IV. SUBSTITUTION & ELIMINATION TEST 1 1 / 20 Which organic bond is formed in the principal substitution pathway? CH₃CH₂Br + NaSCH₃; dry DMSO, 20 C S-Br as the target organic product C-Br retained with an extra C-S bond CH₂-SCH₃ CH₂-OCH₃ CH₂-CH₃ from reagent carbon attack Solution: Sulfur attacks the primary carbon and bromide leaves, giving CH₃CH₂SCH₃. A. Incorrect. That does not connect the nucleophile to the substrate carbon. B. Incorrect. Substitution replaces the leaving group. C. Correct. Sulfur attacks the primary carbon and bromide leaves, giving CH₃CH₂SCH₃. D. Incorrect. The reagent contains sulfur rather than oxygen. E. Incorrect. Thiolate attacks through sulfur. 2 / 20 Which bond-breaking event is essential for replacing the alcohol group? CH₃CH₂OH -> CH₃CH₂Br; PBr₃, dry ether, then isolate Formation of a fifth carbon bond O-H cleavage alone C-H cleavage as the only event C-C cleavage C-O cleavage after oxygen activation Solution: The alcohol oxygen is activated by phosphorus before displacement at carbon. A. Incorrect. Carbon must maintain its octet during displacement. B. Incorrect. Deprotonation alone leaves oxygen attached. C. Incorrect. That does not replace OH. D. Incorrect. The two-carbon framework is preserved. E. Correct. The alcohol oxygen is activated by phosphorus before displacement at carbon. 3 / 20 Which feature makes this a favorable Williamson coupling? CH₃CH₂O⁻Na⁺ + CH₃I; dry acetone, 20 C Methyl electrophile permits SN2 without beta elimination Water is required to protonate ethoxide before attack Iodide attacks the ethoxide oxygen Two ethoxide ions must attack one methyl simultaneously A tertiary carbocation must form first Solution: Ethoxide displaces iodide to give CH₃CH₂OCH₃; methyl iodide has no beta carbon. A. Correct. Ethoxide displaces iodide to give CH₃CH₂OCH₃; methyl iodide has no beta carbon. B. Incorrect. That would consume the nucleophile. C. Incorrect. The alkoxide is the nucleophile. D. Incorrect. SN2 uses one nucleophile in the displacement event. E. Incorrect. Methyl ionization is not the pathway. 4 / 20 Which route best favors substitution over elimination? Target CH₃OCH(CH₃)₂; compare alkoxide/halide choices CH₃OH + propane only (CH₃)₂CHO⁻ + CH₃I CH₃O⁻ + (CH₃)₃CBr HO⁻ + CH₃CH₂Br CH₃I + neutral acetone Solution: The methyl electrophile allows SN2 while avoiding a secondary halide with a strong base. A. Incorrect. No activated electrophile is provided. B. Correct. The methyl electrophile allows SN2 while avoiding a secondary halide with a strong base. C. Incorrect. A tertiary halide cannot give the target by SN2. D. Incorrect. This forms an alcohol with the wrong framework. E. Incorrect. This does not supply the required alkoxide. 5 / 20 Which observation is compatible with an ionization pathway? (CH₃)₃CBr in aqueous ethanol, 20 C; substitution fraction analyzed Solvent capture gives (CH₃)₃COH among substitution products The only possible organic event is nitrile formation Backside SN2 at tertiary carbon is required Every isolated product must be a single enantiomer The carbon skeleton must gain one carbon from bromide Solution: A tertiary carbocation can be captured by water and then deprotonated. A. Correct. A tertiary carbocation can be captured by water and then deprotonated. B. Incorrect. No cyanide is present. C. Incorrect. Steric crowding prevents ordinary tertiary SN2. D. Incorrect. The substrate center is not stereogenic. E. Incorrect. Bromide contains no carbon. 6 / 20 Which is the final organic product of the following sequence? A. B. C. D. E. Solution: CN adds one carbon by SN2; nitrile reduction retains that carbon as CH₂NH₂. A. Incorrect. This loses the cyanide carbon. B. Correct. CN adds one carbon by SN2; nitrile reduction retains that carbon as CH₂NH₂. C. Incorrect. This adds an extra unsupported carbon. D. Incorrect. This is incomplete conversion by a different reagent class. E. Incorrect. Nitrogen is not replaced with oxygen in the stated reduction. 