Organic Chemistry 1 · Exam review
Substitution and elimination: exam review: practice problems
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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
Draw the mechanism 61
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- Anti E2 of (1R,2R)-1-bromo-1,2-diphenylpropaneDraw the anti-periplanar E2 of (1R,2R)-1-bromo-1,2-diphenylpropane with ethoxide and explain why the only alkene formed is (Z)-1,2-diphenylpropene.
- Anti E2 of (1R,2S)-1-bromo-1,2-diphenylpropaneDraw the anti-periplanar E2 of (1R,2S)-1-bromo-1,2-diphenylpropane with ethoxide and explain why the only alkene formed is (E)-1,2-diphenylpropene.
- E2 of (2R,3R)-2-Bromo-3-methylpentaneDraw the anti-periplanar E2 of (2R,3R)-2-bromo-3-methylpentane with ethoxide and show why the only Zaitsev alkene it can give is (E)-3-methylpent-2-ene.
- E2 of (2R,3S)-2-Bromo-3-methylpentaneDraw the anti-periplanar E2 of (2R,3S)-2-bromo-3-methylpentane with ethoxide and show why this diastereomer gives (Z)-3-methylpent-2-ene as the Zaitsev alkene.
- Zaitsev E2 of 1-bromo-1-methylcycloheptaneDraw the E2 in which ethoxide removes a ring C2 hydrogen of 1-bromo-1-methylcycloheptane as bromide leaves, giving the trisubstituted alkene 1-methylcycloheptene.
- Zaitsev E2 of 2-bromo-4-methylpentane with ethoxideDraw the concerted E2 in which ethoxide removes a C3 hydrogen anti to bromine in (S)-2-bromo-4-methylpentane, giving the Zaitsev alkene (E)-4-methylpent-2-ene.
- E2 of 3-Bromo-3-ethylpentane with EthoxideDraw the E2 of 3-bromo-3-ethylpentane with ethoxide, giving 3-ethylpent-2-ene, and explain why a tertiary bromide with a strong base gives a single alkene and no substitution.
- 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.
- Hofmann E2 of 1-bromo-1-methylcycloheptaneDraw the E2 in which tert-butoxide removes a methyl hydrogen of 1-bromo-1-methylcycloheptane as bromide leaves, giving the exocyclic alkene methylenecycloheptane.
- E2 of 1-bromohexane with tert-butoxideDraw the E2 in which tert-butoxide removes a C2 hydrogen of 1-bromohexane as bromide leaves, giving hex-1-ene, and say why this bulky base eliminates rather than substitutes.
- E2 of 1-Chloro-1-methylcyclopentane with tert-ButoxideDraw the E2 in which tert-butoxide removes a methyl hydrogen of 1-chloro-1-methylcyclopentane, giving the exocyclic Hofmann alkene methylenecyclopentane.
- Hofmann E2 of 2-bromo-2-methylpentane with tert-butoxideDraw the E2 in which tert-butoxide removes a methyl hydrogen of 2-bromo-2-methylpentane as bromide leaves, giving the 1,1-disubstituted Hofmann alkene 2-methylpent-1-ene.
- Hofmann E2 of 2-bromo-4-methylpentane with tert-butoxideDraw the E2 in which tert-butoxide removes a C1 methyl hydrogen of (S)-2-bromo-4-methylpentane as bromide leaves, giving the less substituted alkene 4-methylpent-1-ene.
- E2 of Isobutyl Bromide with tert-ButoxideDraw the one-step E2 in which tert-butoxide removes the tertiary β-hydrogen of isobutyl bromide, giving 2-methylpropene, and explain why this primary bromide eliminates instead of substituting.
- 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.
- 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.
- SN1 Ethanolysis of 1-Bromo-1-phenylpropaneDraw the SN1 ethanolysis of 1-bromo-1-phenylpropane: ionization to the benzylic cation, capture by ethanol and deprotonation to racemic (1-ethoxypropyl)benzene.
- SN1 ethanolysis of 2-bromo-2-methylpentaneDraw the SN1 ethanolysis of 2-bromo-2-methylpentane: ionization, capture of the tertiary cation by ethanol and deprotonation to the ether 2-ethoxy-2-methylpentane.
