Why is it that adding the sulphuric acid slowly to a solution of alcohol and carboxylic acid (making an ester) will increase the yield of the product? It seems that not only does the rate at which it is dispensed affects the yield of ester, but really does change the colour of the solution when heated. For example, my team used it slowly and the colour was a more faint peach colour whilst everyone else, who added it quickly, produced pure black solutions!

What is happening? Why does adding the acid slowly increase yield and purity of the ester?

  • 1
    $\begingroup$ Sulfuric acid of higher concentration acts as a dehydrating agent. $\endgroup$
    – andselisk
    Aug 12, 2017 at 5:52
  • $\begingroup$ It would be really nice if this was all documented, for educational purpose, especially for those who are forced to read the theory without the seeing the experiments. $\endgroup$
    – Cheng
    Oct 3, 2023 at 1:52

1 Answer 1


The esterification is not the only thing catalysed by sulphuric acid. Basically, it’s catalytic activity is protonation of whatever feels happy enough to be protonated. In your desired esterification mechanism, we want to protonate the acid:

$$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=\overset{+}{O}H)-OH + HSO4-}\tag{1}$$

This will ease the nucleophilic attack of the alcohol onto the carboxy carbon. However, we also have other atoms that can be protonated. For example, the hydrogen-carrying oxygen may also accept the proton:

$$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=O)-\overset{+}{O}H2 + HSO4-}\tag{2.1}$$

This could then lead to the removal of water and the formation of a very reactive acyl cation:

$$\ce{R-C(=O)-\overset{+}{O}H2 -> R-C#\overset{+}{O} + H2O}\tag{2.2}$$

This in itself isn’t bad as it could also react with the alcohol in the desired fashion. However, the alcohol also can be protonated:

$$\ce{R-CH2-CH2-OH + H2SO4 <=> R-CH2-CH2-\overset{+}{O}H2 + HSO4-}\tag{3.1}$$

And depending on the type of alcohol you have, a number of side reactions may occur. For instance, you could simply eliminate $\ce{H3O+}$ from the protonated alcohol to give an alkene:

$$\ce{R-CH2-CH2-\overset{+}{O}H2 -> R-CH=CH2 + H2O + H+}\tag{3.2}$$

If you have a suitable neighbouring group that can stabilise positive charges, semi-pinacol-type rearrangements could occur. I won’t draw that with the simple primary alcohol I drew because it would be unlikely in e.g. propanol. It could be a problem in other cases, though.

Finally, the intermediates I drew up can react even further and unpredictably with sulphuric acid still present.

If you add sulphuric acid slowly, the local acid concentration will remain rather low and the reaction mixture is rapidly homogenised. This means that side reactions caused by sulphuric acid will be suppressed. If, however, you add it quickly, there will be areas of your solution with very high local sulphuric acid concentrations. In these, these side reactions are accelerated. Furthermore, concentrated sulphuric acid acts as a dehydrating agent which further speeds up the reactions we don’t want (they all include losing a water molecule).

These side reactions — especially the dehydrating ability of sulphuric acid — tend to end up in ‘black tar’ which would give a more or less uniform black solution. What exactly happened is hard to say except that some polymerisation probably took place.


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