Question

(R)-$2$-octyl tosylate is solvolysis in water under ideal ${S_N}1$ conditions. The product(s) will?
a.) R-$2$-octanol and S-$2$-octanol(excess)
b.) R-$2$-octanol and S-$2$-octanol
c.) R-$2$-octanol only
d.) S-$2$-octanol only

Answer 1

To solve this problem we should know about the R and S form of the structures given in question. Here in solvation water will act as a nucleophile during the reaction. We will consider ideal ${S_N}1$ conditions.

Complete step by step answer:
First we will draw the structure of reactant (R)-$2$-octyl tosylate and will understand about the R and S form.

The above structure is tosylate(OTs) but the given reactant is (R)-$2$-octyl tosylate so the reactant’s structure is on right with the octyl group at C-$2$ position.
- Now we will understand the concept of R and S form. First we will number the chemical compounds in the order of higher molecular mass. For example here in (R)-$2$-octyl tosylate we can number as $OTs(1) > {C_6}{H_{13}}(2) > C{H_3}(3) > H(4)$.
- Now first check the position of hydrogen. If it is placed horizontally then when going from $1$ to $4$ in clockwise direction as shown will be R form and if it is anticlockwise then S form. It will be vice-versa when hydrogen is placed vertically.
Now we will proceed the reaction with solvation of (R)-$2$-octyl tosylate.

- During solvolysis tosylate will be removed and planar carbocation is formed. During this process water will act as a nucleophile and product formed may be R and S as under ideal ${S_N}1$ condition there will be (inversion and Retention) both. Therefore, the product will be both R and S form. The product formed will be $2$-octanol in both R and S form.
- Now we will check all options one by one. Option (A) R-$2$-octanol and S-$2$-octanol(excess). In this S form is in excess but under ideal ${S_N}1$ condition both forms are formed in equal amounts.
Option (B) R-$2$-octanol and S-$2$-octanol. This is the correct option as both R and S are formed in equal amounts.
Option (C) R-$2$-octanol only. The product formed is only R but under ideal ${S_N}1$ condition both forms are formed in equal amounts.
Option (D) S-$2$-octanol only. The product formed is only S but under ideal ${S_N}1$ condition both forms are formed in equal amounts. The correct answer is option “B” .

Note: Under ideal ${S_N}1$ condition both retention and inversion occurs. On the other hand ${S_N}2$ condition only inversion occurs.
- A counter clockwise configuration is S(Sinister, left configuration) and clockwise direction is R(rectus, right configuration).