Answer
64.8k+ views
Hint: The extent of enol content is explained on the basis of orifice hydrogen and intermolecular hydrogen bonding.
Complete step by step answer:
Keto and enol form a compound are isomers.
These isomers are called tautomerism and isomerision is known as tautomerism.
Tautomers are isomers of a compound which differ in portion of protons and electrons.
When a reaction involves simple intermolecular proton transfer is called tautomerism.
Example:
![](https://www.vedantu.com/question-sets/a36b1ee5-f66b-499a-a730-b9002f08cc396479839966790804638.png)
Both isomers form by simple transfer of proton within a molecule.
In the following example $C{H_3}COC{H_2}COC{H_3}$enol content is maximum due to intermolecular h-bonding and resonance stabilization.
![](https://www.vedantu.com/question-sets/7d7a0dae-da00-4aeb-9fec-edc12dc2bc805221604809828692605.png)
Therefore, from the above explanation the correct option is (A) $C{H_3}COC{H_2}COC{H_3}.$
Its IUPAC name is pentane-2,4-dione.
Enol content is high in carbon with electron withdrawing group > carbonyl carbon > number of atoms of hydrogen.
Keto is electrophilic and enol is nucleophilic.
![](https://www.vedantu.com/question-sets/4bbc7352-b797-4bb9-b5c5-7f3666f896f45783643349503472359.png)
Under most condition keto form is covered. This form is important for aldehydes and ketones but not so much for carboxylic acid ester and amides under normal condition.
[A] Already draw enol form
![](https://www.vedantu.com/question-sets/263f0663-8588-4154-8fac-4be9022e0adb4988731514992844959.png)
[B]
Cannot draw enol form so $He$ enol content present in acetone. Atomic form is prominent.
[C] $C{H_3} - COC{H_2}CON{H_2}$
![](https://www.vedantu.com/question-sets/9351c0fa-7357-40df-8d67-12dd7afeb2104755378506328882764.png)
$ - N{H_2}$group decreases enol content in compound
[D] $C{H_3}COC{H_2}COO{C_2}{H_5}$
![](https://www.vedantu.com/question-sets/922a7982-977a-4ac5-8edd-a6fd809db5ad6281711984093466577.png)
Ethoxide$( - 06.25)$group decreases enolic content.
Therefore, from the above explanation the correct option is (A) $C{H_3}COC{H_2}COC{H_3}.$
Note: Under normal condition keto form is fevered. Because keto form has $C - H,C - C$ and $C - O$bond whereas enol has $C = C,C = O$ and $O = H$ bond. The sum of first three is about $359lcal/mol$ and second three is $347kcal/mol.$
Complete step by step answer:
Keto and enol form a compound are isomers.
These isomers are called tautomerism and isomerision is known as tautomerism.
Tautomers are isomers of a compound which differ in portion of protons and electrons.
When a reaction involves simple intermolecular proton transfer is called tautomerism.
Example:
![](https://www.vedantu.com/question-sets/a36b1ee5-f66b-499a-a730-b9002f08cc396479839966790804638.png)
Both isomers form by simple transfer of proton within a molecule.
In the following example $C{H_3}COC{H_2}COC{H_3}$enol content is maximum due to intermolecular h-bonding and resonance stabilization.
![](https://www.vedantu.com/question-sets/7d7a0dae-da00-4aeb-9fec-edc12dc2bc805221604809828692605.png)
Therefore, from the above explanation the correct option is (A) $C{H_3}COC{H_2}COC{H_3}.$
Its IUPAC name is pentane-2,4-dione.
Enol content is high in carbon with electron withdrawing group > carbonyl carbon > number of atoms of hydrogen.
Keto is electrophilic and enol is nucleophilic.
![](https://www.vedantu.com/question-sets/4bbc7352-b797-4bb9-b5c5-7f3666f896f45783643349503472359.png)
Electrophilic | Nucleophilic |
Keto form is electrophilic in nature and reacts with carbonyl carbon. It is acidic at -carbon atom and hydrogen bond acceptor. | Enol from is nucleophilic in nature and reacts with electrophiles at -carbon atom. It is acidic at (O-H) and hydrogen bond donor and acceptor. |
Under most condition keto form is covered. This form is important for aldehydes and ketones but not so much for carboxylic acid ester and amides under normal condition.
[A] Already draw enol form
![](https://www.vedantu.com/question-sets/263f0663-8588-4154-8fac-4be9022e0adb4988731514992844959.png)
[B]
Cannot draw enol form so $He$ enol content present in acetone. Atomic form is prominent.
[C] $C{H_3} - COC{H_2}CON{H_2}$
![](https://www.vedantu.com/question-sets/9351c0fa-7357-40df-8d67-12dd7afeb2104755378506328882764.png)
$ - N{H_2}$group decreases enol content in compound
[D] $C{H_3}COC{H_2}COO{C_2}{H_5}$
![](https://www.vedantu.com/question-sets/922a7982-977a-4ac5-8edd-a6fd809db5ad6281711984093466577.png)
Ethoxide$( - 06.25)$group decreases enolic content.
Therefore, from the above explanation the correct option is (A) $C{H_3}COC{H_2}COC{H_3}.$
Note: Under normal condition keto form is fevered. Because keto form has $C - H,C - C$ and $C - O$bond whereas enol has $C = C,C = O$ and $O = H$ bond. The sum of first three is about $359lcal/mol$ and second three is $347kcal/mol.$
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