HW Review w/ pictures Exam 2
B) II - The structure drawn in the box is the half-chair conformation and it is located at position II on the energy diagram. This structure has adopted a conformation that maximizes the steric and torsional strain. Thus, it is at the highest energy point on this diagram.
On the following energy diagram, identify the energy level of the cyclohexane conformation shown in the box. A) I B) II C) III D) IV
C) III - The structure drawn in the box is the twist-boat conformation and it is located at position III on the energy diagram. This structure twists out of the boat conformation to relieve some of the steric and torsional strain of the boat conformation (IV). Thus, this is found in the second lowest valley on this diagram.
On the following energy diagram, identify the energy level of the cyclohexane conformation shown in the box. A) I B) II C) III D) IV
D) IV - The structure drawn in the box is the boat conformation and it is located at position IV on the energy diagram. This structure has steric and torsional strain causing it to be at a peak but it doesn't have as much strain as the half-chair conformation.
On the following energy diagram, identify the energy level of the cyclohexane conformation shown in the box. A) I B) II C) III D) IV
A) reactant favored - This reaction is reactant favored. The amine is a weaker acid (pKa ~38) than the carboxylic acid (pKa ~5). Acid-base reactions favor the protonation of the weaker acid.
Predict the direction of the following reaction. Approximate pKa values in DMSO are provided. A) reactant favored B) a mixture of reactants and products C) product favored
A) reactant favored - This reaction is reactant favored. The benzylic methyl group is a weaker acid (pKa ~41) than the alkyne (pKa ~35). Acid-base reactions favor the protonation of the weaker acid.
Predict the direction of the following reaction. Approximate pKa values in DMSO are provided. A) reactant favored B) a mixture of reactants and products C) product favored
A) reactant favored - This reaction is reactant favored. The alcohol is a weaker acid (pKa ~17) than the carboxylic acid (pKa ~5). Acid-base reactions favor the protonation of the weaker acid.
Predict the direction of the following reaction. Approximate pKa values in water are provided. A) reactant favored B) a mixture of reactants and products C) product favored
C) product favored - This reaction is product favored. The alcohol is a weaker acid (pKa 17) than the diketone (pKa 12). Acid-base reactions favor the protonation of the weaker acid.
Predict the direction of the following reaction. Approximate pKa values in water are provided. A) reactant favored B) a mixture of reactants and products C) product favored
C) product favored - This reaction is product favored. Water is a weaker acid (pKa 16) than the diketone (pKa 12). Acid-base reactions favor the protonation of the weaker acid.
Predict the direction of the following reaction. Approximate pKa values in water are provided. A) reactant favored B) a mixture of reactants and products C) product favored
B) II - Acidity can be correlated to the electronegativity of the atom that the proton is bound to - the more electronegative, the more acidic. Because of the positive charges on the nitrogen atoms, the protons on those sites are the most acidic. The sp2 hybridized nitrogen is considered to be more electronegative than the sp3 hybridized nitrogen because there is more s-character in the sp2 orbital, and s-orbitals lie closer to the nucleus than p-orbitals.
Quinine, found in tonic water, is shown below. Identify the most acidic proton from choices, I-V. A) I B) II C) III D) IV E) V
D) IV - Conformation IV is the anti conformation is the staggered conformation where the steric repulsion is minimized.
A Newman projection looking down C1-C2 bond 1-chloropropane is shown below. Identify which structure is the anti conformation. A) I B) III C) I & III D) IV E) II
E) II - Conformer II is the eclipsed conformation because it is the one where the atoms and bonds on neighboring carbon atoms align with one another.
A Newman projection looking down C1-C2 bond 1-chloropropane is shown below. Identify which structure is the eclipsed conformation. A) I B) III C) I & III D) IV E) II
C) I & III - Conformations I & III are the gauche conformations are the staggered conformations that fail to minimize steric repulsion.
