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ORGANIC CHEMISTRY I – PRACTICE EXERCISEAlkene reactions and mechanismsFOR QUESTIONS 1-24, GIVE THE MAJOR ORGANIC PRODUCT OF THE REACTION,PAYING PARTICULAR ATTENTION TO REGIO- AND STEREOCHEMICAL CH3H3O 8)H3O 9)H3O 10)CH3Hg(OAc)2, H2ONaBH411)Hg(OAc)2, H2ONaBH4
12)Hg(OAc)2, CH3OH13)CH3BH3H2O2THFOH-14)(Z)-3-hexene1) BH3 / THF-?2) H2O2 / OH15)H2Pt16)CH3Br2CH2Cl2 (solvent)17)Cl2CH2Cl2 (solvent)18)Cl2H2O19)CH31) CH3CO3H 2) H3O20)PhCO3HCH2Cl2 (solvent)21)CH3OsO4H2O222)CH31) O32) (CH3)2S23)KMnO4(hot, conc.)NaBH4
24)1) O32) (CH3)2S25) Treatment of cyclopentene with peroxybenzoic acidA) results in oxidative cleavage of the ring to produce an acyclic compoundB) yields a meso epoxideC) yields an equimolar mixture of enantiomeric epoxidesD) gives the same product as treatment of cyclopentene with OsO4E) none of the above26) Provide a detailed, step-by-step mechanism for the reaction shown below.Br2HO OHBrBr27) Provide a detailed, step-by-step mechanism for the reaction shown below. H3OHOO28) Provide the reagents necessary to complete the following transformation. The synthesis may involvemore than one step.BrOH?OH29) Provide the reagents necessary to complete the following transformation. The synthesis may involvemore than one step.Br?OH enantiomerOH30) Provide the reagents necessary to convert 3-methyl-2-butanol to 2-methyl-2-butanol. The synthesismay involve more than one step.31) Both (E)- and (Z)-hex-3-ene are subjected to a hydroboration-oxidation sequence. How are theproducts from these two reactions related to each other?A) The (E)- and (Z)-isomers generate the same products but in differing amounts.B) The (E)- and (Z)-isomers generate the same products in exactly the same amounts.C) The products of the two isomers are related as constitutional isomers.D) The products of the two isomers are related as diastereomers.E) The products of the two isomers are not structurally related.
32) What alkene would yield the following products upon ozonolysis?CH3CH2CH2CH2CHO CH2O33) Addition of Br2 to (E)-hex-3-ene producesA) a meso dibromideB) a mixture of enantiomeric dibromides which is optically activeC) a mixture of enantiomeric dibromides which is optically inactiveD) (Z)-3,4-dibromo-3-hexeneE) (E)-3,4-dibromo-3-hexene34) The mechanism for the acid-catalyzed hydration of alkenes is the reverse of the acid-catalyzeddehydration of alcohols. This illustrates the principle of .35) Which of the following is the best reaction sequence to accomplish a Markovnikov addition of water toan alkene with minimal skeletal rearrangement?A) water dilute acidB) water concentrated acidC) oxymercuration-demercurationD) hydroboration-oxidationE) none of the above36) Which of the following additions to alkenes occur(s) specifically in an anti fashion?A) hydroboration-oxidationB) addition of Br2C) addition of H2D) addition of H2O in dilute acidE) both A and B37) Which of the following additions to alkenes occur(s) specifically in an syn fashion?A) dihydroxylation using OsO4, H2O2B) addition of H2C) hydroborationD) addition of HClE) A, B, and C38) HBr can be added to an alkene in the presence of peroxides (ROOR). What function does the peroxideserve in this reaction?A) nucleophileB) electrophileC) radical chain initiatorD) acid catalystE) solvent
ANSWERS1)OH enantiomer
11)OH12)OCH313)CH3 enantiomerOH14)or simplyHOHHO15)16)BrCH3 Brenantiomer17)HClCH2CH318) HH3CenantiomerClOHCl19)OHCH3 enantiomerOH20)OHH3CCH2CH3H enantiomer
21)OHCH3 enantiomerOH22)OCHOH3C23)OO OH24)OCHO H25) B26) This mechanism is best approached by working backwards. The product shown is an ether-bromide,with the oxygen and the bromine atoms on adjacent carbons. Every time two functional groups are onadjacent carbons it suggests the possibility that they might be formed by an addition to the C C doublebond. This can be represented generically thus: ABABAn addition of bromine in the presence of water produces such result, adding Br to one carbon andOH to the other (section 8-11 in the textbook).Br2H2OOH BrThis suggests the possibility that an alcohol could be used instead of water, with similar results,except that this would add Br to one carbon and RO to the other one.Br2ROHOR BrThe mechanism of this reaction would be similar to that with water. Bromine adds first to form athree memebered ring intermediate, followed by nucleophilic attack by the alcohol from the back.Let’s use an unsymmetrical alkene to illustrate the point that the most highly substituted carbon getsthe RO group preferentially.