7 / 20 Which structure is intermediate X before conversion to 1-iodopropane? A. B. C. D. E. Solution: PBr₃ gives the bromide, which iodide displaces in a second SN2 step. A. Correct. PBr₃ gives the bromide, which iodide displaces in a second SN2 step. B. Incorrect. PBr₃ is not an alcohol oxidant. C. Incorrect. The alcohol carbon is terminal and the framework does not rearrange. D. Incorrect. PBr₃ supplies no carbon. E. Incorrect. The stated sequence does not require elimination. 8 / 20 Where is ¹⁸O immediately after SN2 at the primary carbon? CH₃CH₂-¹⁸O-SO₂CH₃ + NaI; dry acetone, 25 C In the sulfonate leaving fragment In newly formed H₂¹⁸O as the obligatory sole product On iodide as [18I] In CH₃CH₂I On substrate carbon as a carbon isotope Solution: The C-¹⁸O bond breaks; iodide replaces the entire oxygen-bearing leaving group. A. Correct. The C-¹⁸O bond breaks; iodide replaces the entire oxygen-bearing leaving group. B. Incorrect. No water-forming step is required. C. Incorrect. Isotope identity does not transfer between elements. D. Incorrect. The iodide product contains no oxygen. E. Incorrect. An oxygen isotope cannot become carbon. 9 / 20 Which atom in the product comes from the nucleophile? CH₃CH₂CH₂Br + NaOCH₃; dry DMF, cool; substitution product requested A carbonyl oxygen The ether oxygen and its attached methyl carbon Only a hydrogen on the propyl chain All three propyl carbons The bromine atom Solution: Methoxide contributes OCH₃ intact, giving CH₃CH₂CH₂OCH₃. A. Incorrect. No carbonyl forms in this substitution. B. Correct. Methoxide contributes OCH₃ intact, giving CH₃CH₂CH₂OCH₃. C. Incorrect. This is not proton transfer alone. D. Incorrect. Those come from the substrate. E. Incorrect. Bromine is the leaving group from the substrate. 10 / 20 Where is ¹³C in the final acid? BrCH₂CH₂CH₃ + Na¹³CN; dry DMSO; then acidic hydrolysis, heat At the terminal methyl carbon The label is removed during protonation In CO₂ only At both internal carbons At the carboxyl carbon Solution: The cyanide carbon forms the new C-C bond and becomes CO₂H on hydrolysis. A. Incorrect. No carbon rearrangement moves the label there. B. Incorrect. Protonation does not delete a carbon isotope. C. Incorrect. Nitrile hydrolysis retains its carbon in the acid. D. Incorrect. Only one labeled carbon enters. E. Correct. The cyanide carbon forms the new C-C bond and becomes CO₂H on hydrolysis. 11 / 20 Which carbon-number sequence is correct? BrCH₂CH₂CH₃ -> nitrile -> amine; CN⁻, then LiAlH₄ and H₂O 3 -> 4 -> 4 3 -> 4 -> 5 3 -> 3 -> 3 3 -> 4 -> 3 3 -> 2 -> 2 Solution: SN2 cyanation adds one carbon; nitrile reduction adds hydrogens but no carbon. A. Correct. SN2 cyanation adds one carbon; nitrile reduction adds hydrogens but no carbon. B. Incorrect. Reduction does not add carbon. C. Incorrect. This omits cyanide carbon incorporation. D. Incorrect. The nitrile carbon is retained in the amine. E. Incorrect. No carbon is removed by cyanation. 