I. STRUCTURE, BONDING, & RESONANCE TEST 1 1 / 19 Relative to neutral ¹⁶O in the same bond pattern, what changes? ¹⁸O in CH₃-¹⁸O-H: Z=8 Two additional protons Two additional O-H bonds A formal charge of +2 Two additional neutrons Two additional valence electrons Solution: A stays 18 rather than 16 because neutron count rises from 8 to 10; bonding electrons are unchanged. A. Incorrect. Changing proton count would change the element. B. Incorrect. Isotope substitution preserves connectivity. C. Incorrect. Nuclear mass does not specify ionic charge. D. Correct. A stays 18 rather than 16 because neutron count rises from 8 to 10; bonding electrons are unchanged. E. Incorrect. An isotope label does not add valence electrons. 2 / 19 A mass spectrometer distinguishes these molecules. Which pair of atomic properties explains the difference? ¹³CH₃-CH₂-OH and ¹²CH₃-CH₂-OH Different valence, same mass Different element, same electron count Same mass number, different charge Same Z, different neutron number Different Z, same neutron number Solution: Both labeled atoms are carbon with Z=6; ¹³C has seven neutrons and ¹²C has six. A. Incorrect. Carbon valence is unchanged by isotope substitution. B. Incorrect. Isotopes belong to the same element. C. Incorrect. The superscripts specify different masses. D. Correct. Both labeled atoms are carbon with Z=6; ¹³C has seven neutrons and ¹²C has six. E. Incorrect. Both atoms remain carbon. 3 / 19 A proposed carbon 2p orbital has no nodal plane through the nucleus. What is the defect? Isolated carbon: 2s and 2p orbitals Its lobes must carry opposite electric charges It needs three radial nodes All its points must have equal probability It must be spherical A p orbital requires an angular nodal plane Solution: The two p lobes have opposite phase separated by a plane through the nucleus. A. Incorrect. Phase is not electric charge. B. Incorrect. A 2p orbital has no radial node. C. Incorrect. Probability density varies spatially. D. Incorrect. Spherical angular symmetry belongs to s orbitals. E. Correct. The two p lobes have opposite phase separated by a plane through the nucleus. 4 / 19 A drawing rotates 2px into the orientation formerly labeled 2py. Which property remains the same? Isolated atom: 2px, 2py, 2pz; no external field Orbital energy The plane named yz as its node Cartesian orientation The axis named x The direction of its two lobes Solution: The three p orbitals are degenerate in the stated isolated-atom model; orientation changes. A. Correct. The three p orbitals are degenerate in the stated isolated-atom model; orientation changes. B. Incorrect. A y-oriented p orbital has an xz nodal plane. C. Incorrect. That is precisely what the rotation changes. D. Incorrect. The new orientation was defined as y. E. Incorrect. The lobes rotate with the orbital. 5 / 19 Which electron count is used for oxygen before allocating bonds in a Lewis structure? CH₃-O-H; neutral O: 1s2 2s2 2p4 4 8 7 6 2 Solution: The outer n=2 shell contains 2+4 electrons; the 1s core does not enter the valence count. A. Incorrect. Four counts only the 2p electrons. B. Incorrect. Eight includes the two core electrons. C. Incorrect. That is the neutral halogen valence count. D. Correct. The outer n=2 shell contains 2+4 electrons; the 1s core does not enter the valence count. E. Incorrect. Two counts only 2s electrons. 6 / 19 Which occupancy minimizes energy under the usual Hund model? Three degenerate 2p orbitals; two electrons Two electrons in different orbitals, opposite spins One electron shared fractionally across all orbitals as the Lewis occupancy A pair in one orbital, others empty One electron in each of two orbitals, parallel spins Both electrons in one orbital with parallel spins Solution: Separate occupancy reduces pairing before every degenerate orbital is singly occupied. A. Incorrect. This misses the parallel-spin preference. B. Incorrect. This is not the requested integer orbital-box configuration. C. Incorrect. This pairs prematurely. D. Correct. Separate occupancy reduces pairing before every degenerate orbital is singly occupied. E. Incorrect. Pauli forbids identical spins in one orbital. 