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In this example, why is there is a ring formation with the attack of the OH from the intermediate itself rather than a continuation of addition? I know my proposed product is wrong, however I just want an explanation as to why this occurs. enter image description here

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    $\begingroup$ Careful! The connectivity in the diol changed half-way through drawing that mechanism (I see an oxygen connected to a carbon that is all of a sudden connected to 3 carbons). In addition, in the second line, the step between structures 2 and 3 is not a step; those are resonance structures, not separate species. $\endgroup$ – Zhe Nov 6 '18 at 14:25
  • $\begingroup$ In addition, it is likely that the primary hydroxyl group adds to the protonated aldehyde prior to the secondary hydroxyl. $\endgroup$ – user55119 Nov 6 '18 at 19:51
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Intramolecular reactions are always preferred over intermolecular.

Remember that these reactions are reversible so even if the intermolecular attack proceeded, the product would be unstable with respect to intramolecular attack by the free OH that is within the molecule.

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  • $\begingroup$ Why would the product of the intermolecular reaction be unstable? $\endgroup$ – Tan Yong Boon Nov 10 '18 at 2:01
  • $\begingroup$ It is unstable with respect to intramolecular reaction, otherwise it is stable. Under the reaction conditions the oxygens of the acetal will continue to protonate, since the free OH within the molecule is adjacent it is more likely to react than the OH of another molecule. $\endgroup$ – Waylander Nov 10 '18 at 8:02
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I support Waylander's answer above--the reaction proceeds to form a ring because an intramolecular reaction is favored over an intermolecular reaction. However, I would like to clarify that this is mainly due to kinetic reasons and not due to thermodynamic favorability as suggested before ("Intramolecular reactions are always preferred over intermolecular"). The oxygen in the hydroxyl group will attack the positively charged carbon atom at a higher rate than an oxygen atom in a hydroxyl group on another diol molecule because of the former's proximity (intramolecular vs intermolecular proximity).

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  • $\begingroup$ That's not correct; thermodynamic favourability is definitely at work here. Entropy favours a reaction converting two molecules (carbonyl + diol) to one molecule (acetal), rather than one with three molecules (carbonyl + 2 diol) to one molecule (acetal). As was already pointed out, the reactions are reversible, so the outcome is likely controlled by this thermodynamic consideration rather than the kinetic consideration (which you are correct in mentioning; it's just not the crux of the issue here). $\endgroup$ – orthocresol Jan 10 at 22:32

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