VIII. ALDEHYDES & KETONES TEST 1 1 / 20 Which fragment supplies the new C-C bond? CH₃CH₂CHO + CH₃MgBr/dry ether; then H₃O⁺ Bromide adds a carbon The ether solvent bonds to oxygen The original C-C bond must break The aldehyde H becomes a methyl group The reagent methyl bonds to the aldehyde carbon Solution: Addition gives CH₃CH₂CH(OH)CH₃ after protonation. A. Incorrect. Bromide contains none. B. Incorrect. Ether is not the carbon donor. C. Incorrect. Ordinary addition retains it. D. Incorrect. A hydrogen atom cannot become carbon. E. Correct. Addition gives CH₃CH₂CH(OH)CH₃ after protonation. 2 / 20 Which group pair replaces the carbonyl oxygen environment? CH₃COCH₃ + HOCH₂CH₂OH; catalytic pTsOH, toluene, remove H₂O A C#C bond One C-Br bond Two new C-C bonds from the diol A C-N double bond Two C-O bonds to the diol oxygens in a cyclic ketal Solution: Acetalization uses both diol OH groups and eliminates water. A. Incorrect. No deoxygenative alkyne formation occurs. B. Incorrect. No bromine is present. C. Incorrect. The oxygen atoms are the nucleophiles. D. Incorrect. There is no nitrogen reagent. E. Correct. Acetalization uses both diol OH groups and eliminates water. 3 / 20 Which carbon count and functionality identify the organic alkene product? CH₃CHO + preformed Ph3P=CH₂; dry THF Four carbons, one C#C Three carbons, one C=O Three carbons, one C=C Two carbons, one alcohol Two carbons, one C=C Solution: Wittig methylenation gives CH₃CH=CH₂, adding the ylide carbon. A. Incorrect. Neither count nor bond type matches. B. Incorrect. The carbonyl is converted to alkene. C. Correct. Wittig methylenation gives CH₃CH=CH₂, adding the ylide carbon. D. Incorrect. No hydride reduction occurs. E. Incorrect. This omits the methylene donor. 4 / 20 What does protecting one OH allow? HOCH₂CH₂CH₂OH; one end selectively protected; PCC/dry CH₂Cl₂ Conversion of every C-C bond into C=O Addition of one carbon by PCC Making a tertiary alcohol oxidize identically Oxidation of the remaining primary OH without requiring both ends to react Elimination of all oxygen atoms Solution: A compatible protecting group distinguishes the two alcohol functions. A. Incorrect. Only the free alcohol is targeted. B. Incorrect. PCC supplies no carbon to the product. C. Incorrect. Protection does not create a missing carbinol H. D. Correct. A compatible protecting group distinguishes the two alcohol functions. E. Incorrect. Oxidation retains oxygen functionality. 5 / 20 Which linkage defines the condensation product? Cyclohexanone + NH₂OH; mildly acidic buffer, remove water An amide C(=O)-NH₂ Ring carbon C=N-NH₂ Ring carbon C-O-CH₃ Ring carbon C#N with ring opening required Ring carbon C=N-OH Solution: Hydroxylamine forms an oxime at the former carbonyl carbon. A. Incorrect. Oxime formation is not acyl substitution. B. Incorrect. That is a hydrazone-type linkage. C. Incorrect. No methanol alkylation is specified. D. Incorrect. Ordinary oxime formation retains the ring. E. Correct. Hydroxylamine forms an oxime at the former carbonyl carbon. 6 / 20 Why is a ketone obtained rather than a tertiary alcohol? CH₃CH₂CN + CH₃MgBr/dry ether; then acidic hydrolysis Nitriles cannot form C-C bonds Hydrolysis adds a second methyl group The nitrile carbon is deleted The imine salt hydrolyzes only during workup after Grignard is quenched The methyl group is lost as methane in every successful addition Solution: The ketone appears when active organomagnesium reagent is no longer present to add again. A. Incorrect. They do with Grignard reagents. B. Incorrect. Water supplies no methyl. C. Incorrect. It becomes the carbonyl carbon. D. Correct. The ketone appears when active organomagnesium reagent is no longer present to add again. E. Incorrect. It is incorporated in the ketone. 