one step. If the electrophile is a chiral center, its stereochemistry will be inverted by the reaction. Therefore, an electrophile that has an R configuration before substitution adopts an S configuration after the reaction and vice versa, if the nucleophile has the same priority ranking as the leaving group. In this SN2 reaction, the 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. The structure in Choice B correctly shows carbon 2 in the S configuration and carbon 3 in the R configuration. (Choices A, C, and D) The configuration of carbon 2 should be S (wedge in this case) because it is the electrophile, so its configuration inverts in the SN2 reaction. The configuration of carbon 3 should remain R (dashed in this case) because it does not participate in the SN2 reaction. Educational objective: 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.

In the 1H NMR spectrum of mechlorethamine, which of the following statements most accurately describes the relationship between the chemical shift of proton environment A and proton environment B? A.Proton environment A has a higher chemical shift than proton environment B. B.Proton environment B has a higher chemical shift than proton environment A. C.Proton environments A and B have identical chemical shifts. D.Nothing can be inferred about the relative chemical shift of proton environments A and B.

Proton environment A has a higher chemical shift than proton environment B. When electrons surrounding a nucleus enter the spectrometer magnetic field (B0), they generate a magnetic field that opposes B0 (Binduced). The effective magnetic field Beff experienced by a nucleus is impacted by the presence of Binduced through a process known as electron shielding. Chemical shifts (δ) are reported as the ratio of resonance energy frequency (Hz) to the spectrometer frequency (MHz), in units of parts per million (ppm). Protons in alkyl groups are among the most shielded and normally appear between δ 0.9-1.4, with higher-order (eg, tertiary) alkyl groups having higher chemical shifts than lower-order (eg, primary) alkyl groups. Proton chemical shifts are also influenced by the inductive effect, in which nearby electron-rich (ie, electron donating) or electron-deficient (ie, electron withdrawing) groups respectively increase or decrease the effect of shielding. In this question, the chemical shifts of proton environment A and proton environment B in mechlorethamine are being compared. Since environment B (a primary carbon) is less substituted than environment A (a secondary carbon), the chemical shift of environment A is expected to be higher. In addition, the electronegative chlorine atom is closer to environment A, leading to greater inductive deshielding and a higher chemical shift for environment A. Together, these data point toward the conclusion that proton environment A has a higher chemical shift than proton environment B.

Compound 1 and PBr3 react to form Compound 2 as shown below. Which position in Compound 2 corresponds to the hydrogen(s) that appear as a doublet at approximately 4.0 ppm in the 1H NMR spectrum?

Proton nuclear magnetic resonance spectroscopy (1H NMR) detects hydrogen atoms (1H isotope) in a molecule as an external magnetic field and radio frequency pulse are applied to the sample. The hydrogen atoms in different magnetic environments (resulting from molecular structure) are nonequivalent, and neighboring nonequivalent hydrogens within three bonds of each other cause spin-spin splitting of the peaks in the NMR spectrum. The splitting pattern can be determined by n + 1 rule, where n is the number of nonequivalent hydrogen atoms on adjacent carbons. Position IV in Compound 2 contains two hydrogen atoms that are adjacent to a single H atom bonded to an sp2 carbon. According to the n + 1 rule, n = 1 because there is only one coupled hydrogen to the hydrogen atoms at position IV. Therefore, because n + 1 = 2, the peak will be split in two, making the splitting pattern a doublet. (Choice A) Position I contains two hydrogen atoms that are split by three adjacent hydrogen atoms. Although a quartet might be expected, complex, secondary splitting causes overlapping peaks to appear as a multiplet at 2.1 ppm. (Choice B) Position II consists of a methyl group (three hydrogen atoms) that is adjacent to a 4° carbon (no bonds to hydrogen). Because there are no neighboring hydrogen atoms coupled to the methyl hydrogens, the peak is not split and appears as a singlet at 1.8 ppm. (Choice C) Position III contains an alkene hydrogen atom that is adjacent to an alkyl (-CH2-) group. The two alkyl hydrogen atoms are coupled to the alkene hydrogen, causing the peak to be split into a triplet. 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).

