Dear Sangyub Oh

As you mention, in theory, each NCO group will need only one OH group to react. So in your case, the theoretical answer to your question would be yes. But, normally ratios are different in practice due to the "availability" of these reactive groups. As long as the reaction proceeds, the reaction media changes in properties (viscosity for example), and it becomes harder and harder for the functional groups to "meet" and react between each other.

Agustin Forchetti Casarino Thank you for answer. You mean that contrary to theory, in reality, the R ratio may deviate due to changes in the state of the materials. But, the difference won't be that big, right?

Sangyub Oh , exactly. Regarding the difference, it should not, but again, it will depend on the NCO and OH reagents (whether they are big molecules, or small molecules for example), and on the behaviour of the reaction media as well. If some solvent is used, the ratios will be near the theoretical, since the increase in viscosity won't be a very big issue right? But if no solvent is used, then the ratio to achieve a very high conversion should be higher.

Normally, one of the two reagents are used in excess, to ensure that the other one is fully reacted. In the case of polyurethanes I am not sure which, but I tend to imagine that the OH reagent should be added in excess, to ensure the full reaction of the NCO groups. In a case like this, the ratio might be different from 1:1. It might be, I do not know, 2:1 for example.

Agustin Forchetti Casarino

While reading a paper on polyurethane synthesis, I found a case where the molar ratio of the synthesized material greatly deviated from 1:1. What do you think about this?

In table 1 (attached photo), the molar ratio of "di-ol/amine + extender" is much higher than that of isocyanate.

Sangyub Oh I am not fully sure what do you need me to tell you, but the reasons of the deviations are the following:

The a case (hydroxyl-terminated):

In the first stage, the intention is to obtain an NCO terminated polyurethane between the combination of BATPBP and IPDI. To do so, IPDI must be in excess, to ensure the full reaction of all the amine groups of the BATPBP. Even in theory the reaction ratio is 1 to 1, using a 1 to 1 ratio does not ensure that the final molecules will have free NCO groups. That is why IPDI is used in excess. At this point, the NCO-functionalized molecule will be present, and also some remaining IPDI as well.

In the second stage MDEA appears, because the authors want to obtain an hydroxyl terminated polymer at the very end. In this case, a lot of MDEA is needed, in order to ensure the full reaction of the NCO groups of the previous prepolymer. That is why the molar ratio is 1:2:2 (first row, second column of the table).

The b case (amino-terminated):

Here, the intention is to finish with amino groups, rather than hydroxyl groups. So in the first stage, the authors combine NCO (IPDI) with OH (MDEA), and in order to ensure that NCO groups are remaining, the ratio between these two components is 1.33 to 0.67 (second column, third row of the table).

In the second stage, a lot of NCO groups are readily available, but the authors want to finish the reaction with amino-terminated groups, so a lot of amino reagent (BATPBP) will be needed. Now, the molar concentration of NCO groups is not 1.33. Is less than that, because most of them are reacted in the previous stage. That is why the ratio is 1 to 1.33 to 0.67.

If you take the time to convert these ratios into molar concentrations of each reagent, you will find out that even though in this case the reaction seems to have more NCO than amino groups (if you see the third row, second column of the table), in practice the reaction was carried out in two stages, meaning that the NCO groups available when reacted with the amino groups were less than the originals. Same thing apply for the a case as well of course.

See molecular weights and equivalent weights for the NCO:OH ratios

Yes, that's correct. In the synthesis of linear polyurethanes, where an isocyanate with two NCO groups reacts with a diol containing two OH groups, the ideal molar ratio of NCO to OH (R ratio) should be close to 1:1

In practice the NCO/OH ratio should be above 1.2