II. ACIDS, BASES & MOLECULAR PROPERTIES TEST 1 1 / 19 Which description of X preserves the conjugate-pair relationship? CH₃CH₂NH₂ + H₃O⁺ -> X + H₂O CH₃CH₂NH⁻ CH₃CH₂NHCH₃ CH₃NH₃⁺ CH₃CH₂OH CH₃CH₂NH₃⁺, with unchanged C-N connectivity Solution: The N lone pair accepts H⁺; no carbon-containing group is added or removed. A. Incorrect. This is deprotonation rather than protonation. B. Incorrect. A methyl group has been added without a source. C. Incorrect. One carbon has been lost. D. Incorrect. This replaces nitrogen rather than transferring a proton. E. Correct. The N lone pair accepts H⁺; no carbon-containing group is added or removed. 2 / 19 Which two species are a conjugate acid-base pair? CH₃CO₂H + HO⁻ -> CH₃CO₂⁻ + H₂O CH₃CO₂H and CH₃CO₂⁻ CH₃CO₂⁻ and H₂O CH₃CO₂H and H₂O HO⁻ and CH₃CO₂⁻ CH₃CO₂H and HO⁻ Solution: They differ by one proton while retaining the same heavy-atom skeleton. A. Correct. They differ by one proton while retaining the same heavy-atom skeleton. B. Incorrect. Their elemental compositions differ beyond one H⁺. C. Incorrect. These do not share a heavy-atom framework. D. Incorrect. Both are bases, not conjugates of each other. E. Incorrect. They are reaction partners rather than conjugates. 3 / 19 Which change enables selective aqueous extraction? CH₃CO₂H and C₆H₆ in ether; aqueous NaHCO₃ added CH₃CO₂H becomes an ester Both undergo carbon-carbon cleavage C₆H₆ becomes C₆H₅⁻ Both become permanently covalent bicarbonates CH₃CO₂H becomes ionic CH₃CO₂⁻ Solution: Bicarbonate deprotonates the acid; the hydrocarbon remains neutral and preferentially organic. A. Incorrect. No alcohol esterification conditions are supplied. B. Incorrect. No such reaction is required. C. Incorrect. Bicarbonate cannot deprotonate benzene appreciably. D. Incorrect. The extraction uses acid-base ionization. E. Correct. Bicarbonate deprotonates the acid; the hydrocarbon remains neutral and preferentially organic. 4 / 19 Which atom changes from electron-pair acceptor to having a filled valence octet? H₃N: + B(F)3 -> H₃N-BF₃ Hydrogen None, because only proton transfers are acid-base reactions Nitrogen Fluorine Boron Solution: Boron accepts the nitrogen lone pair into its initially vacant orbital. A. Incorrect. No proton transfer is required. B. Incorrect. Lewis acid-base chemistry includes pair donation. C. Incorrect. Nitrogen donates rather than accepts the pair. D. Incorrect. No fluorine is the bonding acceptor here. E. Correct. Boron accepts the nitrogen lone pair into its initially vacant orbital. 5 / 19 Which feature makes this a Lewis acid-base event without a required Bronsted event? CH₃-O-CH₃ + BF₃ -> donor-acceptor adduct Carbon loses a neutron Oxygen donates a lone pair without transferring H⁺ BF₃ must donate H⁺ Ether must become an alkoxide Oxygen accepts the boron lone pair Solution: Ether coordination supplies electron density to boron even though the ether has no O-H proton. A. Incorrect. Nuclear changes are irrelevant. B. Correct. Ether coordination supplies electron density to boron even though the ether has no O-H proton. C. Incorrect. BF₃ contains no hydrogen. D. Incorrect. No ether O-H bond exists to deprotonate. E. Incorrect. The donor and acceptor roles are reversed. 6 / 19 Which base is weakest among their conjugate bases? Same-solvent pKa: CF₃CO₂H 0.2; CH₃CO₂H 4.8; CH₃OH 15.5 CH₃O⁻ CH₃CO₂⁻ The ranking cannot use pKa even within one solvent CF₃CO₂⁻ All are equal because each has charge -1 Solution: The lowest-pKa acid has the most stabilized, weakest conjugate base. A. Incorrect. Its acid is the weakest of the three. B. Incorrect. Its acid is weaker than CF₃CO₂H. C. Incorrect. Comparable pKa values are appropriate. D. Correct. The lowest-pKa acid has the most stabilized, weakest conjugate base. E. Incorrect. Charge alone does not set basicity. 