My text book says that when an ozonide (formed from an alkene) is treated with $\ce{Zn/CH3COOH}$, it cleaves to give carbonyl Compounds.

1) Does $\ce{ Zn/CH3COOH}$ mean Zinc in acetic acid (or) zinc or acetic acid. I at first thought that it should be zinc in acetic acid, but then I found a separate mechanism for acetic acid alone, (given below) which made me confused. So, which one is it?

enter image description here 2) If the above mechanism is correct, then why doesn't the oxygen-oxygen bond (being unstable and weak) cleave in the step I've encircled?

  • $\begingroup$ Acetic acid may facilitate the collapse of the ozonide to a carbonyl cmpd. and a zwitterion. Reduction of the zwitterion with acetate would also form peroxyacetic acid that would be have to be reduced by Zn. Zinc can effect the same reduction of the ozonide directly. You may wish to see: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/ozonolysis/… $\endgroup$ – user55119 Mar 31 '18 at 20:54
  • $\begingroup$ You want to edit your title as teh question is about alkenes not alkynes $\endgroup$ – Waylander Mar 31 '18 at 21:40
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    $\begingroup$ @user55119 So, reduction by zinc directly would follow the same mechanism as dimethyl sulphide? This would mean that Zn would donate its electrons to the ozonide forming Zn+ and then be stabilized by the O- released from the ozonide, resulting in the formation of ZnO, correct? $\endgroup$ – Gokulakrishnan Shankar Apr 1 '18 at 6:56
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    $\begingroup$ You bet! The ozonide is at the same oxidation level as the sum of the zwitterion and the carbonyl cmpd. You need to reduce the ozonide (or the methanol addition product of the zwitterion if methanol is the solvent) by two electrons to obtain two carbonyl cmpds. E.g., the ozonide of 2-butene is at the oxidation level of latent acetaldehyde and acetic acid. The role of the reductant is two get "acetic acid" to acetaldehyde. $\endgroup$ – user55119 Apr 1 '18 at 13:11


Ozone, an allotrope of oxygen, is highly reactive molecule that can undergo a 1,3-dipolar addition across alkenes, forming a 1,2,3-trioxolanes. This compound is unstable and undergoes a cycloreversion to form a carbonyl compound and a carbonyl oxide. These resultant compounds (the carbonyl compound and the carbonyl oxide) then proceed through a second 1,3-dipolar addition to form a 1,2,4-trioxolane, commonly known as an ozonide. To complete the cleavage of the alkene, the ozonolysis (implies that ozone causes to break (-lysis)) is followed by a work-up designed to break apart the ozonide. The work-up, which can be oxidative or reductive, determines the oxidation state of the products. For example, cis- or trans-alkenes gives two corresponding aldehydes with reductive work-up while same alkene gives two corresponding carboxylic with oxidative work-up.

The most common reductive work up procedure uses dimethyl sulphide (DMS) in the presence of methanol. Methanol reacts with the carbonyl oxide to form a hydroperoxy acetal, and this is in turn reduced by the dimethyl sulphide to give relevant carbonyl compound and dimethyl sulfoxide. It is also possible for dimethyl sulphide to reduce the ozonide without the presence of methanol as well, although reation would proceed slowly (A new and convenient method for converting olefins to aldehydes: Tetrahedron Letters, 1966, 7(36), 4273–4278). Other reductive methods: $\ce {Zn/AcOH}$, triphenylphosphine ($\ce {PPh3}$), or thiourea ($\ce {NH2C(=S)NH2}$).

Note: Recently, there is a publications claiming an efficient and convenient method for reductive workup of the ozonolysis reaction using sodium hydrosulfite. However, the final product could be either aldehyde or acid, thus the method is not selective.

Oxidation: The direct conversion of terminal and 1,2-disubstituted alkenes to carboxylic acids is described as oxidative ozonolysis. This can be easily achieved through treatment of initial ozonolysis products with hydrogen peroxide ($\ce {H2O2}$) or other oxidants.

You may also find answers to your Q3 and Q4 from the reference given above.

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