Sulfuric acid is a colorless, odorless, oily liquid.
Esterification, or the combination of an alcohol with an acid to produce an ester, is a form of condensation reaction, since water is removed in the process. The reverse reaction can also occur: the ester can recombine with water to produce the alcohol and acid. In some cases, this “de-esterification” can be avoided by introducing a small amount of sulfuric acid into the reaction vessel. It helps by combining with the produced water and actually binding it. First of all, the benefit of sulfuric acid in esterification is that it acts as a proton donor, increasing the rate of reaction between the acid and the alcohol; when the acid used is a carboxylic acid, the reaction is sometimes called a Fischer-Speier esterification.
A water molecule is removed during esterification.
Carboxylic acids (R-COOH, where R is an organic appendage) may be too weak to use alone in an esterification reaction. A strong proton donor is needed for the carboxylic acid to act as if it were a good proton source. Sulfuric acid in esterification accomplishes the task by injecting a proton into the carboxylic acid structure through the reaction H 2 SO 4 + R-COOH → HSO 4 – + RC + (OH) 2 . The alcohol molecule, R′-OH, with its electron-rich oxygen atom, is attracted to this protonated carboxylic structure and forms a complex conglomerate, RC + (OH) OR ′ + HSO 4 – → RC(O) -R ′ .
This arrangement of atoms and charge is not very stable, so it undergoes a shift of protons (H+), specifically RC(OH) (O(H 2 )+ )-OR′. In this state, it is easy for the clearly identifiable water molecule to split, providing further stabilization and leaving behind the energetically more favorable species, RC+(OH)OR′. Finally, regeneration with sulfuric acid completes the process: RC + (OH) OR ′ + HSO 4 – → RC (O) -R ′. Because sulfuric acid is regenerated but not consumed in the reaction, it is considered a catalyst, not a reactant.
Interestingly, esterification does not require separate molecules of alcohol and acid, but the reaction can in some cases take place within a single molecule containing both moieties and functional molecular groups. Certain conditions must be met: both the hydroxyl and carboxylic groups must be free of spatial impediments and able to pass through each step of the process without being damaged. An example of a molecule that can undergo this type of esterification is 5-hydroxypentanoic acid, HO-CH 2 CH 2 CH 2 CH 2COOH. The ester produced by this form of esterification, which results in ring closure, is called a lactone, in this case δ-valerolactone. The positioning of the oxygen ring (-COC-) compared to the carbonyl group (C=O) is indicated by the Greek letter delta.
Sulfuric acid in esterification is generally not used in relation to tertiary alcohols, those that have their hydroxyl-containing carbon atom bonded to three other carbon atoms. Dehydration without ester formation occurs in tertiary alcohols when they are in the presence of sulfuric acid. As an example, tertiary butyl alcohol, (CH 3 ) 3 C-OH, when combined with sulfuric acid, produces isobutylene, (CH 3 ) 2 = CH 2 + H 2 O. In this case, the alcohol is what is proton, followed by the exit of a water molecule. The use of sulfuric acid in esterification is not a viable methodology for the preparation of tertiary esters.