7 / 19 What is the approximate standard equilibrium constant? HA + B⁻ <-> A⁻ + HB; pKa(HA)=5, pKa(HB)=9 10^14 4 14 10^4 10^-4 Solution: K is approximately 10 raised to pKa(HB)-pKa(HA), giving 10^(9-5). A. Incorrect. The exponent is a difference, not a sum. B. Incorrect. A logarithmic difference is not K itself. C. Incorrect. The two pKa values should not be added. D. Correct. K is approximately 10 raised to pKa(HB)-pKa(HA), giving 10^(9-5). E. Incorrect. The pKa difference was reversed. 8 / 19 At equilibrium, which side is favored in the stated common solvent? CH₃CH₂O⁻ + HA <-> CH₃CH₂OH + A⁻; product acid is 6 pKa units weaker Neither side because charges cancel No conclusion because acid identity never matters Products by approximately 10^6 Products by exactly sixfold Reactants by approximately 10^6 Solution: Proton transfer favors formation of the weaker acid and weaker base. A. Incorrect. Equal net charge does not imply K=1. B. Incorrect. The stated pKa difference directly determines the estimate. C. Correct. Proton transfer favors formation of the weaker acid and weaker base. D. Incorrect. pKa is logarithmic. E. Incorrect. The favored direction is reversed. 9 / 19 Which explanation is consistent with their conjugate-base ranking? HA is stronger than HB in the same solvent A⁻ must be more basic A⁻ must have a larger negative integer charge HA must contain more hydrogens A⁻ is less basic than B⁻ HB cannot donate a proton at all Solution: A stronger acid more readily loses H⁺ because its conjugate base is more stable. A. Incorrect. This reverses the acid-conjugate-base relationship. B. Incorrect. Both conjugate bases can carry the same charge. C. Incorrect. Acid strength is not a hydrogen count. D. Correct. A stronger acid more readily loses H⁺ because its conjugate base is more stable. E. Incorrect. A weaker acid can still donate under suitable conditions. 10 / 19 Why does the chloro substituent generally increase acidity? CH₃CO₂H versus ClCH₂CO₂H Chlorine makes the conjugate base neutral Inductive electron withdrawal stabilizes the carboxylate The acid becomes an aromatic ring Chlorine breaks carboxylate resonance Chlorine supplies an additional acidic O-H bond Solution: The nearby electronegative substituent lowers electron density through sigma bonds. A. Incorrect. Deprotonation still gives charge -1. B. Correct. The nearby electronegative substituent lowers electron density through sigma bonds. C. Incorrect. No ring is present. D. Incorrect. The carboxylate pi system remains. E. Incorrect. Only one carboxyl O-H bond is present. 11 / 19 Which C-H bond is most acidic under the usual comparable-solvent trend? HC#CH, H₂C=CH₂, CH₃CH₃ All equally because every bond is C-H H₂C=CH₂ HC#CH None can ever lose H⁺ CH₃CH₃ Solution: Its conjugate base places charge in an sp orbital with greater s character. A. Incorrect. Hybridization affects acidity. B. Incorrect. sp2 carbon stabilizes charge less than sp. C. Correct. Its conjugate base places charge in an sp orbital with greater s character. D. Incorrect. Very strong bases can deprotonate suitable hydrocarbons. E. Incorrect. sp3 carbon has the least s character here. 12 / 19 Which species increases as pH drops below the carboxylic acid pKa? CH₃CO₂⁻ in water; strong aqueous acid added CH₄ CH₃CO₂(2-) CH₃CH₂OH CH₃CHO CH₃CO₂H Solution: Protonation converts the carboxylate to its neutral conjugate acid. A. Incorrect. Acidification