Question

The most contributing tautomeric enol form of $MeCOC{H}_{2}C{O}_{2}Et$ is?

• A. $H_{2}C=C\left(OH\right)C{H}_{2}C{O}_{2}Et$
• B. $MeC\left(OH\right)=CHC{O}_{2}Et$
• C. 2-Pentanone
• D. 2-Butene

Answer 1

The correct option is B

$MeC\left(OH\right)=CHC{O}_{2}Et$

Let's take a look at the structure:

$Me-\stackrel{\underset{||}{O}}{C}-C{H}_2-\underset{O}{\underset{||}{C}}-OEt$

$C{H}_3-\stackrel{\underset{||}{O}}{C}-C{H}_2-\underset{O}{\underset{||}{C}}-OEt$ [Here we have two $\alpha$-hydrogens from different carbons]

But here,

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