Alcohols (4) Redox reaction with sodium dichromate(VI)

Let’s discuss in this blog the oxidation of simple alcohols.

And contrary to many texts let’s look at alcohol oxidation beginning with tertiary alcohols:—

The reason is that they are usually just tagged on the end of a long discussion of primary and secondary alcohols and then they get forgotten when there is a significant point to make using them.

And it is this:

Let’s make clear that simple tertiary alcohols do not oxidise to an aldehyde, ketone or carboxylic acid because they have no α-hydrogen on the carbon with the —O—H group.

A tertiary alcohol like this then does not oxidise (but it will of course burn in oxygen.):

Second, let’s be clear that a secondary alcohol will oxidise given its α-hydrogen.  The product is a ketone. 

CH₃CH(OH)CH₃ +   [O]   =     CH₃C=OCH₃    +    H₂O

(The α-hydrogen is highlighted in yellow)

We use [O] as an abbreviation for the oxidizing agent. 

You should satisfy yourself that you can build equations like this for other secondary alcohols e.g: butan-2-ol, pentan-3-ol, pentan-2-ol.

The reaction conditions are to heat the secondary alcohol under reflux with orange acidified  sodium dichromate(VI) H₂SO₄/Na₂Cr₂O₇.

There is a colour change with this redox reaction.

The orange acidified dichromate(VI) (oxidation state +6) is reduced to deep green (Cr³⁺) (oxidation state +3).

The apparatus is illustrated below:

Heating under reflux as a practical technique is discussed in depth here

Ketones have a core carbonyl —C=O group to which are attached two identical or two different alkyl groups.

See the named examples here:

 pentan-2-one

pentan-3-one

There is this crazy weird guy (get his hair) discussing oxidation of secondary alcohols here

Third, let’s look closely now at the oxidation of primary alcohols.

First question: do they have the α-hydrogen?

Yes they will oxidise because they have the α-hydrogen attached to the carbon with the —O—H group.

In fact, they have two α-hydrogens so they can be oxidized in two stages as each α-hydrogen is oxidized successively. 

Stage 1 Oxidation: primary alcohol to aldehyde

Stage 2 Oxidation: aldehyde to carboxylic acid

Let’s look at each stage in turn.

Stage 1 Oxidation: primary alcohol to aldehyde

You say what’s an aldehyde?

The name it is suggested comes from a contraction of the English “Alcohol Dehydrogenated” picking up the “Al” and “Dehyd” gives the new name. 

Here’s the equation for the stage 1 oxidation of a typical primary alcohol:

CH₃CH₂CH₂OH   +   [O]   =     CH₃CH₂CHO   +    H₂O

propan-1-ol                               propanal

To name the aldehyde, add –al to the alkane root.

The aldehyde functional group is the –CHO group: that is a carbonyl group with an added hydrogen atom.

The double bond gives this functional group a planar/flat shape.

The reagent is acidified dichromate H₂SO₄/Na₂Cr₂O_7.

The conditions are to heat the mixture of aldehyde and oxidant very gently and distill over the volatile product as it forms.

Practically speaking this is a very tricky experiment to ensure you do not distill over the reaction mixture’s entire contents!!

You might ask why the aldehyde is more volatile than the alcohol from which it is made?

Aldehydes are not molecules that can hydrogen bond to each other. 

With weaker intermolecular forces (that is permanent dipoles and van der Waals forces only) aldehydes have lower boiling points than their corresponding alcohol and the carboxylic acid that they can be oxidized into.

Aldehydes can be identified using Tollen’s test or Fehlings test.

Tollen’s Test:

Sometimes called the silver mirror test, in Tollen’s test you heat the aldehyde with ammoniacal silver nitrate to produce a silver mirror on the side of the test tube (You must use a fresh, clean test tube straight out of the box for best results.)

You are using the aldehyde to reduce the silver (I) ions (Ag+) to silver metal (Ag)

Fehling’s Test:

This test comes, usually and traditionally, with two solutions Fehling’s #1 and #2.  Mixing both with the aldehyde and warming gently gives a red precipitate if the organic liquid is an aldehyde.

The test is very sensitive and any other colour change in the red-orange to green colour range confirms an aldehyde.

The red precipitate is copper(I)oxide, one of the only places this compound appears in general chemistry. 

Its red colour is a feature of its crystalline structure not its atomic electron arrangement since copper (I) salts ought to be colourless, copper(I) ions having a full 3d subshell.

Fehling’s #1 solution is usually blue copper(II) sulphate.

Fehlings #2 solution is a solution of Rochelle salt or sodium potassium tartrate.  

The essential chemistry in this test is also a reduction this time of copper (II) salts (blue) to copper (I) (red).

Stage 2 Oxidation: aldehyde to carboxylic acid

Here is the equation for the Stage 2 oxidation of an aldehyde to a carboxylic acid.

CH₃CH₂CHO    +    [O]      =      CH₃CH₂COOH

Propanal                              propanoic acid

And from the alcohol we have:

CH₃CH₂CH₂OH  +   2[O}    =      CH₃CH₂COOH   +    H₂O

As you can see twice the number of moles of oxidant is required to fully oxidise the alcohol to the carboxylic acid as is needed to produce the aldehyde from the alcohol or the carboxylic acid from the aldehyde.

Carboxylic acids contain the carboxyl group —COOH

We don’t usually write —CO₂H as this would imply that both oxygen atoms were equivalently bonded to the carbon atom.

Rather there is a double-bonded oxygen and a single-bonded oxygen present in the group hence the use of two separate O atom symbols. 

You ought to be aware too that the carboxylate anion is a delocalized system as these diagrams show. (c.f. benzene)

The reaction requires double the amount of oxidant but we can use the same reactant as before H₂SO₄/Na₂Cr₂O₇

Heating the mixture of organic and oxidant under reflux ensures sufficient time and contact between reactants to produce the carboxylic acid.

To confirm that the carboxylic acid has been formed, convert the apparatus to distillation and remove some of the aqueous organic product.

Addition of sodium carbonate solid should produce fizzing of carbon dioxide from reaction with the organic product. (see equation below)

Na₂CO₃(s) +  2RCOOH(aq)  = 2RCOO^—Na+(aq) + H₂O(l) + CO₂ (g)

This fizzing is not from the sulphuric acid reactant because that has too high a boiling point to have distilled over. 

As this equation shows, only one of the hydrogens are acidic and the carboxylic acid behaves as a typical acid. 

However, carboxylic acids are weak acids, typically weaker than mineral acids like hydrochloric acid.

Their weakness is associated with the stability of the carboxylate anion.

The carboxylate anion is stabilised by delocalisation of the negative charge over the whole anion. 

You can see in this diagram the red space that shows the delocalised electron molecular orbitals above and below the plane of the carboxylate anion.

The whole anion carries the negative charge, it is not localised to one of the oxygen atoms.

Thus we can write the dissociation of a carboxylic acid as follows:

RCOOH    ⇋     RCOO^—     +    H+