7 / 19 What spin is allowed for a second electron occupying that same orbital? One spatial orbital: electron 1 has ms=+1/2 +1/2 0 Either sign with no restriction +1 -1/2 Solution: Pauli exclusion requires the second electron to have the opposite spin quantum number. A. Incorrect. This duplicates all four quantum numbers. B. Incorrect. Electron spin projection is not zero. C. Incorrect. Same-orbital occupancy imposes a restriction. D. Incorrect. This is not an allowed electron spin projection. E. Correct. Pauli exclusion requires the second electron to have the opposite spin quantum number. 8 / 19 A student assigns zero formal charge to O. Which bookkeeping correction is needed? CH₃-O⁻; O has three lone pairs Count six nonbonding electrons and one assigned bonding electron Ignore one lone pair Treat the O-C bond as a double bond Assign both bond electrons to O for formal charge Use oxygen atomic number as its valence count Solution: FC(O)=6-6-1=-1. A. Correct. FC(O)=6-6-1=-1. B. Incorrect. All three lone pairs contribute electrons. C. Incorrect. The displayed connectivity specifies one bond. D. Incorrect. Formal-charge bookkeeping splits the bond equally. E. Incorrect. The two core electrons are excluded. 9 / 19 Which accounting supports the displayed charge? (CH₃)₄N⁺; four N-C bonds, no N lone pair 7-0-4=+3 5-0-4=+1 4-0-4=0 5-0-8=-3 5-2-4=-1 Solution: Nitrogen contributes five valence electrons and is assigned one electron from each of four bonds. A. Incorrect. Atomic number was substituted for valence electrons. B. Correct. Nitrogen contributes five valence electrons and is assigned one electron from each of four bonds. C. Incorrect. Four is not nitrogen's neutral valence-electron count. D. Incorrect. Bonding electrons were not divided equally. E. Incorrect. A lone pair was invented. 10 / 19 Which electron-bookkeeping move could complete the central carbon's octet without adding an atom to the species, even if energetically unfavorable? (CH₃)₃C⁺; three C-C bonds Assign two lone pairs while keeping all three bonds Remove another bonding pair Shift an adjacent C-H bond pair and its H to the cationic carbon Replace the plus sign with zero without changing electrons Add a proton as a lone pair Solution: The migrating hydride forms a fourth bond at the original cation; positive charge moves to the donor carbon. A. Incorrect. That would surround carbon with ten electrons. B. Incorrect. This worsens electron deficiency. C. Correct. The migrating hydride forms a fourth bond at the original cation; positive charge moves to the donor carbon. D. Incorrect. A label cannot repair electron accounting. E. Incorrect. A proton supplies no electrons. 11 / 19 An additional H⁺ protonates the carboxylate. What is the product's net charge? H₃N⁺-CH₂-C(=O)-O⁻ +1 -2 -1 +2 0 Solution: The initial zwitterion is neutral; removing the O negative charge leaves the ammonium positive charge. A. Correct. The initial zwitterion is neutral; removing the O negative charge leaves the ammonium positive charge. B. Incorrect. Both the direction and magnitude are incorrect. C. Incorrect. Protonation does not make the species more negative. D. Incorrect. Only one net positive charge is added. E. Incorrect. This ignores the newly added positive charge. 12 / 19 Why is the second carbon invalid as drawn? Proposed CH₃-C(H)(H)(H)-OH with all drawn bonds single Proposed drawing: all bonds shown are single Oxygen must always have a positive charge Every saturated carbon must have a lone pair It has five sigma bonds and ten surrounding electrons Carbon cannot bond to oxygen It has only six surrounding electrons Solution: It is bonded to CH₃, three H atoms and O; ordinary second-row carbon cannot expand its octet. A. Incorrect. An O-H and an O-C bond give neutral oxygen. B. Incorrect. Neutral alkane carbons have none. C. Correct. It is bonded to CH₃, three H atoms and O; ordinary second-row carbon cannot expand its octet. D. Incorrect. Ordinary carbon-oxygen bonds are allowed. E. Incorrect. Five single bonds supply ten. 13 / 19 How many sigma and pi bonds join the three heavy-atom pairs shown? Three heavy-atom bonds 3 sigma and 1 pi 3 sigma and 2 pi 4 sigma and 2 pi 5 sigma and 0 pi 2 sigma and 3 pi Solution: Each connected pair contributes one sigma bond; both double bonds contribute one additional pi bond. A. Incorrect. This overlooks the carbonyl pi bond. B. Correct. Each connected pair contributes one sigma bond; both double bonds contribute one additional pi bond. C. Incorrect. Only three heavy-atom pairs are joined. D. Incorrect. A double bond is not two sigma bonds. E. Incorrect. The single C-C bond is sigma. 