Question

In the enolization of the given molecule, the $H$-atom involved is :

• A. $\alpha$
• B. $\beta$
• C. $\gamma$
• D. The given compound does not enolize.

Answer 1

The correct option is C $\gamma$

In $\alpha ,\beta$-unsaturated carbonyl compound, the $\alpha$-hydrogen and $\beta$-hydrogen are $s{p}^2$ hybridised so they does not involve in delocalisation to exhibit Keto-enol tautomerism. But $\gamma$-hydrogen is $s{p}^3$ hybridised in $\alpha ,\beta$-unsaturated carbonyl compound so they involve in delocalisation and exhibit Keto-enol tautomerism. The enol form of this compound is more stable than its keto form because it gives highly stable 6 membered aromatic benzene ring.