OK there are 3 types of oxidation equations that utilize ozone but I cant understand why they are able to produce very different products.

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O3/CCl4
1. Terminal alkyne --------> 1 carboxylic acid(RCOOH)+ acetic acid(CH3OOH)
H2O

O3/CH2CHl2
2, 2 propene -----------> 2 acetaldehydes (CH3CHO)
ZN/H20

O3/CH2Cl2
3. Alkene (non terminal) -------------> 2 alcohols
NaBH4, CH3OH

My perspective on question three is that the alkene is first oxidized to an carboxylic acid and then reduced by NaBH4 to alcohol (is this correct?)



Another question that is bugging me from Kaplan review.
H+
2,3 methyl propane -------> 2methyl-2-butene as a major product
heat

I understand where the double bond came from but what happened to the other methyl group it doesnt just disappear and it certainly is not a good leaving group right? So why does the product have a hydrogen boned to the alkene where I think a methyl groups should be?

Any help would be greatly appreciated because I can not stop thinking about these problems. Thank you very much

Isn't the first compound really a butane? That's what I drew when I looked at the name, so no carbons are lost.

Thank You...