14 / 19 How does the number of electrons in carbon-carbon pi bonds change? HC#CH -> H₂C=CH₂, one H₂ added One equivalent of hydrogen added 4 to 2 6 to 4 2 to 4 4 to 0 2 to 0 Solution: The triple bond initially has two pi bonds; the alkene retains one. A. Correct. The triple bond initially has two pi bonds; the alkene retains one. B. Incorrect. This counts sigma electrons as pi electrons. C. Incorrect. Hydrogenation removes unsaturation. D. Incorrect. One pi bond remains after adding one H₂. E. Incorrect. This treats the starting triple bond as a double bond. 15 / 19 Which motion most directly destroys their pi overlap? Two parallel p orbitals perpendicular to a C-C axis Two parallel p orbitals on adjacent carbon atoms Reverse both orbital phase signs together Translate the whole molecule Rotate one p orbital 90 degrees about the C-C axis Rotate the whole molecule as one rigid body Bring the carbons slightly closer without changing orientation Solution: The orbitals become orthogonal and lose the required side-on alignment. A. Incorrect. A common phase reversal changes no physical overlap. B. Incorrect. Internal overlap is unchanged. C. Correct. The orbitals become orthogonal and lose the required side-on alignment. D. Incorrect. Relative orbital orientations are unchanged. E. Incorrect. This does not inherently eliminate alignment. 16 / 19 Which local geometry is expected at the carbonyl carbon? Trigonal pyramidal due to a carbon lone pair Octahedral because oxygen has lone pairs Tetrahedral, with four domains because a double bond counts twice Linear, with two domains Approximately planar, with three electron domains Solution: The C=O double bond counts as one domain, alongside two single bonds. A. Incorrect. This carbon has no lone pair. B. Incorrect. Those lone pairs are not domains centered on carbon. C. Incorrect. Multiple bonds count as one VSEPR domain. D. Incorrect. Three atoms are directly attached. E. Correct. The C=O double bond counts as one domain, alongside two single bonds. 17 / 19 What is the approximate C-C-C angle at the middle carbon? 120 degrees 60 degrees 90 degrees 109.5 degrees 180 degrees Solution: The central allene carbon has two sigma domains and sp hybridization. A. Incorrect. That fits three planar sigma domains. B. Incorrect. No three-membered ring is present. C. Incorrect. The orthogonal terminal pi planes do not set a 90-degree backbone angle. D. Incorrect. That is the ideal tetrahedral angle. E. Correct. The central allene carbon has two sigma domains and sp hybridization. 18 / 19 After constructing its sp hybrids, how many p orbitals remain for pi bonding? H-C#N; consider the carbon atom 2 3 0 4 1 Solution: One s and one p orbital form two sp hybrids, leaving two p orbitals. A. Correct. One s and one p orbital form two sp hybrids, leaving two p orbitals. B. Incorrect. One p orbital was used in hybridization. C. Incorrect. All p orbitals are not hybridized. D. Incorrect. Carbon has only three p orbitals in its valence shell. E. Incorrect. That is the sp2 remainder. 19 / 19 How many carbon atoms are sp hybridized? 2 3 1 4 5 Solution: Both carbons in the triple bond are sp; the remaining saturated carbons are sp3. A. Correct. Both carbons in the triple bond are sp; the remaining saturated carbons are sp3. B. Incorrect. An adjacent saturated carbon is not sp. C. Incorrect. Both ends of the triple bond must be counted. D. Incorrect. There is only one triple bond. E. Incorrect. The whole chain does not share one hybridization. Your score isThe average score is 0% 0% Restart quiz I. STRUCTURE, BONDING, & RESONANCE TEST 2 1 / 20 Why is the C-O-C bond angle associated with a bent molecular arrangement at O? CH₃-O-CH₃; two O bonds and two O lone pairs Oxygen is linear because it has two bonds Oxygen has only two valence electrons A double bond joins the methyl groups Only the two bonded atoms define molecular shape; lone pairs occupy the other domains Four domains require four bonded atoms Solution: Four domains give an approximately tetrahedral electron-domain arrangement, but only two are bonds. A. Incorrect. The two lone pairs prevent the two-domain linear model. B. Incorrect. Neutral oxygen has six. C. Incorrect. No such bond is shown. D. Correct. Four domains give an approximately tetrahedral electron-domain arrangement, but only two are bonds. E. Incorrect. Electron domains include lone pairs. 