does not remove the carboxyl carbon. B. Incorrect. Adding H⁺ does not make charge more negative. C. Incorrect. Acid-base workup does not reduce the carbonyl. D. Incorrect. No reduction reagent is present. E. Correct. Protonation converts the carboxylate to its neutral conjugate acid. 13 / 19 Which recovery strategy follows the charge change? CH₃CH₂NH₃⁺ in water; NaOH added to pH well above pKa Add NaCl to create a covalent C-Cl bond Extract neutral CH₃CH₂NH₂ into a suitable organic solvent Oxidize it to a nitrile Add more strong acid Remove the nitrogen atom Solution: Base removes the N proton, reducing the amine's ionic preference for water. A. Incorrect. Salt addition does not perform that substitution. B. Correct. Base removes the N proton, reducing the amine's ionic preference for water. C. Incorrect. No oxidation is needed to recover the amine. D. Incorrect. This retains the ammonium salt. E. Incorrect. The target is the same amine framework. 14 / 19 Which ordering reflects increasing stabilization from s character alone? C-centered anions with otherwise comparable substitution: sp, sp2, sp3 sp3 < sp2 < sp sp < sp2 < sp3 sp3 < sp < sp2 All equal because carbon has fixed electronegativity sp2 < sp < sp3 Solution: The s fractions increase from 25% to 33% to 50%, holding electrons closer to the nucleus. A. Correct. The s fractions increase from 25% to 33% to 50%, holding electrons closer to the nucleus. B. Incorrect. This reverses the s-character trend. C. Incorrect. sp has more s character than sp2. D. Incorrect. Hybridization changes orbital energy. E. Incorrect. sp3 is not most stabilized by this factor. 15 / 19 Why can CH₃SH be the stronger acid despite oxygen’s greater electronegativity? CH₃OH, CH₃SH; analogous conjugate bases Sulfur's larger, more polarizable valence shell stabilizes negative charge CH₃S⁻ has zero charge Methanol cannot hydrogen bond Sulfur has no valence electrons Sulfur must form an S=O bond during deprotonation Solution: Down a group, charge dispersion can outweigh the across-period electronegativity rule. A. Correct. Down a group, charge dispersion can outweigh the across-period electronegativity rule. B. Incorrect. It remains an anion. C. Incorrect. Methanol both donates and accepts hydrogen bonds. D. Incorrect. Sulfur has six valence electrons. E. Incorrect. Only S-H proton loss is required. 16 / 19 Which polarization is chemically reasonable? CH₃-CH₂-OH; consider C-O C delta- and O delta+ Integer C⁺ and O⁻ in every molecule H on carbon must carry charge -1 No polarization because the molecule is neutral C delta+ and O delta- Solution: Oxygen is more electronegative and attracts the bond electron density. A. Incorrect. The electronegativity order is reversed. B. Incorrect. Partial charges are not formal ionization. C. Incorrect. No hydride is present. D. Incorrect. Neutral molecules can contain polar bonds. E. Correct. Oxygen is more electronegative and attracts the bond electron density. 17 / 19 Which attraction must be included for both pure liquids? CH₃CH₂CH₃ and CH₃CH₂OH London dispersion Permanent ion-ion attraction Metallic bonding Covalent polymer crosslinks O-H hydrogen bonding Solution: Fluctuating electron distributions produce dispersion in polar and nonpolar molecules. A. Correct. Fluctuating electron distributions produce dispersion in polar and nonpolar molecules. B. Incorrect. Both compounds are neutral. C. Incorrect. Neither is a metal. D. Incorrect. Neither liquid requires a polymer network. E. Incorrect. The hydrocarbon has no O-H group. 