Question

Compound A, C6H12O2, was found to be optically active, and it was slowly oxidized to an optically active carboxylic acid B, C6H12O3, by Ag(NH3)2. Oxidation of A by anhydrous CrO3 gave an optically inactive compound D that reacted with Zn amalgam/HCl to give 3-methylpentane. With aqueous H2CrO4, compound A was oxidized to an optically inactive dicarboxylic acid C, C6H10O4. Give structures for compounds A, B, and C; do not specify stereochemistry.

Answer 1

kindly check the attach file for the drawing of the chemical structures.

Step-by-step explanation:

So, we are going to start from the compound D, which is stated in the question to be optically active. Therefore, we will have that:

STEP ONE: THE OXIDATION OF COMPOUND A, C6H12O2 TO GIVE COMPOUND C.

The oxidation of compound A,C6H12O2 gives another chemical compound that is chemical compound C which is a optical inactive di-carboxylic acid. The chemical equation is given below:

C6H12O2 + H2Cr2O4 --------------------------------------------> HOOCCH2CHCH3CH2COOH.

STEP TWO: THE OXIDATION OF COMPOUND A, C6H12O2 TO GIVE COMPOUND B.

The oxidation of compound A,C6H12O2 gives another chemical compound that is chemical compound C which is a optically active acid. The chemical equation is given below:

C6H12O2 + Ag(NH3)2^+ -----------------------------> C6H12O3.

Since the question asked us to give the structures of Compound A,B and C there is no need to to show the chemical reaction for compound D.

Kindly check the picture below for the chemical structures.