[moderate]

Hello,

I just have a quick question about identifying hydrogens in different magnetic environments.

Consider ethyl propenoate as a starting point.



The following groups of hydrogens produce distinct peaks, correct?
8
7a
6a
2b and 2a
3c, 3b, and 3a

A total of 5 peaks.

The 2b and 2a hydrogens are enantiotopic. In other words, replacing 2b with a different group and then replacing 2a with the same group would produce enantiomers. And, enantiomers produce one peak in an achiral solution (although I don't really understand why).

Now, let's replace either the 8 or 7a hydrogen with a halogen.



Do the 2b and 2a hydrogens now split into two peaks? They are no longer enantiotopic, but diastereotopic, correct?

[Agathiyar]

Your first part of the discussion is fine but your second part is confusing, replacing hydrogen at 8 by bromine doesn't affect the 2a or 2b protons, they are still enantiotopic.  You don't have another chiral center to discuss about diastereotopic situation.
 

[discodermolide]

Hello,

I just have a quick question about identifying hydrogens in different magnetic environments.

Consider ethyl propenoate as a starting point.



The following groups of hydrogens produce distinct peaks, correct?
8
7a
6a
2b and 2a
3c, 3b, and 3a

A total of 5 peaks.

The 2b and 2a hydrogens are enantiotopic. In other words, replacing 2b with a different group and then replacing 2a with the same group would produce enantiomers. And, enantiomers produce one peak in an achiral solution (although I don't really understand why).

Now, let's replace either the 8 or 7a hydrogen with a halogen.



Do the 2b and 2a hydrogens now split into two peaks? They are no longer enantiotopic, but diastereotopic, correct?

I would not expect the hydrogens 2a and 2b to be different due to free rotation about the single bond, you should see a 2 proton quartet, i.e. a normal ethyl group.