18 / 19 Which hydrogen-bond interaction is available? CH₃-O-CH₃ mixed with H₂O Ether has no electron pairs available Ether O-H donates to water A C-H bond becomes an ionic O-H bond Water O-H donates to an ether O lone pair Every ether molecule permanently ionizes Solution: Ether oxygen is an acceptor, while water supplies the donor H. A. Incorrect. Oxygen has lone pairs. B. Incorrect. The ether has no O-H bond. C. Incorrect. Hydrogen bonding does not change covalent connectivity. D. Correct. Ether oxygen is an acceptor, while water supplies the donor H. E. Incorrect. Hydrogen bonding does not require ionization. 19 / 19 Which difference supports stronger self-association of the alcohol? CH₃CH₂OH and CH₃OCH₃; same molecular formula The ether has permanent ionic bonds Its oxygen has no lone pairs It supplies both H-bond donors and acceptors It has more carbon atoms It has a larger formula mass Solution: The ether accepts hydrogen bonds but lacks a conventional O-H donor in its pure liquid. A. Incorrect. The ether is neutral and covalent. B. Incorrect. Alcohol oxygen has lone pairs. C. Correct. The ether accepts hydrogen bonds but lacks a conventional O-H donor in its pure liquid. D. Incorrect. Both have two carbons. E. Incorrect. They are constitutional isomers with equal formula mass. Your score isThe average score is 0% 0% Restart quiz II. ACIDS, BASES & MOLECULAR PROPERTIES TEST 2 1 / 20 Which factor generally lowers the branched isomer's boiling point? CH₃CH₂CH₂CH₂CH₃ versus C(CH₃)₄ Reduced intermolecular contact area Loss of all dispersion forces Stronger hydrogen bonding in the straight chain Different carbon-carbon bond orders Fewer electrons in the branched isomer Solution: For comparable alkane isomers, compact branching can reduce dispersion contacts. A. Correct. For comparable alkane isomers, compact branching can reduce dispersion contacts. B. Incorrect. Dispersion remains present. C. Incorrect. Neither alkane has conventional hydrogen-bond groups. D. Incorrect. Both contain only single bonds. E. Incorrect. Both share a molecular formula and electron count. 2 / 20 Why does the boiling-point branching trend not by itself predict their melting points? Two C₅H₁₂ isomers; a compact symmetric crystal is observed Every branched alkane has the lowest melting point Melting is determined only by molecular mass Symmetry cannot affect a crystal Covalent C-C bonds must break on melting Crystal packing and symmetry also control the solid's stability Solution: Melting depends on the relative free energies of solid and liquid, including packing. A. Incorrect. This is not a universal rule. B. Incorrect. The isomers have the same mass. C. Incorrect. Symmetry can favor efficient packing. D. Incorrect. Ordinary melting changes intermolecular organization. E. Correct. Melting depends on the relative free energies of solid and liquid, including packing. 3 / 20 What structural change tends to reduce water miscibility? CH₃OH -> CH₃(CH₂)₅OH; one OH in each Loss of every possible hydrogen bond A larger nonpolar portion per OH group Formation of a permanent negative ion A decrease in carbon number Removal of oxygen Solution: The hydrophobic chain grows without adding a compensating polar group. A. Incorrect. The OH group still hydrogen bonds. B. Correct. The hydrophobic chain grows without adding a compensating polar group. C. Incorrect. Both are neutral alcohols as written. D. Incorrect. Carbon number increases. E. Incorrect. Both molecules retain oxygen. 