An organic compound, 2,3-dichlorobutane, exhibits stereoisomerism due to the presence of chiral carbon atoms. What is the number of optically active isomers and meso forms possible for this compound, respectively?

Step-by-step solution

The given compound is 2,3-dichlorobutane, whose structure is CH₃-CH(Cl)-CH(Cl)-CH₃ . It contains two similar chiral carbon atoms ( n = 2 ). Since the molecule can be divided into two equal halves, it is symmetrical. For a symmetrical molecule with an even number of chiral centers ( n ), the number of optically active isomers is given by a = 2^ n-1 . Substituting n = 2 , we get a = 2²⁻¹ = 2 . The number of meso forms is given by m = 2^ n 2 -1 . Substituting n = 2 , we get m = 2^ 2 2 -1 = 2^0 = 1 . Thus, there are 2