Understanding the Formation of Products When KMnO4 is Added to Alkenes

[prasannapakkiam]

Okay. If hot Potassium PERMANGANATE (KMnO4-) with concentrated acid, is added to Ethene, what will form.

If I look at my textbook, I can deduce that first the alkene would oxidise into a di-diol. Then the bond splits, then;

a hydrogen would get replaced by an oxygen
otherwise, the OH would get replaced by an oxygen.

If I follow this pattern for Ethene, I think 2 Methanoic Acids should be formed, instead 2 Carbon Dioxides are formed according to my textbook!

So what are the products?
Also can someone give me a link or some information about the mechanism involved here.

 

[chemisttree]

If I look at my textbook, I can deduce that first the alkene would oxidise into a di-diol. Then the bond splits, then;

What bond splits? Does it split homolytically or is there a reagent that causes the split?

a hydrogen would get replaced by an oxygen otherwise, the OH would get replaced by an oxygen.

Can you draw the product of this replacement?

If I follow this pattern for Ethene, I think 2 Methanoic Acids should be formed, instead 2 Carbon Dioxides are formed according to my textbook!

Can formic acid be further oxidized to CO2 under these conditions?

 

[prasannapakkiam]

H2C=CH2 CH2...

H2OHC-CHOH CH2...

H H
H C-----C CH2...
OH OH

This bond splits (The dash)

In the case above, it should become:

HOOCH and OCHOH CH2...

H H
O=C and O=C CH2...
OH OH

 

[chemisttree]

H2C=CH2 CH2...

How many carbons does 'ethene' (ethylene?) have?
I see three for your example. What is this structure?

Glycols don't spontaneously split as you have shown. What reagent might cause this C-C scission?

 

[Kushal]

oxidation with hot acidified KMnO4 give products depending on the alkene using.
first, let's take ethene,
when ethene is passed into hot acidified KMnO4, the C=C (only) is broken and ethene is oxidised to CO2,

CH2=CH2 + [O] -----> 2CO2 + 2H2O

the C=C breaks and CH2 is oxidised to CO2 and H2O

now, we'll take propene... ethanoic acid and CO2 will be formed...

CH3CH=CH2 + 5[O] -----> CH3COOH + CO2 +H2O

When the C=C breaks, you are left with CH3CH and CH2
the CH3CH is oxidised to CH3COOH while the CH2 is oxidised to CO2 and H2O (just like ethene)

butene will give similar products, but the acid will be propaoic acid,...

there is a third case, when the alkene is branched...so
let's use 2 methyl propene...
CH3C(CH3)=CH2 + 4[O] -----> CO2 + H2O + CH3COCH3

the ketone propanone will also be formed... i.e CH3C=OCH3

as usual the C=C will break, and the CH2 will be oxidised to CO2 and H2O...
the remaining CH3C(CH3) will not be oxidised to a carboxylic acid, but instead to a ketone...


these are the 3 main cases... from these you can deduce the products of other alkenes.


note... the diol is formed with any alkene when using cold KMnO4

 

[chemisttree]

oxidation with hot acidified KMnO4 give products depending on the alkene using.
first, let's take ethene,
when ethene is passed into hot acidified KMnO4, the C=C (only) is broken and ethene is oxidised to CO2,

CH2=CH2 + [O] -----> 2CO2 + 2H2O

the C=C breaks and CH2 is oxidised to CO2 and H2O

Care to give a mechanism for this H2C=CH2 to CH2 to CO2 conversion?

now, we'll take propene... ethanoic acid and CO2 will be formed...

CH3CH=CH2 + 5[O] -----> CH3COOH + CO2 +H2O

When the C=C breaks, you are left with CH3CH and CH2
the CH3CH is oxidised to CH3COOH while the CH2 is oxidised to CO2 and H2O (just like ethene)

Do you really think that the C=C bond breaks first and the resultant fragments (CH2?) are oxidized to CO2?

note... the diol is formed with any alkene when using cold KMnO4

Yes, the reaction stops at the diol under conditions of cold, dilute permanganate. It keeps going when its heated... all the way to CO2. What might the product of the step after the diol production be? What reagent and mechanism are at play? How does the reagent cause the reaction?

 

[prasannapakkiam]

Same question I have been asking. What causes the Ethene to form CO2?? Instead of Methanoic acids..

 

[[Leo]]

How i can get two acids when the unknown hydrocarbon is heated with KMnO4 , when this hydrocarbon reacts with Br2 i get tetrabr combination? So if it reacts with Br2 and gets tetrabr, so it must be alkin, or it can be alkene?
P.s. sorry for my bad english.

 

[Kushal]

okay, i'll correct wot i said... for the sake of simplicity i messed up with evrything...lol

so... about ethene:
as with all alkenes, it is first oxidised to a diol.
since we are using hot acidified KMnO4, it will get oxidised again to HCOOH, methanoic acid.

CH2=CH2 + 2[O] ------> 2HCOOH

now, HCOOH is a reducing agent (the only carboxylic acid), so obviously it will be further oxidised to CO2 and H2O.



HCOOH + [O] -----> CO2 + H2O

in this way, ethene gets maximally oxidised to CO2 and H2O.

now with propene...
it is first oxidised to a diol: propane 1,2 diol. CH3 CH(OH) CH2(OH)

the diol can be considered as consisting of CH3CH(OH) and CH2(OH)

CH2(OH) as in ethene will get oxidised to HCOOH and will further get oxidsed to CO2 and H2O since it is a Reducing agent.

CH3CH(OH) is then a primary alcohol and primary alcohols get oxidised to carboxylic acids, in this case CH3COOH: ethanoic acid.

for the branched alkene now,... using again the 2 methyl propene.
you'll get 2 methyl propane 1,2 diol

CH3 C(CH3)(OH) CH2(OH)

the CH2(OH) will be maximally oxidised to CO2 and H2O.
CH3 C(CH3)(OH) is a secondary alcohol. this type of alcohol is oxidised to the corresponding ketone, here it is the CH3 C=O CH3

one more example... using but 2 ene...

butanne 2,3 diol will be formed first. CH3CH(OH) CH(OH)CH3

CH3CH(OH) is a primary alcohol, it will get oxidised to the corrsponding acid.
ethanoic acid will be formed...

 

[Kushal]

hhmmm... it would most prolly be an alkane if you can get a tetrabromo thing... but wait, can't an alkene llike pent 2 ene (which gives two different carboxylic acids upon oxidation) react with Br2...

won't you get uuummm, 2,2,3,3 tetrabromo pentane...
i know that 2,3 dibromo pentane will be formed upon addition reaction. but then the halogeno alkane formed can undergo free radical substitution reaction and accept 2 more Br to form a tetrabromo thingy, doesn't matter where it places the 2 Br atom (i.e on C1 or on C2 or on C3,...)

hope I'm not wrong in what I'm speculating!