A Newman projection looking down C1-C2 bond 1-chloropropane is shown below. Identify which structure is the gauche conformation. A) I B) III C) I & III D) IV E) II
B) II - The eclipsed conformation is the one where the atoms and bonds on neighboring carbon atoms are aligned. This is projection II. Staggered conformations have dihedral angles of 60°. This is projection I
A Newman projection looking down C2-C3 bond of 2-methylpropane is shown below. Identify which structure is an eclipsed conformation. A) I B) II
A) I - Newman Projection 1 is the staggered conformation. Staggered conformations have dihedral angles of 60°, while the eclipsed conformation has all bonds aligned (dihedral angles of 0°.)
A Newman projection looking down C2-C3 bond of 2-methylpropane is shown below. Identify which structure is the staggered conformation. A) I B) II
A) axial - This is an axial methyl group. The methyl is pointing vertically up from the plane of the cyclohexane structure.
A Newman projection of methylcyclohexane is shown below. Identify the position of the methyl group. A) axial B) equatorial C) cis D) trans
B) equatorial - This is an equatorial methyl group. The methyl is pointing horizontally in the plane of the cyclohexane structure.
A Newman projection of methylcyclohexane is shown below. Identify the position of the methyl group. A) axial B) equatorial C) cis D) trans
A) 1 - The valleys are lower energy and correspond to staggered conformations. There are two possible staggered conformations for this structure. This conformation has two gauche CH₃/CH₃ steric interactions so it will be the second lowest energy conformation.
A graph representing the potential energy of different conformational states of 2-methylbutane is shown below. Identify the point on the graph for the following conformation. A) 1 B) 2 C) 3 D) 4
B) 2 - The peaks are higher energy and correspond to eclipsed conformations. There are two possible eclipsed conformations for this structure. This conformation has a CH₃/CH₃ steric interaction and torsional strain so it will be the highest energy conformation.
A graph representing the potential energy of different conformational states of 2-methylbutane is shown below. Identify the point on the graph for the following conformation. A) 1 B) 2 C) 3 D) 4
C) 3 - The valleys are lower energy and correspond to staggered conformations. There are two possible staggered conformations for this structure. This conformation has the least steric strain with one CH₃/CH₃ gauche interaction so it will be the lowest energy conformation.
A graph representing the potential energy of different conformational states of 2-methylbutane is shown below. Identify the point on the graph for the following conformation. A) 1 B) 2 C) 3 D) 4
D) 4 - The peaks are higher energy and correspond to eclipsed conformations. There are two possible eclipsed conformations for this structure. This conformation has less steric strain as it is all CH₃/H interactions so it will be the second highest energy conformation.
A graph representing the potential energy of different conformational states of 2-methylbutane is shown below. Identify the point on the graph for the following conformation. A) 1 B) 2 C) 3 D) 4
B) unsaturated - This is an unsaturated hydrocarbon because the molecular formula is C₁₀H₂₀. This molecule has 2 fewer hydrogens that would be expected for a saturated alkane (CₓH₂ₓ₊₂).
A saturated acyclic hydrocarbon has the molecular formula of CₓH₂ₓ₊₂. A hydrocarbon can be classified as unsaturated if it has fewer hydrogens than a saturated formula would predict. Classify the following hydrocarbon skeletal structure as saturated or unsaturated. A) saturated B) unsaturated
B) Atom (electronegativity or polarizability) - The explaining factor is atom electronegativity. When Ha is removed, the oxygen becomes negatively charged. However, when Hb is removed, the nitrogen becomes negatively charged. Oxygen is more electronegative than nitrogen so it can stabilize the charge better.
A structure with a more stable conjugate base will be more acidic. Ha is more acidic than Hb. Choose the factor that best explains the increased stability of the first structure's conjugate base. A) Charge B) Atom (electronegativity or polarizability) C) Resonance D) Inductive stabilization E) Orbital hybridization
D) Inductive stabilization - The explaining factor is inductive stabilization. When Ha is removed, the oxygen becomes negatively charged and the fluorine atoms on the molecule pull the electron density toward them in a dipole. There are no more electronegative atoms to stabilize the resultant charge when Hb is removed.
A structure with a more stable conjugate base will be more acidic. Ha is more acidic than Hb. Choose the factor that best explains the increased stability of the first structure's conjugate base. A) Charge B) Atom (electronegativity or polarizability) C) Resonance D) Inductive stabilization E) Orbital hybridization
A) Charge - The explaining factor is charge. When Ha is removed, the structure becomes neutral. However, when Hb is removed, the structure becomes negatively charged. A neutral conjugate base is more stable than a charged one.