12 / 20 Which functional-group sequence can describe the internal-alkene pathway? CH₃CH(Br)CH₂CH₃; NaOEt/EtOH, heat; then mCPBA/CH₂Cl₂ Alkyl bromide -> alcohol -> carboxylate Alkyl bromide -> nitrile -> amine Alkyl bromide -> ketone -> amide Alkyl bromide -> alkyne -> ester Alkyl bromide -> alkene -> epoxide Solution: E2 removes HBr, and peracid transfers oxygen across the resulting C=C bond. A. Incorrect. These reagents do not give that sequence. B. Incorrect. No cyanide or nitrile reductant is present. C. Incorrect. Neither required nitrogen nor carbonyl chemistry is supplied. D. Incorrect. One elimination cannot create that alkyne here. E. Correct. E2 removes HBr, and peracid transfers oxygen across the resulting C=C bond. 13 / 20 Which bond is retained through both stages? CH₃CH₂Br -> CH₃CH₂SCH₃; NaSCH₃/DMSO; then one equivalent oxidant giving sulfoxide An S-S bond A carbonyl C=O bond The original C-Br bond The newly formed ethyl C-S bond A new N-S bond Solution: Substitution forms C-S, and controlled sulfur oxidation adds oxygen without cleaving that bond. A. Incorrect. No disulfide is formed. B. Incorrect. No carbonyl is present. C. Incorrect. It breaks during substitution. D. Correct. Substitution forms C-S, and controlled sulfur oxidation adds oxygen without cleaving that bond. E. Incorrect. No nitrogen is involved. 14 / 20 Which oxygen pattern results from the alkene stage? CH₃CH(Br)CH₃ -> CH₂=CHCH₃; strong base/heat; then OsO₄, NMO, aqueous workup Two OH groups on the same carbon Adjacent OH groups on the former alkene carbons A carboxyl group from mandatory cleavage An O-O bond joining the two carbons One OH on every carbon Solution: Syn dihydroxylation gives CH₃CH(OH)CH₂OH. A. Incorrect. That is a geminal rather than vicinal pattern. B. Correct. Syn dihydroxylation gives CH₃CH(OH)CH₂OH. C. Incorrect. These are dihydroxylation rather than cleavage conditions. D. Incorrect. A diol contains separate OH groups. E. Incorrect. Only the alkene carbons are oxidized here. 15 / 20 Which step follows water attack on the carbocation? (CH₃)₃CCl in H₂O; substitution product isolated Loss of an oxygen atom Formation of a carbanion at tertiary carbon Deprotonation of the oxonium ion CN⁻ attack from water Addition of another carbon to the central carbon Solution: Water capture initially gives a protonated alcohol, then loss of H⁺ yields neutral (CH₃)₃COH. A. Incorrect. That would remove the alcohol group. B. Incorrect. That is not the ordinary SN1 sequence. C. Correct. Water capture initially gives a protonated alcohol, then loss of H⁺ yields neutral (CH₃)₃COH. D. Incorrect. Water contains no cyanide. E. Incorrect. The central carbon already has its required connectivity. 16 / 20 Why must E/Z identity be retained when predicting epoxide stereochemistry? CH₃CH₂CH(Br)CH₃; small strong base, heat; then alkene epoxidation Peracid epoxidation always scrambles every center All epoxides are the same molecule Epoxidation preserves relative substituent geometry across the alkene Epoxidation necessarily cleaves C-C E and Z have different carbon counts Solution: Different alkene configurations can give different epoxide stereoisomer sets. A. Incorrect. The concerted addition is stereospecific. B. Incorrect. Connectivity and stereochemistry vary. C. Correct. Different alkene configurations can give different epoxide stereoisomer sets. D. Incorrect. The alkene C-C connection remains. E. Incorrect. Their carbon counts are identical. 17 / 20 Which intermediate can lead to both substitution and elimination? CH₃CH₂CH₂CH(Br)CH₃; water/ethanol; SN1 and E1 fractions observed An alkoxide formed before ionization A carbon with five permanent sigma bonds A free bromine radical as the only organic intermediate A nitrile anion A carbocation formed after C-Br ionization Solution: Nucleophile capture gives substitution; beta deprotonation gives elimination. A. Incorrect. No alcohol is initially present. B. Incorrect. That violates the ordinary carbon octet. C. Incorrect. That does not give the ordinary shared SN1/E1 intermediate. D. Incorrect. There is no nitrogen source. E. Correct. Nucleophile capture gives substitution; beta deprotonation gives elimination. 