7 / 20 Which product atom arrangement results? A. B. C. D. E. Solution: CN attacks carbonyl carbon and oxygen is protonated, giving a cyanohydrin. A. Incorrect. HCN is not a hydride reagent. B. Incorrect. That requires different connectivity and reduction. C. Incorrect. No nitrile hydrolysis is specified. D. Incorrect. This adds an unsupported carbon and second OH. E. Correct. CN attacks carbonyl carbon and oxygen is protonated, giving a cyanohydrin. 8 / 20 Which carbonyl location follows the usual terminal-alkyne hydration rule? CH₃CH₂C#CH; HgSO₄, aqueous H₂SO₄ Both C1 and C2 as acids C2 of the four-carbon chain C3 with carbon scrambling A new C5 carbonyl C1 as an aldehyde Solution: The product is CH₃COCH₂CH₃ after enol tautomerization. A. Incorrect. No oxidative cleavage is supplied. B. Correct. The product is CH₃COCH₂CH₃ after enol tautomerization. C. Incorrect. The label C2 describes the nearer-end ketone position without scrambling. D. Incorrect. No carbon is added. E. Incorrect. That is the anti-Markovnikov hydration endpoint. 9 / 20 Which original aldehyde atom becomes the ketone carbon? CH₃CHO -> alcohol -> ketone; EtMgBr/H₃O⁺, then PCC The aldehyde oxygen becomes carbon The original aldehydic H A carbon from PCC The terminal reagent methyl carbon The aldehyde carbonyl carbon Solution: It becomes CH(OH) on addition and returns to C=O on oxidation. A. Incorrect. Elements are conserved. B. Incorrect. Hydrogen does not become carbon. C. Incorrect. No carbon is supplied by PCC. D. Incorrect. That remains in the ethyl substituent. E. Correct. It becomes CH(OH) on addition and returns to C=O on oxidation. 10 / 20 Why is the second stage oxidation? CH₃CH₂CHO -> CH₃CH₂CH(OH)CH₃ -> CH₃CH₂COCH₃ It increases C-H bonding at the carbinol carbon It converts oxygen into nitrogen It cleaves both adjacent C-C bonds It adds a carbon atom The carbinol carbon loses C-H and gains C=O bond order Solution: Secondary-alcohol oxidation increases bonding to oxygen while retaining the carbon skeleton. A. Incorrect. That would be reduction. B. Incorrect. No nitrogen is present. C. Incorrect. Those remain intact. D. Incorrect. PCC adds no carbon. E. Correct. Secondary-alcohol oxidation increases bonding to oxygen while retaining the carbon skeleton. 11 / 20 Which net framework change occurs? CH₃CHO -> CH₃CH=CH₂ -> CH₃CH₂CH₂OH; Ph3P=CH₂, then BH₃/H₂O₂ Loss of the original carbonyl carbon One-carbon homologation to a primary alcohol Conversion to an aromatic phenol Formation of an ester without additional steps Two-carbon homologation Solution: Wittig adds CH₂, and anti-Markovnikov hydration places OH at the new terminal carbon. A. Incorrect. It remains in the chain. B. Correct. Wittig adds CH₂, and anti-Markovnikov hydration places OH at the new terminal carbon. C. Incorrect. No aromatic ring exists. D. Incorrect. The endpoint is an alcohol. E. Incorrect. Only one ylide carbon is added. 