A structure with a stable conjugate base will be more acidic. Ha is more acidic than Hb. Choose the factor that best explains the increased stability of the first structure's conjugate base. A) Charge B) Atom (electronegativity or polarizability) C) Resonance D) Inductive stabilization E) Orbital hybridization
C) Resonance - The explaining factor is resonance. When Ha is removed, the negative charge can be delocalized through several resonance structures. However, when Hb is removed, the negative charge resides completely on the oxygen atom. Delocalization of the electrons through resonance stabilizes the conjugate base.
A structure with a stable conjugate base will be more acidic. Ha is more acidic than Hb. Choose the factor that best explains the increased stability of the first structure's conjugate base. A) Charge B) Atom (electronegativity or polarizability) C) Resonance D) Inductive stabilization E) Orbital hybridization
E) Orbital hybridization - The explaining factor is orbital hybridization. When Ha is removed, the lone pair of electrons resides in an sp hybridized orbital. However, when Hb is removed, the lone pair of electrons resides in an sp² hybridized orbital. An sp orbital holds the electrons closer to the nucleus due to its greater s-character, thus stabilizing the charge.
A structure with a stable conjugate base will be more acidic. Ha is more acidic than Hb. Choose the factor that best explains the increased stability of the first structure's conjugate base. A) Charge B) Atom (electronegativity or polarizability) C) Resonance D) Inductive stabilization E) Orbital hybridization
C) Add aqueous NaHCO₃ and extract the benzoic acid into the aqueous layer. - Benzoic acid is a much stronger acid than phenol. Benzoic acid will react with a weak base (NaHCO₃) to form benzoate while the phenol does not react . This charged benzoate anion will be soluble in water. The water layer can then be extracted removing the benzoate from the phenol.
Benzoic acid and phenol (shown below) are dissolved in an ether solution. Choose the best process to separate the two compounds using an acid-base extraction. A) Add aqueous HCl and extract the phenol into the aqueous layer. B) Add aqueous NaOH and extract the phenol into the aqueous layer. C) Add aqueous NaHCO₃ and extract the benzoic acid into the aqueous layer. D) Add aqueous HCl and extract the benzoic acid into the aqueous layer. E) These two compounds cannot be separated using an acid-base extraction.
A) I - Newman projection I is an anti conformation. The ethyl and methyl substituents are positioned opposite each other with a dihedral angle of 180°.
Identify the Newman projection that depicts the anti conformation of 1-bromopropane. A) I B) II C) III D) IV E) V
B) II - Image II depicts a dihedral angle. A dihedral angle is the angle between two bonds originating from neighboring atoms.
Choose the picture that depicts a dihedral angle. A) I B) II C) III D) IV
B) unsaturated - This is an unsaturated hydrocarbon because there is are double bonds present.
Classify the following hydrocarbon skeletal structure as saturated or unsaturated. A) saturated B) unsaturated
Achiral - This molecule is achiral because it has a mirror plane of symmetry. A mirror plane of symmetry is an imaginary plane that bisects a molecule into halves that are mirror images of each other.
Classify the following molecule as chiral or achiral.
Chiral - This molecule is chiral. Chiral molecules have no plane of symmetry (an imaginary plane that bisects a molecule into halves that are mirror images of each other). This molecule has no plane of symmetry and the asymmetric center is circled below.
Classify the following molecule as chiral or achiral.
Chiral - This molecule is chiral. Chiral molecules have no plane of symmetry (an imaginary plane that bisects a molecule into halves that are mirror images of each other). This molecule has no plane of symmetry with two asymmetric centers.
Classify the following molecule as chiral or achiral.
B) II < I < III < IV < V - The pH increases as the basicity of the compound increases. Compounds I and II are Bronsted-Lowry acids with the sulfonic acid II being more acidic, therefore the lowest pH. This can be determined by considering the structural features involved in stabilizing the conjugate bases; the sulfonate anion is stabilized by additional resonance stabilization relative to the carboxylate ion, which makes it a stronger acid. Of the remaining compounds, use their structural features to determine the relative basicity: III is less basic than IV due to resonance stabilization of the oxygen anion; IV is less basic than V due to the negative charge being on the more electronegative oxygen atom compared to the carbon atom in Grignard reagent V.
