Química Orgânica 1 · Exam review
Substituição e eliminação: revisão para a prova
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- E1 dehydration of 3-methylpentan-3-ol
- E1 dehydration of 2,2-dimethylcyclopentanol with a methyl shift
- Ethoxide converts 2-bromopropane mainly to propene
- E2 of 3-bromopentane gives mainly (E)-pent-2-ene
- E2 of 2-bromo-2-methylbutane with ethoxide (Zaitsev product)
- E2 of neomenthyl chloride gives mainly the Zaitsev alkene
- E2 of 2-bromo-2-methylbutane with tert-butoxide (Hofmann product)
- E2 of trans-1-bromo-2-methylcyclohexane gives 3-methylcyclohexene
- HBr converts 2-methylpentan-3-ol into a rearranged tertiary bromide
Ver todos os 102 problemas de substituição e eliminação: revisão para a prova →
O que você vai aprender
- Decide whether an alkyl halide or alcohol reacts by SN1, SN2, E1 or E2 from the substrate, the nucleophile or base, the solvent and the temperature.
- Predict the product of a substitution, including inversion in SN2 and loss of stereochemistry in SN1, and use rate laws to tell the mechanisms apart.
- Predict the major alkene of an elimination: Zaitsev or Hofmann, E or Z, and the anti-periplanar requirement in cyclohexanes.
- Recognise when a carbocation rearranges by a hydride or methyl shift, draw the rearranged product, and choose conditions that avoid the cation.
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Four pathways, four questions
Every substitution/elimination problem is answered by four questions, asked in order. Substrate: methyl and primary carbons can be attacked from the back (SN2) but cannot form carbocations; tertiary carbons block backside attack but ionise easily (SN1/E1, or E2 with a strong base); secondary carbons can do everything, so the reagent decides. Reagent: a strong base (HO⁻, RO⁻, NH₂⁻) favours E2 at secondary and tertiary carbons; a bulky strong base (tert-butoxide, KOtBu) eliminates even at primary carbons; a good nucleophile that is a weak base (I⁻, Br⁻, CN⁻, N₃⁻, RS⁻) substitutes by SN2; a weak nucleophile and weak base (H₂O, ROH) leaves the substrate to ionise. Solvent: polar aprotic solvents (DMSO, DMF, acetone) speed SN2; polar protic solvents (water, alcohols) stabilise ions and support SN1/E1. Temperature: heating favours elimination, because elimination makes more molecules and so has the larger, positive ΔS.
SN2: one step, backside attack, inversion
In SN2 the nucleophile bonds to carbon from the side opposite the leaving group as the leaving group departs, so the rate law is rate = k[RX][Nu] and a stereocentre is inverted. (R)-2-Bromopentane with methanethiolate gives (S)-2-(methylsulfanyl)pentane: the priorities are in the same order before and after, and the three other groups have turned inside out like an umbrella. Rates fall steeply with crowding: methyl > primary > secondary, while tertiary and neopentyl (a CH₂ next to a tert-butyl group) carbons hardly react. In protic solvents, hydrogen bonding cages small anions, so nucleophilicity follows polarizability: I⁻ > Br⁻ > Cl⁻ > F⁻ in methanol.
SN1 and E1: the carbocation decides
SN1 and E1 share a slow first step: the leaving group departs to give a planar carbocation, so rate = k[RX] and the nucleophile or base does not appear in the rate law. Cation stability, tertiary > secondary ≫ primary > methyl, therefore controls the rate. The flat cation is attacked from either face, so an enantiopure substrate gives a nearly racemic substitution product; a small excess of inversion is common because the departing anion briefly shields one face. The same cation can instead lose a β-hydrogen to the solvent (E1), giving mainly the more substituted alkene; heat increases the share of E1. Alcohols enter these pathways only after protonation turns OH into water, a good leaving group.