R BrHOBrROHROBrBr HBrBrBrThe molecule in question has an oxygen (ether group) and a bromine on adjacent carbons. We canmake a similar reasoning as above that such arrangement forms from the reaction between a C Cbond and Br2 in the presence of an alcohol, a group that also happens to be present in the startingmaterial.1Br23HOO4523145The ether and bromine groups are on adjacentcarbons, suggesting that the original double bondwas between C1 and C2. Notice the oxygen on C5The starting material also happens to have5 carbons, with the double bond on C1 and C2,and the oxygen on C5.With this scenario in place, we can now start the mechanism from the first step, which would be theattack of the pi-bond on bromine to form a three membered ring intermediate.HO453 2BrHOBr45Br321 Br1The alcohol group is now poised to attack the three membered ring at the most highly substitutedcarbon. The carbon chain is long enough to allow for flexibility of movement without introducingstrain.324Br3Br15 1O3245BrO1Br5HH24O HBr27) When the starting material gets placed in acid, two (basic) sites can get protonated: the oxygen atomand the pi-bond. The pi-bond is a weak base,. Energy must be expended to break it in order to protonate itand form a carbocation. The oxygen is also basic, but its unshared electrons are not tied up in bonding andare ready to react. Protonation occurs at the oxygen first. O HOH
The pi-bond is now poised to attack the three membered ring from the back at the most highlysubstituted carbon, to open it. At the same time, a tertiary carbocation forms at one of the carbonsoriginally sharing the pi-bond. A new bond forms between carbons 2 and 7, which results information of a new 6-membered ring.23312413o cation455676O7OHHThe tertiary cation undergoes an elimination reaction, losing the adjacent proton to make the newpi-bond present in the product.H342534H2O2 7661517 H3OHOOH28)BrOHNaOCH3 / CH3OHOsO4 / H2O2(E2)or KMnO4 / OH(syn hydroxylation)-OH29)BrOH1) CH3CO3HNaOCH3 / CH3OH (E2) -2) H3O or OH(anti hydroxylation)enantiomerOH30)OHH2SO41) Hg(OAc)2 / H2O(acid-cat. E1)2) NaBH4(oxymerc.-demerc.)Markovnikov alcohol(2-methyl-2-butanol)3-methyl-2-butanol31) B32) CH3CH2CH2CH2CH CH235) C36) BOH33) A37) E34) microscopic reversibility38) C
organic chemistry i - practice exercise alkene reactions and mechanisms for questions 1-24, give the major organic product of the reaction, paying particular attention to regio- and stereochemical outcomes. 1) o hcl ch3oh 2) hcl ch3 3) hcl 4) hcl 5) hbr 6) hcl 7) ch3 h3o 8) h3o 9) h3o 10) hg(oac)2, h2o nabh4 ch3 11) hg(oac)2, h2o nabh4
