V. ALKENES & ALKYNES TEST 1 1 / 20 Which is the principal alcohol product? A. B. C. D. E. Solution: Hydroboration–oxidation places OH at the less substituted terminal alkene carbon without changing the four-carbon skeleton. A. Incorrect. This is the secondary alcohol associated with the opposite hydration regiochemistry. B. Incorrect. This product adds an unsupported fifth carbon. C. Correct. Hydroboration–oxidation places OH at the less substituted terminal alkene carbon without changing the four-carbon skeleton. D. Incorrect. This branched primary alcohol requires a carbon-skeleton rearrangement; hydroboration does not use a freely rearranging carbocation. E. Incorrect. This vicinal diol requires installation of two OH groups rather than hydration with one oxygen. 2 / 20 Which chain-propagation event controls the anti-Markovnikov constitution? CH₂=CHCH₂CH₃ + HBr; peroxide initiator, radical conditions Br⁻ attacks a saturated carbon by SN2 Two Br radicals add to give a dibromide H⁺ adds first to form a free carbocation OH radical becomes the final alcohol oxygen Br radical adds to terminal C, leaving the more substituted radical Solution: The resulting secondary radical abstracts H from HBr and regenerates Br radical. A. Incorrect. No alkyl leaving group is present. B. Incorrect. HBr addition installs H and Br. C. Incorrect. That is the ionic pathway. D. Incorrect. The target is a bromide. E. Correct. The resulting secondary radical abstracts H from HBr and regenerates Br radical. 3 / 20 Which quantity is unchanged by the catalyst alone? Alkene reaction: uncatalyzed Ea=80, catalyzed Ea=50 kJ/mol; same temperature Reverse rate constant Activation pathway Standard equilibrium constant Forward rate constant Time needed to approach equilibrium Solution: The catalyst changes barriers for both directions without changing reactant-product free-energy difference. A. Incorrect. The reverse pathway is also accelerated. B. Incorrect. The catalyst provides an alternative pathway. C. Correct. The catalyst changes barriers for both directions without changing reactant-product free-energy difference. D. Incorrect. Lowering the barrier can increase it. E. Incorrect. Catalysis generally shortens it. 4 / 20 Which organic product has the expected atom balance? A. B. C. D. E. Solution: Dihydroxylation adds two OH groups across C=C, giving HOCH₂CH₂OH. A. Incorrect. No carbon is added. B. Incorrect. That fits an epoxide formula rather than the diol. C. Incorrect. The carbon skeleton is not cleaved. D. Incorrect. That corresponds to one oxygen. E. Correct. Dihydroxylation adds two OH groups across C=C, giving HOCH₂CH₂OH. 5 / 20 Why are both steps propagation? HBr radical addition: Br radical + alkene -> carbon radical; carbon radical + HBr -> product + Br radical Neither step contains a radical Both steps must be proton-transfer equilibria Each consumes one radical and generates one radical Two radicals combine in each step The initiator is consumed in each cycle Solution: The radical carrier is regenerated without net initiation or termination. A. Incorrect. Both explicitly do. B. Incorrect. Radical bond formation and H-atom abstraction are involved. C. Correct. The radical carrier is regenerated without net initiation or termination. D. Incorrect. That would be termination. E. Incorrect. The chain carrier propagates without repeated initiator consumption. 