[id86]

Hi,

New to this forum and wondered if anyone could help. My organic chemistry is not too good so i'm sorry if the answer is obvious.

With the cyclisation of creatine to creatinine under acidic conditions, the only mechanism I can find is the one attached in the image. Would the free H⁺ attack the O double bonded to the carbonyl group first, and if so what would the mechanism be to follow this to get to creatinine?

Also I read that under very acidic conditions, pH <3.5 the creatine would be more stable due to the NH₂ going to NH₃. Can anyone give me help with why this wouldn't happen at a slightly higher pH.

Thanks in advance

[pgk]

1). Yes, but more correctly:
A free electron pair of the carbonyl oxygen attacks on the proton and then, the free electron pair of the primary amine group attacks on the carbonyl carbon, ..etc.., followed by cyclization and final departure of the catalytic proton.
2). It is obvious and easily understood that protonation of the primary amine group, blocks its  free electron pair and inhibits cyclization that way, according to the above mechanism.
On the other hand, creatine is a strong tribasic compound (strongest basic group pka = 12.43, ChemAxon), as being a guanidine derivative and has an order of functional group basicity: imine > tertiary amine > primary amine.
Consequently, the primary amine group can be protonated in high proton concentration, only and which is translated to low pH.