[HughMyron]

So the compound is 2,4 methyl, 3-keto hexane, at least I think that's what you call it.

                 O
                 ||
CH3 - CH - C - CH - CH2 - CH3 
         |             |
         CH3        CH3

I already know how the enol tautomers are created: one of them involves a doublebond between the 2 and 3 carbons. The other involves a doublebond between the 3 and 4 carbons.

My question is: Do E and Z isomers apply here? I don't think so, but I wanted to be sure.



Second Question:
                                    O
                                    ||
        CH2 -- CH -- CH2 -- C - H
      /           |
CH2             |                 O
      \           |                 ||
        CH2 -- CH -- CH2 -- C - CH3

Are there any asymmetric centers here? I said no.

[discodermolide]

Firstly you have a 5 valent carbon in your structure!
You can get E and Z enols, they are in equilibrium with the keto form, which is favoured.
Your second compound has 2 chiral centers. There is an  aldehyde at one end and a ketone at the other. Thus both CH groups are chiral centers.

[HughMyron]

Firstly you have a 5 valent carbon in your structure!
You can get E and Z enols, they are in equilibrium with the keto form, which is favoured.
Your second compound has 2 chiral centers. There is an  aldehyde at one end and a ketone at the other. Thus both CH groups are chiral centers.

Oh, I think I get it now.

Thanks!