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solubility of alkanols in water

 
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Apr21-12, 01:58 AM   #1
 

solubility of alkanols in water


Is propan-1-ol more soluble than propan-2-ol in water?
Propan-2-ol has 2 alkyl group, which is electron donating.
Since the formation of H-bonding is due to the elecrionegativity difference of O atom and H atom is the hydroxl group. I think think that the O atom in propan-2-ol can gain electrons from two alkyl groups while propan-1-ol can gain electrons from one alkyl group only and thus the ability of propan-1-ol to form H-bonding is greater than that of propan-2-ol.

Sorry for my poor English and thx a lot
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Apr21-12, 04:45 AM   #2
 
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But another factor is accessibility to water which is less for propan-2-ol, so that factor is in the opposite direction. Around hydrophobic (-CH3) groups water takes up a more structured form, more like ice and less able to form the H-bonds.

Which effect is more important? My sense is that solvent effects are always more important. I can't justify this sense so let's wait see if anyone who knows what he's talking about happens along.

But these are sorts of things you have to think about or find out about. Not enough to be looking at just one molecule but have a sense of the solution. Or rather, the solution versus the separate bulk liquids. Not easy.
Apr21-12, 08:47 PM   #3

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Both of those propanols are soluble in water in all proportions.
Apr22-12, 04:55 AM   #4
 
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solubility of alkanols in water


Quote by chemisttree View Post
Both of those propanols are soluble in water in all proportions.
Now I remember.

However there would be some number of carbon atoms and some temperature at which the question would arise?
Apr22-12, 02:13 PM   #5

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Probably better to talk about pentanols or 5-carbon alcohols. There are definite trends. More highly branched alkanols are more soluble than the extended chain isomers.
Apr23-12, 07:06 AM   #6
 
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Quote by chemisttree View Post
Probably better to talk about pentanols or 5-carbon alcohols. There are definite trends. More highly branched alkanols are more soluble than the extended chain isomers.
I said it was not easy.

Can you say why that tendency is?
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