0
$\begingroup$

2‐phenyl‐hexahydro‐2H‐pyrano[3,2‐d][1,3]dioxin‐6‐ol to 3‐(5‐hydroxy‐2‐phenyl‐1,3‐dioxan‐4‐yl)propanal

I am given the molecule in black which is treated with $\ce{NaOH}$. The product is in blue.

Could someone explain the mechanism? All I have so far is that the $\ce{-OH}$ group on the bottom right carbon has been deprotonated to $\ce{-O-},$ but am not sure how the rest of the reaction should proceed.

$\endgroup$
1
  • 2
    $\begingroup$ In similar tune @MathewMahindaratne's comment, because many of the reactions with carbonyl groups are ruled by equilibria (e.g., aldol reaction vs. retro-aldol reaction), see if you draw the mechanism starting from the structure depicted in blue toward the one one depicted in black. It might be the opposite direction of what you want, but may help you to identify key intermediates. $\endgroup$ – Buttonwood Mar 23 at 21:13
3
$\begingroup$

The molecule in black consists of a cyclic ketal and a hemiacetal. The hemiacetal is known to be in equilibrium with its aldehyde form, which is the the molecule in blue, in aqueous solutions even without $\ce{NaOH}$. Recall the reaction of glucose (a cyclic hemiacetal) with dilute sodium hydroxide, which is known as Lobry de Bruyn-van Ekentein rearrangement (Ref.1).

Nonetheless, a hemiacetals can be decomposed to the relevant carbonyl and the alcohol under basic conditions as explained here (a reliable mechanism is given). In your case both carbonyl and the alcohol groups are within the same molecule (the molecule in blue):

Reaction of hemiacetal with NaOH

All acetals and ketals are stable to alkaline conditions since they don't carry free hydroxyl group.


References:

  1. Roy L. Whistler, J. N. BeMiller, "Alkaline Degradation of Polysaccharides," Advances in Carbohydrate Chemistry 1958, 13, 289-329 (DOI: https://doi.org/10.1016/S0096-5332(08)60359-8).
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.