Sciencemadness Discussion Board

Alcohol oxidation via chromic acid

coherent - 23-7-2023 at 03:19

I've been having some difficulty with what should be a fairly straightforward lab prac, the oxidation of cyclohexanol into cyclohexanone with acidified potassium dichromate.

The potassium dichromate solution was prepared by the addition of 5g potassium dichromate with 10ml of distilled water, mixed until dissolved and then 20ml of conc. sulfuric acid was added which yielded a dark red solution. A 250ml flask was charged with the above chromic acid solution.

To this solution 25ml of cyclohexanol was carefully added dropwise followed by the attachment of a reflux condenser.

However from this stage onward no reaction has occurred, it's got me stumped honestly. I've tried gently heating and stirring the solution to prompt the oxidation but nothing is happening.

Could someone please help me understand what's going wrong?

[Edited on 23-7-2023 by coherent]

Fery - 23-7-2023 at 03:39

Stirring is necessary as cyclohexanol is very poorly miscible with water and even less miscible with solution of inorganic compounds in water.
I believe you got some product (mixed with excess of reactant). Did the color change from orange to green? You used not enough of dichromate. 1 mol of dichromate (294g of K2Cr2O7) oxidizes 3 mols of alcohol (3x100 = 300 g of cyclohexanol). Using some excess of oxidizer is preferred to complete the oxidation so no reactant remains present. In that case you got mixture of green and orange water phase, do not expect change into green.
http://wwwchem.uwimona.edu.jm/lab_manuals/c10expt21.html

coherent - 23-7-2023 at 03:57

Quote: Originally posted by Fery  
Stirring is necessary as cyclohexanol is very poorly miscible with water and even less miscible with solution of inorganic compounds in water.
I believe you got some product (mixed with excess of reactant). Did the color change from orange to green? You used not enough of dichromate. 1 mol of dichromate (294g of K2Cr2O7) oxidizes 3 mols of alcohol (3x100 = 300 g of cyclohexanol). Using some excess of oxidizer is preferred to complete the oxidation so no reactant remains present. In that case you got mixture of green and orange water phase, do not expect change into green.
http://wwwchem.uwimona.edu.jm/lab_manuals/c10expt21.html


Oddly I haven't observed any colour change at all, on a smaller scale with a much stronger chromic acid preparation I noticed some very small bubbles rising (very few, very easy to miss) which subsided quickly. Perhaps the oxidation did occur but was not enough to reduce the chromate? I've moved both experiments to an ice bath to see if the cyclohexanol solidifies.

I'm hoping that I've simply missed the reaction occur and that I've got the desired product, figured it would be a bit more dramatic.

coherent - 23-7-2023 at 04:29

Nope, definitely not oxidized. Both samples solidified entirely.

I'll have to try again tomorrow. Theoretically, I could perform the oxidation using potassium chromate rather than dichromate, my understanding is that the end product is chromic acid, also, acidification of the chromate yields the dichromate so it seems a bit like needless complexity? Unless I'm mistaken.

I'm quite confident that miscibility isn't the issue as I left the solution vigorously stirring for quite some time to no effect. The quantities in the chromic acid surely must be the problem...

Edit: I'll reattempt using the procedure in the link above, however I'm hoping to be able to substitute potassium dichromate with potassium chromate to simplify the process a little. Should there be any adjustment to the quantity of chromate compared to the dichromate?

[Edited on 23-7-2023 by coherent]

Texium - 23-7-2023 at 06:17

What exactly you mean by “vigorously stirring for quite some time?” Were you using magnetic stirring to get a really good vortex going for a couple hours, or were you just manually beating it with a glass rod until your arm got tired?

I don’t buy that the amount of dichromate you used was the major problem here. Since you used an excess of the alcohol, the solution should have turned green, and your problem would have then been having excess unreacted cyclohexanol. No reaction occurring points to a different experimental error. That being said, you should definitely use an excess of Cr(VI) next time you try it. Just know that there may still be another issue to correct.

