Question

Explain the optical activity of 2 - chloro - butane.

Answer 1

Hint: At first we should know about the compound $2 - chloro - butan e$.Then we would talk about the optical activity in compounds. Then we will examine the optical activity of $2 - chloro - butan e$.

Complete step by step solution:
Step1. $2 - chloro - butan e$ is a chemical compound consisting of chlorine, hydrogen and carbon. It is also called $\sec - butyl{\text{ chloride}}$.It is colorless and volatile at room temperature. Its chemical formula is ${C_4}{H_9}Cl$ .
Step2. Optical activity is the ability of a compound to rotate in the plane of polarization of a beam of light that passes through it. A normal polarimeter consists of a light source, tube, analyzing lens and polarizing lens. For a compound to be optical active one of its carbon must be chiral. It happens when there are 4 different atoms or items attached to it. It must be asymmetrical. If it’s not chiral then it will not be optically active.
Step3. $2 - chloro - butane$ is an optically active molecule. The carbon at position 2 is chiral and it has different substitutes which are $-{C_2}{H_5}$, $-C{H_3}$, $-H$, $-Cl$ attached to it. Now when it has different substitutes so when a plane polarized light is passed then it rotates the light in different directions. This phenomenon gives rise to the two enantiomers of the compound. Those two enantiomers are R and S.

Hence, the answer would be that the $2 - chloro - butane$ is optically active.

Note: Stereoisomers are the molecules that have the same constitution, but different position of its elements in open space. They differ in relative orientation which result in optical activity. Isomers depict different chemical properties.