18 / 20 Which comparison distinguishes two branches from the same cation? (CH₃)₂C(Cl)CH₂CH₃; dilute water, warm Both add a new carbon from chloride E1 consumes two substrate molecules in one elementary elimination H₂O capture forms C-O; beta H loss forms C=C Substitution must increase unsaturation by one Both require backside attack at the original tertiary center Solution: Substitution and E1 differ in whether a nucleophile bonds or a beta proton is removed. A. Incorrect. Chloride supplies no carbon. B. Incorrect. E1 deprotonation occurs on a single carbocation. C. Correct. Substitution and E1 differ in whether a nucleophile bonds or a beta proton is removed. D. Incorrect. Alcohol formation does not require a C=C bond. E. Incorrect. SN1/E1 begin with ionization. 19 / 20 What happens to reaction rate per unit volume? E2 rate law r=k[RX][B]; [RX] tripled, [B] halved, temperature fixed Multiplies by 0.5 Multiplies by 3 Multiplies by 1.5 Remains unchanged Multiplies by 6 Solution: The two concentration factors multiply: 3x0.5=1.5. A. Incorrect. This ignores the substrate increase. B. Incorrect. This ignores the base decrease. C. Correct. The two concentration factors multiply: 3x0.5=1.5. D. Incorrect. The concentration factors do not cancel. E. Incorrect. Halving was treated as doubling. 20 / 20 Which product has the correct atom balance after complete hydrolysis? A. B. C. D. E. Solution: Cyanide contributes its carbon, and hydrolysis converts CN to CO₂H without deleting it. A. Incorrect. Only one carbon enters. B. Correct. Cyanide contributes its carbon, and hydrolysis converts CN to CO₂H without deleting it. C. Incorrect. This omits cyanide carbon. D. Incorrect. Complete acidic hydrolysis goes beyond the amide and retains four carbons. E. Incorrect. Hydrolysis retains oxygen-containing carboxyl functionality. Your score isThe average score is 0% 0% Restart quiz IV. SUBSTITUTION & ELIMINATION TEST 2 1 / 20 Which geometry is directly productive for ordinary anti E2? CH₃CH(Br)CH₂CH₃; reactive conformer H-C-C-Br dihedral=180 degrees; NaOEt, heat The image shows connectivity only. Use the stated H-C-C-Br dihedral angle. C-H perpendicular to C-Br with no rotation A 0-degree H/Br dihedral as the required anti arrangement Any remote methyl H regardless of adjacency The specified anti-periplanar beta H and Br Loss of Br with no beta H removal Solution: Their sigma bonds align for concerted C-H/C-Br cleavage and C=C formation. A. Incorrect. That lacks the ideal anti alignment. B. Incorrect. Zero degrees is syn-periplanar. C. Incorrect. The eliminated H must be beta to the leaving group. D. Correct. Their sigma bonds align for concerted C-H/C-Br cleavage and C=C formation. E. Incorrect. That is not the complete E2 event. 2 / 20 Which spectral change supports X formation? CH₃CH₂CH₂I -> CH₃CH₂CH₂CN -> X; CN⁻/DMSO, then LiAlH₄ and aqueous workup Loss of C#N and appearance of primary-amine N-H features A new ester C=O as the required product Loss of all carbon signals A stronger nitrile band than before A molecular formula with no nitrogen Solution: Nitrile reduction gives CH₃CH₂CH₂CH₂NH₂ while retaining its carbon. A. Correct. Nitrile reduction gives CH₃CH₂CH₂CH₂NH₂ while retaining its carbon. B. Incorrect. No ester is formed. C. Incorrect. The carbon skeleton remains. D. Incorrect. The nitrile is consumed. E. Incorrect. Nitrogen is retained in the amine. 