Regarding chromate vs dichromate, it doesn’t matter. Chromate turns into dichromate in acidic solution. It’s just a matter of what you happen to have on the shelf. If you’re following a procedure that uses chromate though, you’ll have to use half the amount of dichromate (in terms of moles, not mass) to ensure you end up with the same amount of Cr(VI) in solution. If you need help with the stoichiometry, I can help you, but I won’t spoonfeed you the answer.

solo - 23-7-2023 at 07:22

Jones Oxidation
Reference Information


The Jones Oxidation allows a relatively inexpensive conversion of secondary alcohols to ketones and of most primary alcohols to carboxylic acids. The oxidation of primary allylic and benzylic alcohols gives aldehydes. Jones described for the first time a conveniently and safe procedure for a chromium (VI)-based oxidation, that paved the way for some further developments such as Collins Reaction and pyridinium dichromate, which also enabled the oxidation of primary alcohols to aldehydes.

http://www.adichemistry.com/organic/organicreagents/jones/jo...





Attachment: Jones Oxidation.pdf (169kB)
This file has been downloaded 171 times

Fery - 23-7-2023 at 11:32

Yes guys but what's strange that the mixture did not turn green although not enough of dichromate was used and what is even stranger that the product solidified which means almost no ketone present. Even few % of ketone would prevent solidification of the alcohol.

Texium - 23-7-2023 at 11:39

Quote: Originally posted by Fery  
Yes guys but what's strange that the mixture did not turn green although not enough of dichromate was used and what is even stranger that the product solidified which means almost no ketone present. Even few % of ketone would prevent solidification of the alcohol.
That’s more or less what I was saying. Yes, they should use more dichromate, but there’s definitely something else wrong.

coherent - 23-7-2023 at 13:51

Quote: Originally posted by Texium  
What exactly you mean by “vigorously stirring for quite some time?” Were you using magnetic stirring to get a really good vortex going for a couple hours, or were you just manually beating it with a glass rod until your arm got tired?


Regarding the stirring:
The solution was mixed in a 250ml flat bottom round flask with magnetic stirring, initially for 10min at a medium speed and then another 10 minutes at a higher speed. I don't know exactly how fast, but the solution was fairly homegenous with many small bubbles of the aqueous layer distributed throughout.

I'm a bit skeptical about the potassium dichromate which was prepared for this experiment, the chromate I initially prepared was canary yellow and looked like textbook dichromate, this was vacuum filtered, thoroughly dried and then dissolved in 100ml of water to be acidified with concentrated H2SO4. However instead of a brilliant orange solution I ended up with a fairly brown, somewhat orange solution which yielded dark brown crystals which upon drying became orange, but it didn't look right compared to the almost fluorescent orange typical or dichromate.

[Edited on 23-7-2023 by coherent]

coherent - 23-7-2023 at 13:59

Given that the dichromate wasn't off the shelf I'm feeling very skeptical about it.
The potassium chromate was prepared using chromic oxide and the molten alkali + KNO3 method. When I was acidifying the redissolved chromate I noticed that the pH was quite high (12-13) so perhaps there was residual contamination of alkali?

After a few drops of H2SO4 the pH came back down to 7, so I don't think there was a lot however I think that I'll return to this stage to neutralize the chromate before attempting again.

teodor - 24-7-2023 at 00:55

I used Na2Cr2O7 solution in glacial acetic acid, it oxidizes cyclohexanol to cyclohexanone and for the next step I used KMnO4 in water. The procedure is described by Louis F. Fieser in "Organic Experiments" as well as in his "Reagents for Organic Sunthesis".
The original procedure mentions 15 g of Na2Cr2O7 in 25 ml AcOH and I tried to use CrO3 but it has lower solubility than the soldium salt. So, I tried to dissolve 40g CrO3 in 100 ml AcOH but failed to do so. Then I've just added the stoihiometric amount of NaOH and this made to dissolution possible. The rest I did exactly as in Fieser but in 4x scale. I've got a good yield of cyclohexanone as well as adipic acid.
If this information is helpful for you I'll attache the pdf scan of the procedure later here.