LiAlH4 (strong reducing agent)

Reducing agents decrease an atom's oxidation state. The reduction of a carbon atom decreases the number of bonds it has to electronegative atoms (including oxygen, nitrogen, and halogens) and increases the number of bonds to less electronegative atoms (typically hydrogen). Hydride reagents, such as LiAlH4 and NaBH4, are common reducing agents and consist 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. NaBH4 is a reducing agent rather than an oxidizing agent. Furthermore, NaBH4 is a mild reducing agent and cannot reduce carboxylic acids or esters

During the synthesis of a tosylate, which of the following pairs of atoms form a covalent bond? A.C and Cl B.S and O C.S and Cl D.C and O

S and O Substitution reactions require strong nucleophiles and electrophiles, as well as good leaving groups. The nucleophile attacks the electrophile and displaces the leaving group. A bond is formed between the nucleophile and electrophile, whereas the bond between the electrophile and the leaving group is broken. Alcohols are weak electrophiles and poor leaving groups; therefore, a nucleophile is unlikely to displace the hydroxyl group (-OH) of an alcohol. 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 p-toluenesulfonyl chloride (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. 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.

Why does the bond between glycerol and the 2-ethylhexyl group tend to form at glycerol's C1 hydroxyl group instead of the C2 hydroxyl group? A.Secondary alcohols are more likely to undergo SN1 reactions than primary alcohols. B.Secondary alcohols are more sterically hindered than primary alcohols. C.Primary alcohols are less acidic than secondary alcohols. D.Primary alcohols are more electrophilic than secondary alcohols.

Secondary alcohols are more sterically hindered than primary alcohols. Steric hindrance can prevent reactions that would otherwise occur by physically blocking the access of one atom to another. Alcohols may be attached to primary, secondary, or tertiary carbon atoms, which are bonded to one, two, or three other carbons, respectively. These alcohols are referred to as primary, secondary, and tertiary alcohols. 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 passage states that ethylhexyl glycerin forms in an SN2 reaction. The alcohol on carbon 2 of glycerol is a secondary alcohol, whereas carbon 1 is a primary alcohol, and 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. (Choice A) Secondary alcohols are more likely than primary alcohols to undergo SN1 reactions. However, the reaction of interest is an SN2 reaction, for which primary alcohols are more efficient. (Choice C) Primary alcohols are more acidic than secondary alcohols. This attribute contributes to their enhanced ability to react because deprotonation by a base makes the alcohol a better nucleophile. (Choice D) The primary alcohol is acting as a nucleophile, so enhanced electrophilicity would make it a worse reactant. Educational objective: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.

carboxylic acids and extraction

Separation of amides and carboxylic acids can be accomplished by extraction. Carboxylic acids with long hydrocarbon chains can be induced to enter the aqueous layer if they are converted to an ionic salt via deprotonation by either a strong base or a weak base. The amide remains in the organic layer because it cannot be deprotonated by a weak base. Amide hydrolysis requires either a strong acid or strong base and heat.

Prior to adduct formation, DNA was isolated from a cellular mixture. Membrane lipids were first extracted using an organic solvent, which was then evaporated, and DNA was precipitated from the aqueous layer using cold ethanol and sodium acetate. Which of the following mechanisms allowed for DNA precipitation? A.Sodium acetate lowers the pH of the solution, changing the conformation of the DNA structure. B.Ethanol has a greater polarity than water, pulling DNA out of the aqueous solution. C.Acetate ions interact favorably with the phosphate groups in DNA. D.Sodium cations form ionic bonds with DNA, neutralizing its charge.

Sodium cations form ionic bonds with DNA, neutralizing its charge. Extraction is a technique that uses immiscible solvents (ie, an organic solvent and an aqueous solution) to separate molecules in a mixture based on differences in solubility. The principle "like dissolves like" indicates that molecules dissolve in a solvent with a similar polarity. Hydrophilic molecules containing several polar or ionic bonds have a greater affinity for the highly polar aqueous layer. Hydrophobic molecules such as lipids have fewer polar bonds, and therefore have a greater affinity for the less polar organic layer. A molecule's charge or polarity may be modified to change its affinity for the organic or aqueous layer. The question states that DNA remains in aqueous solution after extraction of membrane lipids; this is due to negatively charged phosphate groups (PO3−) on the sugar phosphate backbone. The partially positive hydrogen atoms in water interact with the negatively charged DNA backbone through ion-dipole interactions, creating a hydration shell around the DNA. To precipitate the DNA from aqueous solution, its charge must be neutralized through extraction with ethanol and a salt such as sodium acetate. Mixing the aqueous layer with ethanol disrupts the hydration shell around DNA molecules, allowing for sodium cations to better interact with the negatively charged DNA backbone. 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.

