[Technicalhuman]

Short chain alcohols are soluble because they can have hydrogen bonds with water. But for larger alcohols they aren't. But shouldn't they be even more soluble?

Because they already have the H bond acceptors and donors and the long chain would also make the molecules have greater London dispersion forces. So if they were to form an intermolecular bond, shouldn't the bond be even stronger than a short chain molecule?

[curiouscat]

I think of it as the ratio of hydrophobic to hydrophillic parts of the molecule.

Long chain alcohols have mostly a hydrophobic HC chain and very little of the hydrophillic OH group.

[Technicalhuman]

I think of it as the ratio of hydrophobic to hydrophillic parts of the molecule.

Long chain alcohols have mostly a hydrophobic HC chain and very little of the hydrophillic OH group.

Hi Curiouscat thanks for the reply.

Could you emphasize on that? Because I thought the London dispersion forces will increase the attraction between the two rather than repell them actually. Something like why CS2 is more soluble in acetone than CO2 as it has stronger London dispersion forces than CO2.

Thanks so much for the help

[curiouscat]

Why is say hexane or octane or decane insoluble in water?

[Kate]

Short chain alcohols are soluble because they can have hydrogen bonds with water. But for larger alcohols they aren't. But shouldn't they be even more soluble?

Because they already have the H bond acceptors and donors and the long chain would also make the molecules have greater London dispersion forces. So if they were to form an intermolecular bond, shouldn't the bond be even stronger than a short chain molecule?

In terms of considering the solubility of a molecule in water or any other polar solvent, the London forces are very negligible compared to hydrogen bonding.

Water is a dipole. For solubility to happen, the molecules of water have to become more organized around the incoming solute, which represents a decrease in entropy. To counteract this decrease in entropy, the incoming solute has to be able to establish electrostatic interactions with water. And hydrogen bonding allows that electrostatic interaction to take place.

Hydrogen bonding requires hydrogens covalently linked to either oxygen, fluorine or nitrogen. Since a molecule occupies a certain amount of space/volume, if you have a molecule of decanol, for the same amount of volume you can have, say 5 molecules of methanol and 5 times the number of hydrogen bonds.

[Technicalhuman]

Short chain alcohols are soluble because they can have hydrogen bonds with water. But for larger alcohols they aren't. But shouldn't they be even more soluble?

Because they already have the H bond acceptors and donors and the long chain would also make the molecules have greater London dispersion forces. So if they were to form an intermolecular bond, shouldn't the bond be even stronger than a short chain molecule?

In terms of considering the solubility of a molecule in water or any other polar solvent, the London forces are very negligible compared to hydrogen bonding.

Water is a dipole. For solubility to happen, the molecules of water have to become more organized around the incoming solute, which represents a decrease in entropy. To counteract this decrease in entropy, the incoming solute has to be able to establish electrostatic interactions with water. And hydrogen bonding allows that electrostatic interaction to take place.

Hydrogen bonding requires hydrogens covalently linked to either oxygen, fluorine or nitrogen. Since a molecule occupies a certain amount of space/volume, if you have a molecule of decanol, for the same amount of volume you can have, say 5 molecules of methanol and 5 times the number of hydrogen bonds.

Hi thanks for the reply

But actually I was thinking, if the London dispersion forces are very strong, then shouldn't they be comparable with the polar molecules? So the interaction between those non polar molecules and polar molecules should be quite strong as well? Something like how CS2 is more soluble in acetone than CO2 as CS2 has stronger LDFs.

Thanks so much for the help

[Technicalhuman]

Why is say hexane or octane or decane insoluble in water?

Hmm that is true, but shouldn't it be more soluble than methane still?

Thanks for the help

[Corribus]

Why would you think so?

[Technicalhuman]

Why would you think so?

Hi Corribus thanks for the reply

I thought so because if the London dispersion forces are greater, the attraction between the induced dipole on the non polar molecule and the polar molecule should also be stronger. Isn't this true?