- SN1 hydrolysis of 2-bromo-2-methylpentaneDraw the SN1 hydrolysis of 2-bromo-2-methylpentane: ionization to the tertiary cation, capture by water and deprotonation by a second water to 2-methylpentan-2-ol.
- SN1 methanolysis of 1-bromo-1-methylcycloheptaneDraw the SN1 methanolysis of 1-bromo-1-methylcycloheptane: ionization to the tertiary cation, capture by methanol and deprotonation to 1-methoxy-1-methylcycloheptane.
- SN1 methanolysis of 3-bromocyclohexeneDraw the SN1 methanolysis of 3-bromocyclohexene: ionization to the allylic cation, its resonance form with the charge on the far allylic carbon, capture by methanol at that carbon (either end gives the same product) and deprotonation to 3-methoxycyclohexene.
- 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.
- SN2 of (S)-2-bromo-4-methylpentane with acetateDraw the SN2 in which an acetate oxygen attacks C2 of (S)-2-bromo-4-methylpentane from the back as bromide leaves, giving the inverted ester (R)-4-methylpentan-2-yl acetate.
- SN2 of 1-bromohexane with sodium azideDraw the SN2 in which a terminal azide nitrogen attacks C1 of 1-bromohexane as bromide leaves, giving 1-azidohexane, and say how DMF speeds the reaction.
- SN2 of 2-Bromo-3-methylbutane with Azide (No Shift)Draw the concerted SN2 of azide on 2-bromo-3-methylbutane and explain why the product is 2-azido-3-methylbutane with the carbon skeleton unchanged.
- 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.
- SN2 of (R)-(1-Bromoethyl)benzene with AzideDraw the backside SN2 attack of azide on (R)-(1-bromoethyl)benzene and explain why the (S)-(1-azidoethyl)benzene forms with inversion rather than racemization.
- SN2 of (S)-2-bromo-4-methylpentane with azideDraw the one-step SN2 in which azide attacks C2 of (S)-2-bromo-4-methylpentane from the back as bromide leaves, giving (R)-2-azido-4-methylpentane with inversion.
- SN2 of (2-bromoethyl)benzene with cyanideDraw the SN2 in which cyanide carbon attacks the CH2Br carbon of (2-bromoethyl)benzene as bromide leaves, giving 3-phenylpropanenitrile with one new C–C bond.
- SN2 of 3-bromocyclohexene with cyanideDraw the SN2 in which cyanide carbon attacks the allylic C–Br carbon of 3-bromocyclohexene as bromide leaves, giving cyclohex-2-ene-1-carbonitrile.
- Benzyl cyanideDraw the SN2 step in which the carbon of cyanide attacks the benzylic CH2 of benzyl chloride and chloride leaves, giving phenylacetonitrile.
- SN2 of Isobutyl Bromide with CyanideDraw the SN2 step in which the cyanide carbon attacks the CH2 of isobutyl bromide and bromide leaves, giving 3-methylbutanenitrile, and explain why substitution beats elimination.
- 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).
- Williamson ether synthesis from 1-bromohexaneDraw the SN2 in which ethoxide attacks C1 of 1-bromohexane as bromide leaves, giving the ether 1-ethoxyhexane, and say why E2 is only minor here.
- 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.
- SN2 of (2-bromoethyl)benzene with methanethiolateDraw the SN2 in which methanethiolate sulfur attacks the CH2Br carbon of (2-bromoethyl)benzene as bromide leaves, giving the sulfide [2-(methylsulfanyl)ethyl]benzene.
- SN2 of Bromocyclopentane with MethanethiolateDraw the SN2 of methanethiolate on bromocyclopentane giving cyclopentyl methyl sulfide, and explain why a secondary ring bromide substitutes here instead of eliminating.
- SN2 of cis-4-tert-butylcyclohexyl bromide with PhS⁻Draw the SN2 in which thiophenoxide attacks the C–Br carbon of cis-1-bromo-4-tert-butylcyclohexane from the equatorial side, giving trans-4-tert-butylcyclohexyl phenyl sulfide.
Predict the product 7
- Dehydrate a tertiary ring alcoholPredict the major E1 alkene from a tertiary cyclic alcohol by comparing endocyclic and exocyclic double bonds.
- SN2 on a ring: cis becomes transPredict the cis/trans outcome of SN2 at a ring stereocentre that is not the only stereocentre.
