[Westboarder]

I have tried to deterred the treatment of these compounds with MNO2, but I'm not sure if they react in the same way they would when treated with a reagent such as: KMNO3.

What I think happens, is the OH groups get oxidized to ketone groups and the double bonds are oxidized to OH groups with syn addition to the double bond?

If someone could explain to me what exactly is happening after MnO2 and then affecting the chiral centers that would be fantastic. Thanks.

[Honclbrif]

http://en.wikipedia.org/wiki/Manganese_dioxide#Organic_synthesis

[Dan]

Be careful to capitalise your formulas correctly. MnO₂ is manganese dioxide, KMNO2 is nonsense. KMNO3 is also nonsense, and KMnO₃ does not exist (do you mean KMnO₄?).

The reactivity of MnO₂ in organic synthesis is covered in any textbook, and even in the wikipedia article - it chomoselectively oxidises allylic alcohols.

The easiest way to start this problem is to identify which of those starting alcohols are optically active.

[Westboarder]

Be careful to capitalise your formulas correctly. MnO₂ is manganese dioxide, KMNO2 is nonsense. KMNO3 is also nonsense, and KMnO₃ does not exist (do you mean KMnO₄?).

The reactivity of MnO₂ in organic synthesis is covered in any textbook, and even in the wikipedia article - it chomoselectively oxidises allylic alcohols.

The easiest way to start this problem is to identify which of those starting alcohols are optically active.

So, based on chirality centers. It looks like A and D are achiral/mesocompounds respectively. So, therefore B,C,E are optically active?  

Following the logic posted, the bonds to -OH are going to be oxidized to carbonyl substituents.

[Westboarder]

Here is my attempt for post-oxidation. Please let me know if this is off base, so then I can know if my mistake is in determining chirality. Thanks.

[Dan]

Following the logic posted, the bonds to -OH are going to be oxidized to carbonyl substituents.

As I posted, only the allylic alcohols are oxidised with MnO₂

[Westboarder]

Following the logic posted, the bonds to -OH are going to be oxidized to carbonyl substituents.

As I posted, only the allylic alcohols are oxidised with MnO₂

Ah, thank you. I was misinterpreting allylic.

So, I have the correct answer. However, I still do not know why compound B shifts from being chiral to achiral. I have drawn lines of symmetry and have yet to find a way to make it symmetrical. Anything I might be missing?

[Dan]

The oxidation of B gives cyclopent-2-en-1-one. The molecule is planar - the plane of symmetry is the plane of the paper.