12 / 20 Which group is directly attached to the final OH-bearing carbon? CH₃CN -> ketone -> alcohol; EtMgBr/acidic hydrolysis, then NaBH₄ Two ethyl groups CH₃ and CH₂CH₃ OH and NH₂ as the two carbon substituents Two phenyl groups CH₃ and CN retained together Solution: Nitrile carbon becomes ketone carbon, joined to its original CH₃ and the reagent ethyl. A. Incorrect. The nitrile substituent was methyl. B. Correct. Nitrile carbon becomes ketone carbon, joined to its original CH₃ and the reagent ethyl. C. Incorrect. Those are not carbon groups. D. Incorrect. No aromatic group is present. E. Incorrect. The nitrile has hydrolyzed to carbonyl before reduction. 13 / 20 What site carries ¹³C after both steps? HC#CCH₂CH₃ -> ketone -> alcohol; terminal alkyne carbon labeled ¹³C A fifth carbon The OH-bearing carbon A lost CO₂ molecule necessarily The CH₃ next to CH(OH) The oxygen Solution: Markovnikov hydration makes the terminal alkyne carbon the ketone methyl, retained after reduction. A. Incorrect. No fifth carbon is introduced. B. Incorrect. That originates from the internal alkyne carbon. C. Incorrect. No cleavage is involved. D. Correct. Markovnikov hydration makes the terminal alkyne carbon the ketone methyl, retained after reduction. E. Incorrect. The label remains carbon. 14 / 20 Which product set matches facial addition? HC#CCH₂CH₃ -> CH₃COCH₂CH₃; NaBH₄ in achiral solvent A pair of E/Z alkenes Only the R alcohol Equal R and S CH₃CH(OH)CH₂CH₃ in the ideal achiral model Only the S alcohol Two constitutional alcohols from attack at different carbons Solution: The planar ketone has enantiotopic faces and no chiral influence is specified. A. Incorrect. The product is an alcohol. B. Incorrect. There is no enantioselective source. C. Correct. The planar ketone has enantiotopic faces and no chiral influence is specified. D. Incorrect. There is no enantioselective source. E. Incorrect. Hydride attacks the same carbonyl carbon from two faces. 15 / 20 Which atom is absent from the isolated acyl-derived aldehyde? CH₃CH₂CO₂CH₃ -> acyl alcohol -> aldehyde; LiAlH₄/H₂O, then PCC The original ester methoxy carbon Every oxygen atom The original carbonyl carbon Both original acyl C-C bonds The original acyl methyl carbon Solution: It leaves in methanol during ester reduction and is excluded when tracking the acyl-derived product. A. Correct. It leaves in methanol during ester reduction and is excluded when tracking the acyl-derived product. B. Incorrect. The aldehyde retains one oxygen. C. Incorrect. It remains as CHO carbon. D. Incorrect. They remain. E. Incorrect. It remains in the aldehyde chain. 16 / 20 Which reagent order is suitable? C₆H₅CHO -> C₆H₅CH(OH)CH₃ -> C₆H₅COCH₃ Water followed by NaCl PCC followed by H₂/Pd only NaCN followed by NaOH only HBr/peroxide followed by ozone CH₃MgBr/dry ether then H₃O⁺; PCC/dry CH₂Cl₂ Solution: Addition makes the secondary alcohol, and PCC oxidizes it to ketone. A. Incorrect. Neither step makes the required C-C bond. B. Incorrect. This does not add the methyl carbon. C. Incorrect. That does not give the specified ketone sequence. D. Incorrect. The substrate is not an alkene for that route. E. Correct. Addition makes the secondary alcohol, and PCC oxidizes it to ketone. 17 / 20 Which carbon becomes the CH₂OH carbon? C₆H₅CHO -> C₆H₅CH=CH₂ -> C₆H₅CH₂CH₂OH The methylene carbon supplied by the Wittig ylide A carbon supplied by peroxide The original aldehyde carbon A carbon supplied by borane A ring carbon Solution: Anti-Markovnikov hydration puts OH at the terminal carbon introduced in olefination. A. Correct. Anti-Markovnikov hydration puts OH at the terminal carbon introduced in olefination. B. Incorrect. H₂O₂ contains no carbon. C. Incorrect. That becomes the benzylic CH₂ after hydration. D. Incorrect. BH₃ contains no carbon. E. Incorrect. The aromatic ring is retained. 