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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?

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    $\begingroup$ (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. $\endgroup$ – orthocresol Jul 4 at 8:36
  • $\begingroup$ Electrons transfer occours through (usually double) bonds, protons transfer through space. $\endgroup$ – Karl Jul 4 at 9:06
  • $\begingroup$ @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 $\endgroup$ – 234ff Jul 4 at 14:12
  • $\begingroup$ @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? $\endgroup$ – 234ff Jul 4 at 14:16
  • $\begingroup$ 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. $\endgroup$ – Karl Jul 5 at 16:55

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