Hydrazine analysis

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PRAVEEN PALIWAL
PRAVEEN PALIWAL's picture
Hydrazine analysis

Sir, what is the basic procedure to derivatized hydrazine hydrate with benzaldehyde. AS i derivatized it only with benzaldehyde I got two peaks and when I did the derivatization after addition of sodium borohydrate I got single peaks. That means the pH is very critical in this derivatization. I case of monomethyl Hydrazine and ethyl hydrazine the derivatization get very easy and no other peaks was found. What is the basic chemistry of these derivatization. I am also sending the derivatizatiion of hydrazine with benzaldehyde wthout including sodium borohydrate and one blan run of benzaldehyde without including hydrazine.
Best regards Praveen

Dr. Analytical
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This is an example of "Schiff

This is an example of "Schiff base" formation.  An amino compound (NH2) reacts with a carbonyl compound (C=O) to produce a Schiff base (C=N).

The following picture was obtained from:
http://biochem.siuc.edu/bmb_courses/mbmb451b/lectures/imine.html

Hydrazine has two amino groups (2HN-NH2), so it will react with two molecules of benzaldehyde. So the reaction product will be:

Ph(H)C=N-N=C(H)Ph , where Ph is the phenyl ring on benzaldehyde.

Although I can't draw it here, remember that the C=N bond does not rotate, so the two groups attached to the carbon are fixed in place.  With the two double bonds in this molecule there are actually two different products that are possible: the two Ph groups can be on the "same" side of the bonds, relative to each other, or on "opposite" sides.  If you remember your organic chemistry, this is like "cis- and trans-" but a little more complicated.  So, there are two different products and you should see two different peaks in the chromatogram.

Methyl hydrazine (CH3(H)N-NH2) has only one amino group to react with benzaldehyde, and so only produces one compound (and one peak).

When you say "sodium borohydrate" do you mean sodium borohydride or sodium borate?  The first is a reducing agent; the second is a buffer.  I can't say what it is doing until I know the correct name and reaction conditions.

Sami Tuomivaara
Sami Tuomivaara's picture
Dr. Analytical,

Dr. Analytical,

Looks like I'm missing something here, could you please clarify why there are two possible products of this reaction. It seems to me that the middle N-N bond in

Ph(H)C=N-N=C(H)Ph

is freely rotating. Moreover, there are only three groups bonded to each of the carbon atoms, hence they don't convey any stereochemistry to the molecule either. Hence, there should be only single product. Where I go wrong?

Cheers,

Dr. Analytical
Dr. Analytical's picture
Suola:

Suola:
Good point, and I was thinking most people would not notice.  I suspect that there is a barrier to free rotation around the N-N bond; this is the case for hydrazine itself.  The derivative may have similar restrictions due to steric hindrance from the bulky phenyl group and interactions between the lone pairs on the N atoms.  So, my guess is that the molecule is locked in one orientation around the N-N bond, which would then create two different geometric isomers with the phenyl groups, one cis- and one trans-.  (You are correct that there are no stereoisomers here.)  Even if the molecule is not locked into one position, it is likely to have two more stable configurations rather than free rotation.

I have to admit this is theoretical on my part, and my organic skills have waned a bit over the years, but the idea is consistent with the fact that two peaks appear in the chromatogram for this molecule, and only one peak appears when you substitute on one of the nitrogens (monomethyl-), where only one derivative is possible.

Of course, we haven't even considered the chromatographic issues, and there could be other issues also.  The chromatograms were not very helpful so I had to start here.

My compliments to you on calling me out on this!
 

Sami Tuomivaara
Sami Tuomivaara's picture
Dr. Analytical,

Dr. Analytical,

I still have to disagree about the number or products, even if the phenyl rings prevent total unhindered rotation about the N-N bond by steric clashing, the end groups N=C(H)Ph would still sample almost every relative angle, except for the one where they are fully cis. So I can't imagine any reason why the end groups would be locked in any particular orientation.

And even if the end groups of two different molecules were locked to different orientations, say on with +90 and another with -90 degrees (due to the symmetry requirement, the angles need to be the same but with opposite sign), they would be identical molecules. If you make a molecular two models with different relative orientation of the end groups, and look at the two molecules, you can align the molecules perfectly, because they are identical.

Cheers,

Dr. Analytical
Dr. Analytical's picture
Suola:

Suola:
Yes, I agree that if there is rotation about the bond, you will only get one product.  The non-bonded electrons will be one source of steric hindrance.  Also, I don't think this would be a linear molecule.  The C=N double bond is tilted away from the axis of the N-N bond, so I would expect the "opposite" configuration of the two double bonds to be more stable.  There should be a larger barrier to rotation as a result.  This, of course, could lead to four possible products.

However, since two peaks are seen in the chromatogram, we must have some explanation.  It is possible that one of the peaks is the partially derivatized product.  I would need more informatino on the separation conditions, peak locations, reaction conditions, etc. A nice kinetic experiment (sample the mixture after different reaction times) might tell us something about this.

Interesting discussion.

Sami Tuomivaara
Sami Tuomivaara's picture
Dr. Analytical,

Dr. Analytical,

I agree that extra peaks could come from partially derivatized hydrazine. I did some hydrazine derivative reactions with carbohydrate reducing ends. Without sodium borohydride reduction, the linkage is not stable (I think Praveen ment NaBH4 in the first post). Maybe there is competitive hydrolysis going on and the extra peaks come from partially derivatized hydrazine and maybe the NaBH4 drives the reaction to completion. Good discussion indeed, makes me go back and looks some chemistry papers...

Cheers,

Dr. Analytical
Dr. Analytical's picture
NaBH4 probably reduces the C

NaBH4 probably reduces the C=N bonds to CH-NH, which would be more stable (or at least less reactive).  Of course, there would be only one product! 

PRAVEEN PALIWAL
PRAVEEN PALIWAL's picture
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Arpit Kakkar
Arpit Kakkar's picture
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