Q. Keto-enol tautomerism is not shown by

1427 274 NTA AbhyasNTA Abhyas 2020Organic Chemistry – Some Basic Principles and Techniques Report Error

A

butan-2-one

13%

B

1-phenylbutan-2-one

10%

C

$p - O_{2} N C_{6} H_{4} C H_{2} COC H_{2} P h$

19%

D

PhCOPh

59%

Solution:

The compounds A, B and C show keto-enol tautomerism.

A. $C H_{3} - O ∣ ∣ C - C H_{2} C H_{3} ⇌ (i) C H_{3} - O H ∣ C = C H - C H_{3} ⇌ (ii) H_{2} C = O H ∣ C - C H_{2} C H_{3}$

The first enol is more stable, because it has the more substituted double bond.

B. $P h C H_{2} - O ∣ ∣ C - C H_{2} - C H_{3} ⇌ (i) P h C H = O H ∣ C - C H_{2} - C H_{3} ⇌ (ii) P h C H_{2} - O H ∣ C = C H - C H_{3}$

The first enol is more stable, because $C = C$ and Ph are conjugated.

C. $p - O_{2} N C_{6} H_{4} C H_{2} COC H_{3} ⇌ (i) p - O_{2} N C_{6} H_{4} C H = O H ∣ C - C H_{3} ⇌ (ii) p - O_{2} N C_{6} H_{4} C H_{2} O H ∣ C = C H P h$

The first enol is more stable, because the $p - O_{2} N$ extends the delocalization of the conjugated system.

PhCOPh does not show tautomerism as do not have any $α$ -H-atom.