[mike_302]

I have a test on this stuff tomorrow and I need some urgent help figuring this out...

Our lesson on alcohols tells us that alcohols can undergo combustion with O2, creating water and CO2 as products...

Our lesson on aldehydes ketones says a lot more... Oxidation of a 1st degree alcohol (alcohol + oxidizing agent... O2) results in an aldehyde . Oxidation of a 2nd degree alcohol produces a ketone... And you cannot oxidize a third degree alcohol.

This doesn't seem to add up? Isn't combustion just a form of oxidation? Can someone explain this all to me :S ?

[Borek]

You are right - combustion is oxidation, Just we are talking separately about combustion which is "uncontrolled" oxidation that goes to the end, and about "controlled" oxidation in which we are able to stop the reaction at some intermediate stage.

[mike_302]

OK .. And, the oxidation of alcohol requires some sort of catalyst... In the lab we did, we used HCl ...

So maybe you can help me understand what happened there too...

In the first part of the lab, we added HCl to a first second and third degree alcohol... The third degree turned yellow, the rest stayed clear... Sound right?    Too me, I thought only the 1st and 2nd would react... Could someone explain?

In the second part, we used a 1st 2nd and 3rd degree alcohol again, but this time added an oxidizing agent --> KMnO4, with the HCl catalyst. The KMnO4 is purple to begin with, and simply turned the 1st and 2nd degree purple, but the third one reacted and changed yellowish again... Make any sense? Did I miss something and not wait long enough or what?

[mike_302]

Oh, and the other question, going back to my first post: So, if oxidation is the same as combustion, and we are making "Theoretical" equations when it comes to the reaction, why does the theoretical oxidation of alcohol simply stop at aldehyde without any sort of indication?   If I was given a piece of paper that simply said the following:

CH3-CH2-OH (ethanol... In case I got the formula mixed up) +

• (an oxidizing agent) ---> 

And I was asked to complete the reaction, how do I know whether it is an alcohol oxidation, or an alcohol combustion ?

[Guitarmaniac86]

OK .. And, the oxidation of alcohol requires some sort of catalyst... In the lab we did, we used HCl ...

So maybe you can help me understand what happened there too...

In the first part of the lab, we added HCl to a first second and third degree alcohol... The third degree turned yellow, the rest stayed clear... Sound right?    Too me, I thought only the 1st and 2nd would react... Could someone explain?

In the second part, we used a 1st 2nd and 3rd degree alcohol again, but this time added an oxidizing agent --> KMnO4, with the HCl catalyst. The KMnO4 is purple to begin with, and simply turned the 1st and 2nd degree purple, but the third one reacted and changed yellowish again... Make any sense? Did I miss something and not wait long enough or what?

The KMnO4 would have changed the primary alcohol into carboxylic acid however the OH- group would be replaced by the Cl- ion from the HCl to form an -oyl chloride species, the secondary alcohol would turn into a ketone and the tertiary alcohol would remain unchanged in the presence of the KMnO4 and HCl.

HCl is a bad catalyst to use because the tertiary alcohol cannot be oxidised, however, it can undergo and SN1 reaction where the OH- leaves and leaves behind a carbocation (which is stablised by the R groups around it), and a Cl- ion is then attracted to the carbocation forming a alkylhalide compound. This could account for the yellow colour. Ive always used sulphuric acid for the oxidation of alchols because you do not have the competing Cl- ion that you get from HCl.

Oh, and the other question, going back to my first post: So, if oxidation is the same as combustion, and we are making "Theoretical" equations when it comes to the reaction, why does the theoretical oxidation of alcohol simply stop at aldehyde without any sort of indication?   If I was given a piece of paper that simply said the following:

CH3-CH2-OH (ethanol... In case I got the formula mixed up) +

• (an oxidizing agent) ---> 

And I was asked to complete the reaction, how do I know whether it is an alcohol oxidation, or an alcohol combustion ?

Oxidation is not the same as combustion in this sense. If an organic compound is combusted, it yields CO2 and water. Ethanol when it is being oxidised by an oxidising agent will go to ethanal then ethanoic acid. Combustion is an uncontrolled oxidation that gives off a lot of energy, where as an oxidation using an oxidising agent is controlled and does not yield the same products.

Edit:

Combustion of an organic compound, eg an alcohol gives CO2 and water.

Oxidation of an alcohol gives either a ketone, aldehyde or carboxylic acid depending on the type of alcohol.

[mike_302]

So the combustion is giving off so much energy that, in a sense, you could say that ... It's products get messed up and rearranged in comparison to the slow oxidation in which the atoms can properly arrange themselves... Sound right?


Oh and: Thank you VERY much for the *delete me* I very much appreciate it.

[Guitarmaniac86]

So the combustion is giving off so much energy that, in a sense, you could say that ... It's products get messed up and rearranged in comparison to the slow oxidation in which the atoms can properly arrange themselves... Sound right?


Oh and: Thank you VERY much for the *delete me* I very much appreciate it.

Combustion implies you are igniting the compound, and providing so much energy that the C-C bonds break and the C-H bonds break forming lots of radicals that propagate and form CO2 and water.

Using an oxidising agent is a way to control the out come of a reaction at lower energies and is devised as a specific method of attacking a functional group, since most if not all organic reactions are ways of changing a functional group. So in a sense, oxidising an alcohol is attacking just the functional group and changing that, whereas combustion is a molecule wide reaction.