Question

An organic compound having a carbon attached to four different groups is optically active. But, is the opposite also true? That is, do all optically active organic compounds have chiral carbons? Not necessarily. Presence or absence of chiral centre is not the sufficient criterion for optical activity. The ultimate criterion is the presence or absence of either plane or centre of symmetry. Two compounds which are non-superimposable mirror images of each other are called enantiomers. If a compound contains more than one chiral carbon, new words are required to describe the relationship between various stereo isomers of the compounds. Those words are diastereomers and mesomers.
The optical rotation of a solution of pure natural camphor is found to be $+{5.76}^o$ under the following conditions: concentration $=0.13g/ml$, length of polarimeter $=$ 1dm, wavelength $=$ sodium D line, $T={25}^o$C. The specific rotation of camphor is

• A. $+{44.3}^o$
• B. $+{26.7}^o$
• C. $-{26.7}^o$
• D. $-{44.3}^o$

Answer 1

The correct option is A $+{44.3}^o$

We know that

$[\alpha {]}_{\lambda }^T=\frac{\alpha }{C\times l}$

Where $\left[\alpha \right]=$ Specific rotation

$C=$ concentration

$T=$ temperature

$\lambda =$ Wavelength

$\alpha =$ observed rotation.

On putting the values in the formula we get

$[\alpha {]}_{\lambda }^T=\frac{+5\cdot {76}^{\circ }}{(0\cdot 13)\times 1}$

on simplifying we get

$[\alpha {]}_{\lambda }^T=+{44.307}^{\circ }=+{44.3}^{\circ }$

Hence the correct option is $A$.