Consider equimolar solutions of each of the following compounds; rank them in order of increasing pH. A) I < II < III < IV < V B) II < I < III < IV < V C) II < I < IV < III < V D) V < IV < III < I < II
A) I - Because oxygen is more electronegative than carbon, the O-H bonds will be more acidic than the C-H bonds. Therefore, I and V are more acidic. With proton I, there is a neighboring -CF₃ group, which is highly electron withdrawing and will pull away electron density. Because of the electron withdrawing nature of the -CF₃ group, proton I is more acidic. Note 1: functional group acidity = alcohol > alkyne > alkene > alkane Note 2: The more UNSATURATED a hydrocarbon, the more ACIDIC
Consider the structure below. Identify the most acidic proton from choices, I-V. A) I B) II C) III D) IV E) V
E) V - Newman projection V is an eclipsed conformation with all dihedral angles at 0°.
Identify the Newman projection that depicts the eclipsed conformation of pentane sighting down the C3-C2 bond. A) I B) II C) III D) IV E) V
E) I and III are the same compound. - Notice that I and III have the same connectivities though they are written in opposite directions, while II has a different connectivity than I and III, but the same molecular formula as I and III.
Considering structures below, which of the following statements is true? A) I and II have different molecular formulas. B) I and III are structural isomers of each other. C) II and III are stereoisomers of each other. D) II and III are different conformations of the same compound. E) I and III are the same compound.
A) The axial hydrogen (Ha) becomes equatorial while the equatorial hydrogen (Hb) becomes axial. - The axial hydrogen (Ha) becomes equatorial while the equatorial hydrogen (Hb) becomes axial. In a ring flip, all of the axial hydrogens become equatorial and all of the equatorial hydrogens become axial.
Describe what happens to Ha and Hb during a ring flip. A) The axial hydrogen (Ha) becomes equatorial while the equatorial hydrogen (Hb) becomes axial. B) The axial hydrogen (Hb) becomes equatorial while the equatorial hydrogen (Ha) becomes axial. C) The axial hydrogen (Ha) stays axial the equatorial hydrogen (Hb) stays equatorial. D) The axial hydrogen (Hb) stays axial the equatorial hydrogen (Ha) stays equatorial.
2 - A chiral center is formed at a carbon with four different groups attached to it. The carbon with the nitrogen attached and the carbon with the hydroxy (-OH) attached (carbons 2 and 4) fit that criterion, so there are two chiral centers.
Determine the number of chiral centers in the compound shown here.
1 - A chiral center is formed at a carbon with four different groups attached to it. Only the carbon with the nitrogen attached fits that criterion, so there is only one chiral center.
How many chiral centers are in the compound shown here?
A) The axial bromine is pointing in the wrong direction. - The two bonds that form a corner of the chair point in the direction that the axial group should be drawn. This corner points up and so the axial group should point up, too.
Identify the error in the following chair structure. A) The axial bromine is pointing in the wrong direction. B) The axial fluorine is pointing in the wrong direction. C) The equatorial chlorine is pointing in the wrong direction. D) The equatorial methyl group is pointing in the wrong direction.
B) The axial fluorine is pointing in the wrong direction. - The two bonds that form a corner of the chair point in the direction that the axial group should be drawn. This corner points up and so the axial group should point up, too.
Identify the error in the following chair structure. A) The axial bromine is pointing in the wrong direction. B) The axial fluorine is pointing in the wrong direction. C) The equatorial chlorine is pointing in the wrong direction. D) The equatorial methyl group is pointing in the wrong direction.
A) identical - These images are both depictions of 2-methylbutane.
Identify the relationship between the two structures shown below. A) identical B) constitutional isomers C) conformational isomers D) not isomers
B) constitutional isomers - These two molecules are constitutional isomers. Newman projection I represents pentane while Newman projection II represents 2-methylbutane.
