Post a New Question

Volatility

posted by .

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
   \
    O
    |
    H
    /
    0
  /
  C
(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.

Sheryl

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.

  • Volatility -

    i have no idea

  • Volatility -

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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question