Final answer:
To draw the stereoisomers of 2-bromo-4-methylpentane, one must identify the chiral center and configure the substituents in both R and S configurations since there is one chiral center, resulting in two stereoisomers.
Step-by-step explanation:
To draw all stereoisomers of 2-bromo-4-methylpentane, we need to take into account that the molecule contains a chiral center at the second carbon atom (the one with the bromine attached). A chiral center is an atom that has four different groups attached to it, which is the case for the carbon atom attached to the bromine (Br), methyl (CH3), ethyl (CH2CH3), and hydrogen (H) groups in this molecule. Since there is one chiral center, there will be two stereoisomers: one R-(rectus) and one S-(sinister) configuration.
The R configuration has the groups prioritized by atomic number arranged clockwise around the chiral center when looking from a direction where the lowest priority group (in this case, the hydrogen) is pointed away from you. Conversely, for the S configuration, the same groups are arranged counterclockwise.
Remember that in a Fischer projection, the horizontal lines represent bonds that are coming towards the viewer, and the vertical lines represent bonds extending away from the viewer. In a wedge-dash notation, solid wedges represent bonds protruding out of the page towards the viewer, and dashed lines represent bonds extending behind the page.
Here are the two stereoisomers:
- R-2-bromo-4-methylpentane: The Br, CH3, and CH2CH3 are arranged clockwise on the second carbon when hydrogen is pointed away from you.
- S-2-bromo-4-methylpentane: The Br, CH3, and CH2CH3 are arranged counterclockwise on the second carbon when hydrogen is pointed away from you.