Question

(R)-2-octyl tosylate is solvolysis in water under ideal $S_{N}1$ conditions. The products will be:
[A] R-2-octanol and S-2-octanol in a 1:1 ratio
[B] R-2-octanol and S-2-octanol in a 1:5 ratio
[C] R-2-octanol only
[D] S-2-octanol only

Answer 1

HINT: To solve this, remember that $S_{N}1$ is an abbreviation for substitution nucleophilic unimolecular reaction. It is not a concerted mechanism and it takes place in two differentiable steps. Remember that after the formation of carbocation, there can be two possible products, the inversion as well as the retention product. Consider the absolute configuration to solve this question.

COMPLETE STEP BY STEP SOLUTION: Before answering this question, let us discuss what $S_{N}1$ reaction is. We know that $S_{N}1$ mechanism is nucleophilic substitution unimolecular reaction. In $S_{N}1$ reaction, SN basically stands for nucleophilic substitution and 1 stands for unimolecular. In $S_{N}1$ , the first step i.e. leaving of the leaving group and formation of the carbocation is the slow step and thus the rate determining step and the second step i.e. the attack of the nucleophile is the fast step. Now let us see the question given. Firstly, let us see the structure of (R)-2-octyl tosylate in the fischer form so that we can assign the absolute configuration. So, we know that the tosylate group is denoted as OTs. So, we can draw it as-

We can see that the least priority group is in horizontal position and thus the above structures in R-absolute configuration. Now, let us see its solvolysis via $S_{N}1$ mechanism.

Here, we can see that a planar carbocation is formed. So, the nucleophile can attack from any side and thus inversion as well as retention can take place here. Inversion will give us the S-product and retention will give us the R-product. So, we can write the mechanism as-

We will get the R as well as the S product in equal amounts.

Therefore, the correct answer is option [A] R-2-octanol and S-2-octanol in a 1:1 ratio

NOTE: We can describe the mechanism of $S_{N}1$ as a two-step process. In the first step of the mechanism, the leaving group leaves the substrate and this leads to the formation of a carbocation. We can write the reaction as-

Then, the nucleophile (Nu) attacks the carbocation and gives us the product.

The reactivity of compounds undergoing $S_{N}1$ depends upon 2 factors.
- The first factor is the effectiveness of the removal of the leaving group.
- The second factor is whether or not the carbocation formed is stable.