Q33. - Organic Chemistry

Question

Question: Which carbonyl compound in each pair exhibits the higher percentage of the enol tautomer?

Step-by-Step Solution

Answer

Answer

 

The carbonyl compound in each pair, that exhibits a higher percentage of the enol tautomer is shown below:

The enol tautomer of structure B is more stable due to intramolecular H-bonding.

The enol tautomer of structure B is more stable due to intramolecular H-bonding.

Step 1: Definition of tautomers

Tautomers are structural isomers of chemical compounds that readily interconvert in an acidic or a basic medium in the presence of water. The process of conversion of one tautomer into another is known as tautomerization. One such example is keto-enol tautomers.

Step 2: Bonding pattern in case of a keto-enol tautomerization

Enol and keto forms are tautomers that are present in the carbonyl group and differ in the position of a double bond and a proton. They are constitutional isomers  found to be in equilibrium with each other. A keto has a C=O and an additional C-H bond, while an enol has a C-OH and a C=C bond.

Step 3: Stability of enol in case of a keto-enol tautomerization

In some cases, the stability of enol is more than keto due to the presence of intramolecular H-bonding that includes stabilization by forming a six-membered ring.

These cases are found in $\beta$-dicarbonyl compounds.

Representation of the keto-enol tautomerization

 

$\beta$-dicarbonyl structure is present for structure A but due to the lack of enolizable hydrogen, it does not change into that enol form where intramolecular H-bonding is possible, which is not the case for structure B.

Representation of the keto-enol tautomerization

 

$\beta$ -dicarbonyl structure is present for structure B; hence, it changes into that enol form where intramolecular H-bonding is possible, which is not the case for structure A, which lacks a  $\beta$-dicarbonyl structure.