Compound A,C_5H_10O_5, given a tetraacetate with AC_2O and oxidation of A with Br_2-H_2O gives an acid, C_5H_10O_6. Reduction of A with HI gives…

Question

Compound A,C_5H_10O_5, given a tetraacetate with AC_2O and oxidation of A with Br_2-H_2O gives an acid, C_5H_10O_6. Reduction of A with HI gives isopentane. The possible structure of A is :

MARKS Content Team · Accepted Answer

Correct answer: (C). 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

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 :

Solution

Relevant Questions

Compound A,C5H10O5, given a tetraacetate with AC2O and oxidation of A with Br2-H2O gives an acid, C5H10O6. Reduction of A with HI gives isopentane. The possible structure of A is :

Solution

Relevant Questions

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 :