E2: anti-periplanar, Zaitsev or Hofmann
In E2 the base removes a β-hydrogen while the leaving group departs, and the C–H and C–X bonds must be anti-periplanar (180° dihedral angle). With small bases (EtO⁻, HO⁻) the more substituted, more stable alkene dominates (Zaitsev): 2-bromo-2-methylbutane gives mainly 2-methylbut-2-ene. A bulky base such as tert-butoxide removes the more exposed hydrogen on a CH₃ group and gives mainly the less substituted alkene (Hofmann), 2-methylbut-1-ene. When a single carbon offers two hydrogens, the conformer that places the larger groups anti to each other wins, so E alkenes are favoured. In cyclohexanes the anti-periplanar pair must be trans-diaxial: the leaving group must be axial, and only β-hydrogens that are axial on the neighbouring carbons can leave. This can override Zaitsev, as when menthyl chloride gives only the less substituted alkene.
Carbocation rearrangements
A carbocation next to a carbon that carries a hydrogen or an alkyl group can shift that group, with its bonding pair, onto the cationic carbon whenever the move gives a more stable cation, usually secondary to tertiary; migration of a ring bond can also expand a strained ring, as when a cyclobutyl group next to the cation becomes a cyclopentyl cation. A hydride shifts when the neighbouring carbon has a hydrogen; a methyl shifts when the neighbour is quaternary. Shifts are faster than capture by most nucleophiles, so 2,2-dimethylcyclopentan-1-ol dehydrates mainly to 1,2-dimethylcyclopentene, and 2-methylpentan-3-ol with HBr gives 2-bromo-2-methylpentane. A simple tertiary cation with no strained ring beside it has nowhere better to go, and a shift that only gives an equally stable cation brings no gain, so neither rearranges. To keep the skeleton intact, avoid the cation: PBr₃, SOCl₂ or a tosylate followed by SN2 replace OH without ionisation.
What to carry forward
Classify the substrate, then the reagent (strong base, good nucleophile and weak base, or neither), then the solvent and temperature. SN2 inverts, SN1 scrambles stereochemistry and may rearrange, E2 needs an anti-periplanar hydrogen, and the size of the base decides between Zaitsev and Hofmann alkenes.
Problemas de prática (102)
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Desenhe o mecanismo 61
Elimination: regiochemistry and geometry
- E1 dehydration of 3-methylpentan-3-olDraw the E1 dehydration of 3-methylpentan-3-ol through its tertiary cation and remove the hydrogen that gives the more substituted alkene, 3-methylpent-2-ene.
- E2 of 3-bromopentane gives mainly (E)-pent-2-eneDraw the anti-periplanar E2 of 3-bromopentane with ethoxide and explain why the E isomer of pent-2-ene is the major alkene.
- E2 of 2-bromo-2-methylbutane with ethoxide (Zaitsev product)Draw the concerted E2 in which ethoxide removes a hydrogen from the CH₂ carbon of 2-bromo-2-methylbutane, giving the more substituted alkene 2-methylbut-2-ene.
- E2 of neomenthyl chloride gives mainly the Zaitsev alkeneExplain why neomenthyl chloride, with chlorine axial in its preferred chair, eliminates quickly by E2 toward the isopropyl-bearing carbon to give the trisubstituted alkene.
- E2 of 2-bromo-2-methylbutane with tert-butoxide (Hofmann product)Draw the concerted E2 in which bulky tert-butoxide removes a methyl hydrogen of 2-bromo-2-methylbutane, giving the less substituted alkene 2-methylbut-1-ene.
- E2 of trans-1-bromo-2-methylcyclohexane gives 3-methylcyclohexeneUse the trans-diaxial requirement of E2 to explain why trans-1-bromo-2-methylcyclohexane gives 3-methylcyclohexene rather than the more substituted 1-methylcyclohexene.
- Tertiary ring alcohol dehydrationDraw the three E1 steps for dehydration of 1-methylcyclopentanol: protonation of OH, loss of water to the tertiary cation, and removal of a ring β-hydrogen to give 1-methylcyclopentene.
- Conjugated alkene by E2Draw the E2 in which ethoxide removes a C2 hydrogen of 1-bromo-1-phenylpropane as bromide leaves, giving (E)-1-phenylprop-1-ene conjugated with the ring.
- Tertiary ring chloride + ethoxideDraw the concerted E2 in which ethoxide removes a ring β-hydrogen of 1-chloro-1-methylcyclopentane as chloride leaves, giving 1-methylcyclopentene.