Identify the relationship between the two structures shown below. A) identical B) constitutional isomers C) conformational isomers D) not isomers
A) wedge - This is an equatorial hydrogen that points down from the plane of the ring. Note that the lower edge of the chair is assumed to be nearer to the viewer and is drawn with a heavier line to reinforce that perspective. The ring is tilted towards you. Any hydrogens on bonds angled upwards in the chair are towards you (wedges). Any hydrogens on bonds angled downwards in the chair are away from you (dashes).
It is important to be able to convert between wedge-and-dash representations to chair representations. Would Ha in this chair conformation be represented as a wedge or a dash? A) wedge B) dash C) axial D) equatorial
B) dash - This is an equatorial hydrogen that points down from the plane of the ring. Note that the lower edge of the chair is assumed to be nearer to the viewer and is drawn with a heavier line to reinforce that perspective. The ring is tilted towards you. Any hydrogens on bonds angled upwards in the chair are towards you (wedges). Any hydrogens on bonds angled downwards in the chair are away from you (dashes).
It is important to be able to convert between wedge-and-dash representations to chair representations. Would Ha in this chair conformation be represented as a wedge or a dash? A) wedge B) dash C) axial D) equatorial
D) III < I < II - Strong acids have stable conjugate bases. - In this series, the phenol is less acidic than the carboxylic acids because the conjugate base of the acid has a resonance structure that shares the negative charge with two electronegative oxygen atoms. - The presence of the nitro group in the para-nitrobenzoic acid results in a more acidic substance than benzoic acid due to the inductive effect of the electron-withdrawing group
Rank the compounds (shown below) in order of increasing acidity. A) I < II < III B) III < II < I C) II < I < III D) III < I < II E) II < I < III
D) II < V < III < I < IV - To determine the rank order, consider the stability of the conjugate acid and the general basicity trends for each type of functional group. Compound II contains only an ester functional group (-CO₂R) and is the least basic because the conjugate acid, a protonated carbonyl, will be the least stable. Pyridine, compound IV, is ranked the second least basic, followed by the phenoxide II. Phenoxides and alkyl amines such as compound I can often have similar basicity strength, but the presence of the highly electron-withdrawing nitro group in II reduces the basicity significantly. Potassium tert-butoxide is the most basic of the structures shown; alkoxides are typically more basic than neutral amines because they have a negative charge.
Rank the following compounds in order of increasing basicity. A) II < I < V < III < IV B) II < V < I < III < IV C) IV < I < III < V < II D) II < V < III < I < IV
C) II < I < III - Typically, basicity increases as the inductive effects increases. In tert-butanol, the number of electron-donating groups (such as -CH₃) increases making it a stronger base and more reactive towards H⁺.
Rank the following organic compounds in order of increasing basicity. A) I < II < III B) III < II < I C) II < I < III D) III < I < II E) II < III < I
D) III < I < II - Typically, basicity decreases as the resonance increases. In cyclohexyl amine, the carbons are electron-donating which increases the basicity. In the other two aromatic amines, the electrons are delocalized, leading to a decreased electron density on the nitrogen. These nitrogen atoms have a lower ability to share those electrons with a proton, making it a weaker base.
Rank the following organic compounds in order of increasing basicity. A) I < II < III B) III < II < I C) II < I < III D) III < I < II E) II < III < I
D) III < I < II - Typically, basicity increases with hybridization (sp < sp² < sp³). The sp³ orbital does not hold the pair of electrons as close to the nucleus, making it a stronger base and more reactive towards H⁺ .
Rank the following organic compounds in order of increasing basicity. A) I < II < III B) III < II < I C) II < I < III D) III < I < II E) II < III < I
A) I < III < II - Typically, basicity increases as the electronegativity decreases. When comparing atoms within the same row of the periodic table, the more electronegative the atom donating the electrons, the less willing it is to share those electrons with a proton, so the weaker the base.
Rank the following organic compounds in order of increasing basicity. A) I < III < II B) II < III < I C) II < I < III D) III < I < II E) II < III < I
D) I < III < II - To determine the order of acidity, compare the stabilities of the conjugate bases for each acid. Typically, acidity increases when the conjugate base has a smaller charge. Thus, the acid that becomes neutral is the most acidic whereas the molecule that results in a -2 charge on the conjugate base will not be very acidic.
