Reductive ozonolysis of 1-Methoxy-4-methyl-1,4-cyclohexadiene

My Answer:

According to me, both the left side and the right side double bonds undergo cleavage and hence the product must be, two molecules: Keto aldehyde and Ester aldehyde. But since, such product is not given in options, let us assume that only one bond gets oxidized. But which one? Which bond is the most reactive? I don't know.

Answer given in the book:


How come?


2 Answers 2


There are two types of control, that governs reaction path:

  • kinetic control (path with least activation energy) and
  • thermodynamic control (path with least energy of products).

Given temperature favors kinetic control.

Ozonolysis starts as 2+3 electrocyclic addition (with later regroup of the product), similar to Diels-Alder reaction . Please, read about this reaction in good organic chemistry textbook, especially about what controls geometry of products, it will help to understand what I'm going to say next.

Ozone has low energy of LUMO, so alkene part here provides HOMO, that is still lower then Ozone's LUMO. Oxygen from $OMe$ group donates some electronic density to $C-C$ double bound, so this bond becomes more active, then opposite one.

  • $\begingroup$ I didn't actually understand your last paragraph. To understand what you said, I've read about MO theory. Most of the books I referred to were giving examples of diatomic molecules like $Be_2$, $O_2^+$ etc. none of them discussed polyatomic molecules. Anyway, even now I still don't understand your last paragraph. Could you kindly suggest some good book or online material which explains (with examples) the kind of reasoning that you are giving in your last paragraph $\endgroup$
    – claws
    Jan 20, 2013 at 9:06
  • 1
    $\begingroup$ Basically, electrocyclic addition starts with HOMO of one molecule interacting with LUMO of another molecule. Usually LUMO is pi* orbital of alkene (which is lowered by substitution C-atoms in alkene with N or O atoms) and HOMO is higher pi-orbital of diene. This is why Diels-Alder reaction works well with methoxy-butadien+acetilenedicarboxylic acid. Role-reversion is possible, and first stage of ozonolisys mechanism is an example. As I remember, J. March in his 4-book textbook and later 1-book reedition covered the basics pretty well. $\endgroup$
    – permeakra
    Jan 20, 2013 at 13:06

Any reaction in general will try to attain maximum possible stability.
The reactivity of a molecule or a group is inversely proportional to its stability.
The general order of reactivity of carbonyl compounds is as follows:

$$\small\ce{Aldehyde > Ketone > Acid halide > Anhydride > Ester > Carboxylic Acid > Acid Amide}$$ (Reference)

So if the left bond breaks it results in a ketone which is more reactive(hence less stable) as compared to the product formed by the cleavage of the right bond which is an ester.
Hence the bond on the right would break to form an ester.

  • 1
    $\begingroup$ My answer may seem simple(because it only relies on logic) but @permeakra 's answer actually states the reason why it happens. In my answer some one can argue that how will the reacting molecule know if the formed product is more stable or less, it's like looking into the future. $\endgroup$ Jan 16, 2013 at 3:07

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