- Tetrasubstituted alkeneDraw the E2 in which ethoxide removes the C3 hydrogen of 2-bromo-2,3-dimethylbutane as bromide leaves, giving the tetrasubstituted alkene 2,3-dimethylbut-2-ene.
- Zaitsev with a small baseDraw the E2 in which ethoxide removes the hydrogen on C3 of 2-bromo-3-methylbutane, giving the trisubstituted Zaitsev alkene 2-methylbut-2-ene.
- Hofmann with a bulky baseDraw the E2 in which tert-butoxide removes a C1 methyl hydrogen of 2-bromo-3-methylbutane, giving the less substituted Hofmann alkene 3-methylbut-1-ene.
Carbocation rearrangements
- E1 dehydration of 2,2-dimethylcyclopentanol with a methyl shiftDraw the E1 dehydration of 2,2-dimethylcyclopentan-1-ol: protonation, loss of water, the 1,2-methyl shift to a tertiary cation and loss of a proton giving 1,2-dimethylcyclopentene.
- HBr converts 2-methylpentan-3-ol into a rearranged tertiary bromideDraw the SN1 reaction of 2-methylpentan-3-ol with HBr: protonation, loss of water, the 1,2-hydride shift to the tertiary cation and bromide capture giving 2-bromo-2-methylpentane.
Choosing SN1, SN2, E1 or E2
- Ethoxide converts 2-bromopropane mainly to propeneDecide that a secondary bromide with a strong, unhindered base reacts mainly by E2, and draw the concerted elimination to propene.
- Strong base on a tertiary chlorideDraw the concerted E2 in which hydroxide removes a methyl hydrogen of tert-butyl chloride as chloride leaves, giving 2-methylpropene.
- Branched primary bromide with ethoxideDraw the concerted E2 in which ethoxide removes the tertiary β-hydrogen of 1-bromo-2-methylpropane while bromide leaves, giving 2-methylpropene.
- Benzylic SN1 racemisationDraw the SN1 methanolysis of (S)-1-chloro-1-phenylethane: ionization to the benzylic cation, capture by methanol and deprotonation, and explain why the ether is racemic.
- Tertiary bromide hydrolysisDraw the SN1 hydrolysis of 2-bromo-2-methylbutane: ionization to the tert-pentyl cation, capture by water and deprotonation to 2-methylbutan-2-ol.
- Weak base, good nucleophile on a 2° bromideDraw the SN2 step in which hydrosulfide attacks C2 of 2-bromopropane and bromide leaves, giving propane-2-thiol.
Substitution: substrate, nucleophile and solvent
- SN1 hydrolysis of (S)-3-bromo-3-methylhexaneDraw the SN1 hydrolysis of (S)-3-bromo-3-methylhexane and explain why the planar cation gives nearly racemic 3-methylhexan-3-ol.
- SN1 methanolysis of 1-bromo-1-methylcyclopentaneDraw the SN1 solvolysis of 1-bromo-1-methylcyclopentane in methanol: ionisation to the tertiary cation, capture by methanol and deprotonation by methanol.
- Iodide displaces chloride from 1-chlorobutane (Finkelstein reaction)Draw the one-step SN2 attack of iodide on 1-chlorobutane in acetone that gives 1-iodobutane and chloride.
- Methanethiolate displaces bromide from (R)-2-bromopentane with inversionDraw the backside SN2 attack of methanethiolate on (R)-2-bromopentane and assign the configuration of the (S)-2-(methylsulfanyl)pentane formed.
- Propynide alkylation with bromoethaneDraw the SN2 attack of the propynide carbanion on bromoethane that forms a C–C bond and gives pent-2-yne.
- Williamson synthesis of tert-butyl methyl etherDraw the SN2 attack of tert-butoxide on iodomethane and explain why this pairing, not methoxide with a tert-butyl halide, makes tert-butyl methyl ether.
- Piperidine by intramolecular SN2Draw the intramolecular SN2 in which the amine nitrogen of 5-bromopentan-1-amine attacks C5 to close a six-membered ring, then the deprotonation of the piperidinium ion by carbonate.
- ¹⁸O-ethanol traces the new bondDraw the SN1 ethanolysis of tert-butyl iodide in ¹⁸O-labelled ethanol and show that the label ends up on the ether oxygen bonded to the tert-butyl carbon.
