Solubility depends on polarity or on the number of hydrogen bonds?

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  • #1
pisluca99
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Does solubility depend on the polarity of the molecule or on the number of hydrogen bonds that are formed with water?
For example: CF4 can form hydrogen bonds, but it's not polar.
Glucose Is Polar e can form hydrogen bonds.
How can I predict solubility? Thanks
 

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  • #2
Borek
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It is not "or". It is some of that and some of that.

There are no strong rules that will allow you to predict solubility, more like rules of thumbs, with the main one being "similar dissolves similar". In rare cases where you can assume that everything else stays the same, you can tell which molecule will dissolve better than the other - but it works only for very similar, homological molecules. Once they are different it is just guessing.
 
  • #3
pisluca99
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Ok, thank you. Just one more question: a molecule is polar when it has a dipole moment. It is easy enough to determine when the molecule is simple (such as CHCl3). But when we have a drug (such as chloramphenicol), it is too expensive to determine the resulting dipole moment to determine if the molecule is polar.
So, in this case, is it possible to associate the polarity with the number of hydrogen bonds that the molecule can form with water?

for example: Chloramphenicol has 2 OH groups, an amide group and a nitro group. Since it can form enough hydrogen bonds with water, can it be said that it is polar and, therefore, soluble in water without evaluating the resulting dipole moment?
 
  • #4
TeethWhitener
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Whether or not a molecule has a dipole moment is dependent on whether it is symmetrical about a center of inversion (so chloramphenicol has a dipole moment, because it does not have inversion symmetry). But in reality, some molecules act nonpolar even without centers of inversion, namely asymmetric hydrocarbons. While something like 2,2,4-trimethylpentane technically has a (small) dipole moment, in reality it will not mix with water, it will dissolve nonpolar substances, and it will fail to dissolve polar substances.

Larger molecules are almost never going to have inversion symmetry, and it’s easiest to tell how they interact with solvents by looking at their functional groups. But even then, you can run into trouble if you aren’t careful. Cellulose is chock full of OH groups, but it doesn’t dissolve in water very easily because it’s a large crystalline polymer. Ultimately, if you have a molecule you want to dissolve, it’s often more practical to simply try out solvents to see what dissolves in what, rather than trying to guess based on molecular structure.
 
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  • #5
pisluca99
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Whether or not a molecule has a dipole moment is dependent on whether it is symmetrical about a center of inversion (so chloramphenicol has a dipole moment, because it does not have inversion symmetry). But in reality, some molecules act nonpolar even without centers of inversion, namely asymmetric hydrocarbons. While something like 2,2,4-trimethylpentane technically has a (small) dipole moment, in reality it will not mix with water, it will dissolve nonpolar substances, and it will fail to dissolve polar substances.

Larger molecules are almost never going to have inversion symmetry, and it’s easiest to tell how they interact with solvents by looking at their functional groups. But even then, you can run into trouble if you aren’t careful. Cellulose is chock full of OH groups, but it doesn’t dissolve in water very easily because it’s a large crystalline polymer. Ultimately, if you have a molecule you want to dissolve, it’s often more practical to simply try out solvents to see what dissolves in what, rather than trying to guess based on molecular structure.
Thanks!
 
  • #6
pisluca99
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Whether or not a molecule has a dipole moment is dependent on whether it is symmetrical about a center of inversion (so chloramphenicol has a dipole moment, because it does not have inversion symmetry). But in reality, some molecules act nonpolar even without centers of inversion, namely asymmetric hydrocarbons. While something like 2,2,4-trimethylpentane technically has a (small) dipole moment, in reality it will not mix with water, it will dissolve nonpolar substances, and it will fail to dissolve polar substances.

