What is the mechanism for the following:
1)KOH in alchohol
phenanthrene-9,10-dione ------------->
2) H^+
9-hydroxy-9H-fluorene-9-carboxylic acid
The mechanism for the reactions you mentioned involves the conversion of phenanthrene-9,10-dione to 9-hydroxy-9H-fluorene-9-carboxylic acid. Let's break it down step-by-step:
1) KOH in alcohol:
In this step, the base KOH (potassium hydroxide) reacts with the phenanthrene-9,10-dione in an alcohol solvent. The alcohol acts as a solvent to provide a medium for the reaction to occur. The KOH deprotonates the alpha position next to the carbonyl group in the phenanthrene-9,10-dione to form an enolate intermediate.
The mechanism involves the following steps:
- The hydroxide ion (OH-) from KOH abstracts a proton from the alpha position next to the carbonyl group, leading to the formation of an enolate ion.
- The resulting enolate ion stabilizes itself via resonance by delocalization of the negative charge across the conjugated system.
This enolate intermediate can then react further in subsequent steps.
2) H+:
In this step, an acid, represented as H+, is added. The acid protonates the oxygen atom in the enolate intermediate formed in the previous step, changing it back to a carbonyl group.
The resulting intermediate is 9-hydroxy-9H-fluorene-9-carboxylic acid, as the enolate has been converted back into a carbonyl group, and a hydroxyl group is added at the alpha position.
These two steps collectively convert phenanthrene-9,10-dione to 9-hydroxy-9H-fluorene-9-carboxylic acid.