- Azide on 1-bromo-3-methylbutaneDraw the SN2 step in which a terminal nitrogen of azide attacks C1 of 1-bromo-3-methylbutane and bromide leaves, giving 1-azido-3-methylbutane.
- Benzyl cyanideDraw the SN2 step in which the carbon of cyanide attacks the benzylic CH2 of benzyl chloride and chloride leaves, giving phenylacetonitrile.
- Butyl ethyl ether by SN2Draw the concerted SN2 step in which ethoxide attacks C1 of 1-bromobutane and bromide leaves, giving 1-ethoxybutane (butyl ethyl ether).
Eliminação: regioquímica e geometria
- E2 anti do (1R,2R)-1-bromo-1,2-difenilpropanoDesenhe a E2 antiperiplanar do (1R,2R)-1-bromo-1,2-difenilpropano com etóxido e explique por que o único alceno formado é o (Z)-1,2-difenilpropeno.
- E2 anti do (1R,2S)-1-bromo-1,2-difenilpropanoDesenhe a E2 antiperiplanar do (1R,2S)-1-bromo-1,2-difenilpropano com etóxido e explique por que o único alceno formado é o (E)-1,2-difenilpropeno.
- E2 do (2R,3R)-2-bromo-3-metilpentanoDesenhe a E2 antiperiplanar do (2R,3R)-2-bromo-3-metilpentano com etóxido e mostre por que o único alceno de Zaitsev que ele pode gerar é o (E)-3-metilpent-2-eno.
- E2 do (2R,3S)-2-bromo-3-metilpentanoDesenhe a E2 antiperiplanar do (2R,3S)-2-bromo-3-metilpentano com etóxido e mostre por que esse diastereoisômero gera o (Z)-3-metilpent-2-eno como alceno de Zaitsev.
- E2 de Zaitsev do 1-bromo-1-metilciclo-heptanoDesenhe a E2 em que o etóxido remove um hidrogênio do C2 do anel do 1-bromo-1-metilciclo-heptano enquanto o brometo sai, gerando o alceno trissubstituído 1-metilciclo-hepteno.
- E2 de Hofmann do 1-bromo-1-metilciclo-heptanoDesenhe a E2 em que o terc-butóxido remove um hidrogênio da metila do 1-bromo-1-metilciclo-heptano enquanto o brometo sai, gerando o alceno exocíclico metilenociclo-heptano.
- E2 do 1-cloro-1-metilciclopentano com terc-butóxidoDesenhe a E2 em que o terc-butóxido remove um hidrogênio da metila do 1-cloro-1-metilciclopentano, gerando o alceno exocíclico de Hofmann metilenociclopentano.
- E2 de Hofmann do 2-bromo-2-metilpentano com terc-butóxidoDesenhe a E2 em que o terc-butóxido remove um hidrogênio da metila do 2-bromo-2-metilpentano enquanto o brometo sai, gerando o alceno de Hofmann 1,1-dissubstituído 2-metilpent-1-eno.
Escolhendo entre SN1, SN2, E1 ou E2
- E2 de Zaitsev do 2-bromo-4-metilpentano com etóxidoDesenhe a E2 concertada em que o etóxido remove um hidrogênio do C3 anti ao bromo no (S)-2-bromo-4-metilpentano, gerando o alceno de Zaitsev (E)-4-metilpent-2-eno.
- E2 do 3-bromo-3-etilpentano com etóxidoDesenhe a E2 do 3-bromo-3-etilpentano com etóxido, que gera o 3-etilpent-2-eno, e explique por que um brometo terciário com uma base forte gera um único alceno e nenhuma substituição.
- E2 do 1-bromo-hexano com terc-butóxidoDesenhe a E2 em que o terc-butóxido remove um hidrogênio do C2 do 1-bromo-hexano enquanto o brometo sai, gerando o hex-1-eno, e explique por que essa base volumosa elimina em vez de substituir.
- E2 de Hofmann do 2-bromo-4-metilpentano com terc-butóxidoDesenhe a E2 em que o terc-butóxido remove um hidrogênio da metila C1 do (S)-2-bromo-4-metilpentano enquanto o brometo sai, gerando o alceno menos substituído 4-metilpent-1-eno.
