Inverted mind!!

Chemistry Level 2

Optically pure (+)-A with [ $\alpha$ ]=+40° reacts with I $^-$ to give optically pure (-)-B with [ $\alpha$ ]=-20°. If partially racemic A with [ $\alpha$ ]=+30° reacts with I $^-$ in two different beakers S and T having solvents D and E respectively with polarity of D greater than polarity of E, it gives B with [ $\alpha$ ]=-5° in both the beakers. Find in which beaker will reaction proceed faster.

Details:

A and B are Organic Compounds with no double or triple bonds.
A undergoes substitution reaction to give B.

Data insufficient In beaker T Same in both beakers In beaker S

1 solution

Aryan Sanghi
Feb 14, 2019

Optical purity of A= $\frac{30}{40}$ ×100=75℅

Optical purity of B= $\frac{-5}{-20}$ ×100=25℅

Percentage of inversion= $\frac{Optical purity of B}{Optical purity of A}$ ×100=33.33%

Percentage of racemisation=100-33.33=66.66%

As reaction is taking place with a strong nucleophile, I $^-$ , it is substitution by nucleophile reaction

As percentage of racemisation is greater than percentage of inversion, SN $^1$ reaction will dominate over SN $^2$ reaction.

As rate of SN $^1$ reaction increases with polarity of solvent, therefore reaction in beaker S will be faster as it has more polar solvent D.

Nice question dude!!!