Rank the hydrogens shown in bold in order of increasing acidity. A) I < II < III B) III < II < I C) II < I < III D) I < III < II E) II < III < I
E) II < III < I - To determine the order of acidity, compare the stabilities of the conjugate bases for each acid. Typically, acidity increases when the hydrogen is bonded to a more electronegative atom (from left to right across the periodic table). The conjugate bases are more stable when an electronegative atom accepts the electrons from the bond to the hydrogen.
Rank the hydrogens shown in bold in order of increasing acidity. A) I < II < III B) III < II < I C) II < I < III D) I < III < II E) II < III < I
B) III < IV < I < II - III < IV < I < II. To rank the stability of a conformation of an acyclic alkane, you must consider both the steric strain and torsional strain. The eclipsed conformers (III and IV) are less stable than the staggered conformers (I and II) due to torsional strain. III is less stable than IV because there is more steric strain due to the eclipsing methyl groups. Similarly, conformer I is less stable than II due the higher steric strain of the gauche methyl groups.
Rank the stability of the following conformations of 2-chlorobutane from least stable to most stable. A) I < II < III < IV B) III < IV < I < II C) II < I < IV < III D) IV < III < II < I E) II < III < IV < I
A) I has more angle strain. I has sp² carbons which require 120° bond angles, but the bond angles here are 60°. - The double bonds require sp² carbons. Those carbons require 120° angles. The internal angles of an equilateral triangle are 60°. These bond angles are much smaller than optimum, leading to angle strain.
Select the ring that has the most angle strain (ring strain). Identify the reason for the difference. A) I has more angle strain. I has sp² carbons which require 120° bond angles, but the bond angles here are 60°. B) I has more angle strain. I has all sp³ carbons which require 109° bond angles, and the bond angles here are 120°. C) II has more angle strain. II has all sp³ carbons which require 109° bond angles, and the bond angles here are 60°. D) II has more angle strain. I has sp² carbons which require 120° bond angles, and the bond angles here are 120°.
B) basic - The compound shown, 4-(dimethylamino)pyridine, does not contain acidic hydrogens that can dissociate to increase the hydrogen ion concentration, but it does contain basic nitrogen atoms that can decrease the hydrogen ion concentration. Thus, the solution will be basic.
The compound 4-(dimethylamino)pyridine (DMAP) shown is dissolved in n-butanol to form a homogeneous solution. Consider the structural features of the compound to determine if the solution will be acidic, basic or neutral. A) acidic B) basic C) neutral
A) constitutional - Isomers with the atoms connected in a different way are called constitutional isomers.Stereoisomers, conformational isomers, and geometric isomers have atoms arranged differently in space, but the connections between the atoms is the same. "Molecular" is not an isomer type.
The two structures shown exhibit what type of isomerism? A) constitutional B) stereoisomerism C) conformational D) geometric E) molecular
A) Ha - The most acidic proton is H₁. When this proton is removed the conjugate base is stabilized through two factors: electronegativity of oxygen and the ability to delocalize the charge through resonance. None of the other hydrogen atoms on this molecule would form a conjugate base stabilized by two factors.
Use your understanding of the key factors that influence acidity to choose the most acidic proton on this structure. A) Ha B) Hb C) Hc D) Hd
C) Hc - The most acidic proton is H₃. When this proton is removed the conjugate base is stabilized from the delocalization of charge through resonance over three atoms. Delocalization of the electrons through resonance stabilizes the conjugate base.
Use your understanding of the key factors that influence acidity to choose the most acidic proton on this structure. A) Ha B) Hb C) Hc D) Hd
B) trans - One of the F substituents is shown facing to the front of the ring and one to the rear of the ring, so they are on opposite sides of the ring. This relationship is called trans, from the Latin meaning "across."
What arrangement do the substituents in the following compound exhibit? A) cis B) trans C) anti D) adjacent E) gauche
E) They are not isomers. - The two compounds have different molecular formulas. Isomers have the same molecular formula. These two molecules are not isomers.
What is the relationship between the two molecules shown below? A) They are identical. B) They are constitutional isomers. C) They are isotopes. D) They are enantiomers. E) They are not isomers.