SN1 and SN2 rnx

Substitution reactions occur in one of two ways: an SN1 or an SN2 mechanism. These substitution reactions differ in number of steps and stereospecificity: An SN1 reaction occurs in two steps and is not stereospecific whereas an SN2 reaction occurs in a single step and is stereospecific. Stereospecific reactions occur when stereoisomers that undergo the same reaction each give only one specific stereoisomer of the product. The stereochemistry of the starting material determines the stereochemistry of the product. The first step of an SN1 reaction involves breaking the bond between the leaving group and the adjoining carbon, forming a planar carbocation intermediate. In the second step, the nucleophile can add to either the top or bottom of the planar carbocation, leading to the formation of two products. In contrast, SN2 reactions occur in a single step and the nucleophile always attacks on the side opposite the leaving group (backside attack), causing an inversion at the electrophilic carbon. An SN2 reaction forms only one product. Tertiary alkyl halides form more stable carbocations than primary halides because the additional alkyl groups stabilize the carbocation by the inductive effect. Therefore, tertiary alkyl halides readily undergo SN1 reactions. In contrast, carbocations are less stable on primary alkyl halides; however, the relative lack of steric hindrance allows primary alkyl halides to more readily undergo SN2 reactions. Chiral tertiary alkyl halides form two products (stereoisomers) in SN1 reactions because SN1 reactions are not stereospecific. (Choices A and C) SN2 reactions occur in one step, and primary alkyl halides undergo this reaction more readily than the more sterically hindered tertiary halides. However, SN2 reactions are stereospecific, and only one stereoisomer is formed. (Choice D) The backside of the electrophilic carbon on a tertiary halide is sterically hindered by the bulk of the other substituents. Educational objective: Stereospecific reactions occur when stereoisomers undergo the same reaction, and each gives only one stereoisomer 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 one of two stereoisomers. Primary alkyl halides readily undergo SN2 reactions, which are stereospecific because the nucleophile attacks only from the opposite side, where the leaving group is eliminated (backside attack).

boiling chip use

Superheating occurs when a liquid is heated past its boiling point but does not boil. Surface tension can prevent the formation of bubbles and cause superheating to occur. When bubbles try to form, the surface tension can cause an increase in the local vapor pressure that goes beyond the ambient pressure, allowing the liquid to heat past its boiling point. This can cause large bubbles to form suddenly and erupt from the distillation flask (ie, an effect known as bumping). Boiling chips, which are made of a porous material, provide surfaces with nucleation sites where small bubbles can form. This overcomes the surface tension and allows the liquid to boil evenly, preventing superheating. Boiling chips are used in simple and fractional distillations but cannot be used in vacuum distillations because the vacuum removes the air from the boiling chips. Therefore, magnetic stir bars are used in place of boiling chips in vacuum distillations to ensure even heating and prevent superheating. Therefore, the student used boiling chips in the simple distillation and a stir bar in the vacuum distillation to provide sites for bubble formation, which overcomes the surface tension and allows the liquid to boil evenly.

terpene

Terpenes are natural lipid products derived from the five-carbon isoprene unit building block. Analysis of lipid structure provides clues about whether the lipid can be classified as a terpene.

steroid structure and make

Terpenes are precursors to steroids, which are lipids involved in signaling pathways. Individual terpenes are made up of two or more 5-carbon groups known as isoprenes and are classified based on the number of isoprene units present. Terpenes can be joined together in one of three ways. Those containing two isoprenes are known as monoterpenes, and molecules made up of two monoterpenes (four isoprenes) are called diterpenes. Steroids are made up of triterpenes (three monoterpenes, or six isoprene units), which are synthesized by the cyclization of the triterpene squalene, forming cholesterol. Cholesterol itself contains four fused hydrocarbon rings (three 6-membered and one 5-membered rings). Lipids can be classified based on whether they contain any hydrolyzable linkages (ie, an ester or amide). Lipids containing an ester or amide (eg, fatty acids, phospholipids) are classified as hydrolyzable lipids whereas lipids that do not have an ester or amide (eg, steroids) are known as nonhydrolyzable lipids. Steroids do not contain any hydrolyzable linkages, and as such they are nonhydrolyzable lipids (Number III). nonhydrozyable lipids: steroids, ADEK, prostaglandins (Number I) Steroids contain four rather than only three fused hydrocarbon rings. (Number II) Steroids are made from triterpenes, which in total contain six (not two) isoprene units. Educational objective:Steroids are lipids made from triterpenes, which contain six isoprene units that cyclize to form four fused hydrocarbon rings. Steroids are classified as nonhydrolyzable lipids because they do not contain any hydrolyzable ester or amide linkages.

Thin-layer chromatography (TLC) was used to determine the best solvent system for the mobile phase to purify a mixture by column chromatography. The TLC results using different mobile phase solvent systems are given in the following table. Which statement accurately describes the results? A.100% Hexanes are too nonpolar because the mixture components have a higher affinity for the mobile phase. B.The 1:1 hexanes/ethyl acetate solvent system decreased the compounds' affinity for the mobile phase relative to 100% ethyl acetate. C.100% Ethyl acetate is too polar because the mixture components have a higher affinity for the stationary phase. D.The 1:1 hexanes/ethyl acetate solvent system increased the compounds' affinity for the stationary phase relative to 100% hexanes.