18 / 20 Which product reflects Wolff–Kishner deoxygenation? A. B. C. D. E. Solution: The carbonyl carbon becomes CH₂ while all four carbons remain. A. Incorrect. That is reduction to alcohol, not complete deoxygenation. B. Incorrect. That is an oxidation endpoint. C. Incorrect. This deletes the carbonyl carbon. D. Correct. The carbonyl carbon becomes CH₂ while all four carbons remain. E. Incorrect. Wolff-Kishner gives methylene rather than required alkene. 19 / 20 Which nitrogen fate is consistent with acidic hydrolysis of the Grignard adduct? C₆H₅CN -> C₆H₅COCH₃ -> C₆H₅CH(OH)CH₃ Nitrogen leaves the tracked organic framework during hydrolysis Nitrogen supplies the reagent methyl carbon Nitrogen remains as an NH₂ substituent on the alcohol carbon The nitrile stays unchanged through both stages Nitrogen becomes oxygen Solution: The imine-derived intermediate becomes ketone before reduction to alcohol. A. Correct. The imine-derived intermediate becomes ketone before reduction to alcohol. B. Incorrect. It cannot become carbon. C. Incorrect. The specified product has no nitrogen. D. Incorrect. It is consumed in the first transformation. E. Incorrect. Element identity is conserved. 20 / 20 Why is the ketone itself not a tetrahedral stereocenter at C=O? CH₃CH₂CN -> ketone -> alcohol; carbonyl-stage product is chiral only after reduction Every ketone is optically active Carbonyl carbon is trigonal planar It has no carbon substituents It has five sigma bonds Nitrogen forces tetrahedral geometry after hydrolysis Solution: Hydride addition creates a fourth substituent and tetrahedral geometry. A. Incorrect. Simple achiral ketones need not be. B. Correct. Hydride addition creates a fourth substituent and tetrahedral geometry. C. Incorrect. It has two. D. Incorrect. It has three sigma domains. E. Incorrect. The nitrile nitrogen has left the organic product. Your score isThe average score is 0% 0% Restart quiz VIII. ALDEHYDES & KETONES TEST 2 1 / 19 Which isotope placement discriminates hydride addition from mere O-H exchange? HC#CCH₂CH₃ -> CH₃COCH₂CH₃ -> alcohol; NaBD₄, H₂O D on nitrogen D replacing every methyl H Only O-D after aqueous workup A C-D bond at the OH-bearing carbon D as a new carbon atom Solution: Deuteride forms a nonexchangeable carbon-deuterium bond in the addition step. A. Incorrect. No nitrogen is in the product. B. Incorrect. No such exhaustive exchange is specified. C. Incorrect. That label can exchange and does not establish C attack. D. Correct. Deuteride forms a nonexchangeable carbon-deuterium bond in the addition step. E. Incorrect. Deuterium remains hydrogen. 2 / 19 Which final mass-accounting statement is correct? CH₃CO₂CH₂CH₃ -> ethanol; isolate acyl-derived ethanol; PCC It has four carbons because both ester fragments couple It has one carbon because C=O disappears It has three carbons from hydride The aldehyde has two carbons from the acyl fragment It contains the entire ethoxy group plus the acyl group Solution: Hydride reduction retains CH₃CO as CH₃CH₂OH, then oxidation gives CH₃CHO. A. Incorrect. They separate rather than couple. B. Incorrect. The carbonyl carbon is retained. C. Incorrect. Hydride adds no carbon. D. Correct. Hydride reduction retains CH₃CO as CH₃CH₂OH, then oxidation gives CH₃CHO. E. Incorrect. That is the starting ester, not the isolated aldehyde. 