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In Organic Chemistry (Wade) there is a question:

Which of the following compounds would give a positive Tollens' test? (Remember that the Tollens' test involves mild basic aqueous conditions.)

The answers manual says this compound - $\ce{CH3-(CH2)3-CH(OCH3)2}$ - gives a negative result as it is an "acetal stable in base: no reaction"

Now, I would have thought this is positive. Earlier in the chapter it's taught that since the Tollens reagent removes any aldehyde present, even if at equilibrium the amount of aldehyde is very small, the Tollens reagent will nevertheless predominate due to Le Chatelier's principle (the small amount of aldehyde will be removed slowly but surely, until eventually all of the solution is converted to aldehyde and then to the tollens product).

In this example, whilst I respect that the acetal product may be by far the major product, surely a very, very small percentage of the acetal is nevertheless converted to the aldehyde, via protonation of a methoxy group, which leaves, and then attack by $\ce{H2O}$, followed by deprotonation of $\ce{H2O -> OH}$, and then protonation of the other methoxy group, which leaves, leaving a carbocation, which the $\ce{OH}$ attacks forming a double bond and a positive charge on the $\ce{OH}$, followed by deprotonation on the $\ce{OH}$ giving an aldehyde?

This may be unlikely, but according to Le Chatelier's principle, it just needs to occur at some small quantity in order to be continuously removed by the tollens reagent, and thus ends up predominating and removing all of the acetal product given enough time? Why is this not positive?

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    $\begingroup$ That there is a equilibrium position says nothing about the speed of attaining it, keep that in mind. $\endgroup$ – Jori Mar 6 '15 at 10:43
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The description of the hydrolysis of an acetal is accurate. The trouble is that the initial protonation event happens to such a small amount that the reaction does not proceed. We can estimate where the equilibrium of that protonation lies by looking at the pKa's of the relevant acids [Evans pKa table]. In basic solution, water (pKa 16) is the strongest acid available to protonate the acetal. The protonated acetal will have a pKa close to protonated methanol (pKa -2). So the equilibrium for that protonation will lie to the water side by approximately $\ce{10^18}$. As Jori is alluding to in the comment, the rate moving forward is going to be exceedingly slow because the concentration of the protonated intermediate will be exceedingly low. This is where the rule of thumb that acetals are stable to base comes from. See also: Why are acetals stable to bases and nucleophiles?

If the Tollen's test was performed under acidic conditions, then you would be correct that the acetal would test positive for an aldehyde. A hemi-acetal would also test positive.

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  • $\begingroup$ Really informative answer thanks, and that Pka table is great. Out of curiosity though, given enough time and/or heat and the presence of excess Tollens reagent, would this reaction in theory move to the left completely due to Le Chatelier's principle, even in basic conditions? $\endgroup$ – user4779 Mar 7 '15 at 1:45
  • $\begingroup$ I suppose so, but I wonder if there are other decomposition pathways that would compete if we're allowed to just heat the reaction indefinitely. $\endgroup$ – jerepierre Mar 7 '15 at 3:22

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