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You should know that oxidation is defined as the loss of electrons.
In addition, there are two other definitions of oxidation in organic chemistry:
When alcohols are oxidised, hydrogen is removed and oxygen is added to the carbon attached to the -OH group.
The structure of the alcohol affects the outcome of their oxidation. Regardless, all types of alcohol oxidation involve the same reagents.
See if you can identify these isomers of butanol (molecular formula: C4H10O). Are they primary, secondary, or tertiary alcohols?
'B' is a primary alcohol, 'A' is a secondary alcohol, and 'C' is a tertiary alcohol. 'A' is secondary because it has two R groups, while 'B' is primary because it has only one R group. 'C's three R groups make it tertiary. It is extremely important to understand this to see why primary, secondary, and tertiary alcohols have different outcomes upon oxidation. (Do refer to Alcohols should you struggle with this.)
There are two reagents used in alcohol oxidation - potassium dichromate (VI) and dilute sulfuric acid. Below is a table showing the molecular formulae and functions of these reagents.
|Potassium dichromate (VI) K2Cr2O7||Oxidising agent, facilitates the oxidation of alcohols|
|Dilute sulfuric acid H2SO4||Catalyst, lowers the activation energy to speed up the reaction|
The reaction mixture containing the two reagents is also known as acidified potassium dichromate (VI).
Primary alcohols can undergo partial and full oxidation.
Primary alcohols contain only one R group attached to the C-OH carbon. Methanol, shown below, is an exception - it is a primary alcohol without any R groups.
Butan-1-ol is the example given below, where the R group is highlighted in orange.
Let’s now simplify the butan-1-ol structure further by representing the highlighted part with an R, and compare it with another molecule on the right.
Do you notice that the molecule on the right has one less hydrogen attached to the carbon? Also, instead of an OH group, the molecule on the right now has a double-bonded oxygen? To get from our primary alcohol to the molecule on the right, we use partial oxidation. This results in a product known as an aldehyde. The oxidation process also releases a water molecule.
The general chemical equation for the process is given below:
RCH2OH + Cr2O7 or [O] --------> RCHO + H2O
Remember to include the structural formula in your answers whenever you are asked to write chemical equations. Do also remember to expand on the R group! Also, [O] represents an oxidising agent.
The aldehyde can further oxidise into a carboxylic acid. In this reaction, oxygen is added to the H bonded to the C=O carbon to form an -OH group as circled below. The new molecule now contains the C=O and OH functional groups, which combine to form the carboxylic acid functional group.
In brief, partial oxidation of primary alcohols results in aldehydes, whereas full oxidation results in carboxylic acids.
If we want to produce aldehydes through partial oxidation, we use a technique called distillation with addition. We only gently heat the oxidising agent, potassium dichromate, and add the alcohol slowly to the heated flask. The aldehyde produced evaporates off immediately and travels through a condenser that condenses the aldehyde vapour. Evaporating the aldehyde immediately is important to prevent further oxidation of the aldehyde into a carboxylic acid, which we do not want. We then collect the condensed aldehyde in the receiver.
On the other hand, complete oxidation uses a reflux apparatus. The apparatus consists of a reaction flask with a condenser placed on the neck of the flask. By heating under reflux, the products of oxidation are contained in the vessel as the vapour condenses and drips back into the reaction flask. This prolongs the oxidation process, ensuring all the product is fully oxidised. Without reflux, the alcohol would first oxidise to an aldehyde which would evaporate off immediately. Using reflux traps the aldehyde so that it can oxidise further into a carboxylic acid.
Learn how to draw out the apparatus as exams do test you on that!
Secondary alcohols contain two R groups attached to the C-OH carbon.
Butan-2-ol is the example given below, where the R groups are highlighted in orange.
Let’s now simplify the above structure further by representing the circled parts with R and R’, and compare it with another molecule on the right:
Do you notice that the molecule on the right has no hydrogens attached to the carbon? Also, instead of an OH group, the molecule on the right now has a double-bonded oxygen? Oxidation of a secondary alcohol results in the product on the right, known as a ketone. The oxidation process also releases a water molecule.
Note that ketones and aldehydes are isomers of each other. Aldehydes contain the carbonyl (C=O) group at the end of the carbon chain, as highlighted below. On the other hand, ketones have the carbonyl (C=O) group in the middle of the carbon chain. Furthermore, both functional groups have different names. Aldehydes use the suffix ‘-al’ (eg: propanal) whereas ketones use the suffix ‘-one’ (eg: propanone). This is shown in the example below.