3 / 19 Why does this sequence give an aldehyde rather than a ketone? HCHO + EtMgBr/ether, H₃O⁺; then PCC Formaldehyde addition gives a primary alcohol Ethyl group leaves during workup Every Grignard product is tertiary PCC adds a terminal H to a ketone Formaldehyde has a methyl substituent initially Solution: The intermediate CH₃CH₂CH₂OH has a terminal CH₂OH group that oxidizes to CHO. A. Correct. The intermediate CH₃CH₂CH₂OH has a terminal CH₂OH group that oxidizes to CHO. B. Incorrect. The new C-C bond survives. C. Incorrect. Alcohol class depends on the carbonyl substrate. D. Incorrect. The substrate is a primary alcohol. E. Incorrect. It bears two H atoms. 4 / 19 Which net change in carbon count occurs? CH₃CH₂CHO -> CH₃CH₂CH=CH₂ -> terminal alcohol; Wittig then hydroboration-oxidation 3 to 4 3 to 6 3 to 5 3 to 2 3 to 3 Solution: Wittig methylenation adds one carbon; hydration adds none. A. Correct. Wittig methylenation adds one carbon; hydration adds none. B. Incorrect. The substrate is not dimerized. C. Incorrect. Only one carbon donor enters. D. Incorrect. No carbon cleavage is specified. E. Incorrect. This omits the ylide carbon. 5 / 19 Which final reduction product is consistent with one carbonyl and no carbon loss? A. B. C. D. E. Solution: Nitrile addition gives 2-butanone, which reduces to the four-carbon secondary alcohol. A. Correct. Nitrile addition gives 2-butanone, which reduces to the four-carbon secondary alcohol. B. Incorrect. The carbonyl-derived OH is internal, not terminal. C. Incorrect. NaBH₄ adds H rather than methyl to the ketone. D. Incorrect. Only one oxygenated center is formed. E. Incorrect. This loses the reagent methyl carbon. 6 / 19 Which observation would contradict the proposed completed sequence? HC#CCH₂CH₃; Hg²⁺/H₂O then NaBH₄ A secondary carbinol carbon Four carbons No nitrogen A strong retained C#C feature in a pure final product An OH group Solution: Hydration consumes the triple bond before carbonyl reduction. A. Incorrect. That is expected. B. Incorrect. Both steps preserve the carbon count. C. Incorrect. No nitrogen source is involved. D. Correct. Hydration consumes the triple bond before carbonyl reduction. E. Incorrect. The endpoint is an alcohol. 7 / 19 Which proposed PCC endpoint violates the original carbon mapping? CH₃CH₂CO₂CH₃ -> acyl-derived CH₃CH₂CH₂OH -> aldehyde One carbonyl oxygen One aldehydic H CH₃CH₂CHO CH₃CHO Retention of the original acyl methyl Solution: It has only two carbons rather than the three-carbon acyl-derived chain. A. Incorrect. That is expected. B. Incorrect. That fits the product. C. Incorrect. This is the correct aldehyde. D. Correct. It has only two carbons rather than the three-carbon acyl-derived chain. E. Incorrect. That is expected. 8 / 19 Why is regioisomer selection simplified for the initial enolate? CH₃CH₂COCH₂CH₃; 1 eq LDA/THF, -78 C, then 1 eq CH₃I The ketone is an ester Both carbonyl oxygens compete The two alpha CH₂ sites are symmetry-equivalent Iodide supplies the carbonyl carbon Neither side has alpha H Solution: Deprotonation on either side gives equivalent constitutional enolate sites for single methylation. A. Incorrect. It has two carbon substituents on C=O. B. Incorrect. There is only one oxygen. C. Correct. Deprotonation on either side gives equivalent constitutional enolate sites for single methylation. D. Incorrect. The methyl carbon is the electrophilic donor. E. Incorrect. Both have CH₂ alpha carbons. 