- Secondary bromide with a weakly basic nucleophilePredict that a secondary bromide gives the SN2 product with a good nucleophile that is a weak base.
- E2 of menthyl chloridePredict the only E2 alkene from menthyl chloride by finding the trans-diaxial hydrogen after a ring flip.
- Neopentyl alcohol with HBrPredict the rearranged bromide formed when a neopentyl alcohol ionises with a methyl shift.
- E2 product of meso-2,3-dibromobutanePredict alkene geometry from anti-periplanar E2 on a specific diastereomer.
- Product of 2-bromo-2-methylhexane with KOtBuPredict the major alkene from a tertiary bromide and a bulky base.
Classify 6
- Which cation has no reason to rearrange?Classify carbocations by whether a 1,2-shift or ring expansion would make them more stable.
- 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.
- Pathway for a secondary bromide with ethoxideClassify the major pathway for a secondary alkyl halide with a strong base.
- Pathway for 1-iodopropane with azide in DMFClassify the pathway of a primary iodide with azide in a polar aprotic solvent.
- Pathway for cyclohexyl tosylate with methoxideClassify the major pathway of a secondary tosylate with a strong, small base.
Compare 6
- Rank alkenes by stabilityRank isomeric alkenes by thermodynamic stability using the degree of substitution and E/Z geometry.
- Rank carbocations by stabilityRank simple alkyl carbocations by stability as the basis for predicting SN1/E1 rates and rearrangements.
- Which isomer gives 1-methylcyclohexene?Compare cis and trans 2-substituted cyclohexyl chlorides to decide which can form the Zaitsev alkene by E2.
- 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.
- Rank four bromides by SN1 rateRank alkyl bromides by SN1 solvolysis rate from the stability of the carbocation each forms.
Calculate 7
- Count the E2 alkenes from 3-bromohexaneCount every constitutional and E/Z alkene isomer that E2 can give from a secondary bromide.
- Enantiomeric excess after SN1Calculate the enantiomeric excess of an SN1 product formed with a slight excess of inversion.
- 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.
- Count E2 alkenes from 2-bromo-2-methylpentaneCount the distinct alkenes an E2 reaction can give, including stereoisomers.
- E2 alkene count for 3-bromo-2-methylpentaneCount the E2 alkenes, E/Z isomers included, that 3-bromo-2-methylpentane can give with ethoxide.
- ee after an SN1 with partial inversionCalculate the enantiomeric excess left after an SN1 reaction with ion-pair shielding.
Find the error 2
- 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.
- Find the error: SN2 on neopentyl bromideDiagnose why a primary but β-branched bromide fails to react by SN2.
Interpret the evidence 3
- Read the rate dataDeduce the rate law from initial-rate data and assign the substitution mechanism.
- Read a rate tableIdentify a substitution mechanism from rate data.
- Read solvolysis rates in aqueous ethanolInterpret how the water content of aqueous ethanol changes an SN1 solvolysis rate.
Work backward 2
- Plan a Williamson ether synthesisChoose the alkoxide and alkyl halide pair that makes an unsymmetrical ether by SN2 without elimination.
- Which halide gives only 2-methylbut-1-ene by E2?Choose the alkyl bromide whose E2 gives 2-methylbut-1-ene as the only alkene.
Plan a synthesis 5
- 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.
- Turn 1-bromobutane into but-1-eneChoose a bulky strong base to divert a primary alkyl bromide from SN2 to E2.
- Make an exocyclic alkeneChoose the base that converts a primary alkyl bromide into its terminal alkene without substitution or rearrangement.
- Best Williamson route to tert-butyl methyl etherChoose the alkoxide/halide pair that gives an ether by SN2 without elimination.
- Choose conditions for SN2 on a secondary bromidePlan conditions that favor SN2 over E2 for 2-bromo-4-methylpentane.
Explain why 3
- Why heat favours eliminationExplain with ΔG = ΔH − TΔS why raising the temperature increases the proportion of elimination.
- Why DMSO speeds up SN2Explain why an anionic nucleophile reacts faster by SN2 in a polar aprotic solvent than in a protic one.
- Why tert-butoxide gives the less substituted alkeneExplain the Hofmann selectivity of a bulky base in E2 on 2-bromo-2-methylpentane.
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