Which Tautomeric form of 4-methyl-5-r-Imidazole is more stable and why?

Question

Tautomerism in 4-methyl-5-r-Imidazole. Which of the both tautomers will be more stable and why?

Answer 1

Tautomery is in general:


X=C-Z-H <=> [X=C-Z- <-> -X-C=Z] H+ <=> HX-C=Z

(one bond for C is missing; can't really draw here)

The form that is going to be dominant is the one that corresponds to the less stable/more reactive anion (strong base means weak conjucate acid and thus the H+ stays at that position)

I have no clue what R is, but depending on its nature you can have induction or resonance phenomena favoring either form.
Check the properties (induction, resonance) of your R and you will know. If it is an alkyl group there won't be much difference with the methyl group so you don't care (two forms at the same levels)
If it is acyl it gets complicated because apart from the two phenomena you could have H-bond formation of the H in one tautomer and the C=O. I would say that then the tautomer with the H near R is more abundant

Answer 2

If a quadratic equation has roots x1 and x2 ... then it can be written so (x-x1)(x-x2) = 0 Then as the roots are 4 and -5 it is (x-4)(x+5) = 0 ... the standard form is ax^2 + bx + c = 0 ... To get it just expand the product... it is x^2 +x -20 = 0 OK! There is another way... in the equation of the form x^2 + bx + c = 0... -b is the sum of the roots and c is the product of the roots... Then here 4-5= -1 and 4*(-5) = -20.... then we can write x^2 +1x - 20 = 0 OK!

Answer 3

The methyl group must be on the allylic carbon. This produces a carbocation that benifits most from resonance stability, and hyperconjugation and hence is more stable over all.

Answer 4

I don't really know how to make that structure but i can assure you that whichever form will have the maximum dipole moment will be more stable.