Oxidation of Aldehyde with Potassium Dichromate(VI)

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SUMMARY

The oxidation of aldehydes can be effectively achieved using acidified potassium dichromate(VI) solution, specifically with dilute sulfuric acid. The reaction involves warming the mixture to convert the orange solution to green, indicating the presence of aldehyde. While sulfuric acid is preferred due to its inertness and ability to maintain a high redox potential, other strong acids like phosphoric acid can also be utilized, although they may not dissociate as effectively. The mechanism involves the formation of active chromate species and chromate esters, which facilitate the oxidation process.

PREREQUISITES
  • Understanding of redox reactions and Nernst equation
  • Familiarity with potassium dichromate(VI) and its properties
  • Knowledge of acid-base chemistry, particularly strong acids
  • Basic principles of organic oxidation reactions
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  • Research the mechanism of Jones oxidation in detail
  • Explore alternative oxidizing agents for aldehydes
  • Study the effects of different acids on the oxidation of aldehydes
  • Learn about the safety and handling of potassium dichromate(VI)
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Chemists, organic chemistry students, and laboratory technicians involved in oxidation reactions and synthesis of organic compounds will benefit from this discussion.

DK168
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One method to oxidize aldehyde is by using acidified potassium dichromate(VI) solution

The step: A small amount of potassium dichromate(VI) solution is acidified with dilute sulphuric acid and a few drops of the aldehyde is added. If nothing happens in the cold, the mixture is warmed gently for a couple of minutes - for example, in a beaker of hot water. The presence of aldehyde will turn the orange solution into green solution.

Question. Is sulphuric acid can be replaced with other acid to acidified the solution? Thanks in advance :)
 
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DK168 said:
other acid
Nitric? Perchloric? You are preparing chromic acid (the oxidizer) with a dehydration of chromate/dichromate.
 
Yes, but no.

The reduction half reaction is

Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

so it requires plenty of H+ to keep the redox potential high (think in terms of the Nernst equation). Technically any strong acid will do, but it happens that most common strong acids that can be used are themselves oxidizing. Perhaps with the exception of HCl, but chlorides can get oxidized to Cl2 - and it is not something you want to have produced in the solution, as it can make a mess of your molecule.

SO42- happens to be relatively inert and as such rarely interferes with other reactions, which makes the sulfuric acid a perfect choice whenever you need a highly acidic solution.
 
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Bystander said:
You are preparing chromic acid (the oxidizer) with a dehydration of chromate/dichromate.
There is way too much water in dilute sulfuric acid to dehydrate chromate. You generally hydrate chromium (VI) oxide to get chromic acid. Or you hydrate dichromate to get dichromic acid. See link below.
DK168 said:
oxidize aldehyde
Aldehydes are generally pretty easy to oxidize (sometimes annoyingly so), so you can get away with pretty mild conditions. In fact, many aldehydes will oxidize if you leave them out in air under light for a few days. Also, Borek pointed out:
Borek said:
SO42- happens to be relatively inert and as such rarely interferes with other reactions, which makes the sulfuric acid a perfect choice whenever you need a highly acidic solution.
to which I'll add: 1M H2SO4 = 2N H2SO4, so you get more H+ bang for your H2SO4 buck. Phosphoric acid donates even more protons per mole, so that will probably work as well.

The acid mainly catalyzes the formation of the active chromate species and the chromate ester, which reductively disproportionates to give the carboxyl and a Cr(IV) species. The general mechanism for Jones oxidation can be found here:
http://www.organic-chemistry.org/namedreactions/jones-oxidation.shtm
 
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TeethWhitener said:
to which I'll add: 1M H2SO4 = 2N H2SO4, so you get more H+ bang for your H2SO4 buck. Phosphoric acid donates even more protons per mole, so that will probably work as well.

No and no. HSO4- is a weak acid, so in typical case where we use sulfuric acid for acidification it is almost not dissociated at all (in 1M solution only about 1-2% dissociaties up to SO42-). This is even worse for the phosphoric acid, as H3PO4 is weaker even than HSO4-.
 
Borek said:
No and no. HSO4- is a weak acid, so in typical case where we use sulfuric acid for acidification it is almost not dissociated at all (in 1M solution only about 1-2% dissociaties up to SO42-). This is even worse for the phosphoric acid, as H3PO4 is weaker even than HSO4-.
You're right of course. Thanks for the correction. Though I doubt it matters for the Jones oxidation: you probably don't need very much acid to push the reaction forward.
 
That sounds an awful lot like a qualitative test for the presence of aldehydes. It is a qualitative technique, not a real suitable method for oxidizing aldehydes. I'm sure Morrison and Boyd (or any organic text) has several milder oxidation reactions that will suffice.
 
Mea culpa. Went from a "general" statement regarding acid anhydrides to a specific case without thinking.
 
Dichromate Ions are formed when solid K2Cr2O7 is dissolved in water and the aqueous solution can be represented by an equilibrium :
Cr2O7 2- +H2O <---> CrO42- + 2H+
Now, this equilibrium is pH sensitive and if you want more dichromate ions, you can add more acid (HCl or H2SO4 or HNO3), which will follow Le Chatelier's principle and shift the equilibrium to the left.
I guess all strong acids can work as long as they are not reacting with potassium dichromate.
 
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Thank you all for the helpful information..
 

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