[THECapedCaper]

I have a little confusion as to what determines a chiral center in the larger molecules, namely hydrocodone.



As far as I can tell, anything that is a CH2 or CH3 cannot be considered chiral, but do C=O or C=C bonds fall under the same category because they're not bonded to four groups? What are the chiral centers of hydrocodone?

[discodermolide]

C=O and C=C bonds cannot be chiral.
In your molecule there seem to be 4 chiral centers.
Simply put the same criteria apply for simple and complex molecules, 4 different substituents and non-superimposable on it's mirror image.

[curiouscat]

C=O and C=C bonds cannot be chiral.
In your molecule there seem to be 4 chiral centers.
Simply put the same criteria apply for simple and complex molecules, 4 different substituents and non-superimposable on it's mirror image.

Silly question: Why isn't this (see below) atom considered chiral? Even accounting for resonance aren't there essentially four different substituents?

[THECapedCaper]

So, then these four carbons would be the chiral centers (marked with red stars)?



In that case, we have 16 stereoisomers (2^n) and 14 diasteroisomers (2^n - 2), correct?

[discodermolide]

There are not 4 different substituents on that carbon atom. It's part of a phenyl ring. The double bonds are not static as you said.

[discodermolide]

Yes, that is correct.

[curiouscat]

There are not 4 different substituents on that carbon atom. It's part of a phenyl ring. The double bonds are not static as you said.

I see. Thanks!

Would that position (red box) become a Chiral Centre if I, say, brominated it there?  {Bet it can't be done! But who knows! You wizards have all kinds of synthetic magic up you sleeves!  }

[discodermolide]

Well assuming you could brominate there then it would give a new chiral centre.

[curiouscat]

Well assuming you could brominate there then it would give a new chiral centre.

Thanks again!

[fledarmus]

So, then these four carbons would be the chiral centers (marked with red stars)?



In that case, we have 16 stereoisomers (2^n) and 14 diasteroisomers (2^n - 2), correct?

Not really - build some models. Having ring structures will restrict the total number of possible stereoisomers. You will find that certain combinations of R and S just won't be possible to get the atoms to line up.