Chapter 7, Problem 7.44P

Interpretation Introduction

Interpretation:

The mechanism for the conversion of $\text{propanedial}$ from its keto form to its enol form under basic and acidic conditions is to be drawn. Similarly, the mechanism for the conversion of $\text{propanedial}$ from its enol form to the keto form is to be drawn.

Concept introduction:

In the enol form, the species has a carbon atom that is simultaneously a part of a $\text{C = C}$ bond, characteristic of an alkene, and is bonded to OH, characteristic of an alcohol. In the keto form, there is a hydrogen atom on the carbon atom that is adjacent to the $\text{C = O}$ group, so called alpha carbon, whereas in the enol form, the hydrogen appears on the oxygen atom instead. The proton transfer step serves to add or remove a proton from the particular species. The mechanism of this transformation depends on the presence of a strong acid or strong base. In the basic condition, the strong base removes a proton from the $\text{α}$ carbon of the species in the keto form, producing an enolate anion having a negative charge. In the resonance structure of this anion, the electron-rich atom gains a proton from water, which acts as an acid and forms its enol form. In the acidic condition, the strong acid donates a proton to the O atom of the C = O group of the keto form. Then water removes the proton from the $\text{α}$ carbon to produce enol from of the species. These reactions take place in reverse direction in the transformation of the enol form to the keto form.