6 / 20 Which is the carbonyl product of the following reaction? A. B. C. D. E. Solution: Markovnikov hydration of this terminal alkyne gives a methyl ketone. A. Correct. Markovnikov hydration of this terminal alkyne gives a methyl ketone. B. Incorrect. No oxidative cleavage is specified. C. Incorrect. The ketone is not reduced under these conditions. D. Incorrect. Two terminal alcohols are not formed. E. Incorrect. That is the anti-Markovnikov hydration endpoint. 7 / 20 Where does the nonexchangeable D enter? HC#CCH₂CH₃ -> ketone -> alcohol; Hg²⁺/aqueous acid, then NaBD₄/MeOH and H₂O At the former carbonyl carbon In a newly added carbon At all four carbons Only on oxygen At the terminal methyl carbon only Solution: Deuteride attacks the ketone C, giving CH₃C(D)(OH)CH₂CH₃ after workup. A. Correct. Deuteride attacks the ketone C, giving CH₃C(D)(OH)CH₂CH₃ after workup. B. Incorrect. NaBD₄ supplies no carbon. C. Incorrect. Only one deuteride adds per carbonyl reduction. D. Incorrect. O-H exchange is distinct from C-D formation. E. Incorrect. That is not the electrophilic carbonyl site. 8 / 20 Which product has the expected carbon framework and unsaturation? A. B. C. D. E. Solution: Acetylide alkylation gives HC#CCH₂CH₃, then partial hydrogenation gives CH₂=CHCH₂CH₃. A. Incorrect. Only two carbons are added to acetylene. B. Incorrect. No hydration occurs. C. Correct. Acetylide alkylation gives HC#CCH₂CH₃, then partial hydrogenation gives CH₂=CHCH₂CH₃. D. Incorrect. Lindlar conditions stop at alkene. E. Incorrect. The ethyl group was omitted. 9 / 20 Which geometric alkene is favored in the standard dissolving-metal reduction? A. B. C. D. E. Solution: Stepwise electron/proton addition gives the trans alkene overall. A. Incorrect. Dissolving-metal reduction normally stops at alkene. B. Correct. Stepwise electron/proton addition gives the trans alkene overall. C. Incorrect. The double-bond position is not shifted that way. D. Incorrect. That is favored by Lindlar hydrogenation. E. Incorrect. No oxygen source performs hydration. 10 / 20 Which carbon becomes the aldehyde carbon? CH₂=CHCH₂CH₂CH₃ -> alcohol -> aldehyde; BH₃/oxidation then PCC/dry CH₂Cl₂ The original terminal alkene carbon The original second alkene carbon The remote terminal methyl carbon A carbon introduced by PCC Both alkene carbons become aldehydes Solution: Anti-Markovnikov hydration makes a primary alcohol there; PCC oxidizes that carbon to CHO. A. Correct. Anti-Markovnikov hydration makes a primary alcohol there; PCC oxidizes that carbon to CHO. B. Incorrect. That is the Markovnikov position. C. Incorrect. It was not the alcohol site. D. Incorrect. PCC adds no carbon. E. Incorrect. No cleavage is used. 11 / 20 Which is the final organic product? A. B. C. D. E. Solution: The terminal alcohol formed by hydroboration is replaced by Br without carbon rearrangement. A. Incorrect. No extra carbon enters. B. Correct. The terminal alcohol formed by hydroboration is replaced by Br without carbon rearrangement. C. Incorrect. PBr₃ substitution does not form an acyl bromide here. D. Incorrect. This loses a carbon and introduces unsaturation. E. Incorrect. This uses the wrong hydration regiochemistry. 12 / 20 Which product places the incoming methoxy group at the correct original alkene carbon? A. B. C. D. E. Solution: Methoxide attacks the less substituted terminal epoxide carbon; the original epoxide oxygen becomes OH on the neighboring carbon. A. Incorrect. Opening installs one incoming methoxy group, not two. B. Incorrect. This reverses the normal basic-opening regioselectivity, placing methoxy at the more substituted epoxide carbon. C. Incorrect. The oxygen-bearing sites must correspond to the former adjacent alkene carbons, not both ends of the chain. D. Incorrect. The methoxy group is on a carbon outside the original epoxide. E. Correct. Methoxide attacks the less substituted terminal epoxide carbon; the original epoxide oxygen becomes OH on the neighboring carbon. 