Larger molecules are almost never going to have inversion symmetry, and it’s easiest to tell how they interact with solvents by looking at their functional groups. But even then, you can run into trouble if you aren’t careful. Cellulose is chock full of OH groups, but it doesn’t dissolve in water very easily because it’s a large crystalline polymer. Ultimately, if you have a molecule you want to dissolve, it’s often more practical to simply try out solvents to see what dissolves in what, rather than trying to guess based on molecular structure.
Sorry to reopen the discussion, but I have another small doubt.
We said that like dissolves like: this means that all those molecules that are able to form the same intermolecular interactions that are formed between water molecules (hydrogen bonds, dipole-dipole interactions, mainly, and dispersion forces of London, to a lesser extent), dissolve in water. In particular, as the entity of these interactions increases, solubility tends to increase.
If, at this point, propanol and butanol are considered, butanol is less soluble than propanol in water, even if greater London dispersion forces are formed compared to propanol, being it a larger molecule. The extent of the hydrogen bonds and of the dipole-dipole interactions remains practically unchanged, as the two molecules have only one OH group, so they always form the same number of hydrogen bonds. How do you explain this?
 
  • #7
TeethWhitener
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Sorry to reopen the discussion, but I have another small doubt.
We said that like dissolves like: this means that all those molecules that are able to form the same intermolecular interactions that are formed between water molecules (hydrogen bonds, dipole-dipole interactions, mainly, and dispersion forces of London, to a lesser extent), dissolve in water. In particular, as the entity of these interactions increases, solubility tends to increase.
If, at this point, propanol and butanol are considered, butanol is less soluble than propanol in water, even if greater London dispersion forces are formed compared to propanol, being it a larger molecule. The extent of the hydrogen bonds and of the dipole-dipole interactions remains practically unchanged, as the two molecules have only one OH group, so they always form the same number of hydrogen bonds. How do you explain this?
I think I get what you’re trying to say but correct me if I’m wrong. I read this as you’re wondering why butanol isn’t more soluble in water than propanol because of larger London dispersion forces.

London dispersion forces are usually at least an order of magnitude weaker than hydrogen bonds. Since butanol interacts proportionally more via London forces and proportionally less via hydrogen bonds, it’s interaction with water isn’t as strong as propanol’s interaction with water.
 
  • #8
pisluca99
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I think I get what you’re trying to say but correct me if I’m wrong. I read this as you’re wondering why butanol isn’t more soluble in water than propanol because of larger London dispersion forces.

London dispersion forces are usually at least an order of magnitude weaker than hydrogen bonds. Since butanol interacts proportionally more via London forces and proportionally less via hydrogen bonds, it’s interaction with water isn’t as strong as propanol’s interaction with water.
Yes, that's what i needed. I apologize for my english.

In what sense proportionally?
Overall, both butanol and propanol form the same number of hydrogen bonds, but in butanol the overall magnitude of London forces is higher. Thus, overall, the total interaction force between propanol and water is higher than that between butanol and water. It follows that butanol should be more soluble, regardless of the type of bond. If the amount of interaction is higher, the solubility is, in general, higher. Where am I wrong? Thank you.
 
  • #9
TeethWhitener
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Yes, that's what i needed. I apologize for my english.

In what sense proportionally?
Overall, both butanol and propanol form the same number of hydrogen bonds, but in butanol the overall magnitude of London forces is higher. Thus, overall, the total interaction force between propanol and water is higher than that between butanol and water. It follows that butanol should be more soluble, regardless of the type of bond. If the amount of interaction is higher, the solubility is, in general, higher. Where am I wrong? Thank you.
You are only considering the interactions of water with the alcohol. There are also interactions between water molecules and interactions between alcohols. The strongest of these interactions are water-water interactions: each water can form 4 hydrogen bonds with other waters in a quasi-infinite network. The second strongest of these interactions will be water-alcohol (let’s assume a simple aliphatic primary alcohol, but the concept is general). The OH on alcohol can form 2 hydrogen bonds. The interaction of the water with the part of the alcohol molecule that is not an OH is only a small fraction of the strength of the hydrogen bonds. Because of this, if given a choice, water will choose to interact with another water rather than the non-OH portion of the alcohol. As the non-OH portion becomes a larger fraction of the molecule, the water will interact strongly with proportionally less of the alcohol, because it is more energetically favorable for the water to interact with another water than with an aliphatic group. This means longer chain alcohols will be less soluble than shorter chain alcohols.