The 1:1 hexanes/ethyl acetate solvent system decreased the compounds' affinity for the mobile phase relative to 100% ethyl acetate. Thin-layer chromatography (TLC) is a technique used to separate compounds based on polarity. In normal-phase TLC, the stationary phase is made up of a polar adsorbent material, typically silica (SiO2), and the mobile phase is an organic solvent that travels up the stationary phase via capillary action. The rate at which a compound travels up the plate is a function of the relative polarities of the compound and the solvent. If a mixture remains near the plate's origin, a more polar solvent is needed to increase the compound's affinity for the mobile phase relative to its affinity for the stationary phase. Conversely, if a mixture travels with the solvent front, a more nonpolar solvent is needed to decrease the compound's affinity for the mobile phase relative to its affinity for the stationary phase. The 1:1 hexanes/ethyl acetate solvent system is more polar than 100% hexanes and less polar than 100% ethyl acetate, and it achieved the compound separation essential for column chromatography. This solvent system decreased the mixture components' affinity for the mobile phase relative to 100% ethyl acetate. (Choice A) When using 100% hexanes, the compounds did not separate and remained near the origin. Therefore, 100% hexanes are too nonpolar and result in the mixture components having a higher affinity for the stationary phase than the mobile phase. (Choice C) When 100% ethyl acetate was used alone, both compounds moved with the solvent front. Therefore, ethyl acetate is too polar of a solvent and results in the mixture components having a higher affinity for the mobile phase than for the stationary phase. (Choice D) The 1:1 hexanes/ethyl acetate solvent system increased the mixture components' affinity for the mobile phase, rather than the stationary phase, relative to 100% 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 po

Strecker Syntheis

The Strecker synthesis is used to make α-amino acids from an aldehyde using ammonium chloride (NH4Cl) and potassium cyanide (KCN). The first step of the reaction proceeds with protonation of the carbonyl oxygen by ammonium (NH4+), followed by nucleophilic attack of the carbonyl carbon by ammonia (NH3), resulting in dehydration and imine formation. In the second step, a cyanide anion is added to the imine to form an aminonitrile. Finally, in the third step, the nitrile (R-CN) nitrogen is protonated, and two water molecules add to the nitrile carbon in succession, eliminating ammonia from the nitrile and forming the carboxylic acid of the α-amino acid. Alanine can be synthesized via the Strecker synthesis from acetaldehyde, NH4Cl, and KCN. Acetaldehyde does not have a stereocenter, and the imine intermediate also does not have a stereocenter and is a planar molecule, allowing nucleophilic attack to occur from either above or below the plane. Therefore, the reaction is not stereospecific and the product will be a mixture of L- and D-alanine. The same outcome is obtained for amino acids produced by the Gabriel synthesis because the final step (decarboxylation) is not stereospecific. (Choices A and B) Because the reaction proceeds through the formation of a planar intermediate, the product will be a mixture of L- and D-amino acids rather than just one or the other. (Choice D) The α-carbon of alanine, as well as all amino acids except glycine, contains four different substituents, making it a stereocenter. Educational objective: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.

IR spectroscopy

The absorption of IR radiation by molecules causes intramolecular stretching and bending vibrations of different bond types at different frequencies, allowing for functional group identification. Infrared (IR) spectroscopy is a technique used to identify functional groups in a compound through the analysis of the absorption of IR radiation by different types of bonds present in the molecule. A sample is irradiated with IR light, and the spectrometer detects and records the amount of light that passes through the sample over a range of frequencies. The data is represented on the IR spectrum as the percent transmittance (y-axis) versus wavenumber (x-axis). The IR radiation interacts with the different types of bonds within the molecule, causing stretching vibrations and rotations of the bonds. Functional groups absorb different amounts of IR light (ie, have different absorption intensities at different frequencies). As a result, IR spectroscopy can be used to confirm the formation of a reaction product by identifying the peaks unique to the newly formed functional group(s). Therefore, the structure of Compound 2 can be confirmed using IR spectroscopy because the absorption of IR radiation by molecules causes intramolecular bending and stretching vibrations of different bond types at different frequencies, allowing for functional group identification. Infrared (IR) spectroscopy is a technique used to identify functional groups in a compound by analyzing the types of bonds present. The sample is irradiated with IR light, which interacts with the bonds in the molecule. The amount of light absorbed is recorded by the detector, and the data collected is displayed on the IR spectrum as percent transmittance (y-axis) versus wavenumber (x-axis). Transmittance is inversely related to absorbance (ie, if all the light is absorbed, there is 0% transmittance). The wavenumber (with units of cm−1) is the IR frequency and is defined as the inverse of the wavelength. Because the wavenumber is related to wavelength, it is also directly related to frequency by the equation: where E is energy and h is Planck's constant. IR radiation ranges from 700 nm to 1 mm and causes bond stretching vibrations and rotations at differ

axial hydroxyl groups in a chair are not in the correct geometry for a cyclic structure so cannot react

