An optically active compound A, C6H10O2, when dissolved in NaOH solution, consumed one equivalent of base. On acidification, compound A was slowly regenerated. Treatment of A with LiAlH4 in ether followed by protonolysis gave an optically inactive compound B that reacted with acetic anhydride to give an acetate diester derivative C. Compound B was oxidized by aqueous chromic acid to β-methylglutaric acid (3-methylpentanedioic acid), D. Identify compounds A, B, and C; do not specify stereochemistry. (The absolute stereochemical configurations of chiral substances cannot be determined from the data.)

A = β-methyl-δ-valerolactone; B = 3-methylpentane-1,5- diol;

C = 3-methylpentane-1,5- diol diacetate

Explanation:

1. Calculate the unsaturation value

The formula for a 6-carbon would be C₆H₁₄.

U = (14 - 10)/2 = 4/2 = 2

The compound has two rings and/or double bonds.

2. Compound B

Treatment of Compound B  with acetic anhydride gave a diacetate.  

B must be a diol.

Oxidation of compound B gives β-methylglutaric acid.

Compound B must be 3-methylpentane-1,5-diol.

3. Compound A

Compound A reacted slowly with NaOH to form a compound that slowly regenerated Compound A on acidification.

It could be an ester. However, it must be an internal ester (a lactone), because all six carbons are in the diol.

The most likely structure for A is β-methyl-δ-valerolactone.

there are two significant figures.