DAT OC odyssey Ch4. Stereochemistry #28, #40, need answer to this questions.

[qkrgkssk]

Need help with this two question from Organic Chemistry Odyssey.
I have attached a picture of these questions.
#28, I am confused why C=O has priority over CH-BR, giving the answer as D (2S, 3R).
I was thinking that C=O should be higher priority since it has no Hydrogen attached making it (2R, 3R).
#40 I was thinking the answer should be active racemic mixture for Br2/light reaction. The answer was B. I need explanation how it is a inactive racemic mixture.

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[HopeStyle]

#28 Chiral center priorities are ranked by atomic number not by how many H's are attached

#40 Br2/Light -> Radical rxn -> always attacks carbon where it will form the most stable radical -> you will get a mixture of R & S products (racemic) which cancels each other out (inactive).

수고하셔요~

 

[HopeStyle]

oops repeated

 

[Queen Beach]

40.) Ok, so it's inactive because you took the H away and pretty much anything denatured becomes inactive.

 

[HopeStyle]

40.) Ok, so it's inactive because you took the H away and pretty much anything denatured becomes inactive.

It has nothing to do with denaturing lol
It's literally meaning "optically" inactive.

If you just form a single product, it will be active as there IS a chiral center.
However, if you look as a whole R & S will cancel out. (as they are racemic = 50:50)

 

[Queen Beach]

Well maybe I am wrong, but thought of it like this.... if you took the H away now the methyl isnt rigid, so inactive. So binding would be a different orientation as before.

 

[qkrgkssk]

#28 Chiral center priorities are ranked by atomic number not by how many H's are attached

#40 Br2/Light -> Radical rxn -> always attacks carbon where it will form the most stable radical -> you will get a mixture of R & S products (racemic) which cancels each other out (inactive).

수고하셔요~

Thank you for the reply~ 감사해요~