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Oxidation of Salycylic Acid

Salicylic acid is a monohydroxybenzoic acid, a type of phenolic acid and a beta hydroxy acid. This colorless crystalline organic acid is widely used in organic synthesis and functions as a plane hormone. It is derived from the metabolism of salicin. In addition to being an important active metabolism of aspirin (acetylsalicylic acid), which acts in part as a prodrug to salicylic acid, it is probably best known for its use in anti-acne treatments.

I have a question: Can Salycylic Acid be oxidized by Potassium Permanganate in a Neutral Solution?

The answer is yes. It gives brownish black precipitate and brown solution as the result.

Now the problem is, how it can be?
How the mechanism reaction of this oxidation?

From internet, I found that only very strong oxidizing reagents, such as permanganate, dichromate or Fremy's salt (KSO₃)₂NO, oxidize simple phenols to 2,5-cyclohexadien-1,4-ones (benzoquinones).

From the reaction above, I propose a mechanism of the oxidation of phenol as follows:
First, hydroxy group will resonance to form a carbonyl.

 

In carbonyl form, the aromatic property will be lost so that the benzene ring will act as an alkane. This compound will be attacked by permanganate ion to form an inorganic ester (syn-addition reaction).

 

hypomanate ion (in the form of dihydrohypomanganic acid) is an unstable acid which will be disproportionated to mangan dioxide and permanganate ion (concluded from the result given by the reaction).

For the last, the compound will release water molecule to form a derivate of quinone.

 

The weaknesses of my proposed mechanism are: 

1. Can we break the aromatic property in benzene easily like that?

2. Why water molecule released in this reaction?

If you have another proposed mechanism, please share with me. 

Thank You

Bibliography:

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch24/ch24-3-5.html

http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/oxi_phenol.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/entfernen_wasserstoff/oxidation_phenole.vscml.html

http://users.humboldt.edu/rpaselk/C328.Su04/C328Notes/C328nLec15.htm

Synthesis of 2,4-dimethyl-3,5-dicarboethoxypyrrole

In organic chemistry, this compound is popular enough to be synthesized. The procedure are written as follows:

In a 3-l. three-necked, round-bottomed flask, fitted with a liquid-sealed mechanical stirrer and dropping funnel, are placed 390 g. (3 moles) of ethyl acetoacetate and 900 cc. of glacial acetic acid. The solution is cooled in an efficient freezing mixture to 5°, and a cold solution of 107 g. (1.47 moles) of 95 per cent sodium nitrite in 150 cc. of water is added dropwise with vigorous stirring at such a rate that the temperature remains between 5° and 7°. With efficient cooling about one-half hour is required to add the nitrite. The mixture is stirred for one-half hour longer and then allowed to stand for four hours, during which time it warms up to room temperature.

The separatory funnel is replaced by a wide-bore condenser, and the third neck of the flask is fitted with a stopper. The solution is stirred and portions of 196 g. (3 gram atoms) of zinc dust are added quickly through the third neck of the flask until the liquid boils and then frequently enough to keep it boiling. After the addition has been completed, the mixture is heated by a burner and refluxed for one hour. While still hot the contents of the flask are decanted from the remaining zinc into a crock containing 10 l. of water which is being vigorously stirred. The zinc residue is washed with two 50-cc. portions of hot glacial acetic acid which are also decanted into the water. After standing overnight, the crude product is filtered by suction, washed on the filter with two 500-cc. portions of water, and dried in air to constant weight. The yield is 205–230 g. (57–64 per cent of the theoretical amount) of material melting at 126–130°. On recrystallizing a 50-g. portion from 100 cc. of 95 per cent alcohol and washing twice with 20-cc. portions of cold alcohol, there is obtained 38.5 g. of pale yellow crystals melting at 136–137° .

Here's the mechanism:
1. Step One. Oxime Forming

HNO₂ which is formed by the in situ reaction of NaNO₂ and HCl at 0 ^oC reacts with the alpha hydrogen to form Nitrosonium Ion (NO⁺). Nitrosonium ion will be attacked by ethylacetoacetate in concerted way. See the mechanism as follows:

The nitrosonyl group will resonance to form oxime which is a more stable group.

2. Step Two: Reduction of Oxime to Form Amine

You have to know that this is not Clemmensen Reduction, it's just a usual reduction.

3. Step Three: Forming the Target Compound

Another ethylacetoacetate will be attacked (Nucleophilic addition and Elimination) by the compound.

Finally, the compound above arrange itself to form the target compound (controlled by reaction environment).

 

Glacial Acetic Acid act as catalyst and solvent in this reaction. The heat we use will release H₂O compound to form 2,4-dimethyl-3,5-dicarboxyethoxypyrrole.

Total Reaction:

This mechanism is approved by a Strata 2 student.
Thank You

Bibliography:

Clayden. Advanced in Organic Chemistry.
Fessenden. Organic Chemistry Ed. 3rd

http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv2p0202

Normalization Factor of Hydrogen Atom's Orbital Wave Function

In this post, We will only discuss 1 problem about how to Normalize an orbital wave function in Hydrogen Atom. For further reference, I will try to post it in the next posts.

There's an unnormalized Hydrogen Atom's Wave Function:

Now, let's normalize that function!


Problem 1

First, you have to know about these mathematical operations:

 

 Where x is a variable, e is the euler number, and a is a constant (In this case: Bohr's Atomic Radius [53 pm])

Second, you also have to know that in Hydrogen Atom, we use Spherical Coordinate where the Normalization Factor must be determined by the following operation:

Where N is the normalization factor, Ψ and Ψ* is the wave function and dτ is the derivative for spherical coordinate that can be expanded as follows:

 

Now, Let's calculate it.

To make an easier calculation, let's separate those integrals.

First Integral:

 

Second Integral:

By using Substitution Integral Method, (let x = cos θ so that dx = - sin θ d θ), we will get:

Third Integral:

 

We already calculated those three integrals, now let's finish this problem

Thank You
 

Olimpiade Kimia ITS