The most contributing tautomeric enol form of MeCOCH 2 CO 2 Et is-A- H 2 C - C OH CH 2 CO 2 EtB- MeC OH - CHCO 2 EtC- 2-PentanoneD- 2 -Butene

The correct option is B MeC(OH) = CHCO2Et Let's take a look at the structure: Me ||OC CH_2 O||C OEt CH_3 ||OC CH_2 O||C OEt [Here we have two α Hydrogen from ...

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Standard XII

Chemistry

Aromatic Compounds

The most cont...

Question

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

H₂C=C(OH)CH₂CO₂Et

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MeC(OH) = CHCO₂Et

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2-Pentanone

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2-Butene

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Solution

The correct option is B
MeC(OH) = CHCO₂Et

Let's take a look at the structure:

$M e - O | | C - C H_{2} - C | | O - O E t$

$C H_{3} - O | | C - C H_{2} - C | | O - O E t$ [Here we have two $α$ -Hydrogen from different carbons]

But here,

Because it will form hydrogen bonding and make five membered ring.

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The most contributing tautomeric enol form of MeCOCH2CO2Et is:

The correct option is B MeC(OH) = CHCO2Et Let's take a look at the structure: Me−O||C−CH2−C||O−OEt CH3−O||C−CH2−C||O−OEt [Here we have two α-Hydrogen from different carbons] But here, Because it will form hydrogen bonding and make five membered ring.

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