9 / 19 Which carbon is added relative to the starting aldehyde? CH₃CH₂CHO -> secondary alcohol by CH₃MgBr; PCC -> ketone The original carbonyl carbon The methyl substituent on the ketone carbonyl carbon A carbon inside the original ethyl fragment No carbon is added Two carbons from ether solvent Solution: The Grignard methyl remains bonded after alcohol oxidation. A. Incorrect. It was not newly added. B. Correct. The Grignard methyl remains bonded after alcohol oxidation. C. Incorrect. That fragment was already present. D. Incorrect. The reagent methyl is incorporated. E. Incorrect. Ether is not the donor. 10 / 19 Which final product class follows? HC#CCH₂CH₃ -> ketone -> alcohol -> bromide; Hg²⁺/H₂O, NaBH₄, PBr₃ Vinyl bromide with retained C=C Aryl bromide Primary bromide with carbon migration Acyl bromide Secondary alkyl bromide Solution: The carbonyl site becomes secondary alcohol, then its OH is replaced by Br. A. Incorrect. The specified endpoint is saturated at the former carbonyl site. B. Incorrect. There is no aromatic ring. C. Incorrect. No such migration is required. D. Incorrect. No carbonyl is restored after alcohol formation. E. Correct. The carbonyl site becomes secondary alcohol, then its OH is replaced by Br. 11 / 19 Which step introduces nitrogen? CH₃CHO -> CH₃CH(OH)CH₂CH₃ -> CH₃COCH₂CH₃ -> oxime Every stage introduces one N PCC oxidation Condensation of the ketone with NH₂OH Grignard addition Acid workup alone Solution: Grignard addition and PCC oxidation involve no nitrogen installation. A. Incorrect. Only hydroxylamine supplies retained N. B. Incorrect. PCC does not supply the oxime nitrogen to the target. C. Correct. Grignard addition and PCC oxidation involve no nitrogen installation. D. Incorrect. The reagent supplies carbon and magnesium, not nitrogen. E. Incorrect. H₃O⁺ contains no nitrogen. 12 / 19 What is the net carbonyl-to-carbonyl change for Wittig/BH₃-oxidation/PCC? CH₃CHO -> terminal alkene -> primary alcohol -> aldehyde CH₃CHO becomes CH₃CH₂CHO CH₃CHO is regenerated unchanged CH₃CHO loses one carbon CH₃CHO becomes CH₃COCH₃ CH₃CHO becomes an amide Solution: The ylide adds one carbon, which becomes the terminal alcohol and then aldehyde carbon. A. Correct. The ylide adds one carbon, which becomes the terminal alcohol and then aldehyde carbon. B. Incorrect. The added carbon remains. C. Incorrect. No cleavage occurs. D. Incorrect. Anti-Markovnikov hydration gives a terminal alcohol. E. Incorrect. No nitrogen source is supplied. 13 / 19 Where would a carbon-count check fail if the final drawing were CH₃CH₂CH₂Br? CH₃CH₂C#CH -> CH₃CH₂COCH₃ -> alcohol -> bromide At hydration because water adds carbon The final drawing has lost one original carbon Nowhere, because functional-group changes ignore carbon balance At reduction because NaBH₄ removes carbon At PBr₃ because it must homologate Solution: All stated steps retain the four-carbon skeleton. A. Incorrect. Water adds none. B. Correct. All stated steps retain the four-carbon skeleton. C. Incorrect. Carbon conservation remains necessary. D. Incorrect. It does not. E. Incorrect. It supplies no carbon. 14 / 19 Which structural connection persists through the final condensation? C₆H₅CHO -> C₆H₅CH(OH)CH₃ -> C₆H₅COCH₃ -> C₆H₅C(=NOH)CH₃ Phenyl-O-CH₃ Phenyl-NH-COCH₃ Phenyl-C(oxime)-CH₃ A newly closed three-membered carbon ring Phenyl-CH₂-CH₂NH₂ Solution: Oxime formation replaces carbonyl O functionality while retaining both C-C bonds. A. Incorrect. That would be an ether with different connectivity. B. Incorrect. That is an amide arrangement. C. Correct. Oxime formation replaces carbonyl O functionality while retaining both C-C bonds. D. Incorrect. No cyclization occurs. E. Incorrect. That is a different reduction product. 