4 / 20 Why is this N a poor hydrogen-bond acceptor compared with an ordinary amine N? CH₃CONH₂; consider the N lone pair Its lone pair is delocalized into C=O An N-H bond can never donate a hydrogen bond Nitrogen has no electrons The whole molecule cannot hydrogen bond The nitrogen is an alkane carbon Solution: Amide conjugation reduces availability of the nitrogen lone pair. A. Correct. Amide conjugation reduces availability of the nitrogen lone pair. B. Incorrect. Amide N-H can be a donor. C. Incorrect. It still has valence electrons. D. Incorrect. The carbonyl oxygen remains an acceptor. E. Incorrect. The atom identity remains nitrogen. 5 / 20 Which side is favored after proton transfer? CH₃CO₂H + CH₃O⁻; pKa values 4.8 and 15.5 CH₃CO₂⁻ + CH₃OH A 1:1 equilibrium regardless of pKa CH₃COCH₃ + HO⁻ Unchanged reactants because both contain oxygen CH₄ + CO₂ only Solution: The product acid has the higher pKa, so K is approximately 10^10.7. A. Correct. The product acid has the higher pKa, so K is approximately 10^10.7. B. Incorrect. pKa differences set equilibrium bias. C. Incorrect. This invents a carbon-carbon coupling. D. Incorrect. The pKa gap strongly favors products. E. Incorrect. No decarboxylation conditions are supplied. 6 / 20 Which atom of methanol accepts H⁺ in the ordinary first acid-base step? CH₃OH + HCl -> protonated methanol + Cl⁻ Oxygen The chlorine already in methanol A methyl hydrogen Carbon No atom because neutral molecules cannot be bases Solution: The oxygen lone pair forms O-H, giving CH₃OH₂⁺. A. Correct. The oxygen lone pair forms O-H, giving CH₃OH₂⁺. B. Incorrect. Methanol contains no chlorine. C. Incorrect. That does not use methanol's basic lone pair. D. Incorrect. Carbon has no lone pair in neutral methanol. E. Incorrect. Neutral lone-pair donors can accept protons. 7 / 20 Which operation restores the original neutral functional group? Aqueous CH₃CH₂CO₂⁻; neutral acid needed for isolation Add NaBH₄ Heat with excess oxygen Add NaOH Add ethyl bromide as the only operation Add aqueous acid until pH is well below its pKa Solution: Protonation restores CO₂H without changing the carbon skeleton. A. Incorrect. This is not the necessary proton-transfer workup. B. Incorrect. Combustion destroys the target framework. C. Incorrect. This favors the carboxylate. D. Incorrect. That risks ester alkylation rather than neutral acid recovery. E. Correct. Protonation restores CO₂H without changing the carbon skeleton. 8 / 20 Which dominant net-charge state follows? H₃N⁺-CH₂-CO₂⁻; pKa values 2.3 and 9.6; pH 6 -1 +1 -2 0 +2 Solution: CO₂H is mostly deprotonated and NH₃⁺ mostly protonated, giving a zwitterion. A. Incorrect. That would require mostly neutral NH₂ at pH 6. B. Incorrect. That would require mostly neutral CO₂H at pH 6. C. Incorrect. The amine does not become a dianion. D. Correct. CO₂H is mostly deprotonated and NH₃⁺ mostly protonated, giving a zwitterion. E. Incorrect. There are not two positively ionized groups here. 9 / 20 What is the dominant net charge after equilibration? H₂N-CH₂-CO₂H; pKa values 2.3 and 9.6; pH 1 +2 +1 -1 -2 0 Solution: The carboxyl group is mostly CO₂H and the amine mostly NH₃⁺. A. Incorrect. CO₂H is neutral, not positively charged. B. Correct. The carboxyl group is mostly CO₂H and the amine mostly NH₃⁺. C. Incorrect. The carboxylate is not dominant below pKa 2.3. D. Incorrect. Acidic water does not produce two negative groups. E. Incorrect. This overlooks amine protonation in strong acid. 10 / 20 Which form dominates? H₃N⁺-CH(CH₃)-CO₂⁻; pKa values 2.4 and 9.7; pH 12 HN⁻-CH(CH₃)-CO₂⁻ H₂N-CH(CH₃)-CO₂H H₃N⁺-CH(CH₃)-CO₂H H₂N-CH(CH₃)-CO₂⁻ H₃N⁺-CH(CH₃)-CO₂⁻ Solution: Both relevant proton-loss equilibria favor the deprotonated states at pH 12. A. Incorrect. Ordinary aqueous pH 12 does not remove a proton from neutral NH₂. B. Incorrect. The carboxyl group is not mostly neutral at pH 12. C. Incorrect. This is the low-pH form. D. Correct. Both relevant proton-loss equilibria favor the deprotonated states at pH 12. E. Incorrect. The ammonium group is mostly deprotonated. 