Question

The total number of tautomers possible for I and II together is .......... 

Hint:

When dealing with tautomers, always consider the potential positions for protons and double bonds that can shift, affecting the molecular structure.

Answer 1

Correct Answer: 13

Analysis of tautomers:

Compound I: Symmetric diketone with two Ph-NH groups

Possible tautomers:

1. Diketo form (as shown)
2. Mono-enol form (left C=O → C-OH, C=C shifts) - 2 positions
3. Mono-enol form (right C=O → C-OH, C=C shifts) - equivalent to #2 due to symmetry
4. Di-enol form (both C=O → C-OH)
5. Imino-enol forms: NH can tautomerize to N with H moving to oxygen
   • Left NH → N, left C=O → C-OH (2 forms)
   • Right NH → N, right C=O → C-OH (equivalent by symmetry)
   • Both NH → N, both C=O → C-OH forms

Compound I total: 4 distinct tautomers (accounting for symmetry)

Compound II: Asymmetric - one Ph-NH, one Me-NH

Possible tautomers:

1. Diketo form (as shown)
2. Mono-enol (left C=O → C-OH)
3. Mono-enol (right C=O → C-OH)
4. Di-enol form
5. Imino-enol forms:
   • Ph-NH → N with left enol
   • Ph-NH → N with right enol
   • Me-NH → N with left enol
   • Me-NH → N with right enol
   • Both NH → N with various enol combinations

Compound II total: 9 distinct tautomers (no symmetry)

Total: 4 + 9 = 13 tautomers 

Answer: 13