3 / 20 Which stage breaks the alcohol C-O bond? HOCH₂CH₂CH₃ -> sulfonate; TsCl, pyridine, 0-20 C; then NaI/acetone The initial O-S bond formation Neither stage can replace oxygen Iodide displacement of the sulfonate Pyridine removal of the O-H proton alone Solvent evaporation alone Solution: Tosylation forms O-S while retaining C-O; subsequent SN2 breaks C-O. A. Incorrect. Tosylation does not replace oxygen at carbon. B. Incorrect. The displacement removes the oxygen-bearing leaving group. C. Correct. Tosylation forms O-S while retaining C-O; subsequent SN2 breaks C-O. D. Incorrect. That breaks O-H rather than C-O. E. Incorrect. Drying is not a substitution step. 4 / 20 Which pair can arise from attack at the two allylic cation termini, ignoring stereochemistry? CH₂=CH-CH(Cl)-CH₃; ionization followed by H₂O capture Two identical drawings of CH₂=CH-CH(OH)-CH₃ only CH₃COCH₃ and methanol CH₃CH(OH)CH(OH)CH₃ and CH₃CH₂CH₂CH₃ CH₂=CH-CH(OH)-CH₃ and HOCH₂-CH=CH-CH₃ CH₃CH₂CH₂CH₂OH and CH₃CH₂CH₂CHO Solution: Allylic resonance places electrophilic character at both termini of the three-carbon pi system. A. Incorrect. This omits the distinct allylic connectivity. B. Incorrect. That requires carbon-carbon cleavage. C. Incorrect. Capture does not give dihydroxylation or reduction. D. Correct. Allylic resonance places electrophilic character at both termini of the three-carbon pi system. E. Incorrect. These require hydrogenation or oxidation beyond capture. 5 / 20 Which bond formed in the substitution must survive ordinary nitrile hydrolysis? CH₃CH₂Cl + CN⁻; nucleophile carbon marked C* CH₂-C* C*-N as a triple bond CH₂-Cl A new C*-C* bond N-Cl Solution: The new substrate-cyanide carbon bond becomes the bond to the carboxyl carbon. A. Correct. The new substrate-cyanide carbon bond becomes the bond to the carboxyl carbon. B. Incorrect. Hydrolysis replaces the nitrile nitrogen functionality. C. Incorrect. The leaving-group bond breaks in substitution. D. Incorrect. Only one cyanide carbon is introduced per substitution. E. Incorrect. This is not the coupling bond. 6 / 20 Which proposed correction to a three-carbon amine drawing is unnecessary? CH₃CH₂Cl -> CH₃CH₂CN -> CH₃CH₂CH₂NH₂ Giving neutral N three bonds after workup Converting nitrile C to CH₂ Removing the C#N triple-bond depiction Adding another carbon during hydride reduction Retaining the original ethyl fragment Solution: The nitrile carbon already supplies the third carbon; reduction changes bonding to nitrogen without carbon addition. A. Incorrect. That is correct for the primary amine. B. Incorrect. That is the correct reduction endpoint. C. Incorrect. The final amine no longer contains a nitrile. D. Correct. The nitrile carbon already supplies the third carbon; reduction changes bonding to nitrogen without carbon addition. E. Incorrect. That fragment should remain. 7 / 20 Which pair of operations avoids using unactivated OH as an SN2 leaving group? CH₃CH₂CH₂OH; route to CH₃CH₂CH₂OCH₃ NaH, then CH₃I in dry solvent H₂/Pd, then NaCl NaI added to neutral alcohol only Br₂/light, then oxygen NaBH₄, then water Solution: Deprotonation creates the alkoxide, which attacks methyl iodide. A. Correct. Deprotonation creates the alkoxide, which attacks methyl iodide. B. Incorrect. No suitable ether-forming electrophile is created. C. Incorrect. Unactivated OH is a poor leaving group. D. Incorrect. These conditions do not selectively make the ether. E. Incorrect. This does not install OCH₃. 8 / 20 Which bromide comes from the original acyl fragment? A. B. C. D. E. Solution: The acyl CH₃CO fragment reduces to ethanol, then converts to bromoethane. A. Incorrect. No additional carbon is introduced. B. Incorrect. This loses the former carbonyl carbon. C. Incorrect. This changes both carbon count and connectivity. D. Correct. The acyl CH₃CO fragment reduces to ethanol, then converts to bromoethane. E. Incorrect. PBr₃ does not restore a carbonyl to ethanol. 