Compound A , C 5 H 10 O 5 , given a tetraacetate with AC 2 O and oxidation of A with Br 2 - H 2 O gives an acid, C 5 H 10 O 6 . Reduction of A with HI gives isopentane. The possible structure of A is :

Step-by-step solution

Formation of tetraacetate with acetic anhydride means compound A has four - OH linkage. Reduction of A with HI gives isopentane i.e. molecule contains five carbon atoms. So, C is the correct option