[Edited on 24-7-2023 by teodor]

coherent - 24-7-2023 at 03:26

Quote: Originally posted by teodor  
I've got a good yield of cyclohexanone as well as adipic acid.
If this information is helpful for you I'll attache the pdf scan of the procedure later here.


That would be great if you could pass on the pdf :)

I recrystallized some of my dichromate after dissolving it in water and adjusting the pH down to 4.9, this seemed to help a great deal and yielded bright orange crystals.
I figured I'd test it with a more straightforward test, a small clump in a test tube followed by two droppers of concentrated H2SO4 and finally a drop of ethanol, it reacted quite violently and immediately turned green, bingo!

I think I thoroughly underestimated the quantity of solid chromate / dichromate needed per mol of alcohol too, I've spent the afternoon preparing another batch of chromate from chromic oxide so I can have a go again soon.

Regarding the stoichiometry of dichromate VS chromate, would this be as simple as 1:2? Perhaps I'm being naive... but I figure that the dichromate would yield two chromates per molecule, so twice molar quantity is required in the case of chromate? If this is the case I can see why the dichromate is preferable as the amount of solid material must get a bit out of hand.

Texium - 24-7-2023 at 04:45

Quote: Originally posted by coherent  
I think I thoroughly underestimated the quantity of solid chromate / dichromate needed per mol of alcohol too, I've spent the afternoon preparing another batch of chromate from chromic oxide so I can have a go again soon.

Regarding the stoichiometry of dichromate VS chromate, would this be as simple as 1:2? Perhaps I'm being naive... but I figure that the dichromate would yield two chromates per molecule, so twice molar quantity is required in the case of chromate? If this is the case I can see why the dichromate is preferable as the amount of solid material must get a bit out of hand.
Can you write a balanced equation for the reaction? That will be a good starting point. I have a sinking feeling that you’re playing somewhat of a guessing game with your reagents rather than doing proper stoichiometry, which can be quite dangerous when strong oxidizers are involved…

Please, show a balanced equation for the reaction and how you would calculate the stoichiometric amount of dichromate needed to oxidize 25 mL of cyclohexanol. If you don’t know how to do that, I’m more than happy to help you learn, but it’s something you should know before setting up more reactions with seriously hazardous reagents.

teodor - 24-7-2023 at 05:06

coherent, I believe my procedure on 4x scale was:

Put 40g CrO3 into 100ml of glacial acetic acid, add 16g NaOH, heat to dissolve everything and proceed as in the attached pdf.

The equation to calculate NaOH amount is:
2 CrO3 + 2 NaOH = Na2Cr2O7 + H2O

Attachment: img-230724130702.pdf (152kB)
This file has been downloaded 187 times

[Edited on 24-7-2023 by teodor]

Texium - 24-7-2023 at 08:57

teodor, I would appreciate it if you would please refrain from further spoonfeeding until we’ve established that OP knows how to do basic stoichiometry. I want to make sure people understand what they’re doing, rather than just following recipes.

teodor - 24-7-2023 at 11:23

Quote: Originally posted by Texium  
teodor, I would appreciate it if you would please refrain from further spoonfeeding until we’ve established that OP knows how to do basic stoichiometry. I want to make sure people understand what they’re doing, rather than just following recipes.


OK.
But those alcohol/ketone oxidations sometimes are tough staff. I myself experimenting with camphor oxidation to camphoric acid and I am puzzled why some methods work well and other don't work at all. Temperature/concentration/solvent/time of the reaction is hard to predict just by stoihiometry. At least for me.

coherent - 24-7-2023 at 16:39

Quote: Originally posted by Texium  
I’m more than happy to help you learn, but it’s something you should know before setting up more reactions with seriously hazardous reagents.