- E2 do brometo de isobutila com terc-butóxidoDesenhe a E2 de uma etapa em que o terc-butóxido remove o hidrogênio β terciário do brometo de isobutila, gerando 2-metilpropeno, e explique por que esse brometo primário sofre eliminação em vez de substituição.
- Etanólise SN1 do 1-bromo-1-fenilpropanoDesenhe a etanólise SN1 do 1-bromo-1-fenilpropano: ionização ao cátion benzílico, captura pelo etanol e desprotonação ao (1-etoxipropil)benzeno racêmico.
- Etanólise SN1 do 2-bromo-2-metilpentanoDesenhe a etanólise SN1 do 2-bromo-2-metilpentano: ionização, captura do cátion terciário pelo etanol e desprotonação ao éter 2-etoxi-2-metilpentano.
- Hidrólise SN1 do 2-bromo-2-metilpentanoDesenhe a hidrólise SN1 do 2-bromo-2-metilpentano: ionização ao cátion terciário, captura pela água e desprotonação por uma segunda água ao 2-metilpentan-2-ol.
- Metanólise SN1 do 1-bromo-1-metilciclo-heptanoDesenhe a metanólise SN1 do 1-bromo-1-metilciclo-heptano: ionização ao cátion terciário, captura pelo metanol e desprotonação ao 1-metoxi-1-metilciclo-heptano.
- Metanólise SN1 do 3-bromociclo-hexenoDesenhe a metanólise SN1 do 3-bromociclo-hexeno: ionização ao cátion alílico, sua forma de ressonância com a carga no carbono alílico da outra ponta, captura pelo metanol nesse carbono (qualquer uma das pontas dá o mesmo produto) e desprotonação ao 3-metoxiciclo-hexeno.
- SN2 do (S)-2-bromo-4-metilpentano com acetatoDesenhe a SN2 em que um oxigênio do acetato ataca o C2 do (S)-2-bromo-4-metilpentano por trás enquanto o brometo sai, gerando o éster invertido acetato de (R)-4-metilpentan-2-ila.
- SN2 do 2-bromo-3-metilbutano com azida (sem rearranjo)Desenhe a SN2 concertada da azida no 2-bromo-3-metilbutano e explique por que o produto é o 2-azido-3-metilbutano, com o esqueleto carbônico inalterado.
- SN2 do (S)-2-bromo-4-metilpentano com azidaDesenhe a SN2 de uma etapa em que a azida ataca o C2 do (S)-2-bromo-4-metilpentano por trás enquanto o brometo sai, gerando o (R)-2-azido-4-metilpentano com inversão.
- Síntese de éteres de Williamson a partir do 1-bromo-hexanoDesenhe a SN2 em que o etóxido ataca o C1 do 1-bromo-hexano enquanto o brometo sai, gerando o éter 1-etoxi-hexano, e explique por que a E2 é apenas minoritária aqui.
Substituição: substrato, nucleófilo e solvente
- SN2 do 1-bromo-hexano com azida de sódioDesenhe a SN2 em que um nitrogênio terminal da azida ataca o C1 do 1-bromo-hexano enquanto o brometo sai, gerando o 1-azido-hexano, e explique como o DMF acelera a reação.
- SN2 do (R)-(1-bromoetil)benzeno com azidaDesenhe o ataque SN2 pelo lado oposto da azida ao (R)-(1-bromoetil)benzeno e explique por que o (S)-(1-azidoetil)benzeno se forma com inversão, e não com racemização.
- SN2 do (2-bromoetil)benzeno com cianetoDesenhe a SN2 em que o carbono do cianeto ataca o carbono CH2Br do (2-bromoetil)benzeno enquanto o brometo sai, gerando a 3-fenilpropanonitrila com uma nova ligação C–C.
- SN2 do 3-bromociclo-hexeno com cianetoDesenhe a SN2 em que o carbono do cianeto ataca o carbono alílico ligado ao Br do 3-bromociclo-hexeno enquanto o brometo sai, gerando a ciclo-hex-2-eno-1-carbonitrila.
- SN2 do brometo de isobutila com cianetoDesenhe a etapa SN2 em que o carbono do cianeto ataca o CH2 do brometo de isobutila e o brometo sai, gerando a 3-metilbutanonitrila, e explique por que a substituição prevalece sobre a eliminação.