Thanks for the help here

[Corribus]

By now you may know that instead of speculating or guessing, I like to look at real data, which for compounds like this are widely available.  Here is some real solubility data. I've converted it all into the same units - hopefully no mistakes made. 

n	n-Alkane	n-Alcohol
1	7.06E5	miscible
2	2.32E5	miscible
3	5.83E4	miscible
4	1.47E4	7.3E7
5	3.85E4	2.2E7
6	9.50E3	6.23E6
7	2.93E3	1.80E6
8	660	8.385E5
9	171	n/a
10	46	n/a
11	9	n/a
12	1	4.285E4
13	0.4	n/a
14	n/a	300
15	0.04	102.6
16	n/a	41.2

References: Alkane, n = 1 to 8, McAuliffe, Journal of Physical Chemistry, 1966, 70, 1267; Science, 163, 1969, Alkane, n 8 to 15, Tolls, et al, Journal of Physical Chemistry A, 2002, 106, 2760. n-Alcohol, Bell et al, Chemistry and Physics of Lipids, 1973, 1-10.  NOTE: For alkanes, n = 1 to 4, value is predicted based on experimental measurement and extrapolation to account for high vapor pressure.  Solubility data for n-alkanes have been predicted for n values into the twenties, see Ferguson et al, Journal of PHysical Chemistry B, 2009, 113 (18), pp 6405–6414. In most cases, branched alkanes with the same carbon number have a higher solubility than linear isomers.

Now, seeing all that data, do you still have questions?

[Technicalhuman]

By now you may know that instead of speculating or guessing, I like to look at real data, which for compounds like this are widely available.  Here is some real solubility data. I've converted it all into the same units - hopefully no mistakes made. 

n	n-Alkane	n-Alcohol
1	7.06E5	miscible
2	2.32E5	miscible
3	5.83E4	miscible
4	1.47E4	7.3E7
5	3.85E4	2.2E7
6	9.50E3	6.23E6
7	2.93E3	1.80E6
8	660	8.385E5
9	171	n/a
10	46	n/a
11	9	n/a
12	1	4.285E4
13	0.4	n/a
14	n/a	300
15	0.04	102.6
16	n/a	41.2

References: Alkane, n = 1 to 8, McAuliffe, Journal of Physical Chemistry, 1966, 70, 1267; Science, 163, 1969, Alkane, n 8 to 15, Tolls, et al, Journal of Physical Chemistry A, 2002, 106, 2760. n-Alcohol, Bell et al, Chemistry and Physics of Lipids, 1973, 1-10.  NOTE: For alkanes, n = 1 to 4, value is predicted based on experimental measurement and extrapolation to account for high vapor pressure.  Solubility data for n-alkanes have been predicted for n values into the twenties, see Ferguson et al, Journal of PHysical Chemistry B, 2009, 113 (18), pp 6405–6414. In most cases, branched alkanes with the same carbon number have a higher solubility than linear isomers.

Now, seeing all that data, do you still have questions?

Hi thanks for the reply Corribus

This shows me that the greater the London dispersion forces, the less soluble it is. But why is this so? Because I always thought we want the forces of attraction to be as strong as possible to ensure that the bond forming process is enough to overcome the bond breaking part. So if the LDFs are stronger shouldnt the bond Also be stronger?

Thanks again for the help

[Technicalhuman]

By now you may know that instead of speculating or guessing, I like to look at real data, which for compounds like this are widely available.  Here is some real solubility data. I've converted it all into the same units - hopefully no mistakes made. 

n	n-Alkane	n-Alcohol
1	7.06E5	miscible
2	2.32E5	miscible
3	5.83E4	miscible
4	1.47E4	7.3E7
5	3.85E4	2.2E7
6	9.50E3	6.23E6
7	2.93E3	1.80E6
8	660	8.385E5
9	171	n/a
10	46	n/a
11	9	n/a
12	1	4.285E4
13	0.4	n/a
14	n/a	300
15	0.04	102.6
16	n/a	41.2

References: Alkane, n = 1 to 8, McAuliffe, Journal of Physical Chemistry, 1966, 70, 1267; Science, 163, 1969, Alkane, n 8 to 15, Tolls, et al, Journal of Physical Chemistry A, 2002, 106, 2760. n-Alcohol, Bell et al, Chemistry and Physics of Lipids, 1973, 1-10.  NOTE: For alkanes, n = 1 to 4, value is predicted based on experimental measurement and extrapolation to account for high vapor pressure.  Solubility data for n-alkanes have been predicted for n values into the twenties, see Ferguson et al, Journal of PHysical Chemistry B, 2009, 113 (18), pp 6405–6414. In most cases, branched alkanes with the same carbon number have a higher solubility than linear isomers.

Now, seeing all that data, do you still have questions?

Hi Corribus, from this link http://www.chemguide.co.uk/organicprops/alkanes/background.html

They said that to be soluble the bond forming between the molecules but release as much heat as the bind breaking process between the pure components. So if the alkane is larger won't the bond forming process be greater since the bond between water and the alkane would be stronger?

Thanks in advance for the help