A) Torsional - There are a number of eclipsed bonds leading to torsional strain. In the circled portion of the molecule, there are no atoms positioned too closely in space so there is no steric strain. The bond angles are not distorted.
What type of strain is present in the circled portion of this molecule? A) Torsional B) Angle C) Steric
C) Steric - A methyl substituent in the axial position will be too close to atoms in the other axial positions on the chair and experiences steric strain. This is sometimes called 1,3-diaxial strain.
What type of strain is present in the circled portion of this molecule? A) Torsional B) Angle C) Steric
C) Steric - These two hydrogens on neighboring methyl groups are forced too closely in space leading to steric strain.
What type of strain is present in the circled portion of this molecule? A) Torsional B) Angle C) Steric
B) II - Structure II has both substituents pointing up and located in the correct location on the ring.
Which of the following chair conformations represents the molecule as it is drawn below? A) I B) II C) III D) IV
C) III - Structure III has the correction arrangement and stereochemistry of the three methyl groups.
Which of the following chair conformations represents the molecule as it is drawn below? A) I B) II C) III D) IV
C) III - Molecule III is an amine. This functional group will react with the acidic solution to form a charged species . This charged species is soluble in water and can separated from the organic layer.
Which of the following compounds could be extracted into an acidic aqueous solution from an organic solvent? A) I B) II C) III D) I and II E) II and III
A) I - Molecule I is a strong acid. Molecule III is a weak acid. Only a strong acid (I) will react with the weakly basic, NaHCO₃, This reaction results in a charged species is soluble in water and can separated from the organic layer.
Which of the following compounds could be extracted into an aqueous solution of NaHCO₃ from an organic solvent? A) I B) II C) III D) I and II E) II and III
E) I and III - Molecule I is a strong acid. Molecule III is a weak acid. Both structures will react with NaOH These reactions result in charged species that are soluble in water and can separated from any neutral compounds in the organic layer.
Which of the following compounds could be extracted into an aqueous solution of NaOH from an organic solvent? A) I B) II C) III D) I and II E) I and III
D) I and II - Molecule I and II are both strong acids and will react with NaOH These reactions result in charged species is soluble in water and can separated from the organic layer.
Which of the following compounds could be extracted into an aqueous solution of NaOH from an organic solvent? A) I B) II C) III D) I and II E) II and III
A) C₅H₁₀ - There are five carbons, three of which are singly bonded to a carbon and require three hydrogens each. One of the middle carbons has four bonds, but the other has only three bonds and requires a hydrogen, for a total of 10 hydrogens. Therefore the molecular formula is C₅H₁₀.
Which of the following is the correct molecular formula for the molecule below? A) C₅H₁₀ B) C₅H₃ C) C₂H₃ D) C₅H₈
B) II - This Newman projection depicts (R)-2-pentanol.
Which of the following line drawings represents the correct stereochemistry of molecule shown in the Newman projection below? A) I B) II C) III D) IV
A) I - This structure has both substituents pointing up and located in the correct location on the ring.
Which of the following wedge-and-dash structures represents the chair structure shown below? A) I B) II C) III D) IV
C) III - Structure III has both substituents trans and located in the correct location on the ring.
Which of the following wedge-and-dash structures represents the chair structure shown below? A) I B) II C) III D) IV
B) C2 only - A chiral center is formed at a carbon with four different groups attached to it. The carbon labeled C2 fits that criterion. (The carbon labeled C1 has only three groups attached to it.)
Which of the labeled carbons is chiral? A) C1 only B) C2 only C) C1 and C2 D) neither is chiral
A) Ha < Hb < Hc - When Hb or Hc are removed, the conjugate base is stabilized by delocalization. In addition, the conjugate base formed when H₃ is removed allows the charge to be delocalized onto another electronegative oxygen so it has two stabilizing factors.
You can identify relative strengths of acidity of molecules using key factors such as those in the mnemonic CARIO (Charge, Atom, Resonance, Inductivity, and Orbitals). Using this mnemonic, rank these structures in order of acidity of the indicated protons (least acidic to most acidic). A) Ha < Hb < Hc B) Hb < Ha < Hc C) Hc < Ha < Hb D) Hc < Hb < Ha E) Ha < Hc < Hb