- SN2 do (2-bromoetil)benzeno com metanotiolatoDesenhe a SN2 em que o enxofre do metanotiolato ataca o carbono CH2Br do (2-bromoetil)benzeno enquanto o brometo sai, gerando o sulfeto [2-(metilsulfanil)etil]benzeno.
- SN2 do bromociclopentano com metanotiolatoDesenhe a SN2 do metanotiolato no bromociclopentano, que gera o sulfeto de ciclopentila e metila, e explique por que, aqui, um brometo secundário de anel sofre substituição em vez de eliminação.
- SN2 do brometo de cis-4-terc-butilciclo-hexila com PhS⁻Desenhe a SN2 em que o tiofenóxido ataca o carbono ligado ao Br do cis-1-bromo-4-terc-butilciclo-hexano pelo lado equatorial, gerando o sulfeto de trans-4-terc-butilciclo-hexila e fenila.
Preveja o produto 7
Elimination: regiochemistry and geometry
- Dehydrate a tertiary ring alcoholPredict the major E1 alkene from a tertiary cyclic alcohol by comparing endocyclic and exocyclic double bonds.
- E2 of menthyl chloridePredict the only E2 alkene from menthyl chloride by finding the trans-diaxial hydrogen after a ring flip.
Substitution: substrate, nucleophile and solvent
- SN2 on a ring: cis becomes transPredict the cis/trans outcome of SN2 at a ring stereocentre that is not the only stereocentre.
Choosing SN1, SN2, E1 or E2
- Secondary bromide with a weakly basic nucleophilePredict that a secondary bromide gives the SN2 product with a good nucleophile that is a weak base.
Carbocation rearrangements
- Neopentyl alcohol with HBrPredict the rearranged bromide formed when a neopentyl alcohol ionises with a methyl shift.
E2 stereochemistry
- E2 product of meso-2,3-dibromobutanePredict alkene geometry from anti-periplanar E2 on a specific diastereomer.
Eliminação: regioquímica e geometria
- Produto do 2-bromo-2-metil-hexano com KOtBuPreveja o alceno principal formado a partir de um brometo terciário e uma base volumosa.
Classificar 6
Carbocation rearrangements
- Which cation has no reason to rearrange?Classify carbocations by whether a 1,2-shift or ring expansion would make them more stable.
Choosing SN1, SN2, E1 or E2
- 1-Bromopentane with cyanide in DMSOClassify the reaction of an unhindered primary bromide with cyanide in a polar aprotic solvent as SN2.
- tert-Butyl chloride in methanolClassify the solvolysis of a tertiary alkyl chloride in neutral methanol as SN1.
SN1/SN2/E1/E2 decisions
- Pathway for a secondary bromide with ethoxideClassify the major pathway for a secondary alkyl halide with a strong base.
Escolhendo entre SN1, SN2, E1 ou E2
- Mecanismo do 1-iodopropano com azida em DMFClassifique o mecanismo de um iodeto primário com azida em um solvente polar aprótico.
- Mecanismo do tosilato de ciclo-hexila com metóxidoClassifique o mecanismo principal de um tosilato secundário com uma base forte e pequena.
Comparar 6
Elimination: regiochemistry and geometry
- Rank alkenes by stabilityRank isomeric alkenes by thermodynamic stability using the degree of substitution and E/Z geometry.
- Which isomer gives 1-methylcyclohexene?Compare cis and trans 2-substituted cyclohexyl chlorides to decide which can form the Zaitsev alkene by E2.
Carbocation rearrangements
- Rank carbocations by stabilityRank simple alkyl carbocations by stability as the basis for predicting SN1/E1 rates and rearrangements.
Substitution: substrate, nucleophile and solvent
- Rank halides as nucleophiles in methanolRank the halide ions by SN2 nucleophilicity in a protic solvent using solvation and polarizability.
- Rank bromides by SN2 rateRank alkyl bromides by SN2 rate, including the very slow neopentyl case where crowding is one carbon away.
Substituição: substrato, nucleófilo e solvente
- Ordene quatro brometos pela velocidade de SN1Ordene brometos de alquila pela velocidade de solvólise SN1 a partir da estabilidade do carbocátion que cada um forma.