2 / 20 Which explanation predicts the C-C length trend? CH₃-CH₃; CH₂=CH₂; HC#CH A pi bond replaces the sigma bond entirely All three have equal lengths because both atoms are carbon Higher bond order generally gives shorter C-C bonds Higher bond order always gives longer bonds Only total molecular mass controls bond length Solution: More bonding electron density strengthens and shortens comparable carbon-carbon bonds. A. Incorrect. The sigma component remains. B. Incorrect. Bond order distinguishes them. C. Correct. More bonding electron density strengthens and shortens comparable carbon-carbon bonds. D. Incorrect. The trend is reversed. E. Incorrect. Local bonding dominates this comparison. 3 / 20 In a direct thermal rotation model, which energetic cost is central? CH₃-CH=CH-CH₃, E form -> Z form Rotation about the central double bond Ionizing every C-H bond Pairing two previously unpaired alkene electrons Removing both methyl groups Breaking the C-C sigma bond completely Loss of pi overlap during twisting Solution: The p orbitals must lose alignment before the geometric configuration changes. A. Incorrect. That is unrelated to double-bond rotation. B. Incorrect. The ground-state alkene already has paired bonding electrons. C. Incorrect. The carbon framework is retained. D. Incorrect. Geometric isomerization need not cleave the sigma bond. E. Correct. The p orbitals must lose alignment before the geometric configuration changes. 4 / 20 Where should the arrow for formation of the O-H bond begin? CH₃-O⁻ + H-Br O-H bond formation: reactants At bromine toward carbon At the minus sign as a physical electron orbital At carbon because every arrow starts on carbon At an oxygen lone pair At H⁺ because protons supply electrons Solution: A full curved arrow traces an electron pair from the base to the electrophilic proton. A. Incorrect. That does not form the requested O-H bond. B. Incorrect. A charge symbol is not an electron-pair location. C. Incorrect. The reacting pair is on oxygen. D. Correct. A full curved arrow traces an electron pair from the base to the electrophilic proton. E. Incorrect. The proton accepts an electron pair. 5 / 20 Which electron allocation describes the bond cleavage? Br-Br –light–> 2 Br radicals Homolytic cleavage Both electrons go to one Br One bond electron goes to each Br Both electrons are destroyed Each Br gains two electrons A proton moves from one Br to the other Solution: Homolysis divides the bond pair equally and gives two radicals. A. Incorrect. That is heterolysis. B. Correct. Homolysis divides the bond pair equally and gives two radicals. C. Incorrect. Electrons are conserved. D. Incorrect. Only two bonding electrons are available. E. Incorrect. No hydrogen is present. 6 / 20 What is conserved in this heterolytic bookkeeping step? CH₃-Br -> CH₃⁺ + Br⁻, idealized cleavage Idealized heterolytic cleavage Total charge and total electron number The charge on each fragment An octet at the methyl carbon The number of covalent bonds One electron on each fragment from the broken bond Solution: A neutral starting species gives oppositely charged fragments without creating electrons. A. Correct. A neutral starting species gives oppositely charged fragments without creating electrons. B. Incorrect. The fragments acquire opposite charges. C. Incorrect. The carbocation has only six electrons. D. Incorrect. One covalent bond is broken. E. Incorrect. Both bond electrons go to bromine. 7 / 20 What remains fixed between these resonance contributors? CH₃-C(=O)-O⁻ <-> CH₃-C(-O⁻)=O Location of each pi bond Formal charge on each oxygen Every electron's assignment to one atom Nuclear positions and sigma connectivity Which oxygen has the double bond Solution: Only electron placement and pi bond order change; neither atom migration nor proton transfer occurs. A. Incorrect. The pi bond changes location. B. Incorrect. The negative charge alternates between oxygens. C. Incorrect. Electron distribution changes. D. Correct. Only electron placement and pi bond order change; neither atom migration nor proton transfer occurs. E. Incorrect. That is the contributor difference. 8 / 20 A prediction assigns two distinct C-O bond lengths to the actual isolated ion solely from one contributor. What is the correction? Two equivalent resonance drawings of CH₃CO₂⁻ One drawing is the only real molecule The negative charge disappears by resonance The methyl group migrates between the oxygens The ion switches between two isolable constitutional isomers Both C-O bonds are equivalent in the resonance hybrid Solution: Equivalent contributors describe one delocalized ion rather than separate rapidly switching structures. A. Incorrect. No single equivalent contributor is the full description. B. Incorrect. The total charge stays -1. C. Incorrect. Nuclei do not move in resonance. D. Incorrect. Resonance contributors are not distinct isomers. E. Correct. Equivalent contributors describe one delocalized ion rather than separate rapidly switching structures. 