Mod 7 Oxidation of Alcohols (1 Viewer)

[okayokay123]

So with the oxidation of a primary alcohol --> aldehyde --> carboxylic acid, what is the need for an oxidising agent such as Cr2O7(2-) and dilute acid such as H2SO4?

Also are the acid and dichromate actually involved in the reaction or are they just catalysts?

 

[jazz519]

Oxidation is one type of redox reaction (a reaction in which electrons are transferred). Reduction is another type of redox reaction.

In oxidation electrons are lost and in reduction electrons are gained.

These two reactions happen together normally and don't occur by themselves.

For oxidation of alcohols, the oxidising agent is the chemical which causes the alcohol to be oxidised (i.e. lose electrons). In the process the oxidising agent is reduced as it gains the electrons that were lost by the alcohol.

If you look on the HSC data sheet towards the bottom you will see an equation which shows the reduction of the Cr2O7 2- and within that equation there is also H+ (this is why acid is needed it is a part of the reaction).

They are not catalysts as the dichromate and acid are not in the same form at the end of the reaction, which would be needed for a catalyst as it reforms.

 

[CM_Tutor]

Adding to what Jazz has said...

One reason that redox reactions can be confusing is that it is not obvious that electrons are being lost or gained. This is one advantage of looking at half-equations that make this clearer. Also, the gain or loss of electrons can occur by transferring atoms.

Consider the oxidation of ethanol by dichromate. The oxidation reaction is:

CH₃CH₂OH + H₂O -----> CH₃COOH + 4H⁺ + 4e^-​

and so we can see that the gain of an O atom and the loss of two H atoms from ethanol (net) has led to electrons as a product and confirming that the process is an oxidation. Now, we can't go and get a bottle of electrons nor can we have electrons just hanging around unbound in a reaction, so we need a second process (reduction) that gains electrons to balance out the process. In this case, that process is:

Cr₂O₇^2- + 14H⁺ + 6e^- -----> 2Cr³⁺+ 7H₂O​

The chromium centre in the dichromate ion Cr₂O₇^2- has chromium in a formal Cr⁶⁺ state that is gaining electrons in becoming the Cr³⁺, taking up the electrons lost in the oxidation. The stoichiometry of the overall process can be found by adding these half-equations by balancing the electrons. So, by looking at 2 x RED + 3 x OX:

2Cr₂O₇^2- + 28H⁺ + 12e^- + 3CH₃CH₂OH + 3H₂O -----> 4Cr³⁺+ 14H₂O + 3CH₃COOH + 12H⁺ + 12e^-​

and then cancelling the duplicates:

3CH₃CH₂OH + 2Cr₂O₇^2- + 16H⁺ -----> 3CH₃COOH + 4Cr³⁺+ 11H₂O​

We see that the reaction requires two dichromate ions for every three ethanol molecules plus quite a bit of acid.

As Jazz said, the dichromate ion and the acid are consumed / transformed in the reaction and so are not catalysts. Furthermore, they are directly involved in the reaction as the intermediates in the process include chromate esters of ethanol with structures like CH₃CH₂O-CrO₃^2-. These are formed as part of transferring the extra O atoms to make the carboxyl groups.

 

[okayokay123]

Adding to what Jazz has said...

Thanks, the ionic equations helped