Arenes-oxidation by KMnO4
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My friends and teachers taught me that part of the side chain is broken to become oxidised like alkenes while the other part remains to become carboxylic acid.
CO2+H20However,what I fail to understand is, the by-products I get.
1st:ok,so i broke benzene-CH2 that CH2 becomes H--- O ---C=O
By the oxidation reaction in alkenes,it would have been 2(OH) -----C=O
2nd:The other part broken,gets me -CH3 CO2 and H20 2(OH)---C=O
but I got 3H---C=O and was stucked.
OH
Next,Alkaline oxidation results in one OH added to a broken C=, so its C----H
Can't seem to understand.
Last but not least,hot alkaline oxidation and its difference between hot acidified oxidation by KMnO4
Seems like oxidation is alright for me in alkenes,but a huge headache in arenes.I could memorise,but that defeats the point of understanding the alkylbenzene oxidation reaction.
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To fully understand oxidation of aromatic species, necessarily involves the full mechanism (including the MnO4- species and multiple intermediates), which is not just beyond H2 & H3 A levels, but is also beyond much of Uni level Chemistry as well (in which you will have to choose to focus on your specialties and subspecialties; this holds true for all sciences and all disciplines).
As such, even your school JC teachers, and even some of the Cambridge examiners, themselves would not know the mechanism for this reaction (*particularly* for this aromatic oxidation reaction, which is significantly more complicated than say, oxidation of alkenes, alcohols, aldehydes, etc).
Bottomline : as far as A levels H2 Chemistry is concerned, Cambridge only requires you to appreciate that the alkyl side chain (containing at least 1 benzylic H atom) of a benzene ring, can be oxidized by heating under reflux with acidified KMnO4 to generate benzoic acid, and other by-products (usually including CO2).
If the benzene ring's side chain contains a C=C group, you have to treat it as an alkene and oxidize it accordingly to generate the expected products on the other side of the C=C goup. Any remaining C atoms between the benzylic C atom and the double bond, you may assume to be oxidized to CO2.
Last but not least,hot alkaline oxidation and its difference between hot acidified oxidation by KMnO4
For A level purposes, temperature together with pH matters only for oxidation of alkenes, ie. cold alkaline KMnO4 generates vicinal diol, while hot acidified KMnO4 results in oxidative cleavage.Other than alkenes, the H2 syllabus ignores any differences between acidified vs alkaline KMnO4 (but "heat under reflux" must always be specified), other than (obviously!) the required protonation or deprotonation of the oxidation products.
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how about the formation of H20,still a little edgy for me, so all Ch2Ch3 becomes 2 c02 but why does ch2ch2ch3 becomes ch3cooh?
I can't understand because my school notes did not clearly mention the rationale behind this concept,but only shows the reaction and products only.
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Originally posted by a mugger:
how about the formation of H20,still a little edgy for me, so all Ch2Ch3 becomes 2 c02 but why does ch2ch2ch3 becomes ch3cooh?
I can't understand because my school notes did not clearly mention the rationale behind this concept,but only shows the reaction and products only.
To fully understand the reaction, you have to appreciate its mechanism. Google "mechanism for oxidative cleavage of alkenes".For A level purposes, it will suffice to note the products based on the cleavage of the C=C group, and apply this accordingly to any alkene.
Ethene undergoes oxidative cleavage by KMnO4 to generate 2 molecules of carbonic(IV) acid, each of which exists in equilibrium with, and hence can decompose into CO2 and H2O, with favourable positive entropy change.
A molecule with the group R=CH2CH2=R will undergo oxidative cleavage by KMnO4 to generate 2 molecules of R=O (in which the R groups may or may not be the same), and ethandial (OS of C increases from -2 to +1), which is further oxidized to ethandioic acid (OS of C increases from +1 to +3), which is further oxidized to (2 molecules of) carbonic(IV) acid (OS of C increases from +3 to +4), which exists in equilibrium with, and hence can decompose into CO2 and H2O (OS of C remains unchanged, from to +4 to +4), with favourable positive entropy change.