Chemistry: the reactivity of Aldehydes and Ketones

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#1
Report Thread starter 4 years ago
#1
Hi,
I was wondering if someone could help me understand the differences in reactivity between aldehydes and ketones when mixed with acidified potassium dichromate (K2Cr2O7/H+) under reflux.

I understand the basic principles of the reaction, where by the aldehyde is oxidised by the potassium dichromate which is reduced, to produce a carboxylic acid (propanoic acid) turning the orange solution, green.

And that the Ketone (Propanone) can not be further oxidised and remains orange.

But what I don't understand, is the mechanism... what is actually happening to the aldehyde in order for it to be oxidised by the potassium dichromate?

And why is the 'lack of hydrogen' the reason why the ketone can not be further oxidised?

If anyone can help, I'd greatly appreciate it.
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#2
Report 4 years ago
#2
This page explains it in excellent detail.

https://www.masterorganicchemistry.c...ol-oxidations/

Firstly, OILRIGS.

This when applied to Organic Chemistry; the definition of 'loss of electrons' becomes equivalent to the 'loss of Hydrogen'.

So removing Hydrogen results in oxidation. This definition strictly only works for Organic Compounds. It doesn't actually work with Metals e.g NaH is actually the oxidised form of Na, because Na+ (+1) -----------> Na (0) (i.e Losing Hydrogen reduces it!). The 'Organic Definition' actually arises from electronegativity.

A good explanation of how this works, if you're interested, lies here: https://chemistry.stackexchange.com/...ecome-oxidised

So because this doesn't work for Metals, OILRIGS with respect to electrons, is taught instead. Nevertheless we are using the 'Organic redox definition' for this case,


If you are familiar with E2 eliminations, we need a leaving group and a Beta-hydrogen.

Essentially these reactions place an oxaphilic lewis-acid, leaving group in coordination with a carbonyl and then use a base to remove an alpha Hydrogen (which is acting essentially as a beta hydrogen if you compare the mechanisms to E2 eliminations).


Example: A good 3 step mechanism is shown 1/4 down the website below, with CrO3, after 'Tertiary Alcohols'.

https://chem.libretexts.org/LibreTex...on_of_Alcohols

The first two steps are proton transfers, the key step is the third, you can see the Hydrogen being removed and the carbonyl reforming. That is really the essence of the majority of Organic Oxidation reactions.

The lack of what really is an alpha-Hydrogen attached to the carbonyl group is what prevents Ketones oxidising further.

The key with oxidation of Aldehydes, they undergo a further 'hydrate' step involving H20, shown in the 'Master Organic Chemistry' page above. That turns them into Carboxylic Acids. Without water, Aldehydes won't oxidise to Acids.


The Colour change can explained by 'Crystal Field Theory', as a result of the change in Oxidation state from Cr6+ and Cr3+. This actually is a change in the number of electrons in the D-orbitals of the Chromium ion, along with the ligands (=O, OH). These variables change the wavelength of photons that are absorbed and reflected. As a result, the observed colour changes. There's a lot more to it though and it's beyond A level.
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#3
Report Thread starter 4 years ago
#3
Wow, Thank you so much for your intricately detailed answer. You've answered my question in full, which I truly appreciate and i've gained some extra foundation knowledge aswell.

Thanks again.
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#4
Report Thread starter 4 years ago
#4
I'm sorry, but I do have one more question, if you do have time.

As I just want to make sure that I defiantly understand.

During the redox reaction between:

Propanal (mixed with acidified potassium dichromate) to form propanoic acid.

I assume that the hydride H- ion comes from propanal and that the additional proton needed to make H2 comes from the acidified potassium dichromate?

Therefore the propanal becomes Dehydrogenated (or oxidised by the loss of electrons) and the acidified potassium dichromate becomes hydrogenated (or reduced by the gain of electrons).

But then i'm still not sure how propanoic acid is formed?
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