How todraw all stereoisomers of 2-bromo-4-methylpentane?

2 Answers
Feb 24, 2018

Make a model....

Explanation:

You gots...H_3C-stackrel(2)CHBrCH_2stackrel(4)CH(CH_3)stackrel(5)CH_3

AS written stackrel(2)C is CHIRAL, and can generate left and right-handed isomers depending on the geometry of the stackrel(2)C centre.

And once you have drawn one geometry, the interchange of ANY 2 substituents around stackrel2C gives the enantiomer....interchange the substituents again, and you get the enantiomer of an enantiomer, i.e. the original stereoisomer.

Is stackrel(4)C chiral? Why or why not?

Feb 24, 2018

Here's how to do it.

Explanation:

The structure of 2-bromo-4-methylpentane is

2Br4Me2Br4Me

There is only one chiral centre, "C2", so there are only two stereoisomers.
Draw a wedge-dash structure. Put the "Br" atom on a wedge in one structure and on a dash in the other isomer,

RR

SS