15 / 19 Which carbon is the final CHO carbon? C₆H₅CHO -> C₆H₅CH=CH₂ -> C₆H₅CH₂CH₂OH -> C₆H₅CH₂CHO A carbon supplied by PCC The carbon supplied by Ph3P=CH₂ The original aldehyde carbon A random ring carbon A carbon supplied by H₂O₂ Solution: It is the terminal alkene carbon, then CH₂OH, then aldehyde carbon. A. Incorrect. PCC adds none. B. Correct. It is the terminal alkene carbon, then CH₂OH, then aldehyde carbon. C. Incorrect. It becomes the benzylic CH₂. D. Incorrect. The ring is retained. E. Incorrect. Peroxide has no carbon. 16 / 19 Which endpoint follows replacement of the alcohol group? A. B. C. D. E. Solution: The benzylic carbon retains the methyl and phenyl groups while OH is replaced. A. Incorrect. That would be an acyl bromide with a different carbon count. B. Incorrect. Only one OH group is replaced; no second Br addition is specified. C. Correct. The benzylic carbon retains the methyl and phenyl groups while OH is replaced. D. Incorrect. That replaces a ring H rather than alcohol OH. E. Incorrect. That moves the functional site to a different carbon. 17 / 19 Which species is the electrophile in oxime formation? CH₃CH₂CH₂OH -> CH₃CH₂CHO -> CH₃CH₂CH=NOH The terminal methyl carbon The PCC chromium as the final oxime center The aldehyde carbonyl carbon Hydroxylamine oxygen as an alkyl halide carbon The oxime OH proton as the carbon donor Solution: Hydroxylamine N attacks the polarized C=O carbon before dehydration. A. Incorrect. It lacks the carbonyl electrophilicity. B. Incorrect. The organic carbonyl carbon is retained. C. Correct. Hydroxylamine N attacks the polarized C=O carbon before dehydration. D. Incorrect. It is neither carbon nor an alkyl halide. E. Incorrect. It supplies no carbon. 18 / 19 Which change distinguishes the nitrile stage from ester Grignard overaddition? CH₃CN -> CH₃COCH₂CH₃ -> alcohol; EtMgBr/acid workup, then NaBH₄ The nitrile carbon leaves as cyanide Two ethyl groups necessarily add before hydrolysis The ketone is released only on aqueous hydrolysis The final alcohol contains no reagent carbon NaBH₄ adds a second ethyl group Solution: The imine-salt intermediate prevents immediate ketone availability during the dry addition stage. A. Incorrect. It is retained as carbonyl carbon. B. Incorrect. Ordinary nitrile addition gives one added organic group in this model. C. Correct. The imine-salt intermediate prevents immediate ketone availability during the dry addition stage. D. Incorrect. The ethyl group is incorporated. E. Incorrect. It adds hydride. 19 / 19 Where is the oxygen label after NaBH₄/H₂O workup? HC#CCH₂CH₃ -> ketone -> alcohol; hydrated in H₂¹⁸O, ideal no later O exchange On every hydrogen In the alcohol oxygen Necessarily lost during hydride attack In a C-C bond Only in a new nitrile Solution: Hydration supplies the carbonyl O, and hydride reduction retains that O as OH. A. Incorrect. Oxygen isotope labeling does not become hydrogen labeling. B. Correct. Hydration supplies the carbonyl O, and hydride reduction retains that O as OH. C. Incorrect. Carbonyl reduction retains O. D. Incorrect. Oxygen is not carbon. E. Incorrect. No nitrile forms. Your score isThe average score is 0% 0% Restart quiz GO TO THE NEXT CHAPTER
