Kevin B. answered • 09/08/22
Enthusiastic Orgo Professor Can Help You Unlock Organic Chemistry
No problem! Detailed is my specialty... ;-) As you will soon see....
Solubility as a qualitative test used to determine the identity of organic unknowns usually uses three solutions -- dilute NaOH, dilute NaHCO3 (bicarbonate), and dilute HCl. (The "dilute" could be a variety of actual concentrations, the specific concentration does not matter much.) The NaOH and NaHCO3 are bases and can deprotonate certain functional groups, forming organic anions which are then soluble in the aqueous solution. Conversely, HCl can protonate certain functional groups to form organic cations, which dissolve. So the solubility of the compound depends on its acid/base reactivity -- whether it can be protonated or deprotonated by the solution.
These "tests" are predicated on the fact that most organic compounds are not soluble in neutral water itself. For small compounds (say, 4-5 carbons max) with a polar functional group, this may not be true. For example, small alcohols (like methanol, ethanol, and isopropanol) are all soluble in neutral water, as are small carboxylic acids, amines, and even ketones (like acetone). Any organic compound which is small and polar enough to dissolve in neutral water will dissolve in ALL THREE of the above solutions, because they are all water-based. That is not useful information -- since these kinds of molecules don't need to be protonated or deprotonated in order to dissolve, this does not really say anything about the presence or absence of any specific functional groups.
As a result, these solubility tests are only meaningful if the unknown is first determined to be INSOLUBLE in neutral water. If that is the case, then these solubility tests do tell you something useful!
The NaOH solution should dissolve any molecules with acidic functional groups -- "acidic" being defined as having a pKa of less than 14. The only two FGs which fit in that category are carboxylic acids (pKa usually 4-5 or even lower) and phenols (pKa 10 or lower). Theoretically, any carboxylic acid or phenol should dissolve in dilute NaOH, though this process sometimes is slow and requires a little patience. Any unknown which does NOT dissolve in dilute NaOH should be assumed to NOT be a carboxylic acid or a phenol.
You can theoretically distinguish a carboxylic acid from a phenol using the dilute NaHCO3 solution. This is a weaker base (sodium bicarbonate, while basic, is a weaker base than NaOH), and because of that, it is not strong enough to deprotonate most phenols. As a result, most phenols are not soluble in dilute NaHCO3. This allows you to distinguish carboxylic acids from phenols -- carboxylic acids are usually soluble in BOTH NaOH AND NaHCO3, while phenols are soluble in NaOH but NOT NaHCO3. (There are some phenols with lots of strong electron withdrawing groups, like -NO2, which are acidic enough to dissolve in NaHCO3, but they are the exception rather than the rule.)
Again, any other functional groups (of which there are very many) should not be soluble in EITHER NaOH or NaHCO3. This includes alcohols. While the O-H of an alcohol can theoretically be deprotonated, the pKa of most alcohols is too high (typically ~16 or even higher), and while a small percentage will be deprotonated at equilibrium, it is not enough to bring the alcohol into aqueous solution, even with the strong base NaOH. In order to cause the compound to dissolve, the deprotonation would need to be favorable at equilibrium (pKa less than about 14) in order to cause the unknown to dissolve. So alcohols, like the majority of organic FGs (anything besides carboxylic acids and phenols) would be insoluble in NaOH (and, obviously, NaHCO3 as well).
The HCl solution will protonate any basic functional group, which is essentially only amines. While many other functional groups can be reversibly protonated to a small extent (say, ~1%, plus or minus 1%) by HCl, that is not sufficient to bring the compound into aqueous solution. Amines will be completely deprotonated (~99%, plus or minus 1%) by HCl, and will thus dissolve. Any organic compound which is insoluble in water but soluble in dilute HCl can be assumed to be an amine.
