How may optical isomers are possible for C H 3 C | | H — O H C | | H — O H C O 2 H

Step-by-step solution

C H 3 C * | | H — O H C * | | H — O H C O 2 H It has two chiral centres so optical isomers = 2 n = 2 2 = 4 Structure (I) and (II) form a pair of enantiomers. Similarly, structure (III) and (IV) also form a pair entiomers. Structure (I) or (II) does not have a plane of symmetry bisecting the bond between the two asymmetric carbon atoms. Hence, this compound exists in four isomeric forms, which are optically active.