13 / 20 What is the alcohol class at the former carbonyl carbon? HC#CCH₂CH₃ -> ketone by Hg²⁺/H₂O; then CH₃MgBr/dry ether, H₃O⁺ Primary Secondary No alcohol can form because ketones cannot accept Grignards Tertiary Phenolic Solution: The ketone already has two carbon substituents; Grignard addition installs a third. A. Incorrect. There are three carbon neighbors, not one. B. Incorrect. This overlooks the newly added methyl group. C. Incorrect. Ketones normally undergo Grignard addition. D. Correct. The ketone already has two carbon substituents; Grignard addition installs a third. E. Incorrect. No aromatic C-OH bond is present. 14 / 20 Which mass balance is correct for organic carbon? Terminal alkyne -> ketone -> alcohol; Hg²⁺/aqueous acid then NaBH₄ Carbon count is unchanged through both stages NaBH₄ adds one carbon Hydration adds one carbon Tautomerization removes the terminal carbon Reduction halves the carbon count Solution: Hydration and hydride reduction supply no carbon atoms. A. Correct. Hydration and hydride reduction supply no carbon atoms. B. Incorrect. It contains no carbon. C. Incorrect. It adds water-derived atoms. D. Incorrect. Tautomerization retains all atoms. E. Incorrect. No carbon-carbon cleavage occurs. 15 / 20 Which feature distinguishes the final product from full hydrogenation? HC#CCH₃; NaNH₂, then CH₃CH₂Br; then H₂/Lindlar A retained C#C bond as the required endpoint A lost ethyl group A nitrogen atom in the product A new C=O group A retained C=C bond Solution: Alkylation gives a five-carbon internal alkyne; poisoned-catalyst reduction gives a cis alkene. A. Incorrect. The partial reduction consumes the triple bond. B. Incorrect. The alkylation C-C bond remains. C. Incorrect. Amide base does not install nitrogen into the organic framework. D. Incorrect. No oxygen chemistry is supplied. E. Correct. Alkylation gives a five-carbon internal alkyne; poisoned-catalyst reduction gives a cis alkene. 16 / 20 Which comparison is expected? CH₃C#CCH₂CH₃; H₂/Lindlar versus Na/NH₃ Both necessarily give the same Z alkene Z alkene versus E alkene Lindlar gives ketone; sodium gives alcohol Both necessarily give the alkane Carbon count differs by one Solution: Catalytic syn delivery favors cis, whereas dissolving-metal reduction gives trans overall. A. Incorrect. Their stereochemical outcomes differ. B. Correct. Catalytic syn delivery favors cis, whereas dissolving-metal reduction gives trans overall. C. Incorrect. Neither is hydration. D. Incorrect. Both are partial-reduction methods. E. Incorrect. Neither reduction adds carbon. 17 / 20 Which distinction from dry PCC oxidation matters? CH₂=CHCH₂CH₃ -> primary alcohol -> acid; BH₃/H₂O₂ then aqueous Cr(VI) Cr(VI) adds a methyl group Aqueous strong oxidation can carry the primary alcohol through to CO₂H Both must stop at aldehyde regardless of water The terminal carbon disappears PCC always cleaves every C-C bond Solution: Water permits aldehyde hydration and further oxidation. A. Incorrect. It is an oxidant, not a carbon donor. B. Correct. Water permits aldehyde hydration and further oxidation. C. Incorrect. The endpoint depends on oxidant and conditions. D. Incorrect. It becomes the carboxyl carbon. E. Incorrect. It does not normally do so here. 18 / 20 Which statement describes the second stage? CH₂=CHCH₂CH₃ -> primary alcohol; then TsCl/pyridine -> sulfonate It