[number2jazzfan]

I'm given many equations.
1. write an equation that represents the reaction of 2% aqueous KMnO₄ solution with 3-hexene, CH₃-CH₂CH=CH-CH₂-CH₃

I think the answer is:

CH₃-CH₂-CH=CH-CH₂-CH + KMnO₄ + H₂O

But it doesn't seem right. Can I take two other Hydrogen's from the far left 'CH₃' or even from any of the 'CH₂'s? I'm told that "because KMnO₄ oxidizes the unsaturated compound, we call the reaction an oxidation-reduction reaction. As a result of this reaction, an OH group is added to each carbon at each end of a double bond in the alkene." So this makes me think that the answer would be:

CH₃-CH₂CH-CH-CH₂-CH₃ + KMnO₂
             l    l
           OH OH

But the end CH₃'s would have to lose a Hydrogen


2. Write an equation that represents the reaction of 2% ethanolic PBP solution with 4-methylcyclo-hexene and name the product

The thing I'm confused about this is that both are rings, though the 4-methylcyclo-hexene only has one double bond. We have never learned how to react two rings. Do I just bond the two together through a carbon?... There is a H₃C-- on the left and the double bond is 2 C's clockwise away.


There is another question that asks practically the same thing as the first two (there is an a and b), but with a benzene ring with a H₃C-- single bonded on the left, and a   
                                                                  CH₃
                                                                   /
                                                               --C
                                                                  \\
                                                                  CH₂
single bonded on the right. If I get an answer to the first two, I think I can figure this out on my own.
Thanks!

[Dan]

But the end CH₃'s would have to lose a Hydrogen

Why/how?

2. Write an equation that represents the reaction of 2% ethanolic PBP solution with 4-methylcyclo-hexene and name the product

What is PBP?