Carboxylic Acids (3)

We have seen in a previous blog that carboxylic acids react with alcohols to produce esters and water.

The question I want to deal with in this blog is this: does the alcohol –OH group end up in the water molecule or the ester?

In other words which of these two reactions takes place on esterification:

In this reaction the oxygen of the alcohol –OH group is found in the water molecule.

But in the second reaction the oxygen of the ethanol –OH group is found in the ester.

Which is it and how can we find out which it is?

In a famous experiment in 1938 two American chemists Irving Roberts and Harold Urey found out the destination of the alcoholic oxygen using isotopic labelling of the oxygen atom in the alcohol

In the reaction below, they used isotopically labelled oxygen ¹⁸O in the methanol to see which of the two reactions below actually took place. 

C₆H₅CO | OH     +    CH₃¹⁸OH     ⇋   C₆H₅CO¹⁸OCH₃    +  H₂O ……..A

Or

C₆H₅COO | H    +    CH₃¹⁸OH     ⇋    C₆H₅COOCH₃    +  H₂¹⁸O ……..B

They used a mass spectrometer to determine the masses of the products and found that the labelled oxygen appeared in the ester and not in the water molecule.

This meant that equation A was correct not equation B.

The oxygen of the alcohol appears in the ester.

The mechanism involves the loss of an –OH group from the acid. 

This means that the mechanism begins with the alcohol making a nucleophilic attack on the acid as the mechanism shows below.

Step 1 involves the protonation of the carbonyl group, the strong acid catalyst facilitates this.

Step 2 involves the nucleophilic addition of methanol to the protonated carboxylic acid. 

Methanol is a nucleophile because of the lone pair of electrons on the oxygen atom.

Step 3 involves the elimination of a water molecule due to proximity of the -H and -OH groups.

At this point, the oxygen of the acids -OH group ends up in the water molecule not the ester. 

Step 4 is the final step in which the species on the right above deprotonates to form the ester.

The product in this example is methyl benzoate.

Examination of the reactants and products IR spectra reveal the change in major functional groups.

Here is benzoic acid’s IR spectrum:

Note the large absorption around 1700cm^-1 due to the carbonyl group

Here next is methanol’s IR spectrum

This spectrum is characterised by the large absorption at around 3300cm^-1 due to the H bonded OH group.

Now contrast those two spectra with that of methyl benzoate:

The strong absorption at 3300cm^-1 is no longer present because the ester contains no H bonded OH group. 

The strong carbonyl absorption is still present at about 1700 cm^-1 because this is central to the ester functional group

We have seen then that the mechanism for the esterification of a carboxylic acid involves an addition elimination reaction.

In this reaction substitution of the –OH group of the acid by the CH₃O- group of the methanol occurs.

This reaction mechanism was first confirmed when Irving Roberts and Harold Urey used isotopic labelling to follow the reaction.