immediately replaces oxygen with carbon It adds OH to the remote methyl It oxidizes the alcohol to an acid The alcohol C-O bond remains during O-sulfonylation It changes four carbons into three Solution: A new O-S bond activates the oxygen group for later displacement. A. Incorrect. Tosylation does not perform that substitution. B. Incorrect. No such hydroxylation occurs. C. Incorrect. TsCl is not the stated oxidant. D. Correct. A new O-S bond activates the oxygen group for later displacement. E. Incorrect. The carbon framework is retained. 19 / 20 Which relative diol geometry follows ordinary backside ring opening? Cyclohexene -> epoxide, mCPBA; then aqueous H₃O⁺ 1,3-diol cis-1,2-diol 1,1-diol trans-1,2-diol Aromatic phenol Solution: The epoxide opens anti, placing the two OH groups on opposite faces. A. Incorrect. No migration moves an OH to C3. B. Incorrect. That is the syn-dihydroxylation relationship. C. Incorrect. The oxygens remain on adjacent carbons. D. Correct. The epoxide opens anti, placing the two OH groups on opposite faces. E. Incorrect. No aromatization is specified. 20 / 20 In an achiral environment, what stereochemical outcome is expected from NaBH₄ reduction of the ketone? HC#CCH₂CH₃ -> CH₃COCH₂CH₃ -> CH₃CH(OH)CH₂CH₃ A meso alcohol with two centers Only R alcohol by definition An E/Z alkene pair Only S alcohol because hydride is negative Racemic R/S alcohol Solution: Attack on the two enantiotopic carbonyl faces creates equal enantiomers in the ideal achiral model. A. Incorrect. Only one stereocenter is created. B. Incorrect. No chiral reagent or catalyst is present. C. Incorrect. The product is an alcohol. D. Incorrect. Charge does not select one enantiotopic face. E. Correct. Attack on the two enantiotopic carbonyl faces creates equal enantiomers in the ideal achiral model. Your score isThe average score is 0% 0% Restart quiz V. ALKENES & ALKYNES TEST 2 1 / 20 Where is the label in the resulting methyl-carbinol fragment? Terminal alkyne carbon is ¹³C; Hg²⁺/aqueous hydration then NaBH₄ Lost as CO₂ necessarily Duplicated at two carbons At the OH-bearing carbon In the CH₃ attached to CH(OH) On oxygen Solution: Markovnikov hydration makes the terminal alkyne carbon the methyl group of the ketone; reduction preserves it. A. Incorrect. No oxidative cleavage occurs. B. Incorrect. Only one labeled atom was present. C. Incorrect. That is the internal alkyne carbon in this pathway. D. Correct. Markovnikov hydration makes the terminal alkyne carbon the methyl group of the ketone; reduction preserves it. E. Incorrect. A carbon isotope cannot label oxygen. 2 / 20 Which error would produce an incorrect final skeleton? CH₃C#C⁻ + CH₃CH₂Br; dry solvent; then Na/NH₃ Removing one pi bond during partial reduction Keeping five carbons during reduction Counting the bromide carbon chain as a leaving fragment Forming a new C-C bond in alkylation Retaining the original three alkyne carbons Solution: Bromide leaves, but its ethyl carbons join the acetylide to form a five-carbon framework. A. Incorrect. That is correct. B. Incorrect. Reduction adds no carbon. C. Correct. Bromide leaves, but its ethyl carbons join the acetylide to form a five-carbon framework. D. Incorrect. That is the intended reaction. E. Incorrect. Those are retained. 3 / 20 Which intermediate must exist before reduction? HC#CH; deprotonate once, add CH₃Br; repeat deprotonation and CH₃Br; then Lindlar CH₃CH=CHCH₃ CH₃C#CH only CH₃CH₂CH₂CH₃ CH₃COCH₃ CH₃C#CCH₃ Solution: Two successive terminal acetylide alkylations install two methyl groups on acetylene. A. Incorrect. That is the reduction product, not the alkyne intermediate. B. Incorrect. This omits the second alkylation. C. Incorrect. This is fully reduced. D. Incorrect. No hydration stage is present. E. Correct. Two successive terminal acetylide alkylations install two methyl groups on acetylene. 