London forces are basically a rounding error compared to hydrogen bonds, and also they’re always present. So aliphatic groups, which only interact via weak London forces (or induced forces) will be strongly outmatched by the possibility of forming hydrogen bonds (which are maximized in water-water interactions), and water-water interactions will also exhibit London forces in addition to hydrogen bonding.
 
  • #10
snorkack
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Indeed. If you compare butanol and propanol in terms of vapour pressure, then butanol-butanol bonds are stronger (because of more London forces) so butanol has lower vapour pressure and higher boiling point. If you compare the solubility in water, the solubility of butanol in water is lower because butanol breaks the hydrogen bonds in water. Indeed, if you compare the vapour pressures of water-methanol, water-ethanol, water-propanol and water-butanol mixtures, you will find that the ethanol, propanol and butanol solutions have higher vapour pressure and lower boiling point than either component, and the effect increases from ethanol to butanol. Because they break the water hydrogen bonds.
To the contrary, if you add to water a solute which has stronger bonds with water than water or solute with itself, you will find that the solution has lower vapour pressure and higher boiling point than either component. Like most strong acids.
 
  • #11
pisluca99
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You are only considering the interactions of water with the alcohol. There are also interactions between water molecules and interactions between alcohols. The strongest of these interactions are water-water interactions: each water can form 4 hydrogen bonds with other waters in a quasi-infinite network. The second strongest of these interactions will be water-alcohol (let’s assume a simple aliphatic primary alcohol, but the concept is general). The OH on alcohol can form 2 hydrogen bonds. The interaction of the water with the part of the alcohol molecule that is not an OH is only a small fraction of the strength of the hydrogen bonds. Because of this, if given a choice, water will choose to interact with another water rather than the non-OH portion of the alcohol. As the non-OH portion becomes a larger fraction of the molecule, the water will interact strongly with proportionally less of the alcohol, because it is more energetically favorable for the water to interact with another water than with an aliphatic group. This means longer chain alcohols will be less soluble than shorter chain alcohols.

London forces are basically a rounding error compared to hydrogen bonds, and also they’re always present. So aliphatic groups, which only interact via weak London forces (or induced forces) will be strongly outmatched by the possibility of forming hydrogen bonds (which are maximized in water-water interactions), and water-water interactions will also exhibit London forces in addition to hydrogen bonding.

so, what you want to tell me is this: water-alcohol interactions are very similar to water-water and alcohol-alcohol interactions, since, in all these systems, dipole-dipole interactions, hydrogen bonds, dipole-induced dipole and London forces can form.
However, as the number of carbon atoms in the alcohol increases, the apolar character of the molecule increases, as the various CH2 groups are much less polarized than the C-OH bond. In particular, these various CH2 groups, being very poorly polarized, tend to interact mainly through London interactions which, overall, are weaker than the prevailing interactions between water molecules, such as hydrogen bonds and dipole-dipole interactions, therefore the water molecules will prefer to interact with each other, rather than with the apolar portion of the alcohol. Obviously all this is more accentuated as the number of carbon atoms in the alcohol increases, so as the number of carbon atoms increases, it is as if the alcohol is proportionally less and less 'surrounded' by molecules of water (which always interact only with the strongly polarized C-OH moiety of alcohol, as only with this moiety water can form dipole-dipole interactions and hydrogen bonds, which are the predominant interactions between water molecules) , and therefore, less soluble.

Therefore, it is precisely for this reason that we say 'like dissolves like', because the more polar portions there are in a molecule, the more this molecule will predominantly form the same interactions that are prevalent between water molecules and will be more soluble.
Right?
 
