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A) 1-Bromo-3-methylbutane
B) 2-Cyclohexen-1-ol

Pls explain your choice. Thank you

2007-08-31 20:46:26 · 2 answers · asked by sky_blue 1 in Science & Mathematics Chemistry

2 answers

The ring system in compound B forces the presence of a couple of enantiomers. The -OH group has to stick either "up" or "down".

2007-08-31 20:59:54 · answer #1 · answered by Gervald F 7 · 0 5

Enantiomers are stereoisomers that are nonsuperimposable. mirror images.of each other.

A) 1-Bromo-3-methyl butane (or Isoamyl Bromide) has no optically active Carbon center (no enantiomers).

B) (R)-2-Cyclohexen-1-ol and (S)-2-Cyclohexen-1-ol are enantiomers (Carbon 1 is optically active and has 4 "different groups").

2007-09-01 05:04:23 · answer #2 · answered by Richard 7 · 8 0

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