4 / 20 Which observation supports completion of stage 3? CH₂=CHCH₂CH₃; 1) BH₃/THF 2) H₂O₂/NaOH 3) PCC/dry CH₂Cl₂ Appearance of an aldehydic proton near 9-10 ppm Persistence of the original terminal alkene as the target An increase from four to five carbons A new nitrile stretch Two new aromatic proton sets Solution: The terminal primary alcohol is oxidized to CH₃CH₂CH₂CHO. A. Correct. The terminal primary alcohol is oxidized to CH₃CH₂CH₂CHO. B. Incorrect. Hydroboration consumes it. C. Incorrect. No carbon is added. D. Incorrect. No nitrogen source is present. E. Incorrect. No aromatic ring is formed. 5 / 20 Which reagent order fits the specified intermediates? CH₂=CHCH₂CH₃ -> HOCH₂CH₂CH₂CH₃ -> BrCH₂CH₂CH₂CH₃ BH₃/THF, H₂O₂/NaOH; then PBr₃ HBr/peroxide; then NaBH₄ O₃/Zn; then PBr₃ PCC; then NH₂OH HgSO₄/H₂O; then H₂/Pd Solution: Anti-Markovnikov hydration creates the primary alcohol, followed by OH replacement. A. Correct. Anti-Markovnikov hydration creates the primary alcohol, followed by OH replacement. B. Incorrect. The first step gives bromide directly, not the listed alcohol. C. Incorrect. Cleavage changes the skeleton. D. Incorrect. These do not hydrate an alkene. E. Incorrect. The starting alkene is not the stated terminal alkyne hydration substrate. 6 / 20 Where does sulfur enter under basic epoxide opening? CH₂=CH-CH₂-OCH₃; mCPBA, then NaSCH₃ in dry polar solvent, H₂O At a newly created carbon On the epoxide oxygen by forming an O-S target bond At both epoxide carbons with two S groups The ether methyl carbon necessarily The less substituted former alkene carbon Solution: Thiolate attacks terminal epoxide carbon; the epoxide O becomes OH at the neighboring carbon. A. Incorrect. No carbon is created by the opening step. B. Incorrect. Opening makes a C-S bond. C. Incorrect. One opening installs one nucleophile. D. Incorrect. The strained epoxide is the specified electrophile. E. Correct. Thiolate attacks terminal epoxide carbon; the epoxide O becomes OH at the neighboring carbon. 7 / 20 Which carbon bears Br in the final constitution? HC#CCH₂CH₃ -> ketone -> alcohol -> bromide; Hg²⁺/H₂O, NaBH₄, PBr₃ A fifth carbon from hydride The original internal alkyne carbon The original terminal alkyne carbon Both original alkyne carbons The remote terminal methyl carbon Solution: Hydration places C=O there; reduction gives OH there, and PBr₃ replaces OH. A. Incorrect. Hydride supplies no carbon. B. Correct. Hydration places C=O there; reduction gives OH there, and PBr₃ replaces OH. C. Incorrect. That becomes CH₃ under Markovnikov hydration. D. Incorrect. Only one alcohol group is formed and replaced. E. Incorrect. No functional-group migration occurs. 8 / 20 Which atoms supply the ester's OCH₂CH₃ group? CH₂=CHCH₂CH₃ -> alcohol -> acid -> ethyl ester; BH₃/H₂O₂, aqueous Cr(VI), EtOH/H⁺ Air nitrogen The original alkene's terminal two carbons exclusively Borane The ethanol added in the last stage The chromium reagent Solution: Fischer esterification retains the acid skeleton and incorporates the alcohol-derived ethoxy group. A. Incorrect. No nitrogen enters the ester. B. Incorrect. Those remain in the acid-derived chain. C. Incorrect. BH₃ supplies no carbon or oxygen. D. Correct. Fischer esterification retains the acid skeleton and incorporates the alcohol-derived ethoxy group. E. Incorrect. It contains no ethoxy group. 