Solubility Question from TPRH Orgo Passage 19 in SWB

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ap594

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Passage 19 describes the reaction of methyl salicylate -> salicylic acid -> aspirin. I am running into a problem with question 5: it says that only the latter two would be soluble in an aqueous NaHCO3 solution because methyl salicylate does not have a carboxylic acid functional group, and thus does not have an acidic enough proton. Why does this matter? If the molecule has a polar O-H bond, wouldn't that deem it soluble in an aqueous solution?

Here's the methyl salicylate molecule:

http://en.wikipedia.org/wiki/Methyl_salicylate

Thanks.
 
So, if the molecule had a carboxylic acid group, it would donate a proton, and thus exhibit an overall negative charge, increasing its solubility in the solution? Is that the right train of thought?
 
Passage 19 describes the reaction of methyl salicylate -> salicylic acid -> aspirin. I am running into a problem with question 5: it says that only the latter two would be soluble in an aqueous NaHCO3 solution because methyl salicylate does not have a carboxylic acid functional group, and thus does not have an acidic enough proton. Why does this matter? If the molecule has a polar O-H bond, wouldn't that deem it soluble in an aqueous solution?

Here's the methyl salicylate molecule:

http://en.wikipedia.org/wiki/Methyl_salicylate

Thanks.


NAHC03 will almost always only deprotonate carb acids, because the HC03 is only a weak base, so it its not strong enough to deprotonate an alcohol...you need a strong base (NAOH for ex)
 
bicarbonate actually has more 'basic strength' than it seems, because the carbonic acid that forms upon deprotonation can decompose to CO2 and leave solution, driving the deprotonation forward, so I could see it "unfavorably" deprotonating things, as long as the newly formed base is weaker than carbonate. Idk if that's MCAT relevant or not.
 
NAHC03 will almost always only deprotonate carb acids, because the HC03 is only a weak base, so it its not strong enough to deprotonate an alcohol...you need a strong base (NAOH for ex)

That is exactly correct. In salicylic acid and aspirin, the bicarb (NaHCO3) deprotonates the carboxylic acid, and the resulting (charged) carboxylate is now highly soluble in aqueous solution. NaHCO3 is not strong enough to deprotonate the OH in methyl salicylate completely.

pKa of carboxylic acid = 4
pKa of sodium bicarbonate = 6.3
pka of phenol (aromatic O-H) = 10

-James
 
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