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
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
single bonded on the right. If I get an answer to the first two, I think I can figure this out on my own.