9 / 20 Which final thioether linkage follows this sequence? A. B. C. D. E. Solution: The alkene becomes a terminal alcohol, then bromide, then C-S substitution product. A. Correct. The alkene becomes a terminal alcohol, then bromide, then C-S substitution product. B. Incorrect. This uses the wrong hydration regiochemistry. C. Incorrect. The alcohol oxygen leaves during bromide formation. D. Incorrect. The nucleophile supplies SCH₃ rather than SH. E. Incorrect. No disulfide-forming step is included. 10 / 20 Which final constitution is supported without assigning stereochemistry? A. B. C. D. E. Solution: All stages retain the skeleton and transfer the functional site through carbonyl, alcohol and chloride. A. Incorrect. No cyclization occurs. B. Incorrect. Alcohol chlorination replaces oxygen. C. Correct. All stages retain the skeleton and transfer the functional site through carbonyl, alcohol and chloride. D. Incorrect. No carbon is added. E. Incorrect. No cyanide is present. 11 / 20 Which atom replaces carbonyl oxygen in the final C=N bond? CH₂=CHCH₂CH₃ -> terminal alcohol -> aldehyde; then NH₂OH/buffer, remove water Hydroxylamine oxygen Boron A carbon from PCC Hydroxylamine nitrogen A hydrogen from solvent Solution: Oxime formation gives CH₃CH₂CH₂CH=NOH after condensation. A. Incorrect. Oximes have C=N-OH, not C=O-NH₂. B. Incorrect. Borane does not supply the final N atom. C. Incorrect. PCC adds no carbon. D. Correct. Oxime formation gives CH₃CH₂CH₂CH=NOH after condensation. E. Incorrect. Hydrogen cannot replace nitrogen in a C=N bond. 12 / 20 Which proposed result improperly retains a leaving-group atom? CH₂=CHCH₂CH₃ -> alcohol -> bromide -> CH₃S-linked product Sulfur with two carbon neighbors CH₃CH₂CH₂CH₂OSCH₃ Bromide in the inorganic fraction CH₃CH₂CH₂CH₂SCH₃ Five organic carbons in the thioether Solution: The original alcohol O leaves in bromide formation; thiolate makes C-S directly. A. Incorrect. That is the sulfide connectivity. B. Correct. The original alcohol O leaves in bromide formation; thiolate makes C-S directly. C. Incorrect. Bromide is expelled on thiolate attack. D. Incorrect. This is the intended connectivity. E. Incorrect. Four substrate carbons plus thiolate methyl give five. 13 / 20 What happens to the oxygen introduced in hydration? HC#CCH₂CH₃ -> ketone -> alcohol -> bromide It remains as a hidden ether in the bromide It is removed during alcohol-to-bromide conversion It becomes an extra carbon It must be lost during NaBH₄ reduction It becomes bromine Solution: The carbonyl oxygen becomes alcohol O on reduction, then leaves during substitution. A. Incorrect. The specified bromide has no oxygen. B. Correct. The carbonyl oxygen becomes alcohol O on reduction, then leaves during substitution. C. Incorrect. Oxygen cannot become carbon. D. Incorrect. Reduction retains carbonyl oxygen as alcohol oxygen. E. Incorrect. Elements do not transmute. 14 / 20 Which bond-count change characterizes the cycloaddition? CH₂=CH-CH=CH₂ + CH₂=CH-CO₂CH₃; thermal Diels-Alder Three new pi bonds and no ring Two new sigma bonds and one retained C=C in a six-membered ring A four-membered ring only Loss of the ester carbonyl as CO₂ Cleavage of the diene's central C-C bond Solution: A diene and dienophile form a cyclohexene framework in the concerted cycloaddition. A. Incorrect. That reverses the cycloaddition pattern. B. Correct. A diene and dienophile form a cyclohexene framework in the concerted cycloaddition. C. Incorrect. The [4+2] framework has six ring atoms. D. Incorrect. Ordinary Diels-Alder does not require decarboxylation. E. Incorrect. That sigma connection is retained. 