# Maximum distance for intramolecular proton transfer

In this reaction (formation of ester from ketone / aldehyde), intramolecular proton transfer happens, since it can stabilize the formal positive charge by resonance after that. This transfer happens across 3 atoms. ($$\ce{O-O-C}$$)

However, in this step (formation of acetal from $$\ce{C=O}$$) using alcohol, the book says it does not occur, since the $$\ce{O}$$ and $$\ce{H}$$ are 'too far' apart. This 'transfer' happens only across 2 atoms ($$\ce{C-O}$$). Which is less than the above reaction ($$\ce{O-O-C}$$). If that can occur, why can't the proton transfer here not occur?

What is the real factor that determines whether an intramolecular proton transfer can happen?

• (1) What book is this? (2) Consider looking into Baldwin's rules. FWIW I would be inclined to draw the first one as two separate steps. – orthocresol Jul 4 '20 at 8:36
• Electrons transfer occours through (usually double) bonds, protons transfer through space. – Karl Jul 4 '20 at 9:06
• @orthocresol 'organic chemistry as a second language', I am struggling to figure out Baldwin's rules. It seems to apply in a ring formation (I don't know whether I am correct), is this applicable to the explanation for the second one? Thanks – 234ff Jul 4 '20 at 14:12
• @Karl sorry if I am not mistaken, if any protons can transfer through space, this means actually the transfer in the second one is workable? – 234ff Jul 4 '20 at 14:16
• What do you mean by "any"? I only said that if there is a viable (especially short enough) trajectory, a proton does not care from what (part of a) molecule it comes. If the reaction happens of course depends on a lot of additional factors. – Karl Jul 5 '20 at 16:55