11 / 20 Which average-charge tendency is expected? Amino-acid sample: pI 6.0; buffer pH 8.5 Exactly zero at every pH Neutral because all amino acids are always zwitterions Negative Permanent +2 independent of structure Positive Solution: Above pI, the population is on average more deprotonated and carries net negative charge. A. Incorrect. Zero average charge occurs near pI, not all pH values. B. Incorrect. Zwitterions are not the only protonation forms. C. Correct. Above pI, the population is on average more deprotonated and carries net negative charge. D. Incorrect. Protonation depends on ionizable groups and pH. E. Incorrect. This is the tendency below pI. 12 / 20 Which quantitative implication follows if activities are used? HA + B⁻ <-> A⁻ + HB; K=10^3 The forward activation barrier is exactly 3 kJ/mol Every initial mixture is 99.9% converted The forward rate is 1000 times the reverse rate at equilibrium pKa(HB)-pKa(HA) is approximately +3 The pKa difference is -3 Solution: Taking log10 of the proton-transfer K gives the compatible pKa difference. A. Incorrect. Equilibrium does not directly set an activation barrier. B. Incorrect. Conversion also depends on starting amounts and equilibrium mass balances. C. Incorrect. Forward and reverse rates are equal at equilibrium. D. Correct. Taking log10 of the proton-transfer K gives the compatible pKa difference. E. Incorrect. The sign is reversed. 13 / 20 Why is a direct one-unit strength comparison unsupported? Acid X: pKa 10 in DMSO; acid Y: pKa 11 in water DMSO has no molecules Solvation and reference scales differ Larger pKa always means stronger acid pKa values never measure equilibrium Water makes every acid equally strong in all ranges Solution: Acidity constants are medium-dependent; compatible measurements are needed for a close ranking. A. Incorrect. It is a molecular solvent. B. Correct. Acidity constants are medium-dependent; compatible measurements are needed for a close ranking. C. Incorrect. The relation is reversed on a common scale. D. Incorrect. They describe acid dissociation equilibrium. E. Incorrect. Leveling has limits and does not justify this comparison. 14 / 20 Which conjugate-base feature explains enhanced acidity of the nitro derivative? p-NO₂-C₆H₄-OH versus C₆H₅-OH It removes aromatic conjugation from the ring Electron withdrawal stabilizes the phenoxide anion It requires deprotonation at a ring carbon It adds a second phenolic OH It makes oxygen less electronegative as an element Solution: The para nitro group supports resonance and inductive stabilization of the conjugate base. A. Incorrect. The aromatic framework is retained. B. Correct. The para nitro group supports resonance and inductive stabilization of the conjugate base. C. Incorrect. The comparison concerns phenolic O-H loss. D. Incorrect. No second OH is shown. E. Incorrect. Substitution does not change elemental identity. 15 / 20 Which carbon product and coefficient balance the three carbon atoms? CH₃CH₂CH₃ + excess O₂; complete combustion 1 C₃H₆ 3 CO 3 CH₃OH 3 CO₂ 1 CO₂ Solution: Complete combustion converts each substrate carbon to CO₂. A. Incorrect. That is dehydrogenation, not complete combustion. B. Incorrect. CO is an incomplete-combustion product. C. Incorrect. Complete combustion does not stop at methanol. D. Correct. Complete combustion converts each substrate carbon to CO₂. E. Incorrect. This loses two carbons. 