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Find the major product of the following reaction:

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My approach

From my knowledge of reagents, I figured out that that the acid will protonate the oxygen(s) in the reactant and after that $\ce{C-O \sigma}$ bond should break, but I am not able to decide which $\ce{C-O}$ bonds to break in the reaction after protonation by the aqueous acid.

Am I thinking in the right direction or are there some other concepts involved?

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The actual mechanism

Waylander has given an accurate reaction since this is a THP ether

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The actual mechanism of the de-protection using alcoholysis would be as follows:

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The actual mechanism of the de-protection using acidic hydrolysis would be as follows:

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A place to start from if you didn't know the mechanism

Seeing that the compound given resembles a carbonyl group that has been protected using ethylene glycol, a mechanism using a reaction similar to that of protection of carbonyl groups was used here to hypothesis what would happen(not knowing the original mechanism) and to decide the products formed.

protection of ketone

Hence, taking a reversed approach here, the first question I ask is:

What could have formed this acetal in the first place?

One part of the acetal seems to be connected to the carbonyl group and is part of the main molecule. The other side, however, is a different molecule. So, this is what the acetal formation looks like in the mechanism. Here, I assume $\ce{R}-$ is just an inert part of the molecule and so is equivalent to $\ce{CH3-CH2}-$. This may not be the right mechanism but it would give us a good starting point.

Formation of the hemiacetal takes place as follows:

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Formation of acetal using propanol can happen as follows:

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Which is the THP ether in the question.


What are the final products?

Therefore, the products of acid catalysed hydrolysis of the acetal would be $\ce{R-CH2-OH}$ and 5-hydroxypentanal which exists in equilibrium with its cyclic form dominating in mildly acidic conditions.

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The hydrolysis of the hydroxyderivative of THP would happen as follows. Both alcohols will get protonated but protonating the oxygen in $\ce{-OH}$ results in the same molecule being regenerated.

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Therefore, the final products would give $\ce{R-CH2-OH}$ and a hydroxy derivative of THP.

However, if $\ce{HCl}$ or $\ce{NaHSO3}$ is used instead of a weak lewis acid, we see the dominating compound in the equilibrium being 5-hydroxypentanal as shown in this paper here:

[...] As a result of the acid-catalyzed hydration of 1, the 2-hydroxytetrahydropyran (2) was formed primarily which equilibrated with its open chain form, the 5-hydroxypentanal (3). The acid-catalyst (conc. hydrochloric acid), however, promotes not only the addition of water to the C-C double bond of 1, but also that of the OH groups in 1 and 2, thus resulting in the protected compounds 2-(2-tetrahydropyranyloxy) tetrahydropyran 0, and 5-(2-tetrahydropyranyloxy)pentanal (4). The ratio of 4 + 5 to 1 + 2: depends on the amount of acid used.[...]

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Reference:

  1. Peter Vinczer , Zoltan Juvancz , Lajos Novak & Csaba Szantay (1989) AN IMPROVED SYNTHESIS OF 5-HYDROXYPENTANAL, Organic Preparations and Procedures International: The New Journal for Organic Synthesis, 21:3, 344-346, DOI: 10.1080/00304948909356391
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    $\begingroup$ The THP ether is usually formed from dihydropyran which is regenerated when the group is removed. $\endgroup$
    – Waylander
    Aug 17 '20 at 10:25
  • $\begingroup$ @user55119 It is SN1 (the reaction takes place via alcholysis). I'll modify that. It's the same mechanism as aditya's answer, with the nucleophilic attack on the other side. I couldn't manually draw it for a generic compound since chemdoodle doesn't support it( currently do not have access to anything else), so I added the closest one. $\endgroup$ Aug 17 '20 at 16:08
  • $\begingroup$ @safdar: Back in pre-history we used the Fieser triangle. sigmaaldrich.com/catalog/product/aldrich/… Alas, it has been discontinued. $\endgroup$
    – user55119
    Aug 17 '20 at 22:53
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enter image description here 5-hydroxypentan-1-al is the product.

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    $\begingroup$ That SN2 reaction is ugly as sin $\endgroup$
    – PCK
    Aug 17 '20 at 13:50
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    $\begingroup$ Indeed, you made a bad shortcut here, as this C atom isn't nearly as electrophilic as the other one. $\endgroup$
    – Mithoron
    Aug 17 '20 at 14:04
  • $\begingroup$ @Safdar You are correct that under aq acid conditions the 5-hydroxypentanal is formed en.wikipedia.org/wiki/3,4-Dihydropyran $\endgroup$
    – Waylander
    Aug 17 '20 at 14:07
  • $\begingroup$ @Waylander however, according to Greene's Protective Groups in Organic Synthesis, Fourth Edition, it depends on the conditions as you said. Using TsOH/MeOH yields a hydroxy derivative as shown in my answer (there are 30+ conditions giving different reactions). $\endgroup$ Aug 17 '20 at 14:11
  • $\begingroup$ @PCK: If the distasteful SN2 mechanism in the last step were operative, you couldn't form RCH2OTHP in the first place from THP and RCH2OH/H+? $\endgroup$
    – user55119
    Aug 17 '20 at 15:09

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