The most contributing tautomeric enol from of $M e C O C H_{2} C O_{2} E t$ is :

C H_{2} = C (O H) C H_{2} C O_{2} E t

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M e C (O H) = C H C O_{2} E t

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M e C O C H = C (O H) O E t

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C H_{2} = C (O H) C H = C (O H) O E t

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Solution

The correct option is B $M e C (O H) = C H C O_{2} E t$
It is the most contributing tautomeric form.

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The most contributing tautomeric enol form of $M e C O C H_{2} C O_{2} E t$ is:

The most contributing tautomeric enol form of $M e C O C H_{2} C O_{2} E t$ is?

M e C O C H_{2} C O_{2} E t

C H_{2} = O H | C - C H_{2} - O | | C - C H_{3} ⇌ C H_{3} - O | | C - C H_{2} - O | | C - C H_{3} ⇌ C H_{3} - O H | C = C H - O | | C - C H_{3}

I

I I

I I I

C H_{2} = O H | C - C H_{2} (I) - O | | C - C H_{3} ⇌ C H_{3} - O | | C - C H_{2} (I I) - O | | C - C H_{3}

⇌ C H_{3} - O H | C = C H (I I I) - O | | C - C H_{3}

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