15 / 20 How many carbons are in the final alcohol? CH₂=CHCH₂CH₃ -> terminal alcohol -> aldehyde -> secondary alcohol; BH₃/H₂O₂, PCC, CH₃MgBr/H₃O⁺ 8 4 5 6 3 Solution: Hydration and oxidation retain four carbons; Grignard addition supplies one methyl carbon. A. Incorrect. The substrate does not couple with another whole substrate molecule. B. Incorrect. This omits the Grignard carbon. C. Correct. Hydration and oxidation retain four carbons; Grignard addition supplies one methyl carbon. D. Incorrect. Only one carbon is added. E. Incorrect. No cleavage removes a carbon. 16 / 20 Which property must X have? HC#CCH₂CH₃ -> X -> CH₃CH(OH)CH₂CH₃; Hg²⁺/H₂O then NaBH₄ A carboxylate salt only A terminal aldehyde carbonyl Two ether oxygens A C-Br bond An internal ketone carbonyl Solution: The Markovnikov alkyne hydration endpoint is CH₃COCH₂CH₃, which reduces to the alcohol. A. Incorrect. NaBH₄ does not ordinarily reduce it to this product. B. Incorrect. That would reduce to a primary alcohol. C. Incorrect. The sequence involves one carbonyl oxygen. D. Incorrect. No bromine reagent is present. E. Correct. The Markovnikov alkyne hydration endpoint is CH₃COCH₂CH₃, which reduces to the alcohol. 17 / 20 Which intermediate X forms before reduction to (Z)-2-pentene? Intermediate X after alkylation; before H₂/Lindlar A. B. C. D. E. Solution: Ethylation gives the five-carbon internal alkyne, then syn partial hydrogenation gives Z alkene. A. Incorrect. This has four carbons. B. Incorrect. This has six carbons. C. Incorrect. The saturated alkane is not the required intermediate. D. Correct. Ethylation gives the five-carbon internal alkyne, then syn partial hydrogenation gives Z alkene. E. Incorrect. A ketone does not give this Lindlar product. 18 / 20 Which distinction identifies the final reduction method? CH₃C#CH -> X -> E-CH₃CH=CHCH₂CH₃; acetylide ethylation, then Na/NH₃ Retention of both alkyne pi bonds Removal of an ethyl group Addition of an oxygen atom Opposite-face overall addition giving E geometry Formation of a single tetrahedral stereocenter as the only change Solution: Dissolving-metal reduction differs from poisoned-catalyst syn hydrogenation. A. Incorrect. One pi bond is reduced. B. Incorrect. The C-C bond made in alkylation remains. C. Incorrect. The reagents do not hydrate the alkyne. D. Correct. Dissolving-metal reduction differs from poisoned-catalyst syn hydrogenation. E. Incorrect. The requested product has an E double bond. 19 / 20 Which product set accounts for all six starting carbons? A. B. C. D. E. Solution: Cleavage converts the ring alkene carbon to ketone and the exocyclic CH₂ to formaldehyde. A. Correct. Cleavage converts the ring alkene carbon to ketone and the exocyclic CH₂ to formaldehyde. B. Incorrect. Reductive ozonolysis does not give that pair. C. Incorrect. This changes the carbon connectivity incorrectly. D. Incorrect. The ring carbon framework is not opened at a ring bond. E. Incorrect. This omits the exocyclic carbon. 20 / 20 Which structure is intermediate X? A. B. C. D. E. Solution: The anti-Markovnikov intermediate is primary and oxidizes to an aldehyde. A. Incorrect. No nitrogen enters. B. Incorrect. Hydroboration-oxidation supplies only one OH per alkene. C. Correct. The anti-Markovnikov intermediate is primary and oxidizes to an aldehyde. D. Incorrect. No carbon is added. E. Incorrect. The first stage gives an alcohol. Your score isThe average score is 0% 0% Restart quiz GO TO THE NEXT CHAPTER