  • #12
snorkack
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Therefore, it is precisely for this reason that we say 'like dissolves like', because the more polar portions there are in a molecule, the more this molecule will predominantly form the same interactions that are prevalent between water molecules and will be more soluble.
Compare a conspicuous exception.
Hydrogen halides of heavy halides (chlorine to iodine) are very unlike water. They boil at a very low temperature (HCl an -85, HBr at -67, HI at -35 degrees) and when condensed, the liquids are low polarity and also very poor solvents for water and salts. Yet they are extremely soluble in water!
 
  • #13
pisluca99
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Compare a conspicuous exception.
Hydrogen halides of heavy halides (chlorine to iodine) are very unlike water. They boil at a very low temperature (HCl an -85, HBr at -67, HI at -35 degrees) and when condensed, the liquids are low polarity and also very poor solvents for water and salts. Yet they are extremely soluble in water!
But how can liquids be low polarity if H-Cl, H-Br, exc, are polarized bonds?
 
  • #14
TeethWhitener
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You basically have the right idea, but a few technical things are worth pointing out:
In particular, these various CH2 groups, being very poorly polarized, tend to interact mainly through London interactions which, overall, are weaker than the prevailing interactions between water molecules, such as hydrogen bonds and dipole-dipole interactions, therefore the water molecules will prefer to interact with each other, rather than with the apolar portion of the alcohol.
The water can still induce a dipole moment in the CH2 groups, but to a lesser extent than in the hydroxyl.
, it is as if the alcohol is proportionally less and less 'surrounded' by molecules of water (which always interact only with the strongly polarized C-OH moiety of alcohol, as only with this moiety water can form dipole-dipole interactions and hydrogen bonds, which are the predominant interactions between water molecules)
The molecules of water interact with the entire molecule of alcohol, but they interact very strongly with the hydroxyl and only very weakly with the rest of the molecule.

One fact that surprises a lot of students is that the interaction of water with the non-hydroxyl portion of the alcohol is stronger than the interactions between the non-hydroxyl portions on two alcohol molecules. Students make the mistake that "like dissolves like" implies that non-polar groups interact more strongly with other non-polar groups than with polar groups. In reality, non-polar groups interact more strongly with polar groups, but polar groups interact so much more strongly with other polar groups that the non-polar groups tend to get excluded from the interaction.

Here's an analogy: water is like the cool kids in high school, and nonpolar molecules are like the uncool kids. The cool kids interact strongly with each other and don't really want to let uncool kids into their crowd. The uncool kids really want to interact with the cool kids, but most of the time they're shunned and have to interact with other uncool kids, which isn't their first choice. Maybe if the uncool kid is wearing a cool shirt (like maybe it has an OH group on it), the cool kids will let them hang out for a bit. But for the most part, the uncool kids are stuck with other uncool kids and excluded by the cool kids.
 
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  • #15
snorkack
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But how can liquids be low polarity if H-Cl, H-Br, exc, are polarized bonds?
1) Not that polar to begin with. The dipole moments in D are:
  1. HF - 1,82 D for 0,92 A°, that is 1,98 D/A° or 0,41 e
  2. HCl - 1,08 D for 1,27 A°, that is 0,85 D/A° or 0,18 e
  3. HBr - 0,82 D for 1,41 A°, that is 0,58 D/A° or 0,12 e
  4. HI - 0,44 D for 1,61 A°, that is 0,27 D/A° or 0,056 e
  5. H2O - 1,84 D between 2 bonds at 104 degrees and 0,96 A° each. Each bond´s dipole moment is 1,84 D/(2*cos(104°/2))=1,84/1,23=1,50 D, so 1,56 D/A° or 0,32 e
As you can see, hydrogen partial positive charge in these bonds is much smaller than in water
2) What does the interaction energy consist of? Well, monopole-monopole attractive force grows with product of charges and shrinks with square of distance. So the attraction energy shrinks with distance. Now dipole-dipole attraction energy increases with product of moments and shrinks with cube of distance - but since moment is itself a product of partial charge and distance, and atom sizes also hold the other molecules at distance, the energy grows with product of partial charges and shrinks with distance. So when you compare water and HCl for polarity, look at partial charge more than moment.
 