16 / 20 What can dominate their different aqueous behavior? C₆H₅CO₂H versus C₆H₅CO₂⁻Na⁺; same temperature and water volume The neutral acid must be more ionic A change in carbon count Ionic hydration of the salt Formation of a covalent Na-C bond Removal of both oxygens Solution: Ionization can strongly change water affinity even when the carbon framework is unchanged. A. Incorrect. The salt is the ionic form. B. Incorrect. Both organic species have seven carbons. C. Correct. Ionization can strongly change water affinity even when the carbon framework is unchanged. D. Incorrect. The salt is not an organosodium compound. E. Incorrect. The carboxylate retains both oxygens. 17 / 20 Which trend is consistent with electronegativity, other factors held comparable? CH₃CH₂⁻, CH₃NH⁻, CH₃O⁻; compare second-row charge-bearing atoms Carbon always stabilizes charge best Oxygen cannot carry a negative formal charge Negative charge is better stabilized toward oxygen Atomic size increases strongly from C to O All three anions have identical basicity Solution: Across C, N, O, increasing electronegativity generally stabilizes anionic charge. A. Incorrect. This reverses the across-row trend. B. Incorrect. Alkoxides are ordinary anions. C. Correct. Across C, N, O, increasing electronegativity generally stabilizes anionic charge. D. Incorrect. Across the period, size generally decreases. E. Incorrect. The charge-bearing atom matters. 18 / 20 Which interaction uses the carbonyl oxygen without changing covalent structure? CH₃-C(=O)-CH₃ in water It becomes a permanent hydronium ion It donates an O-H hydrogen bond O accepts a hydrogen bond from H₂O Carbon accepts through an oxygen lone pair located on carbon It must cleave a C-C bond Solution: A carbonyl lone pair can accept a water O-H hydrogen bond. A. Incorrect. Hydrogen bonding does not require full proton transfer. B. Incorrect. The ketone has no O-H bond. C. Correct. A carbonyl lone pair can accept a water O-H hydrogen bond. D. Incorrect. The relevant lone pairs are on oxygen. E. Incorrect. Hydrogen bonding preserves connectivity. 19 / 20 After removing the hydrocarbon layer, what sequence recovers free amine? C₆H₅CH₂NH₃⁺Cl⁻ in water, mixed with neutral hydrocarbon Add an acyl chloride to neutralize N Basify above pKa, then extract with a suitable organic solvent Acidify further, then expect all salt to enter hexane Evaporate water and call the ammonium salt free amine Reduce the aromatic ring first Solution: Deprotonation yields neutral C₆H₅CH₂NH₂, which can partition into the organic phase. A. Incorrect. That changes the amine into an amide. B. Correct. Deprotonation yields neutral C₆H₅CH₂NH₂, which can partition into the organic phase. C. Incorrect. The charged salt remains water-favored. D. Incorrect. Drying alone does not remove the extra proton. E. Incorrect. This changes the target and is unnecessary. 20 / 20 Which pair of pKa values belongs in log K=pKa(product acid)-pKa(reactant acid)? NH₄⁺ + CH₃O⁻ <-> NH₃ + CH₃OH CH₃O⁻ and CH₃OH H₂O and H₃O⁺ NH₄⁺ and NH₃ NH₃ and CH₃O⁻ CH₃OH and NH₄⁺ Solution: These are the acids on the product and reactant sides, respectively. A. Incorrect. This uses the methanol pair only. B. Incorrect. Neither is a stated acid in this reaction equation. C. Incorrect. This mixes one conjugate pair instead of the two acids. D. Incorrect. These are the conjugate bases in the stated transfer. E. Correct. These are the acids on the product and reactant sides, respectively. Your score isThe average score is 0% 0% Restart quiz GO TO THE NEXT CHAPTER
