# Why is keto is more stable than enol?

Dec 30, 2015

The main reason is that $\text{C=O}$ bond is more stable than a $\text{C=C}$ bond.

#### Explanation:

The essential bond changes in a keto-enol tautomerism are

$\text{H-C-C=O ⇌ C=C-O-H}$

The $\text{C-H}$, $\text{C-C}$, and $\text{C=O}$ bonds in the keto form become $\text{O-H}$, $\text{C-O}$, and $\text{C=C}$ bonds in the enol.

For the keto form, the bond energies in kilojoules per mole are

$\text{C-H} \textcolor{w h i t e}{l l} = \textcolor{w h i t e}{l l} 413$
$\text{C-C} \textcolor{w h i t e}{l l} = \textcolor{w h i t e}{l l} 347$
$\text{C=O} = \textcolor{w h i t e}{l l} 745$
"Tot".color(white)(l) = stackrel(——)(1505)

For the enol, the bond energies are

$\text{O-H} \textcolor{w h i t e}{l} = \textcolor{w h i t e}{l l} 467$
$\text{C-O} \textcolor{w h i t e}{l} = \textcolor{w h i t e}{l l} 358$
$\text{C=C} = \textcolor{w h i t e}{l l} 614$
"Tot." color(white)(l)= stackrel(——)(1439)

The difference in the sums of the bond energies is

${\sum}_{\text{keto"("BE") – sum_"enol"("BE") = "(1505-1439) kJ/mol" = "66 kJ/mol}}$

The keto form is thermodynamically more stable by about 66 kJ/mol.

Hence the position of equilibrium strongly favours the keto form.