The acetonide protecting group can form when adjacent -OH groups have an accommodating geometric relationship. In carbohydrates, acetonides can only form between hydroxyl groups that are in an equatorial position—axial hydroxyl groups do not have the correct geometry to form the cyclic structure. The -OR group on carbon 1 cannot react to form an acetonide because it is not a hydroxyl group. The question asks why a glucose derivative would not be a good substrate for acetonide protection if only one product is desired. The given glucose derivative has carbons 2, 3, and 4 with equatorial -OH groups. As a result, acetonides could form between -OH groups on carbons 2 and 3 OR carbons 3 and 4.

In the ultraviolet (UV) range, pyrimidine has maximum absorptions at 240 nm due to π electron excitation and at 280 nm due to nonbonding electron excitation. The nonbonding electrons on nitrogen in pyrimidine undergo which of the following transitions when absorbing UV light? A.π → π* B.n → π* C.π* → π D.π* → n

The electromagnetic spectrum consists of waves of varying wavelengths, energy, and frequency. Ultraviolet (UV) light, which corresponds to wavelengths between 200 nm and 400 nm, is used in UV spectroscopy. 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 (ie, lone electron pairs) are in the n nonbonding molecular orbital. The nonbonding molecular orbitals are higher in energy than the π bonding molecular orbitals. Upon interactions with UV light of sufficient energy, π and nonbonding electrons are excited to the Lowest Unoccupied Molecular Orbital (LUMO), called the π* antibonding orbital. These excitations are described as a transition from the lower energy bonding molecular orbital to the higher energy antibonding orbital in the form of π → π* or n → π*, depending on whether a π or n electron is excited. The lone electron pairs on nitrogen in pyrimidine are nonbonding (n) electrons. When absorbing UV light, these electrons are excited to the antibonding (π*) orbital and therefore undergo a n → π* transition.

Compounds 1 and 2 are both intermediates that could form by a reaction with AgNO3 in ethanol, but Compound 1 is less stable than Compound 2 because: A.Compound 1 contains an electron donating group that destabilizes the carbocation. B.there is an inductive effect from the fluorine atoms that destabilizes the carbocation. C.the methyl groups on the structure withdraw electrons from the positively charged carbon. D.the electronegative fluorine atoms cause the carbocation to rearrange into a secondary carbocation.

The inductive effect occurs with the donation of electron density through sigma bonds. Electronegative atoms or electron withdrawing groups (EWG) pull electrons away from an adjacent atom, creating a dipole with a partial negative charge on the electronegative atom and a partial positive charge on the adjacent 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. (Choices A and C) The -CH3 groups in Compound 1 are EDG, which donate electrons to the carbocation and help stabilize the carbocation. It is the -CF3 group that withdraws electrons and destabilizes the carbocation. (Choice D) The fluorine atoms are highly electronegative and do not cause the carbocation to rearrange. 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.

difference of polarity between acetaminophen and 4-nitrophenol

The only difference between the two compounds is the change from amine in 4-aminophenol to amide in acetaminophen. An amide is a hydrogen bond donor and acceptor via two electronegative atoms (oxygen and nitrogen) whereas an amine (primary and secondary) is a hydrogen bond donor and acceptor via only one electronegative atom (nitrogen). Because an amide has more hydrogen bond acceptors, it is more polar; therefore, spot B must be acetaminophen.

In mass spectrometry, which of the following statements most accurately describes the definition of the base peak?

