ozonolysis

In ozonlysis, an alkene is firstly treated with ozone to form an ozonide intermediate, and different products can be obtained by different work-up, i.e. treating the ozonide intermediate with different reagents. I think you are asking the difference between using zinc dust in H2O2 and using H2O2 alone to treat the ozonide intermediate.

When the ozonide intermediate is treated with H2O2 alone, work-up is oxidative which leads to carboxylic acid(s) and/or ketone(s).
http://i320.photobucket.com/albums/nn347/old-master/100225_1a.jpg?t=1267080798

圖片參考：http://i320.photobucket.com/albums/nn347/old-master/100225_1a.jpg?t=1267080798


When the ozonide intermediate is treated with zinc dust in H2O2, zinc is an reducing agent which lower the oxidizing power of ketone and thus the products are aldehyde(s) and/or ketone(s). Aldehyde(s) is/are formed instead of carboxylic acid(s).
http://i320.photobucket.com/albums/nn347/old-master/100225_1b.jpg?t=1267080893

圖片參考：http://i320.photobucket.com/albums/nn347/old-master/100225_1b.jpg?t=1267080893


For you reference, two more different work-ups are shown below:
http://i320.photobucket.com/albums/nn347/old-master/100225_1c.jpg?t=1267080982

圖片參考：http://i320.photobucket.com/albums/nn347/old-master/100225_1c.jpg?t=1267080982


2010-02-25 15:02:10 補充：
參考資料：
老爺子 +　http://www.organic-chemistry.org/namedreactions/ozonolysis-criegee-mechanism.shtm

2010-02-25 23:38:58 補充：
In the 3rd paragraph, "zinc is an reducing agent which lower the oxidizing power of ketone and ......" should read "zinc is an reducing agent which lower the oxidizing power of H2O2 and ......".

When u reacted alkene with ozone, an intermediate would be generated, which was known as ozonide. It consisted of C-O-C-O-O, which was an ether-like structure. The mechanism was shown as below. The use of this picture was protected by the two lines.


圖片參考：http://imgcld.yimg.com/8/n/HA00031906/o/701002240011613873411640.jpg


This intermediate was stable and could be decomposed in presences of reducing and oxidation agent. For example, if oxidising agent was used, like H2O2, this would give carboxylic acid, strong reducing agent such as NaBH4 would give alcohol, while mild reducing agent such as zine dust and triphenylphosphine or S(Me)2 would give aldehyde and ketone.

Therefore, the answer for the first part of the question would be to reduce the intermediate into ketone or aldehyde. If strong reducing agent was used, this would cause alcohol to form instead.

Sometimes, we might not need zinc dust. This might because we might carry out an oxidative work up instead of a reductive work up.

Hope this helps :)