Exercício de Química Sintética e Organica Mista (183)

Prévia do material em texto

Which of these form diastereomeric dibromides on anti addition of bromine?
We are told in the problem that two diastereomeric bromides were formed, thus the compound
must be 3-methylcyclopentene.
7.44 Dehydration of this tertiary alcohol can yield 2,3-dimethyl-1-pentene or 2,3-dimethyl-2-pentene.
Only the terminal alkene in this case is chiral.
The 2,3-dimethyl-1-pentene formed in the dehydration reaction must be optically pure because it
arises from optically pure alcohol by a reaction that does not involve any of the bonds to the stereo-
genic center. When optically pure 2,3-dimethyl-1-pentene is hydrogenated, it must yield optically
pure 2,3-dimethylpentane—again, no bonds to the stereogenic center are involved in this step.
CH3CH2CHC(CH3)2
H3C OH
2,3-Dimethyl-2-pentanol
(chiral, optically pure)
2,3-Dimethyl-1-pentene
(chiral, optically pure)
*
CH3CH2CHC
CH3
*
CH2
CH3
2,3-Dimethylpentane
(chiral, optically pure)
CH3CH2CHCH(CH3)2
CH3
*
2,3-Dimethyl-2-pentene
(achiral, optically inactive)
� CH3CH2C
CH3
CH3
CH3
C
H2, Pt
2,3-Dimethylpentane
(chiral, optically inactive)
CH3CH2CHCH(CH3)2
CH3
*
H2, Pt
CH3H
Br BrBrBr
CH3H CH3H
Br2
4-Methylcyclopentene
� (enantiomers)
CH3H
Br
Br
CH3H CH3H
Br
Br
Br2
3-Methylcyclopentene
� (diastereomers)
CH3 CH3Br
Br
Br
Br2
1-Methylcyclopentene
� (enantiomers)
CH3
Br
CH2BrBr
Methylenecyclopentane
CH2
Br2
(only product)
178 STEREOCHEMISTRY