  • #16
pisluca99
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Here's an analogy: water is like the cool kids in high school, and nonpolar molecules are like the uncool kids. The cool kids interact strongly with each other and don't really want to let uncool kids into their crowd. The uncool kids really want to interact with the cool kids, but most of the time they're shunned and have to interact with other uncool kids, which isn't their first choice. Maybe if the uncool kid is wearing a cool shirt (like maybe it has an OH group on it), the cool kids will let them hang out for a bit. But for the most part, the uncool kids are stuck with other uncool kids and excluded by the cool kids.
Nice One!

therefore, simply, a molecule is more soluble in water if it forms a greater number of stronger interactions with it (hydrogen bonds and dipole-dipole interactions). In the case of butanol, the carbon chain is longer, so that only a small portion of the molecule can interact with stronger interactions (compared to, for example, methanol), while most of the molecule interacts with water through dipole-induced dipole interactions, which are very weak and 'instantaneous', so much so that water molecules prefer, more than anything else, to interact with each other. Therefore, moment by moment, butanol is 'surrounded' by a reduced number of water molecules, proportionally, compared to what happens in alcohols with shorter chains. This reduces solubility. The effect increases as the number of carbon atoms increases.

or, more simply, as the size of the alkyl chain increases, the amount of dipole-induced dipole interactions between alcohol and water increases proportionally, which are weaker than the hydrogen bonds formed between alcohol and water. Consequently, there is, proportionally, a greater quantity of weaker interactions, with consequent lower solubility, because the weaker interactions lead to a more unstable system.

I hope I caught what you meant.
 
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  • #17
pisluca99
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Indeed. If you compare butanol and propanol in terms of vapour pressure, then butanol-butanol bonds are stronger (because of more London forces) so butanol has lower vapour pressure and higher boiling point. If you compare the solubility in water, the solubility of butanol in water is lower because butanol breaks the hydrogen bonds in water. Indeed, if you compare the vapour pressures of water-methanol, water-ethanol, water-propanol and water-butanol mixtures, you will find that the ethanol, propanol and butanol solutions have higher vapour pressure and lower boiling point than either component, and the effect increases from ethanol to butanol. Because they break the water hydrogen bonds.
To the contrary, if you add to water a solute which has stronger bonds with water than water or solute with itself, you will find that the solution has lower vapour pressure and higher boiling point than either component. Like most strong acids.
so you consider a more energetic point of view: as the size of the alkyl chain increases, the number of hydrogen bonds between the water molecules which must break increases, so that they can interact with dipole-induced dipole interactions with the alcohol . Obviously this is an unfavorable process from a thermodynamic point of view, being the hydrogen bonds much stronger, therefore, solubility is reduced, because the final system Is more unstable.
This also has repercussions on the boiling point/vapour pressure: if the alcohol is 'longer', more hydrogen bonds must be broken, so the system is more unstable, so vapor pressure of the solution will be higher and boiling point lower.
 
  • #18
TeethWhitener
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therefore, simply, a molecule is more soluble in water if it forms a greater number of stronger interactions with it (hydrogen bonds and dipole-dipole interactions).
To be precise, it’s an overall balance of interaction energies between different molecular species that determines solubility behavior, but I think you have the basic gist of the idea.
 
  • #19
pisluca99
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To be precise, it’s an overall balance of interaction energies between different molecular species that determines solubility behavior, but I think you have the basic gist of the idea.
Yeah, perhaps what I wrote in response to @snorkack is more accurate
 

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