The general chemical equation for the oxidation process is stated below:
RCH(OH)R’ + Cr2O7 or [O] --------> RCOR’ + H2O
Remember to include the structural formulae in your answers whenever you are asked to write chemical equations. Do also remember to expand on the R group!
Unlike aldehydes, which can be oxidised again into carboxylic acids, ketones cannot be oxidised further. This is because there are no carbon-hydrogen bonds on the carbonyl carbon left in ketones for oxidation to take place. Propanone is shown below.
As secondary alcohols will only be fully oxidised, the technique used would be heating under reflux.
The products of oxidation are contained in the vessel as the vapour condenses and drips back into the reaction flask. This prolongs the oxidation process, ensuring all the product is fully oxidised. The oxidation of secondary alcohols does not require immediate distillation.
Tertiary alcohols contain three R groups attached to the C-OH carbon. The example below is a molecule of 2-methylpropan-2-ol with the R groups circled in blue.
Let’s now simplify this molecule by replacing the circled R groups with R, R’ and R’’.
Since there are three R groups, there are no C-H bonds in tertiary alcohols. Thus, the lack of C-H bonds meant that tertiary alcohols can’t undergo oxidation. This is similar to why the ketones from the oxidation of secondary alcohols cannot be oxidised further. The enthalpy of the remaining C-C bonds is too high for oxidation to take place - breaking the C-C bonds for oxidation requires too much energy.
Even when 2-methylpropan-2-ol is heated with acidified potassium dichromate, the alcohol will remain unchanged.
An alcohol can have a primary, secondary, or tertiary structure. To classify alcohols, we use a two-step process.
This step is based on the principle that tertiary alcohols cannot be oxidised.
When heated with orange acidified potassium dichromate, a solution containing primary or secondary alcohols turns green, whereas a solution containing tertiary alcohols remains orange. The colour change is due to the dichromate ion getting reduced into Cr3+ as it acts as an oxidising agent.
If we know from step 1 that an unknown alcohol is not a tertiary alcohol, we can carry out step 2. As we learnt, ketones cannot be oxidised further whereas aldehydes can. So, we use another colour change to identify any aldehydes present in our solution. If there is a colour change, there must be an aldehyde present, hence our alcohol must have been a primary alcohol.
There are two approaches to this step. One is to heat under Fehling’s solution. The solution containing the primary alcohol would turn from blue to brick red.
Another approach is to heat the solution under Tollen’s reagent. A silver mirror is formed in the solution containing the primary alcohol.
(The detailed mechanism behind the colour changes seen in both Fehling’s solutions and Tollen’s reagent is explored at A2 level.)
To help make better sense of the principles underlying alcohol oxidation, here's a table to compare the oxidation of primary, secondary, and tertiary alcohols.
Alcohol name and structure
Oxidation reaction techniques
Products upon oxidation
When heated under K2CrO4
When heated under Fehling’s or Tollen’s
Partial - butanal and water
Full - butanoic acid and water
Colour change from orange to green
Fehling’s - blue to brick red
Tollen’s - silver mirror
Butanone and water
Colour change from orange to green
Fehling’s - remains blue
Tollen’s - remains clear
Irrelevant as cannot be oxidised
No oxidation products formed, hence irrelevant
The oxidation of alcohol is a type of reaction where alcohol loses hydrogens or gains oxygens in the presence of an oxidising agent such as acidified potassium dichromate.
Ketones are formed by the oxidation of secondary alcohols. Tertiary alcohols cannot be oxidised.
Yes, reflux is needed to oxidise secondary alcohols to the final product, ketone.
Only primary and secondary alcohols can be oxidised. Both alcohols will lose a proton, though only primary alcohols can gain extra oxygen to form carboxylic acids.
Based on the OILRIG mnemonic, oxidation is the gain of electrons. True or False?
False - oxidation is the LOSS of electrons.
Outline two other definitions of oxidation specific to organic chemistry.
Loss of hydrogens, and gain of oxygens, all on the same carbon.
Name and describe the technique for oxidising ethanol into ethanal.
Distillation with addition. The reagent mixture is heated with the ethanol being added slowly to immediately distill the ethanal vapour. Ethanal is collected in the receiver.
Name and describe the technique that is responsible for oxidising propan-2-ol into propanone.
Heating with reflux. The opening reflux apparatus is sealed to stop the vapour from escaping the vessel. There is a condenser surrounding the neck of the apparatus for the vapour to condense back into the reaction vessel.
What are the two steps to identifying the class of alcohols?
1. Heating the unknown solution under acidified potassium dichromate (VI).
2. Heating the unknown solution under Fehling’s solution or Tollen’s reagent (if NOT tertiary).
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