9 / 20 Which bookkeeping statement is correct? CH₃-¹⁸O-H -> CH₃Br -> CH₃CN; PBr₃, then NaCN/DMSO The final nitrile retains ¹⁸O on nitrogen The final nitrile must contain an O-N bond The final nitrile has one carbon The final nitrile has two carbons and no ¹⁸O The final nitrile has three carbons Solution: Substitution removes the alcohol oxygen; cyanide adds one carbon. A. Incorrect. Elemental isotope identity cannot transfer that way. B. Incorrect. No oxygen remains in the target. C. Incorrect. Cyanide adds a carbon. D. Correct. Substitution removes the alcohol oxygen; cyanide adds one carbon. E. Incorrect. Only one new carbon enters. 10 / 20 Which intermediate is present immediately before the subsequent LiAlH₄ reduction and aqueous workup? A. B. C. D. E. Solution: Alcohol substitution preserves two carbons; cyanation produces the three-carbon nitrile. A. Incorrect. No hydrolysis stage was specified. B. Correct. Alcohol substitution preserves two carbons; cyanation produces the three-carbon nitrile. C. Incorrect. Cyanation does not extend the chain as another bromide. D. Incorrect. One original carbon is missing. E. Incorrect. That is not the isolated nitrile intermediate. 11 / 20 Which nitrile comes from the tracked acyl-derived fragment? A. B. C. D. E. Solution: The acid-derived three-carbon fragment becomes 1-propanol, then gains one carbon by cyanation. A. Incorrect. The carbonyl is removed by LiAlH₄. B. Incorrect. This loses the original carbonyl carbon during reduction. C. Incorrect. This adds one extra carbon. D. Correct. The acid-derived three-carbon fragment becomes 1-propanol, then gains one carbon by cyanation. E. Incorrect. This incorrectly tracks and couples the alkoxy fragment. 12 / 20 Which final acid structure locates the cyanide carbon label correctly? A. B. C. D. E. Solution: The cyanide carbon becomes the carboxyl carbon after complete hydrolysis. A. Incorrect. The cyanide ¹³C label is retained, rather than lost from the acid. B. Incorrect. This places ¹³C at the alpha methylene, which came from the original alcohol chain. C. Incorrect. This places ¹³C at the beta methylene, which came from the original alcohol chain. D. Incorrect. This places ¹³C at the terminal methyl carbon, which came from the original alcohol chain. E. Correct. The cyanide carbon becomes the carboxyl carbon after complete hydrolysis. 13 / 20 Why can the bulky alkoxide increase the elimination fraction? CH₃CH₂Br with CH₃O⁻ versus (CH₃)₃CO⁻; same solvent and temperature It must add a carbonyl to the substrate It cannot accept a proton It eliminates the need for a beta hydrogen It turns the primary carbon into a tertiary carbon Steric hindrance impedes attack at carbon more than beta-H abstraction Solution: Basicity and nucleophilicity do not respond identically to steric crowding. A. Incorrect. No such reagent function exists. B. Incorrect. tert-Butoxide is a strong base. C. Incorrect. E2 still requires a beta H. D. Incorrect. Reagent size does not reclassify substrate substitution. E. Correct. Basicity and nucleophilicity do not respond identically to steric crowding. 14 / 20 Which experiment best tests carbon incorporation at X? CH₃CH₂Br -> X -> CH₃CH₂CO₂H; NaCN/DMSO then aqueous acid/heat Omit cyanide and expect identical homologation Use D₂O only and count one extra carbon Measure only the solvent boiling point Use ¹³CN⁻ and locate ¹³C at the final carboxyl carbon Label bromine and expect labeled carbon Solution: The cyanide carbon is retained through both substitution and hydrolysis. A. Incorrect. The carbon donor is necessary. B. Incorrect. D₂O adds no carbon. C. Incorrect. That does not track atom origin. D. Correct. The cyanide carbon is retained through both substitution and hydrolysis. E. Incorrect. An isotope cannot change element. 