The peak with the highest abundance An electron-impact mass spectrometer bombards a sample with electrons to remove an electron from a molecule (ionization). This creates the molecular ion, a radical cation with a mass-to-charge ratio (m/z) equal to the molecular weight of the molecule. Normally, the peak with the highest m/z value corresponds to the molecular ion. The molecular ion can also break into a radical and a cation to produce smaller molecular fragments. Ionized material enters a magnetic field, which curves the path of the particles by their m/z value. Only cations with m/z values that match the curvature of the instrument tube reach the detector. Finally, a mass spectrum is generated. The abundance of cations detected at each m/z value is variable and depends on the frequency (likelihood) of fragmentation patterns in a population, the stability of cations produced, or a combination of factors. The y-axis in a mass spectrum (abundance) is reported on an arbitrary 0 to 100% scale, where the highest abundance peak is given a relative abundance of 100%. The highest abundance peak (called the base peak) corresponds to the cation with the greatest stability and/or relative rate of formation, and it can provide structural clues toward the identity of the original molecule. Therefore, the statement that most accurately describes the definition of base peak in mass spectrometry is the peak with the highest abundance.

lactams

The properties of lactams (cyclic amides) result from the delocalization of electrons. Delocalization of the nitrogen lone pair into the carbonyl generates a resonance form with both charge separation and a different pattern of single and double bond character. The hybrid resonance structure of a lactam reflects the extent of electron delocalization. Infrared (IR) spectroscopy is an analytical technique in which a sample is irradiated with IR light and a spectrometer detects and records the percentage of radiation that passes through the sample over a range of frequencies. Absorption of IR radiation causes vibrational and rotational motion in a molecule. Different bonds and functional groups absorb IR radiation at different intensities and different frequencies, and therefore the signal from each bonding environment appears in a characteristic region of the IR spectrum. The data shows that a smaller lactam ring size leads to a higher wavenumber carbonyl stretch in their IR spectrum. Ring strain includes both angle strain and torsional strain and is greater for smaller ring sizes. In a lactam, the partial double bond character of the C-N bond leads to angle strain when bond angles significantly deviate from the preferred 120° bond angle of a planar sp2 carbon atom. To minimize ring strain, smaller lactam rings demonstrate less resonance delocalization and less C-N double bond character. Smaller lactam rings also have angle strain for the lactam carbon atom. Together, these features strengthen the C-O bond and lead to a higher wavenumber for the carbonyl stretching vibration. Consequently, ring strain is responsible for the observed trend in IR stretching vibrations among lactams. (Choice A) By Hooke's Law, atoms act as weights on either side of a spring (bond), where heavier atoms lead to slower frequencies of oscillation. The atoms contributing to the C-O bonds do not vary between the three molecules. (Choice B) The inductive effect describes how electron density can be donated or removed through sigma bonds. The sigma bond environment around the three lactam functional groups is unchanged. (Choice C) The N-H protons of a lactam are only weakly acidic and do not vary significantly between lactam molecul

formation of cyanohydrin

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.

base hydrolysis of triacylglycerol

Triacylglycerols are lipids made up of three fatty acyl groups and a molecule of glycerol. Each fatty acyl group is bonded to glycerol through an ester linkage. Hydrolysis of the triacylglycerol can occur under acidic or basic conditions and breaks the ester bonds, releasing glycerol and three fatty acids under acidic conditions or three fatty acid salts under basic conditions. Base-mediated ester hydrolysis is known as saponification and begins by the addition of hydroxide to the carbonyl, creating a tetrahedral intermediate. The fatty acid salt is released from the glycerol molecule when the carbonyl is reformed. The number of unique fatty acid salts released through hydrolysis depends on the composition of the hydrocarbon tails: All hydrocarbon tails are the same—only one unique fatty acid salt is released Two hydrocarbon tails are the same and one is different—two unique fatty acid salts are released All the hydrocarbon tails are different—three unique fatty acid salts are released The hydrocarbon tails on the triacylglycerol in this question are represented by R, R′, and R′′. Because each R group is different, the triacylglycerol has three different hydrocarbon tails. Therefore, three unique fatty acid salts will be released along with glycerol upon basic hydrolysis of the triacylglycerol. (Choice A) In the basic hydrolysis of a triacylglycerol, no change in the number of C-O bonds occurs. Therefore, the reaction is not a reduction. (Choice B) If all of the R groups (hydrocarbon tails) on the three fatty acyl groups were the identical, then basic hydrolysis of the triacylglycerol in this question would give 3 moles of the same fatty acid salt. (Choice C) SN2 reactions are a concerted process. Because the hydrolysis of a triacylglycerol occurs in more than one step as a nucleophilic acyl substitution, it cannot be an SN2 reaction. Educational objective:Triacylglycerols are composed of three fatty acyl groups bonded to glycerol through ester linkages. Basic hydrolysis of the ester linkage releases three fatty acid salts and glycerol. The number of unique fatty acid salts released depends on the composition of the hydrocarbon tails on the fatty acyl groups.