Calcular 7
Elimination: regiochemistry and geometry
- Count the E2 alkenes from 3-bromohexaneCount every constitutional and E/Z alkene isomer that E2 can give from a secondary bromide.
Substitution: substrate, nucleophile and solvent
- Enantiomeric excess after SN1Calculate the enantiomeric excess of an SN1 product formed with a slight excess of inversion.
Choosing SN1, SN2, E1 or E2
- Change concentrations in an SN1 reactionUse the first-order SN1 rate law to show that the nucleophile concentration does not affect the rate.
- Change concentrations in an SN2 reactionUse the second-order SN2 rate law to find how the rate changes when both concentrations change.
SN1/SN2/E1/E2 decisions
- Count E2 alkenes from 2-bromo-2-methylpentaneCount the distinct alkenes an E2 reaction can give, including stereoisomers.
Eliminação: regioquímica e geometria
- Número de alcenos de E2 do 3-bromo-2-metilpentanoConte os alcenos de E2, incluindo isômeros E/Z, que o 3-bromo-2-metilpentano pode gerar com etóxido.
SN1 stereochemistry
- ee after an SN1 with partial inversionCalculate the enantiomeric excess left after an SN1 reaction with ion-pair shielding.
Encontre o erro 2
Choosing SN1, SN2, E1 or E2
- Find the error: SN2 at a tertiary carbonIdentify why a proposed SN2 reaction of a tertiary bromide with methoxide is wrong and what happens instead.
Substituição: substrato, nucleófilo e solvente
- Encontre o erro: SN2 no brometo de neopentilaDiagnostique por que um brometo primário, mas ramificado no carbono β, não reage por SN2.
Interprete as evidências 3
Choosing SN1, SN2, E1 or E2
- Read the rate dataDeduce the rate law from initial-rate data and assign the substitution mechanism.
Rates and mechanism evidence
- Read a rate tableIdentify a substitution mechanism from rate data.
Substituição: substrato, nucleófilo e solvente
- Interprete velocidades de solvólise em etanol aquosoInterprete como o teor de água do etanol aquoso altera a velocidade de uma solvólise SN1.
Trabalhe de trás para frente 2
Substitution: substrate, nucleophile and solvent
- Plan a Williamson ether synthesisChoose the alkoxide and alkyl halide pair that makes an unsymmetrical ether by SN2 without elimination.
Eliminação: regioquímica e geometria
- Qual haleto gera apenas 2-metilbut-1-eno por E2?Escolha o brometo de alquila cuja E2 gera o 2-metilbut-1-eno como único alceno.
Planeje uma síntese 5
Carbocation rearrangements
- Replace OH by Br without a shiftChoose a reagent that converts a secondary alcohol into the alkyl bromide without forming a carbocation that would rearrange.
Choosing SN1, SN2, E1 or E2
- Turn 1-bromobutane into but-1-eneChoose a bulky strong base to divert a primary alkyl bromide from SN2 to E2.
Elimination: regiochemistry and geometry
- Make an exocyclic alkeneChoose the base that converts a primary alkyl bromide into its terminal alkene without substitution or rearrangement.
Williamson ether planning
- Best Williamson route to tert-butyl methyl etherChoose the alkoxide/halide pair that gives an ether by SN2 without elimination.
Escolhendo entre SN1, SN2, E1 ou E2
- Escolha condições para SN2 em um brometo secundárioPlaneje condições que favoreçam a SN2 em relação à E2 para o 2-bromo-4-metilpentano.
Explique por quê 3
Choosing SN1, SN2, E1 or E2
- Why heat favours eliminationExplain with ΔG = ΔH − TΔS why raising the temperature increases the proportion of elimination.
Substitution: substrate, nucleophile and solvent
- Why DMSO speeds up SN2Explain why an anionic nucleophile reacts faster by SN2 in a polar aprotic solvent than in a protic one.
Eliminação: regioquímica e geometria
- Por que o terc-butóxido gera o alceno menos substituídoExplique a seletividade de Hofmann de uma base volumosa na E2 do 2-bromo-2-metilpentano.
Travou em substituição e eliminação: revisão para a prova? Escaneie o problema.
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