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Which of the following compounds will show the maximum enol content?
(A) $C{H_3}COC{H_2}COC{H_3}$
(B) $C{H_3}COC{H_3}$
(C) $C{H_3}COC{H_2}CON{H_2}$
(D) $C{H_3}COC{H_2}COO{C_2}{H_5}$

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Last updated date: 13th Jun 2024
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Answer
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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:

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.

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.

ElectrophilicNucleophilic
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

[B]
Cannot draw enol form so $He$ enol content present in acetone. Atomic form is prominent.
[C] $C{H_3} - COC{H_2}CON{H_2}$

$ - N{H_2}$group decreases enol content in compound
[D] $C{H_3}COC{H_2}COO{C_2}{H_5}$

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 $347kcal/mol.$