Question

Explain the optical activity of 2 - Chloro - butane.

Answer 1

Solutions of some compounds that can rotate the plane of polarized light are known to be optically active. This property of rotating the plane of polarization of plane-polarized light is known as optical activity and the rotation is assigned positive or negative values accordingly.

Complete answer:
Given to us is an organic compound $\rm{2 - Chloro - butane}$, first we have to find out whether this compound is optically active or not.
For a compound to be optically active, it has to have an asymmetric chiral carbon centre. In order to find the chiral centre in our given compound, we have to draw its structure.
The structure of $\rm{2 - Chloro - butane}$ is given as $\rm{C{H_3} - CH(Cl) - C{H_2} - C{H_3}}$, in this compound the second carbon is chiral since its four bonds are with different substituents. Hence, we can say that the given molecule $\rm{2 - Chloro - butane}$ is optically active meaning that it can rotate the plane of polarized light when passed through it. This gives rise to two enantiomers of the compound.

Additional information:
An optically active solution that rotates the plane of polarized right in clockwise direction is said to be dextrorotatory and the solution that rotates the plane of polarized light in counterclockwise direction is known as levorotatory.

Note:
We prove the optical activity of the given compound $\rm{2 - Chloro - butane}$ by finding a chiral carbon in its molecular structure. A carbon is said to be chiral when it has different substituents attached to it. In our given compound, the second carbon has chirality so the compound is optically active.