Question

U DU CILUUUU CO DH L2 achieved by ensuring that the reactants are in the correct orientation as they approach each other. A good example of rate enhancement by this means is pro- vided by the rates of lactone (internal ester) formation in compounds (I) and (II).? The presence of the two methyl groups on the carbon atom adjacent to the aro- matic ring in (II) increases the rate of lactone formation by a factor of 4 x 109. The restrictions placed on rotation about single bonds in the carboxylate side-chain of (II) by the bulky substituents ensure that the preferred orientation of the react- ing groups closely resembles that of the transition state of the reaction. A simple way of looking at this is to say that the molecule will spend a greater proportion of the time in the conformation that leads to reaction, and therefore less rota- tional entropy (fewer degrees of freedom) will be lost as the reaction proceeds, towards the transition state. The smaller negative entropy of activation will ica to an increase in the rate of reaction. Estimates of the effects of orientation and proximity on the entropy of activ CO H CH, CH,
What is the meaning of the statement "less rotational entropy (fewer degrees of freedom )will be lost as the reaction proceeds towards the transition state the smaller negative entropy increases the rate of the reaction"?

Answer 1

First of all, it is important to bear in mind that is the entropy of activation is negative, the smaller it is, the better, due to the fact that we wan't a delta G of activation as negative as possible, and:

$\Delta G_{act}=\Delta H_{act}-T\Delta S_{act}$

So a more negative delta S means a more positive delta G, which is unfavorable.

Now, regarding the rotational entropy. Entropy is directly related to the number of microstates a molecule can exist in. In the particular case of rotations, it means the different positions a molecule (or a part of a molecule) can exist during a rotation. If a group can do a full rotation, it will be able to reach much more positions than a group that is impeded to rotate. This means that the group that can rotate more has a higher rotational entropy and, when the group reacts and limits its rotation in the transition state, the entropy change will be negative and large. The group that has an impeded rotation will have a smaller rotational entropy, and when it reaches the transition state, the change will be negative but smaller than in the other case.

This is what is meant by the text. Compound II has a rather bulky side chain bearing the CO₂H group, which means that it won't be able to rotate freely due to sterical hindrance. In compound I, on the other hand, the side chain bearing the CO₂H group isn't that bulky (there are no methyl groups that increase its volume, and there are isn't a methyl group in the next carbon in the aromatic ring, so the rotation is easier.

This is why it will be much easier for compound II to reach the required geometry to form the transition state (since the -CO₂H group is "locked" in a position close to the -OH). For compound I, the required approach of the -CO₂H to the -OH group involved in the reaction is less efficient, since the group is in constant rotation.

I hope I cleared that up a bit. If not, please write a comment and we can expand this!