distillations

Vacuum distillation is performed under reduced system pressure, thereby lowering a compound's boiling point (bp) relative to its boiling point at atmospheric pressure (Number IV). Compounds with high boiling points (>150 °C) tend to decompose at or near their boiling point. Therefore, vacuum distillation is ideal for compounds with a boiling point over 150 °C to prevent degradation (Number I). Simple distillation (boiling points <150 °C and >25 °C apart) Fractional distillation (boiling points <150 °C and <25 °C apart)

how to make a wax

Waxes are hydrophobic lipids structurally composed of a long-chain carboxylic acid (fatty acid) condensed with a long-chain alcohol through an ester linkage. An esterification reaction results in the formation of a new ester linkage. Direct esterification reactions between a carboxylic acid and an alcohol occur slowly at room temperature and in the absence of a catalyst. The Fischer esterification requires an acid catalyst (usually sulfuric acid, H2SO4) and heat, and is a condensation reaction that eliminates a molecule of water from the reactants alongside ester formation. This question asks which statement correctly describes a method to prepare the depicted wax. The half of the wax containing the carbonyl (C=O) must be derived from a reactant carboxylic acid. Including carbons in the terminal methyl and carbonyl, a total of twelve carbons are in this half of the wax. The IUPAC prefix for twelve is dodeca-, and the reactant carboxylic acid is dodecanoic acid. The other half of the wax is derived from a reactant alcohol. A total of sixteen carbons are in this half of the wax, arranged in a linear chain with the oxygen attached at carbon 1. The IUPAC prefix for sixteen is hexadec-; thus, the reactant alcohol is 1-hexadecanol. For an effective Fischer esterification, an acid catalyst (H2SO4) is required. Choice D lists all the necessary components to prepare the given wax molecule through a Fischer esterification.

Phenylketonuria (PKU)

a disorder related to a defective recessive gene on chromosome 12 that prevents metabolism of phenylalanine Phenylketonuria is characterized by a lack of activity of phenylalanine hydroxylase, which catalyzes the hydroxylation of phenylalanine to tyrosine. Because the enzyme is not functioning properly, tyrosine, which is characterized as an aromatic amino acid with a phenol functional group, is not formed.

effect of esterification on shielding

addition of an ester leads to downfield dishielding because oxygen is electronegative and EWD Because an ester is electronegative, it withdraws electrons and deshields neighboring protons; therefore, the signals for He and Hf must shift downfield

crotyl chloride

can react in SN1 reaction because it is allylic even though it is primary alkyl halide

NP-HPLC and fractional distillation are two methods that can be employed to separate mixtures. Different separation methods employ different chemical principles to facilitate separation, which can lead to differences in elution order.

cannot assume the elution order will be the same for both

signal intensity mass spec

decreasing PH would increase number of multiply charged ions and yields smaller m/z ratio 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 smaller m/z ratio; therefore, an increase in the number of multiply charged molecular ions increases signal intensity for smaller m/z ratios relative to the m/z peak where z = 1.

Acid helps catalyze the Fischer esterification reaction by doing all of the following EXCEPT: A.enhancing the nucleophilicity of alcohols. B.forming a resonance-stabilized intermediate. C.creating a good leaving group. D.increasing carbonyl electrophilicity.

enhancing the nucleophilicity of alcohols. Acids help by forming a resonance stabilized intermediate, creating a good leaving group, and increasing carbonyl electrophilicity A Fischer esterification is the acid-catalyzed condensation of a carboxylic acid and an alcohol into an ester. Strong acids can donate their protons to other molecules to catalyze a reaction. In the Fischer esterification, acidic protonation enhances the rate of esterification. In the mechanism for the Fischer esterification, protonation of the carbonyl oxygen leads to the formation of a resonance-stabilized intermediate (Choice B). A resonance form of this intermediate has a positive formal charge on the carbonyl carbon, which enhances its electrophilicity (Choice D). The alcohol nucleophile attacks the electrophilic carbon. Subsequent deprotonation regenerates the acid catalyst. The protonation of a hydroxyl converts it into a good leaving group (Choice C), which is eliminated as a molecule of water. The resulting cation is a resonance-stabilized intermediate. Deprotonation of the carbonyl oxygen again regenerates the acid catalyst and forms the ester product. Acids do not generally increase the nucleophilicity of a compound, as a nucleophile lone pair is attracted to acidic protons. Consequently, acids tend to protonate nucleophiles and make them less nucleophilic. Educational objective:Acids catalyze many reactions by donating protons to a reactant. Protonation enhances electrophiles and increases the stability of leaving groups. Acids generally decrease the nucleophilicity of molecules.

