Number of optical isomers possible for 2 – chlorobutane ........

The number of optical isomers possible for a compound is ${2}^{n}$ where $n =$ number of asymmetric carbon atoms. 

As $n =1$ for $2 - chlorobutane$, it is an optically active molecule with one chiral carbon atom and it forms dextro and laevo isomers.

${2}^{n} = {2}^{1} = 2$. 

Hence, it has two optical isomers.