9 / 20 With both contributors satisfying octets, why is the O-negative contributor often weighted more? Enolate contributors: CH₂=C(O⁻)-CH₃ and (-)CH₂-C(=O)-CH₃ Carbon is more electronegative than oxygen Oxygen better accommodates negative charge The O-negative contributor has fewer atoms It has a different total charge The contributors are different tautomers Solution: Electronegativity favors charge on oxygen when octet and charge-separation comparisons are otherwise similar. A. Incorrect. This reverses the electronegativity order. B. Correct. Electronegativity favors charge on oxygen when octet and charge-separation comparisons are otherwise similar. C. Incorrect. Both have identical nuclei. D. Incorrect. Both total -1. E. Incorrect. No proton has moved between them. 10 / 20 Which structural feature explains greater charge delocalization in the first ion? CH₃-C(=O)-O⁻ and CH₃-CH₂-O⁻ The carboxylate contains an O-O bond The O lone pair is adjacent to a carbonyl pi system The alkoxide has an aromatic ring Both ions have identical resonance because both contain O The carboxylate has no formal charge Solution: Carboxylate has two equivalent O-centered contributors; the saturated alkoxide lacks that adjacent pi system. A. Incorrect. Its oxygens both attach to carbon. B. Correct. Carboxylate has two equivalent O-centered contributors; the saturated alkoxide lacks that adjacent pi system. C. Incorrect. No ring is present. D. Incorrect. Oxygen alone does not guarantee conjugation. E. Incorrect. Its net charge is -1. 11 / 20 Which proposed observation follows from N-lone-pair conjugation? Free C-N rotation identical to an alkane Independent aromaticity of the three-atom C-O-N chain A permanently tetrahedral carbonyl carbon A full positive net charge on the neutral amide Restricted rotation about C-N Solution: Donation gives C-N partial double-bond character, favoring approximate planarity. A. Incorrect. This neglects conjugation. B. Incorrect. An acyclic chain is not an aromatic loop. C. Incorrect. The carbonyl carbon is planar. D. Incorrect. Charge-separated contributors do not change net charge. E. Correct. Donation gives C-N partial double-bond character, favoring approximate planarity. 12 / 20 Under the ordinary nonconjugated carbocation model, what stabilizes the first cation? (CH₃)₃C⁺ compared with CH₃CH₂⁺ A lone pair on its positive carbon A complete central carbon octet in the first structure Resonance into a carbonyl oxygen A negative net charge More adjacent alkyl sigma bonds can donate by hyperconjugation Solution: The tertiary center has greater alkyl substitution than the primary center. A. Incorrect. No such lone pair is present. B. Incorrect. Both carbocations remain electron deficient. C. Incorrect. No carbonyl exists. D. Incorrect. Both species have charge +1. E. Correct. The tertiary center has greater alkyl substitution than the primary center. 13 / 20 Which contributor is obtained by moving only the pi pair toward the cationic end? CH₂=CH-CH₂⁺ CH₂=CH-CH₂ radical (+)CH₂-CH=CH₂ CH₃-CH=CH(+ ) CH₃-C#CH CH₂=CH-CH₂⁻ Solution: Moving the pi pair creates the adjacent double bond and relocates positive charge to the other terminal carbon. A. Incorrect. This replaces the cation with a neutral radical and changes electron count. B. Correct. Moving the pi pair creates the adjacent double bond and relocates positive charge to the other terminal carbon. C. Incorrect. This changes hydrogen connectivity. D. Incorrect. That changes hydrogen positions and charge. E. Incorrect. This changes total charge. 