Magpie
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quaternary ammonium salt reaction
I have a generic question, based on a JACS paper* I have been reading.
The paper states that S-benzylthiuronic chloride, a quaternary ammonium salt, will react with sodium benzenesulfonate, a salt, to form a 3rd salt.
This derivative then can be obtained by crystallization. It has a melting point that can be used in the identification of the benzenesulfonate (and
other arylsulfonates, for that matter).
This is something new to me, ie, two organic salts reacting to form an apparently less soluble, 3rd organic salt. I wasn't sure just what the
product formula was, the paper not specifying same. But after doing a little math I have come to these conclusions:
1. All 3 hydrogens of the quaternary amine functional group (R1-NH3+) are retained, as well as its positive charge.
2. The R2-SO3- functional group also retains its negative charge.
3. The dried salt, of course, is electrically neutral.
Is this interpretation correct, or does it sound reasonable?
*"The Identification of Sulfonic Acids" by Chambers & Watt, November 8, 1940
[Edited on 6-7-2011 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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Paddywhacker
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Quote: Originally posted by Magpie  | ... two organic salts reacting to form an apparently less soluble, 3rd organic salt...
[Edited on 6-7-2011 by Magpie] |
and sodium chloride.
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497
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I don't see how S-Benzyltiuronium chloride which is what I'm assuming they meant is a quaternary ammonium salt.
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ScienceSquirrel
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That is easy, the proton should be attached to the nitrogen and not the chlorine.
The salt is similar to urea nitrate.
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Magpie
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Yes, that's my understanding:
The single most important condition for a successful synthesis is good mixing - Nicodem
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497
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I always thought they were more along the lines of this:
http://en.wikipedia.org/wiki/Quaternary_ammonium_cation
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Magpie
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That's what SS and I are saying. The graphic I show (copypasta) is an inaccurate representation as it shows the proton as part of HCl.
The single most important condition for a successful synthesis is good mixing - Nicodem
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not_important
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| Quote: Originally posted by Magpie |
That's what SS and I are saying. The graphic I show (copypasta) is an inaccurate representation as it shows the proton as part of HCl.
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But subject compound is RNH3(+), not RR'R''R'''N(+) - which is a quant salt. S-Benzyltiuronium chloride is a substituted ammonium chloride, used to
prepare derivatives of acids (carboxylic, sulfonic, ...) for identification purposes. When reacting with a sulfonic acid, HCl is liberated as the
benzyltiuronium salt of the sulfonic acid is formed.
http://pubs.acs.org/doi/abs/10.1021/ja01260a501
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Magpie
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Yes, I see. So the subject compound is a primary ammonium salt, correct? And it would lose its proton under sufficiently basic conditions, yielding
an amine?
The single most important condition for a successful synthesis is good mixing - Nicodem
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not_important
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I assume so, but ...
the S-benzyltiuronium R-S-C(=NH)(NH3) is a derivative of thiourea S=C(NH2)2. I'm not at all sure what the C=NH will do on neutralisation. Note
that salts such as Me-S(+)=C(NMe2)2 can be formed, or compounds that have resonance structures like that plus Me-S-C(=NMe2)(NMe2) (+).
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Magpie
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The JACS paper I referenced lead me to believe that only the -NH3+ group is affected. I deduced this by calculating %N and %S in agreement with the
values in the paper.
Here's the paper:
http://pubs.acs.org/doi/abs/10.1021/jo01203a002
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Correction: This reagent is covered in Vogel. The form represented there does not define where the proton attaches, ie, at the -NH2 or the =NH. In
fact, I wonder if this is the reason for the two isomers that have different melting points.
[Edited on 7-7-2011 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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