Plzz plzz help. How can we draw the diastereomers of 2,3-dibromobutane.??? Is it follow the #2^n rule for stereoisomers????

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Plzz plzz help. How can we draw the diastereomers of 2,3-dibromobutane.??? Is it follow the #2^n rule for stereoisomers????

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Answer 1 · 2018-01-20T14:40:11

Make a model.....you gots a symmetric disubstituted butane derivative....

Explanation:

As written... $H_{3} 1 C - C H B r C H B r 4 C H_{3}$ $H_3stackrel(1)C-CHBrCHBrstackrel(4)CH_3$ (I had trouble representing this on the editor...the chiral carbons, $2 C$ $stackrel2C$ and $3 C$ $stackrel3C$ are not labelled but their substitution is evident).

Anyway, the beast has two chiral centres... In principle it has $2^{n}$ $2^n$ isomers where $n = 2$ $n=2$ , i.e. $n = the number of chiral carbons$ $n="the number of chiral carbons"$ ; POTENTIALLY 4 diastereomers....

Now $RR$ $"RR"$ is the mirror image of $SS$ $"SS"$ ...and these are non-superposable, and I presume that only the ONE of these enantiomers is biochemically active....and the industrial synthesis would produce this isomer. But the $RS$ $"RS"$ disastereomer is THE SAME as the $SR$ $"SR"$ disastereomer.....given the symmetry of the molecule. With me? You will see this relationship if you make a model.

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Plzz plzz help. How can we draw the diastereomers of 2,3-dibromobutane.??? Is it follow the #2^n rule for stereoisomers????

1 Answer

Explanation:

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