What about activated benzene ring with KMnO4 and heat? In my notes o cresol decomposes So will every activated ring be decomposed? Anything with an OH directly attached to the ring will be oxidized to a quinone and from there it will produce a polymerized mess. Since this substitute has alpha carbon but also this is a ketone group. I have two methyl groups at pyrazole ring at 1,3 position. I want to convert both groups to CA. Please hewlp me or suggest me some other route.
I would suggest doing benzylic bromination or similar reaction first, followed by conversion to alcohol and then mild oxidation to the carboxylic acd.
Benzaldehyde would be the first product formed after oxidative cleavage of the glycol. With acid, as chromic acid H2CrO4, it can be used in the oxidation of toluene and other aliphatic compounds. See here. Tetrahedron , 20, Hi James, what if side chain have one double bond in example 2. Will the main product be same or change?? The double bond will be dihydroxylated by KMnO4. At low temperatures with base this can be isolated as a diol. With heating or acid oxidative cleavage will occur to give an aldehyde which can be further oxidized to give a carboxylic acid.
Hi there… How about this compound.. Could it be oxidised to benzoic acid as well? Even though it is a tertiary alcohols? Such as CH2Cl? That would be fine, it should still oxidize as it will still form a resonance stabilized free radical.
The alpha position of acetophenone is reasonably susceptible to radical C-H abstraction. Given enough heating I would not be surprised to see oxidation. Color change is a part-per-million phenomenon that is often not diagnostic. Methylbenzene toluene is colorless, and benzoic acid is also colorless.
But trace impurities can lead to color. What happens if diphenyl methane is used as reactant? Do 2 molecules of benzoic acid form, or 1 benzoic acid and 1 benzene. Do products differ if oxidant is L. R first and then in excess? I had to look up the structure of cardanol. Find another way to do it. I have an off track doubt. Is there any reaction? Thank you, Sir. Maybe phenol reacts in a different way unlike the alkenes. I would be grateful if you enlightened me a bit.
If you heat naphthalene up enough with KMnO4, you will get phthalic acid. I had a question. If the presence of a handheld hydrogen is necessary, then why does acetophenone get oxidized to Benzoic acid upon subsequent oxidation? Acetophenone is in equilibrium with its enol tautomer, which reacts with KMnO4. Benzophenone would not react. What happens when 3 phenyl propanal is oxidised with kmno4? Does it convert to 3 phenyl propanoic acid or benzoic acid?
I think it would the first one. Sangepu, J. Xie, Synthesis , , 50 , An efficient and economic procedure for the dihydroxylation of various olefin derivatives with commercial KMnO 4 as the oxidant in the presence of a quaternary ammonium salt is suitable for large-scale production of cis -dihydroxy compounds, even those bearing primary and secondary alcohol groups, without overoxidation.
Luo, C. Zhao, J. Xie, H. Lu, Synthesis , , 48 , In a preparation of N -alkyl-substituted sulfoximines from sulfides, in situ generated N -bromoalkylamines serve as readily accessible imidating agents leading to N -alkylsulfiliminium bromides that are subsequently oxidized providing the desired products. Dannenberg, V. That means that the reaction will not sto p at this point unless the potassium manganate VII solution is very dilute, very cold, and preferably not under acidic conditions.
If you are using hot concentrated acidified potassium manganate VII solution, what you finally end up with depends on the arrangement of groups around the carbon-carbon double bond. The formula below represents a general alkene. In organic chemistry, the symbol R is used to represent hydrocarbon groups or hydrogen in a formula when you don't want to talk about specific compounds. If you use the symbol more than once in a formula as here , the various groups are written as R 1 , R 2 , etc. In this particular case, the double bond is surrounded by four such groups, and these can be any combination of same or different - so they could be 2 hydrogens, a methyl and an ethyl, or 1 hydrogen and 3 methyls, or 1 hydrogen and 1 methyl and 1 ethyl and 1 propyl, or any other combination you can think of.
In other words, this formula represents every possible simple alkene:. The acidified potassium manganate VII solution oxidizes the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds. Carbonyl compounds can also react with potassium manganate VII , but how they react depends on what is attached to the carbon-oxygen double bond.
So we need to work through all the possible combinations. Carbonyl compounds which have two hydrocarbon groups attached to the carbonyl group are called ketones. Ketones aren't that easy to oxidize, and so there is no further action.
But see note in red below. If the groups attached either side of the original carbon-carbon double bond were the same, then you would end up with a single ketone.
If they were different, then you would end up with a mixture of two. For example:. In this case, you would end up with two identical molecules called propanone. On the other hand, if one of the methyl groups in the original molecule was replaced by an ethyl group, you would get a mixture of two different ketones - propanone and butanone.
What would you get if there was a methyl and an ethyl group on both sides of the original carbon-carbon double bond? Again, you would get a single ketone formed - in this case, butanone. If you aren't sure about this, draw the structures and see. This last section is a gross over-simplification.
In practice, ketones are oxidized by potassium manganate VII solution under these conditions. The reaction is untidy and results in breaking carbon-carbon bonds either side of the carbonyl group. Using the principles above, we expect KMno 4 to react with alkenes, alkynes, alcohols, aldehydes and aromatic side chains. Examples are provided below. It is easiest to start at the top. Unless great efforts are taken to maintain a neutral pH, KMnO 4 oxidations tend to occur under basic conditions.
Balancing the reactions would involve using the methods learned in general chemistry, requiring half reactions for all processes. Primary alcohols such as octanol can be oxidized efficiently by KMnO 4 , in the presence of basic copper salts.
Although overoxidation is less of a problem with secondary alcohols, KMnO 4 is still not considered generally well-suited for conversions of alcohols to aldehydes or ketones. Under mild conditions, potassium permanganate can effect conversion of alkenes to glycols. It is, however, capable of further oxidizing the glycol with cleavage of the carbon-carbon bond, so careful control of the reaction conditions is necessary.
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