Stereoselective and stereospecific reactions. CH3CH=CHCH3 + Br2 2-butene
Stereoselective and stereospecific reactions. CH3CH=CHCH3 + Br2 2-butene 2 geometric isomers cis- and trans- * * CH3CHCHCH3 Br Br 2,3-dibromobutane 3 stereoisomers (S,S)-, (R,R)-, and (R,S)meso- H CH3 \ / C=C / \ CH3 H trans-2-butene CH3 CH3
\ / C=C / \ H H cis-2-butene H CH3 \ / C=C only product / \ CH3 H + Br2 meso-2,3-dibromobutane trans-2-butene A reaction that yields predominately one stereoisomer (or one pair of enantiomers) of several diastereomers is called a stereoselective reaction. In this case the meso- product is produced and not the other two diastereomers.
H H \ / C = C + Br2 (S,S)- & (R,R)-2,3-dibromobutane / CH3 \ racemic modification only products CH3 cis-2-butene A reaction in which stereochemically different molecules react differently is called a stereospecific reaction. In this case the cis- and trans- stereoisomers give different products. The fact that the addition of halogens to alkenes is both stereoselective and stereospecific gives us additional information about the stereochemistry of the addition and the mechanism for the reaction.
C C C C anti-addition C C C C syn-addition CH3 H3C C C H H H3C C H H C CH3 H
Br Br H CH3 Br2 Br2 CH3 Is the addition of Br2 syn or anti? + CH3 Br H H Br CH3 CH3 H Br H Br CH3 H X
H \ / C=C / \ CH3 X CH3 X | H H C C CH3 CH3 | X anti-addition of X2 to the cis-isomer H X CH3 \ / C=C / \
CH3 X H X | CH3 H CC H CH3 | X anti-addition of X2 to the trans-isomer Br2 H H3C H3C anti- H3C H H3C Br2 CH3
H H Br H H3C H3C H3C Br Br H CH3 H Br Br H CH3 H H3C
Br H Br H3C H H Br anti- H CH3 Br H3C H Br rotate about C2-C3 to get to Fischer projection! CH3 H Br
H Br CH3 In determining whether a stereoselective addition is syn- or anti- you cannot simply look at the Fischer projection. Remember it is often necessary to rotate about a carboncarbon bond to get a molecule into the conformation that corresponds to the Fischer projection! Use your model kit to verify! What does the stereochemistry tell us about the mechanism of addition of halogens to alkenes? RDS 1) C C + X X C C X + "halonium ion" 2)
X + C C X anti-addition X C C X X Hydroxylation of alkenes: CH3CH=CHCH3 + KMnO4 2-butene 2 geometric isomers * * CH3CH-CHCH3 OH OH 2,3-butanediol 3 stereoisomers
cis-2-butene + KMnO4 2,3-butanediol mp 34oC trans-2-butene + KMnO4 2,3-butanediol mp 19oC 2,3-butanediol ( mp 19oC ) is separable into enantiomers. CH3 CH3 H OH HO HO H H CH3 H H
CH3 C C H H H OH H OH CH3 trans-2-butene + KMnO4 (S,S) & (R,R)-2,3-dihydroxybutane mp 19o H3C H C C H CH3 CH3 CH3 H OH
HO H CH3 stereoselective and stereospecific + HO H H OH CH3 Is hydroxylation with KMnO4 syn- or anti-? H O O CH3 \ / C=C / \ CH3 H
OH OH | | H C C CH3 CH3 H syn-oxidation of the trans-isomer H O O H \ / C=C / \ CH3 CH3 OH OH | | H CC H CH3 CH3 syn-oxidation of the cis-isomer
cis-2-butene + HCO3H 2,3-butanediol mp 19oC trans-2-butene + HCO3H 2,3-butanediol mp 34oC 2,3-butanediol mp 19oC is separable into enantiomers. CH3 CH3 H OH HO HO H H CH3 H
H OH OH H OH CH3 CH3 CH3 (S,S) (R,R) meso mp 19oC mp 34oC Oxidation with KMnO4 syn-oxidation
cis-2-butene meso-2,3-dihydroxybutane trans-butene (S,S)- & (R,R)-2,3-dihydroxybutane Oxidation with HCO2OH gives the opposite cis-2-butene (S,S)- & (R,R)-2,3-dihydroxybutane trans-2-butene meso-2,3-dihydroxybutane Oxidation with HCO2OH is anti-oxidation. | | CC | | O O Mn O O | | CC O hydroxylation with KMnO4 is syn- because of an intermediate
permanganate addition product. hydroxylation with HCO2OH is anti- because of an intermediate epoxide. * * CH2-CH-CH-CH=O | | | OH OH OH Four carbon sugar, an aldotetrose. Two chiral centers, four stereoisomers CHO CHO H OH HO H H OH
HO H CH2OH D-erythrose CH2OH L-erythrose CHO HO H H OH CH2OH D-threose CHO H HO OH H CH2OH
L-threose X X X X erythro- X X X X threo- * * C6H5CHCHC6H5 + KOH(alc) C6H5CH=CC6H5 Br CH3 CH3 1-bromo-1,2-diphenylpropane 4 stereoisomers
1,2-diphenylpropene 2 stereoisomers (E)- & (Z)- dehydrohalogenation of an alkyl halide via E2 mechanism C6H5 C6H5 C6H5 CH3 H H CH3 Br H H Br
C6H5 C6H5 CH3 H (E)- H H CH3 Br Br H C6H5 erythroC6H5 CH3 \ / C=C /
\ H C6H5 C6H5 C6H5 threo- C6H5 \ C6H5 / C=C / \ H CH3 (Z)- C6H5 C6H5 CH3 H
H CH3 Br H H Br C6H5 KOH(alc) C6H5 erythro- C6H5 \ H / C=C
(Z)- C6H5 / \ CH3 C6H5 C6H5 CH3 H H CH3 H Br Br H C6H5
KOH(alc) C6H5 threo- C6H5 \ H / C=C (E)- CH3 / \ C6H5 E2 is both stereoselective and stereospecific. 100% anti-elimination of the H & Br: C6H5 CH3 H
CH3 C6H5 Br H C6H5 erythro- HO- Br | CC H | C6H5 H CH3 \ H / C=C / \ C6H5
C6H5 (Z)- C6H5 CH3 H H Br C6H5 threo- CH3 C6H5 HO- Br | C C C6H5 | H H CH3 C6H5
\ / C=C / \ C6H5 CH3 (E)- Once again, you must rotate about the CC bond in the Fischer projection to get the H & Br anti to one another. E2 is an anti-elimination. The hydrogen and the halogen must be on opposite sides of the molecule before the E2 elimination can take place. This makes sense as both the base and the leaving group are negatively charged. Therefore they would try to be as far apart as possible. In addition, the leaving group is large and there is more room for the removal of the adjacent proton if it is on the opposite side from the leaving group. Mechanism = elimination, bimolecular E2 X C C
RDS H base: 100% anti-elimination! C C + H:base + :X CH3 H D H Br CH3 CH3 D H Br H CH3 CH3 D
H H Br CH3 CH3 H D Br H CH3 base H3C H base H3C H C C C C H
D + CH3 H D H + CH3 H C C C C CH3 CH3 CH3 CH3
Addition of halogens anti-addition to alkenes Hydoxylation with KMnO4 syn-oxidation Hydroxylation with HCO2OH anti-oxidation Dehydrohalogenation anti-elimination of alkyl halides E2 stereospecific and stereoselective problems http://chemistry2.csudh.edu/organic/synanti/startsynanti.html http://chemistry2.csudh.edu/organic/synanti/startsynanti.html
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