14 / 20 A proposed hydride shift moves H from the adjacent CH carbon. Which test supports the proposal? CH₃-CH⁺-CH(CH₃)-CH₃ Any hydrogen anywhere in the molecule can shift in one 1,2 step It must create a primary cation It converts a secondary cation into a tertiary cation without changing carbon count The shift adds one carbon from solvent A hydrogen atom leaves with no electrons Solution: The migrating C-H pair forms a bond to the cation while the donor carbon becomes positive. A. Incorrect. The donor must be adjacent. B. Incorrect. That is not the stability improvement here. C. Correct. The migrating C-H pair forms a bond to the cation while the donor carbon becomes positive. D. Incorrect. No carbon enters this rearrangement. E. Incorrect. A hydride shift transfers the C-H bonding pair. 15 / 20 Why is direct benzylic conjugation available only to the first radical? C₆H₅-CH₂ radical and C₆H₅-CH₂-CH₂ radical The first radical has a different benzene formula The second has no C-H bonds Its radical p orbital is adjacent to the aromatic pi system Both radicals are identical by resonance Benzene must lose all sigma bonds to stabilize a radical Solution: The extra saturated CH₂ in the second structure interrupts direct p-orbital overlap. A. Incorrect. Both contain the same phenyl unit. B. Incorrect. It has several. C. Correct. The extra saturated CH₂ in the second structure interrupts direct p-orbital overlap. D. Incorrect. Moving the radical across a saturated bond is not resonance. E. Incorrect. Delocalization uses the pi framework. 16 / 20 What fraction of the modeled monohalogenation events occurs at type b? Model substrate: 9 type-a H and 1 type-b H; per-H rates 1 and 9 a: 9 H total (three methyl groups); b: 1 H. Per-H rates: a = 1, b = 9. 10% 1% 50% 90% 100% Solution: Weighted contributions are 9x1 and 1x9, so b contributes 9/(9+9). A. Incorrect. This uses hydrogen counts without reactivity. B. Incorrect. Neither weighted contribution gives this fraction. C. Correct. Weighted contributions are 9x1 and 1x9, so b contributes 9/(9+9). D. Incorrect. This uses per-H reactivity without degeneracy. E. Incorrect. Type-a abstraction has nonzero modeled reactivity. 17 / 20 Which two electron-pair moves avoid exceeding carbon's octet? CH₃-C(=O)-CH₃ + H⁻, first addition step First nucleophilic addition step O lone pair to H while H pair goes to methyl C H pair to carbonyl C; C=O pi pair to O H pair to C with no other arrow H pair to O; C=O pi pair to C C-O sigma pair to methyl carbon Solution: The incoming bond is accompanied by conversion of C=O to C-O⁻. A. Incorrect. This gives the wrong reacting site and electron flow. B. Correct. The incoming bond is accompanied by conversion of C=O to C-O⁻. C. Incorrect. That would overfill carbon's octet. D. Incorrect. This does not describe hydride addition to the electrophilic carbon. E. Incorrect. The carbonyl sigma bond is retained. 18 / 20 Why can these drawings not be resonance contributors? CH₃-C(=O)-CH₃ and CH₂=C(OH)-CH₃ A hydrogen changes its bonded atom Both must be geometric isomers Resonance never changes pi bonds Their carbon counts differ Their molecular formulas differ Solution: Keto-enol tautomerization moves a proton and changes sigma connectivity. A. Correct. Keto-enol tautomerization moves a proton and changes sigma connectivity. B. Incorrect. They differ in atom connectivity. C. Incorrect. Pi-bond relocation is allowed in resonance. D. Incorrect. Both have three carbons. E. Incorrect. Both are C₃H₆O. 19 / 20 What changes when a base removes only the O-H proton? CH₃-O⁺(H)-CH₂CH₃; O has one lone pair O-H proton removal Oxygen gains a fourth bond The methyl carbon becomes a carbocation Oxygen becomes neutral while retaining two C-O bonds Oxygen becomes -2 Both carbon chains detach Solution: O goes from three bonds and one lone pair to two bonds and two lone pairs. A. Incorrect. The O-H bond is removed. B. Incorrect. The O-H bond pair stays on oxygen. C. Correct. O goes from three bonds and one lone pair to two bonds and two lone pairs. D. Incorrect. One proton removal changes net charge by only -1. E. Incorrect. Deprotonation need not cleave C-O bonds. 20 / 20 From left to right, what are the hybridizations of the three carbon atoms? CH₃-C(=O)-C#N Read the three carbon atoms from left to right sp3, sp3, sp2 sp2, sp2, sp2 sp, sp2, sp3 sp3, sp2, sp sp3, sp, sp2 Solution: The methyl has four sigma domains, the carbonyl has three, and the nitrile carbon has two. A. Incorrect. The multiple-bond centers have been undercounted. B. Incorrect. The terminal methyl and nitrile are not both trigonal. C. Incorrect. The end assignments are reversed. D. Correct. The methyl has four sigma domains, the carbonyl has three, and the nitrile carbon has two. E. Incorrect. The carbonyl and nitrile assignments are reversed. Your score isThe average score is 0% 0% Restart quiz GO TO THE NEXT CHAPTER