Help with understanding the stoichiometry would be much appreciated, I've found it a bit difficult to understand the electron half equations honestly. A big part of why I've come to these forums is to learn more about this subject, it's been a long time since I studied chemistry so please be patient with me.

coherent - 24-7-2023 at 17:32

I'll have a go at this equation:

K2Cr2O7 + H2SO4 + HOCH(CH2)5 -> 2Cr3 + K2SO4 + (CH2)5CO

I'm sure I've made an error as I can't really wrap my head around what's happening with the reduction of dichromate, the dichromate separates as potassium sulfate is formed, but why does Cr2O7 become 2Cr3??

If someone could help explain the half electron equation i'd greatly appreciate it.

[Edited on 25-7-2023 by coherent]

teodor - 25-7-2023 at 00:33

Coherent, how many oxidation states Cr can have? How to get those states in water solution? I think what you lack is basic understanding of metal chemistry and oxidation/reduction with Cr metal as an example.

clearly_not_atara - 25-7-2023 at 06:02

Usually we speak only of Cr3+ and Cr6+, although with a little effort you can make Cr2+ (and these compounds are supposedly good reducing agents, though rarely used for this) and there is also Cr4+ in CrF4 and Cr5+ in K3Cr(O2)4. So, it's complicated :D

coherent - 25-7-2023 at 07:21

Quote: Originally posted by clearly_not_atara  
Usually we speak only of Cr3+ and Cr6+, although with a little effort you can make Cr2+ (and these compounds are supposedly good reducing agents, though rarely used for this) and there is also Cr4+ in CrF4 and Cr5+ in K3Cr(O2)4. So, it's complicated :D


Attachment: Potassium_chromate_xtals (1.1MB)
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Interesting! I'm only familiar with the trivalent, hexavalent and elemental (0) states personally. The actual mechanics of oxidation are still quite confusing to me, got plenty of reading ahead of me...

Texium - 25-7-2023 at 08:06

This is a good place to start: Balancing Redox Reactions by Half-Reaction Method

DraconicAcid - 25-7-2023 at 08:28

Quote: Originally posted by Texium  
This is a good place to start: Balancing Redox Reactions by Half-Reaction Method


And just to reiterate what it says in the first step- use net ionic equations. You can add spectator ions back in at the end if you really want a molecular equation. The potassium ions and the sulphate ions in your example are just going to make your reactions more cluttered during the hard parts.

chornedsnorkack - 26-7-2023 at 03:05

Tracking electrons is one option. Another could be tracking oxygens. Take reaction
K2Cr2O7+?H2SO4+??=2Cr3++???
and break it down into steps:
1) K2Cr2O7+H2SO4=K2SO4+H2O+2CrO3
2) 2CrO3=Cr2O3+3"O"
3) Cr2O3+3H2SO4=Cr2(SO4)3+3H2O
4) (C5H10)CHOH+"O"=(C5H10)CO+H2O

It gets much easier to track oxygens if you substract the spectators like the "K2O" in a salt (neither oxygen nor potassium change oxidation state) and so find out the identity and amount of oxide that actually does get reduced. After all, this way you can recalculate your stoichiometry for a different reagent.

Texium - 26-7-2023 at 06:39

That "oxygen" method works well in this example, but is less general, since it only works with oxidizer species that are traditional oxygen-donors. Doesn't work in situations such as reduction of iron(III) to iron(II) using iron(0), when electrons are the only thing moving around (though that is a very easy one to figure out).

Boffis - 26-7-2023 at 08:32

I have very recently used sodium dichromate and dilute sulphuric acid to oxidizes 2-pentanol to 2-pentanone on a roughly 0.5 molar scale. I obtained a 70% yield after distillation and encountered no particular difficulty. Pleased with the result. 2-pentanol was not completely miscible with the aqueous phase but with vigorous stirring this doesn't appear to be a big problem. For less water miscible materials acetic acid a suggested above is probably the way to go but remember to check the Bps are sufficiently different to allow isolation of the product.

chornedsnorkack - 27-7-2023 at 01:05

Tracking oxygen is actually the definition of Chemical Oxygen Demand. Which has been moved from KMnO4 to K2Cr2O7 as oxidant.