Question: A mixture of p-nitrophenol and o-nitrophenol can be separated by steam distillation. The o-nitrophenol is steam volatile, and the para iosmer is not volatile. Explain. Base your answer on the ability of the isomers to form hydrogen bonds internally.

My answer: Phenols are compounds with an OH group attached to an aromatic carbon. Intra-molecular bonding is possible in some ortho- phenols. This is due to the polar nature of the O-H bonds which can result in the formation of hydrogen bonds within the same molecule. This intramolecular H bonding reduces water solubility and increases solubility. Thus o-nitrophenol is steam distillable while p-nitrophenol is not.

Is this answer sufficient?

Thanks from Sheryl

I don't think your answer will get it. Here is what I found on pp 958-959 of Organic Chemistry by Morrison and Boyd, Fourth Edition,1983. (Old book but a great book.) "An important point emergences from a comparison of the physical properties of the isomeric nitrophenols of Table 24.2. [Table 24.2 shows b.p. ortho of 100, meta of 194 and para decomposes (all at a reduced pressure of 70 mm.) Solubility g/100 g H2O is ortho 0.2 and volatile in steam; meta 1.35g/100 g H2O and not volatile in steam; para 1.69 g/100 g H2O and not volatile in steam.] "We notice that o-nitrophenol has a much lower boiling point and much lower solubility in water than its isomers; it is the only one of the three that is readily steam-distillable. How can these difference be accounted for?
"Let us consider first the m and p isomers. They have very high boiling points because of intermolecular hydrogen bonding. [Then the book shows the N=O of the nitro group hydrogen bonding to the -OH of ANOTHER molecule.] Their solubility in water is due to hydrogen bonding with water molecules. [Then the book shows N=O AND -OH on different ends of the p molcule hydrogen bonding with HOH.]
Steam distillation depends upon a substance having an appreciable vapor pressure at the boiling point of water; by lowering the vapor pressure, intermolecular hydrogn bonding inhibits steam distillatioin of the m and p isomers.
"What is the situation for the o-isomer? Examination of models shows that the -NO2 and -OH groups are located exactly right for the formation of a hydrogen bond within the molecule. This intramolecular hydrogen bonding takes the place of intermolecular with other phenol molecules and with water molecules; therefore, o-nitrophenol does not have the low volatility of an associated liquid, nor does it have the solubility characteristic of a compound that forms hydrogen bonds with water. We recognize this as an example of chelation. [Then the book shows the
C -N-O
&nbsp&nbsp&nbsp O
&nbsp&nbsp&nbsp |
&nbsp&nbsp&nbsp H
&nbsp&nbsp&nbsp /
&nbsp&nbsp&nbsp 0
(I don't know how this drawing will show but it goes from the nitro group at the apex of the ring to the OH in the ortho position and you make a 6 membered ring counting the carbon to which the nitro group is attached at #1, then N is 2, O from NO2 group is 3, H from OH is 4, O from OH is 5 and back to the carbon in the ortho position as 6. That is the chelation part.
A summary of this with the drawing should make this a "sharp" answer. I will try to correct (one way or another) if the drawing doesn't show up very well OR if the bold and italic tags don't do well. I hope this helps.

If you draw a line from the top C to the bottom C that will be the C-C bond on the side of the ring. Actually, the drawing doesn't look all that bad. The C atoms are in the ortho position.

What does it mean that para decomposes rather than having a boiling point? I've never run into that before.


Many organic compounds don't boil. They decompose. For some there is a temperature at which they decompose listed; others simply list dec (which means 'don't look for a boiling point). And that part of the table doesn't have anything to do with the discussion about the three nitrophenols. I once had an organic compound unknown; the melting point was sharp I thought I would do the boiling point, too. Most compounds can be identified just on physical properties, alone, if we have a couple of things like that. Mine decomposed at a specific temperature every time. I looked through my tables for melting points AND decomposition points and found just one that fit. Went to Beilstein, looked up the compound, found solubilities, etc etc and I had it identified. The prof wouldn't let us turn in our unknowns without making two derivatives, BUT I knew what it was and I knew what derivatives to make. So I slopped through a couple of derivatives and turned them in. Got it right, too.

Some compounds decompose instead of melting, also.

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asked by Sheryl
  1. i have no idea

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  2. That answer kicks butt! its the only one that has made any sense! thanks Sheryl!

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    posted by Ally

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