Question

A) In the process of forming an enolate, a ketone like acetone reacts with sodium hydroxide and acts like a nucleophile, how is the negative charge stabilized? Draw the reaction.

B) Given 100 cc of acetone that contains sodium hydroxide, and 10 cc of benzaldehyde, how much dibenzalacetone is produced, show calculations.

Answer 1

The negative charge is stablized through resonance involving the migration of electron between methylene carbon and Oxygen group as was shown in the mechanism below.

10 mL of benzaldehyde is equal to 10 g

no. of moles = weight/mol.wt = 10/106 = 0.094 mol

density of acetone = 0.78

100 ml of acetone is equal to (mass = density x volume) = 78 g

no. of moles = 78/58 = 1.3 moles

As per the equation 2 eq of benzaldehyde reacts with 1 equivalent of acetone to give 1 equivalent of dibenzalacetone

Therefore 0.094 mol should give rise to 0.047 mol of dibenzalacetone

mol. wt of dibenzalacetone = 234g/mol

weight produced = no. of moles x mol.wt = 0.047 mol x 234 g/mol= 10.998 g