Apr 5, 2010
60
1
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Pre-Dental
Guys this would be an example of a diastereomer right ? (R,R) (S,R)

This question ask which produces two diastereomers, to me it seems like it would produce enantiamers, could someone help me out?
 

rmm30

7+ Year Member
Jul 28, 2010
276
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Pre-Dental
Guys this would be an example of a diastereomer right ? (R,R) (S,R)

This question ask which produces two diastereomers, to me it seems like it would produce enantiamers, could someone help me out?
Enantiomers = Opposite configurations at EVERY chiral center. So if a stereoisomer is not an enantiomer, it is a diastereomer. The textbook def of diastereomer is objects that stereoisomers not related by mirror image. But if you remember that one rule you should be ok. I don't get the 2nd part of the question. I think I'd have to see it. But you're right about the 1st.
 

rmm30

7+ Year Member
Jul 28, 2010
276
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Pre-Dental
Ah here we go. I remember this problem. It's a tricky one. First you have to know a little something about this reaction. It is Bromination / addition across a double bond that adds anti. So when Br2 adds and no chiral centers are created you have 1 product. When Br2 adds across a double bond generating 1 chiral center you have 2 products every time. When 2 chiral centers are created you can have at most 2 products; 2 enantiomers. ("At most" 2 products means you could have created a meso compound and in that case you would have 1 product). Anyway adding Br2 to choices A, B, D, will generate 0, 1, and 2 chiral centers respectively. However, a Br2 to compound C will result in a product with 3 stereocenters. Check out what it looks like in the solutions and convince yourself that choice C has 3 stereocenters: 2 that contain Br and a 3rd containing the ETHYL substituent. Now becasuse you have 3 chiral centers, you can have diastereomers. Rem enantiomer = opp config at EVERY chiral center. Look at the eg in the solution. You'll see those are diastereomers.


This problem doesn't give you the configuration of the ethyl group in choice C, which makes it a little awkward but it does let you practice a bunch of different concepts at once. Which is why destroyer is so badass. keep it up.
 
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Apr 5, 2010
60
1
Status
Pre-Dental
Ah here we go. I remember this problem. It's a tricky one. First you have to know a little something about this reaction. It is Bromination / addition across a double bond that adds anti. So when Br2 adds and no chiral centers are created you have 1 product. When Br2 adds across a double bond generating 1 chiral center you have 2 products every time. When 2 chiral centers are created you can have at most 2 products; 2 enantiomers. ("At most" 2 products means you could have created a meso compound and in that case you would have 1 product). Anyway adding Br2 to choices A, B, D, will generate 0, 1, and 2 chiral centers respectively. However, a Br2 to compound C will result in a product with 3 stereocenters. Check out what it looks like in the solutions and convince yourself that choice C has 3 stereocenters: 2 that contain Br and a 3rd containing the ETHYL substituent. Now becasuse you have 3 chiral centers, you can have diastereomers. Rem enantiomer = opp config at EVERY chiral center. Look at the eg in the solution. You'll see those are diastereomers.


This problem doesn't give you the configuration of the ethyl group in choice C, which makes it a little awkward but it does let you practice a bunch of different concepts at once. Which is why destroyer is so badass. keep it up.

Hey! Dang destroyer is tricky, that makes sense now, thank you!
 

jwnichols21

7+ Year Member
Aug 11, 2010
272
2
Status
Pre-Dental
Ah here we go. I remember this problem. It's a tricky one. First you have to know a little something about this reaction. It is Bromination / addition across a double bond that adds anti. So when Br2 adds and no chiral centers are created you have 1 product. When Br2 adds across a double bond generating 1 chiral center you have 2 products every time. When 2 chiral centers are created you can have at most 2 products; 2 enantiomers. ("At most" 2 products means you could have created a meso compound and in that case you would have 1 product). Anyway adding Br2 to choices A, B, D, will generate 0, 1, and 2 chiral centers respectively. However, a Br2 to compound C will result in a product with 3 stereocenters. Check out what it looks like in the solutions and convince yourself that choice C has 3 stereocenters: 2 that contain Br and a 3rd containing the ETHYL substituent. Now becasuse you have 3 chiral centers, you can have diastereomers. Rem enantiomer = opp config at EVERY chiral center. Look at the eg in the solution. You'll see those are diastereomers.


This problem doesn't give you the configuration of the ethyl group in choice C, which makes it a little awkward but it does let you practice a bunch of different concepts at once. Which is why destroyer is so badass. keep it up.
Ah! So tricky. Thank you for the explanation.
 
Aug 14, 2011
48
0
Status
Pre-Dental
My honest opinion is keep studying and don't get so caught up on a single question. I took one look at the DAT Destroyer Orgo section, closed the book, and just read my college text. If you have the time I would suggest this to you. Also, as a side note, I did the best on the orgo portion of the exam.
 
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