15 / 20 Which functional-group change occurs only in stage 2? CH₃CH₂I -> X -> CH₃CH₂CH₂NH₂; CN⁻, then LiAlH₄/H₂O Carbon count rises from two to three C#N becomes CH₂NH₂ Iodide becomes a carbon atom C-C framework cleaves in half C-I becomes C-C Solution: Stage 1 creates the nitrile by substitution; hydride reduction removes its pi bonds. A. Incorrect. That occurs on cyanide incorporation. B. Correct. Stage 1 creates the nitrile by substitution; hydride reduction removes its pi bonds. C. Incorrect. Elements are conserved. D. Incorrect. Neither stage requires this. E. Incorrect. That occurs in stage 1. 16 / 20 Which two neighboring atoms must supply the eliminated fragments for propene formation? BrCH₂CH₂CH₃; strong bulky base, heat Two H atoms on C3 only Br on C1 and H on C3 Oxygen and H from solvent only Br and carbon C1 entirely Br on C1 and H on C2 Solution: Beta elimination removes leaving group and adjacent H, forming C1=C2. A. Incorrect. The leaving group must depart. B. Incorrect. C3 is not beta to C1. C. Incorrect. That cannot account for loss of substrate bromide. D. Incorrect. Ordinary E2 retains all carbons. E. Correct. Beta elimination removes leaving group and adjacent H, forming C1=C2. 17 / 20 Which product indicates formation and deuterolysis of the organomagnesium reagent? A. B. C. D. E. Solution: The C-Mg bond is quenched with D+, replacing the original C-Br position by D. A. Incorrect. No carbon donor is present. B. Incorrect. No oxygen is installed in the hydrocarbon product. C. Correct. The C-Mg bond is quenched with D+, replacing the original C-Br position by D. D. Incorrect. Alkene formation does not specifically demonstrate deuterolysis of a Grignard reagent. E. Incorrect. The leaving group should already have been replaced. 18 / 20 Why can the product still be labeled R when the reactant was R? One-step SN2 at a stereogenic carbon; spatial inversion observed R always means positive optical rotation The entering group can change the CIP priority order SN2 actually retained all bond directions CIP ignores the entering atom Every inverted product must be achiral Solution: Geometric inversion and R/S-label change are distinct questions. A. Incorrect. R does not encode rotation sign. B. Correct. Geometric inversion and R/S-label change are distinct questions. C. Incorrect. The premise specifies inversion. D. Incorrect. The entering atom affects priorities. E. Incorrect. Inversion can preserve a stereogenic center. 19 / 20 Which restriction follows for ordinary E2 from this bromide? (CH₃)₃C-CH₂Br; beta carbon is quaternary E2 must give a terminal alkene rapidly The bromide is tertiary A methyl carbon two bonds farther away counts as beta Bromine cannot leave in any reaction No beta C-H bond is available Solution: The only adjacent carbon has four carbon bonds, so the required beta proton is absent. A. Incorrect. No beta H supports that event. B. Incorrect. The Br-bearing carbon is primary. C. Incorrect. Those hydrogens are not on the adjacent carbon. D. Incorrect. The E2 restriction does not forbid all pathways. E. Correct. The only adjacent carbon has four carbon bonds, so the required beta proton is absent. 20 / 20 Which oxygen-origin statement is correct? CH₃CH₂OH -> CH₃CH₂Br -> CH₃CH₂OCH₃; PBr₃ then NaOCH₃/DMF Ether oxygen comes from bromide Ether oxygen comes from PBr₃ Ether oxygen comes from methoxide Ether oxygen must be the original alcohol oxygen Both original and methoxide oxygens remain in the ether Solution: PBr₃ substitution removes the original alcohol oxygen; methoxide supplies the ether O. A. Incorrect. Bromide contains no oxygen. B. Incorrect. PBr₃ contains no oxygen. C. Correct. PBr₃ substitution removes the original alcohol oxygen; methoxide supplies the ether O. D. Incorrect. That atom leaves before ether formation. E. Incorrect. The product contains only one oxygen. Your score isThe average score is 0% 0% Restart quiz GO TO THE NEXT CHAPTER
