Question

The enantiomeric excess and observed rotation of a mixture containing $6gm$ butanol and $4gm$ of $(-)-2-$ butanol are respectively. (If the specific rotation of enantiomerically pure $(+)-2-$ butanol is $+13.5$ unit).
$A.$ $80\mathrm{\%},+2.7$ Unit
$B.$ $20\mathrm{\%},-27$ Unit
$C.$ $20\mathrm{\%},+2.7$ Unit
$D.$ $80\mathrm{\%},-27$ Unit

Answer 1

Hint: Enantiomers are chiral molecules that are mirror images of one another. Furthermore, the molecules are non-superimposable on one another. It means that they cannot be placed on the top of one another and give the same molecules. Also called enantiomorph, optical isomer.

Complete step by step answer:
An optically compound is the one which rotates plane polarized light. The enantiomer which rotates light clockwise is called dextrorotatory $(+)$ and the one which rotates in anticlockwise is called laevorotatory. This is absolute configuration. We have another system which is the relative configuration known as $R$ and $S$ configuration. Both $R$ and $S$ can be $d$ or $l$ that is $R$ can be $d$ or $l$ and $S$ can also be $d$ or $l$ . The $R$ and $S$ configuration is determined by using $CIP$ Rules.
Optical activity is the ability of a chiral molecule to rotate the plane polarized light. The amount of rotation can be measured by using a polarimeter. When the rotation is measured by using the polarimeter then, it is called the observed rotation. It depends on wavelength $\left(\lambda \right)$ of light used and Temperature $\left(T\right)$ . When $T={25}^{\circ }C$ and $\lambda =589nm$ it is called “SPECIFIC ROTATION”.
${\left[\alpha \right]}_{\lambda }^T={\alpha }_{\circ }÷l\times c$
Where,$\alpha$ is observed rotation using polarimeter,$l$ is length of sample tube, $c$ is concentration, $T={25}^{\circ }C$ and $\lambda =589nm$
Dextrorotatory and laevorotatory, that is, two enantiomers have the same value of ${\left[\alpha \right]}_{\lambda }^T$ but opposite signs because the direction of rotation is opposite. If there is $50:50$ mixture of $d\mathrm{\&}l$ they will cancel each other and net rotation would be $0$ .
Enantiomeric excess $\left(ee\right)$ is the excess of the major enantiomer over the minor enantiomer. Let the mixture contain $75\mathrm{\%}$ of $(+)$ and $25\mathrm{\%}$ of $(-)$ .then,
$\left(ee\right)=75\mathrm{\%}-25\mathrm{\%}=50\mathrm{\%}$
For a mixture it is also given by,
$ee\mathrm{\%}={\displaystyle \frac{{\alpha }_{mixture}}{\left[\alpha \right]}}\times 100$ -
Where, $ee\mathrm{\%}$ is percentage of enantiomeric excess, ${\alpha }_{mixture}$ is observed rotation of mixture and $\left[\alpha \right]$ is the specific rotation of sample
Also, $ee\mathrm{\%}={\displaystyle \frac{R-S}{R+S}}\times 100$
Now, according to question we have given $R=6gm$ , $S=4gm$ and ${\alpha }_{pure-enantiomer}=+13.5$ put in above equation, we get
$ee\mathrm{\%}={\displaystyle \frac{6-4}{6+4}}\times 100$
$=20\mathrm{\%}$
And, by using above equation, $\mathrm{\%}$ of${\alpha }_{observed}$ $=$ ${\displaystyle \frac{ee\mathrm{\%}\times {\alpha }_{pure-enantiomer}}{100}}$
$\mathrm{\%}$ Of ${\alpha }_{observed}$ $={\displaystyle \frac{20\times +13.5}{100}}$
$=+2.7$ Unit
Thus, the correct option is $\left(C\right)$.

Note: A chiral molecule is that molecule which contains four different groups attached to it. They are asymmetric. Enantiomer ratio is extremely important because while one enantiomer is beneficial to the body and the other enantiomer highly toxic. A well known example is Thalidomide.