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.

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

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.

the temperature of the gas, in k, not celsius.

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).