stereoselectivity vs regioselectivity

stereoselective is when a rxn preferentially forms one stereoisomer regioseletive is when a rxn preferentially forms one constitutional isomer Organic transformations can demonstrate several types of selectivity. A stereoselective reaction results in the preferential formation of a stereoisomer. For example, a reaction that prefers either a cis or trans outcome would be stereoselective. A regioselective reaction results in the preferential reaction at one location within a molecule, often resulting in constitutional isomers. In Step 3, the alcohol in Compound 2 is protected as a silyl ether (Compound 3). Without this step, only two possible alcohol groups would react with the acid chloride in Step 4. By protecting one of the alcohols as a silyl ether, the other alcohol could be regioselectively acylated in Step 4

The structure of guaiazulene with annotated protons is shown below. The chemical shift of HB (7.22 ppm) is larger than the chemical shift of HA (3.08 ppm) as a result of: A.the inductive effect. B.steric hindrance. C.resonance. D.the anisotropy of the aromatic ring.

the anisotropy of the aromatic ring. In proton (1H) nuclear magnetic resonance (NMR) spectroscopy, protons in alkyl groups are among the most shielded and normally have chemical shifts (δ) of δ 0.9-1.4. Higher-order (eg, tertiary) alkyl groups have greater chemical shifts than lower-order (eg, primary) alkyl groups. Proton chemical shifts are also influenced by the anisotropic effect, where the circulation of nearby pi electrons generates an induced magnetic field Binduced that either increases or decreases the effect of shielding (depending on the location of the nuclei relative to the vector of Binduced). Protons directly attached to aromatic rings are positioned in a region where Binduced is aligned with the external magnetic field B0, leading to a much larger effective magnetic field Beff and greater chemical shift range (δ 7-8 ppm). In the annotated structure of guaiazulene provided in this question, proton HA is attached to the central sp3 carbon of the isopropyl group—an alkyl environment. Being bonded to a more-substituted tertiary carbon atom, the chemical shift of HA should be near the higher end of the alkyl range (δ 1.4-1.8 ppm). The nearby proximity of the aromatic further contributes to deshielding proton HA. In contrast, proton HB is bonded directly to an sp2 carbon within an aromatic ring. Due to the anisotropy of the aromatic ring, the relative location of proton HB leads to Binduced reinforcing B0, a much larger Beff, and a greater chemical shift for HB.

The experimental data obtained when a chiral molecule interacts with plane-polarized light is called:

the specific rotation. Normal light is made of light waves that oscillate in all planes perpendicular to the direction of propagation. When light encounters a polarizer, only waves that are aligned with the polarizer can pass. Therefore, all the light that passes through the polarizer oscillates in one direction. This resulting uniform light is known as plane-polarized (linear) light. 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. (Choice A) A polarimeter is the instrument used to measure the rotation of polarized light. (Choice B) The R or S configurations are based on the priority ranking of substituents on a stereocenter, and are not correlated to the direction chiral molecules rotate plane-polarized light. Therefore, stereocenter configuration cannot be used to predict specific rotation, or vice versa. (Choice D) An infrared spectrum is obtained from irradiation of a molecule with infrared light, causing bond vibrations and stretching at different frequencies. The amount of light absorbed by the molecule is recorded and displayed as the percent transmittance versus wavenumber on the infrared spectrum. Educational objective: 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.

During conversion of an amino acid from the ʟ to the ᴅ configuration in a PLP-dependent enzyme, a base removes a proton from the α-carbon. Re-protonation of the resulting planar molecule involves the nucleophilic attack of a proton donor: A.in a nonstereospecific manner. B.on either side of the planar intermediate with equal likelihood. C.with an orientation opposite that of the proton that was removed. D.with the same orientation as the proton that was removed.

with an orientation opposite that of the proton that was removed. During a stereospecific reaction, the configuration of the substrate completely determines the configuration of the product. When one chiral molecule is converted to its enantiomer or to a different chiral molecule, the mechanism often includes formation of a planar intermediate such as an alkene or carbocation. The reaction is stereospecific if a new reactant is preferentially added to one side of this intermediate. For example, the racemase-catalyzed conversion of ʟ-glutamate to ᴅ-glutamate is stereospecific. The planar intermediate (imine) is formed in step 3 of the PLP-dependent reactions (Figure 2) after hydrogen is removed from the chiral carbon atom in ʟ-glutamate. In step 4, the enzyme catalyzes the addition of hydrogen to the opposite side, which allows for the conversion to ᴅ-glutamate. (Choices A and B) Addition of hydrogen to either side of the planar intermediate would result in the production of a mixture of ʟ- and ᴅ-amino acids. The question asks specifically for the mechanism involved in converting ʟ-amino acids to ᴅ-amino acids. (Choice D) The addition of hydrogen to the same side as the hydrogen atom that was removed in step 3 would result in reformation of an ʟ-amino acid rather than formation of a ᴅ-amino acid. Educational objective: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.