Now back to your specific question! At this point, I'm tempted to let you answer it on your own.... But given that this particular medium ("Ask the Expert") is non-interactive, I'll sum up. The carboxylic acid can be distinguished from any alcohol or alkene by solubility (assuming all three compounds are large enough to be insoluble in neutral water), since it would be soluble in NaOH and NaHCO3, while the alcohol and alkene would not. But solubility would struggle (well, fail) to distinguish the alcohol from the alkene! The vast majority of compounds containing each of those functional groups would be insoluble in all aqueous solutions -- neutral water, dilute NaOH, dilute NaHCO3, and dilute HCl. There are, of course, exceptions -- as noted above, small alcohols can be completely or partially soluble in water, in which case they would dissolve in ALL of those aqueous solutions. (In contrast, an organic molecule containing only an alkene FG would be insoluble regardless of size, since an alkene is not a polar FG.) Alternatively, it is possible (though very unlikely, in my opinion) to have an alcohol with LOTS of nearby electronegative substituents, whose pKa is lowered by those e'neg substituents sufficiently that it is actually soluble in NaOH. This is very unlikely, but not impossible -- that alcohol would then "appear" to be a phenol, as it would be soluble in NaOH, but not NaHCO3. But again, that is extremely unlikely -- enough so that I would not suggest worrying about that at all.
As such, my summary is that "NO!" solubility is NOT sufficient to distinguish those three FGs! It would easily distinguish the carboxylic acid from the other two, since it would be soluble in both NaHCO3 and NaOH, but would almost certainly fail to distinguish the alcohol and alkene from each other, as they would both be insoluble in all three aqueous solutions.
[There are other solutions occasionally used in solubility tests, like diethyl ether and concentrated H2SO4. But these would also fail to distinguish an alcohol from an alkene. Diethyl ether dissolves the vast majority of organic unknowns, and only those with several polar, H-bonding FGs will not dissolve. The vast majority of alcohols and all alkenes would be expected to dissolve in ether. The only exception would be alcohols with approximately as many -OH groups as carbons, which would not dissolve in ether -- any normal-sized alcohol with a single alcohol FG will dissolve. H2SO4 will dissolve compounds with any FG which can be protonated anywhere on the molecule at all. As a result, any compound containing any type of O or N atom (including alcohols) will dissolve in H2SO4. But it turns out that even alkenes (at least, C=C bonds which are not part of an aromatic ring) can be protonated by H2SO4, and are expected to dissolve (or react/polymerize, which counts as "soluble") in H2SO4. The only classes of compounds which will NOT dissolve in concentrated H2SO4 are alkanes, alkyl halides, or aromatic rings with alkyl and/or halide substituents. Everything else will dissolve or react with H2SO4. So these solubility tests are not likely to help you to distinguish an alcohol from an alkene.]
This was not the question posed here (although it's part of the other question you posted), but there are good chemical tests to distinguish alcohols from alkenes -- they just aren't solubility. Dilute KMnO4 will react with most alkenes to form brown MnO2 precipitate, but will not react with most alcohols. (That's true for neutral KMnO4 -- acidified KMnO4, as you asked about in your other question, is more reactive and may not distinguish an alcohol from an alkene well.) Also, most alkenes will decolorize Bromine in CCl4 solution, while alcohols generally will not. There are both false negatives and false positives for both of these tests, but any unknown which gives a positive test for BOTH KMnO4 and Br2/CCl4 is almost certainly an alkene, and any unknown which gives a negative test for both is almost certainly not an alkene. Negative tests on both does not necessarily mean the unknown is an alcohol, as many FGs would test negative for both, but it would be consistent with the unknown being an alcohol.
There are also chemical tests which give positive results for alcohols. Most alcohols react exothermically with acetyl chloride (although so do all amines and some phenols, and solid alcohols frequently give a very mild exotherm). Primary and secondary alcohols are easily oxidized by CrO3 in acetone ("Jones reagent") to form blue precipitate, but so do most aldehydes, and tertiary alcohols will not. Most alcohols will give a positive (red solution) with ceric ammonium nitrate. These tests would all help distinguish an alcohol from an alkene. But I would not do so based solely on solubility.
