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I was looking at a question which asked to predict the product of reductive ozonolysis of benzene using $\ce{Zn/O3/H2O2}$ and I predicted them correctly using the cleaving of the double bond shortcut but then I realized that benzene is losing its stable aromatic state to form not as stable products.

So I would like to know why benzene proceeds through this reaction even if it doesn't seem as favourable? Also in that case, why don't reactions like hydrohalogenation or hydroxylation occur on benzene? As I see it, they would involve loss of stability too so what's so special about ozonolysis?

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    $\begingroup$ See this . The conditions required are quite vigorous $\endgroup$
    – Yusuf Hasan
    Jun 11, 2019 at 5:04
  • $\begingroup$ Then under such conditions, can the other reactions I mentioned also occur? $\endgroup$
    – Sameer Thakur
    Jun 11, 2019 at 5:10
  • $\begingroup$ As per my knowledge, these other two reactions can't be performed under the conditions mentioned in the answer above. Different conditions may exist for hydrohalogenation and hydroxylation of benzene,but I am not aware of them (if they exist) $\endgroup$
    – Yusuf Hasan
    Jun 11, 2019 at 5:20

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