The H moves from the para carbon to the oxygen: cyclohexa-2,5-dienone becomes phenol, and the aromatic ring makes phenol the clear winner. Likewise 4-pyridone ⇌ 4-hydroxypyridine.
The atom carrying the moving H is lit.A 1,3-shift, just like keto-enol.
Nitro ⇌ aci-nitro
H moves from C to O
The α-H of nitroethane moves onto an oxygen of the nitro group, giving the aci form CH₃CH=N(O)OH.
Imine ⇌ enamine
The nitrogen keto-enol
CH₃CH₂CH=NH ⇌ CH₃CH=CH–NH₂: the H moves from carbon to nitrogen and the double bond shifts.
Imine and enamine.Acetamide and its imidol.
Amide ⇌ imidol
H from N to O
R–CO–NH₂ ⇌ R–C(OH)=NH. The amide form dominates.
JEE-style question
Your turn
Which pair are tautomers?
(a)CH₃CH₂NO₂ and CH₃CH₂–O–N=O
(b)CH₃CH=NOH and CH₃CONH₂
(c)CH₃CH₂CH=NH and CH₃CH=CH–NH₂
(d)CH₃CN and CH₃NC
Show the answer and the traps
In (c), one H moves from carbon to nitrogen and the double bond shifts: imine and enamine. Option (c).
In (a), a whole ethyl group is attached differently, not just an H: functional isomers. (b) and (d) are functional isomers from Part 2 that don't interconvert.
Watch out
Common mistakes
Calling nitroethane and ethyl nitrite tautomers.An ethyl group, not an H, is attached differently: they are functional isomers.
Forgetting the double bond must shift with the H.A tautomer pair differs only by one moved H and a matching shift of the double bond.
Practice
Try these
1. Name the tautomer of nitromethane.
Aci-nitromethane, CH₂=N(O)OH.
2. For acetamide, which form dominates, amide or imidol?