Understanding Enantiomers of 3-Bromo-1-Butanol

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The discussion focuses on the challenges of accurately representing stereochemistry, particularly in drawing enantiomers and converting between different projection types. The main concern is how to correctly assign solid and dashed lines to bonds at the stereogenic center, with the understanding that the lowest priority group should receive the dashed line. There is confusion regarding the placement of atoms, such as bromine, and whether their representation affects the validity of the enantiomers. Consistency in drawing is emphasized, as the orientation of bonds can lead to different representations of the same molecule. Additionally, there is a request for guidance on converting Fischer projections to Newman projections, indicating a broader struggle with visualizing molecular structures in three dimensions. The importance of clarity and accuracy in stereochemical drawings is highlighted, as errors can lead to incorrect interpretations of the compounds.
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We have hit stereochemistry , and I realize I will be screwed unless I can get my drawings down-pat. Okay, so, let's say the question is:

Draw clearly structures for the pair of enantiomers of each:

a) 3-bromo-1-butanol

To start, I drew out the stick-diagram, and identified the stereogenic center. I then realized that I'd have to make a proper 3D representation of it to clearly show the structure in relation to its enantiomer. I hit a road block. At the stereogenic center, you'll have two bonds in the plain of the paper, one going away (dashed) and one towards (wedged).

My question is: How are you to distinguish which gets which type of line? The only hint I have in my head is that you'd give the dashed line to lowest priority at the center (in this case, hydrogen). But, how would you determine if the bromine, say, gets a solid line or a wedged line?

Here's an image of my professors answer:
http://i1120.photobucket.com/albums/l485/Marconis/Picture1.png

He represented the Br with a solid line...why? Why couldn't it be drawn with a wedged line, and be in the place where the methyl group is? If I didn't draw it like that, would my enantiomers be completely false? Please, explain clearly...I'm perishing over here.

Thank you so much in advance.
 
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You can select any two bonds to be on the plane, it doesn't matter which ones you selected - as long as you are consistent and draw the other enantiomer the same way.
 
That's what I was hoping for, it just gets confusing when you draw something one way and then the answer key shows something totally different. I know that in stereochemistry, compounds are very sensitive in how they are to be drawn or else you can totally mess it up, so I wanted to be sure this wasn't the case here. Thank you!
 
Marconis said:
That's what I was hoping for, it just gets confusing when you draw something one way and then the answer key shows something totally different.

I can only sympathize, but as you will get experience you will see when they are the same. It is just a matter of time.
 
Could you perhaps help me with the following? It's not so much a homework question as it is general guidance, so I apologize if I am asking in the wrong forum.

I am having trouble converting Fisher projections into Newman projections, and vice versa. I have gotten several correct, but the method I am using has failed me on others so I think I am doing it incorrectly. Any advice on the following is greatly appreciated:

For (b), the question tells you do draw a correct Newman Projection that correctly shows the stereogenic centers. I am so confused on how to do this:

http://i1120.photobucket.com/albums/l485/Marconis/Picture1-1.png

So it pretty much wants me to finish up the Newman with correct placements of atoms.

The same goes for this:
http://i1120.photobucket.com/albums/l485/Marconis/Picture2.png

Thanks in advance.
 
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