What Are the Products of Reductive Ozonolysis of 1,4-Cyclohexadiene?

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SUMMARY

The discussion centers on the products of reductive ozonolysis of 1,4-cyclohexadiene, specifically addressing the confusion between glyoxal and propanedial. Participants clarify that glyoxal is indeed a product, while propanedial is not, despite initial assumptions. The use of zinc dust or dimethyl sulfide (DMS) in the reductive workup yields aldehydes or ketones, while hydrogen peroxide (H2O2) produces carboxylic acids. The conversation highlights the importance of using methanol as a solvent to facilitate the reaction and the potential for rearrangements affecting product outcomes.

PREREQUISITES
  • Understanding of ozonolysis reactions
  • Familiarity with reductive workup conditions
  • Knowledge of organic compounds such as glyoxal and propanedial
  • Basic principles of nucleophilic attack in organic chemistry
NEXT STEPS
  • Study the mechanism of ozonolysis in organic chemistry
  • Learn about the role of solvents in chemical reactions, specifically methanol
  • Research the effects of different reductive agents like zinc dust and sodium borohydride
  • Explore the concept of rearrangements in organic reactions
USEFUL FOR

Chemistry students, organic chemists, and educators seeking to clarify the products of ozonolysis reactions and the implications of solvent choice and reductive workup conditions.

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Homework Statement


What are the products formed on reductive ozonolysis of 1,4-cyclohexadiene?

Homework Equations


[/B]

The Attempt at a Solution


shouldn't it give propandial?? but the answer says that glyoxal is one of the products, how?
 
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yes. do you know its structure?
Also, propandial is not a product.
 
Suraj M said:

Homework Statement


What are the products formed on reductive ozonolysis of 1,4-cyclohexadiene?

Homework Equations


[/B]

The Attempt at a Solution


shouldn't it give propandial?? but the answer says that glyoxal is one of the products, how?
What are the reductive work up conditions?
If you use zinc dust or DMS you will get aldehydes or ketones.
Use of H2O2 gives you carboxylic acids while sodium borohydride gives you alcohols.
 
Raghav Gupta said:
What are the reductive work up conditions?
If you use zinc dust or DMS you will get aldehydes or ketones.
Use of H2O2 gives you carboxylic acids while sodium borohydride gives you alcohols.
Yes. DMS takes an oxygen atom and gets converted to dimethyl sulphoxide. I think H2O2 gives nascent oxygen which oxidises the products.
Also, these compounds are also formed as minor products:
Untitled.png

its unstable.
Also, if methanol is used as a solvent, then one more product will be formed. Try figuring it out.
 
AdityaDev said:
Also, if methanol is used as a solvent, then one more product will be formed. Try figuring it out.
In the ozonolysis process the methanol is used as a solvent with O3 at -78°C to form ozonide. Using methanol is a necessary condition. We can then use any reagent to get aldehydes, ketones, alcohols or carboxylic acids.

You might be wanting to say that if methanol is not used then what is the product?
In that case we will not get ozonide.
 
I think I used the wrong words. (Typo)? I will check my book and tell you the question later. But what I wanted to say is that, during the reaction, nucleophilic attack of methanol takes place on one of the intermediates To give a different compound.
[This was given by my chemistry teacher Last month.]
 
Got it. Actually, this is the question. I was only able to remember some parts before.
Alcohols react with the carbonyl oxide to give hydroperoxy hemiacetals:
C__Data_Users_DefApps_AppData_INTERNETEXPLORER_Temp_Saved Images_ozonol10.gif
 
Hi you two, thanks for replying :).
the thing is, I just got into 12th, just finished, alcohols, phenols... i didn't understand a lot of what you guys said, my original question was this
Screenshot (48).png

I guess all the things you guys were talking about will be taught to me later(wont it? :frown: ) ,
could you help me solve this question for now? I selected answer D, which was correct because i thought it would give propandial and not glyoxal.. but the question says somethings else..
PS i hope you did well in your exams Aditya :smile:
 
D gives 2 moles of a compound other than glyoxal. All others give only glyoxal on ozonolysis. Waite for a few months. Soon you will learn it.
 
  • #10
Exams did not go as planned. Atleast I can write jee advanced. The FIITJEE topper here scored 320 in mains.
 
  • #11
AdityaDev said:
D gives 2 moles of a compound other than glyoxal. All others give only glyoxal on ozonolysis.
But the first option benzene gives H2O2 other then glyoxal, so why it is not the answer for question?

And how come 4th will give glyoxal? To me it is looking propanedial. Is there some rearrangement here that 4th option is giving glyoxal?
 
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  • #12
Why H2O2? It gives three moles of glyoxal.
Also, propandial is not a proper name.
 
  • #13
AdityaDev said:
Why H2O2? It gives three moles of glyoxal.
Also, propandial is not a proper name.
Sorry Benzene gives glyoxal as well as ZnO2 in reductive work up.

Propanedial is a proper name, see
http://en.m.wikipedia.org/wiki/Malondialdehyde
 
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  • #14
But the question remains unanswered, that are we getting propanedial in 1-4 cyclohexanediene?
 
  • #15
yes. 2 moles of propanedial.
 
  • #16
AdityaDev said:
yes. 2 moles of propanedial.
And how glyoxal?
 
  • #17
question is wrong?
 
  • #18
AdityaDev said:
question is wrong?
It may be right, some rearrangement can be possible.
Also benzene is giving glyoxal as well as ZnO2.
Is there a online calculator to see products?
 
  • #19
Why do you think rearrangement is possible?
 
  • #20
AdityaDev said:
Why do you think rearrangement is possible?
What can the other reason be and how you can directly say question is wrong?
Earlier in some posts you were saying that propanedial is not a product.

In organic chem. there are many rearrangements. Reason could also be because of torsional strain but I am not sure.

Who can help three of us now?
 
  • #21
Thats because I got confused with the structure. I did not get the correct structure using google. @Borek can help us.
 
  • #22
You need an organiker, not me.
 
  • #23
Now, I don't know a decent organiker here in PF.
Would this problem be solved?:rolleyes:
 
  • #24
Please Borek, or anyone else, could you help us out.
I don't understand how glyoxal could be a product from option D,
even by rearrangement how can you get glyoxal,
 
  • #25
Im sure. The question is wrong.
 
  • #26
Ok then, if you're that sure,
thank you
 
  • #27
Suraj M said:
Ok then, if you're that sure,
thank you
Isn't something given in hints and solution section?
 
  • #28
a friend gave me the question, when i asked for the solution, she said her friend asked her, don't know the exact source or the solution! sorry.
but what were you saying about rearrangement?
 
  • #29
Suraj M said:
a friend gave me the question, when i asked for the solution, she said her friend asked her, don't know the exact source or the solution! sorry.
but what were you saying about rearrangement?
The file you have attached was not of a book?
Rearrangement based on torsional strain and something but I am not sure.
 
  • #30
No rearrangement here. I am sure. Take chair form and do the mechanism.
 
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