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If glucose can react with form phenyl hydrazine to form osazone why can't it react with 2,4-DNP?

Yes I know about cyclic form of glucose and its equilibrium with straight chain and all …

I read this, this and this Stack Exchange one too Does glucose react with Brady's reagent?

Still, didn't get an answer to my question. So requesting for clarification …

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Glucose itself keeps interconverting between alpha beta glucopyranose and the chain form but it is more stable in this ring forms. A reagent should be strong enough to convert cyclic structure to the chain form so as to react with aldehyde group. I think due to electron withdrawing action of the Nitro groups in DNP they are less active/weaker than phenyl hydrazine and so DNP shouldnt convert glucose to chain form and react with the aldehyde. That's my explanation though.Someone might post a better answer.

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  • $\begingroup$ Glucose converts itself to the chain form in appropriate solvents, doesn’t it? $\endgroup$ – Jan Dec 24 '15 at 23:25
  • $\begingroup$ @Jan I guess. When the reagent /solvent is highly reactive towards aldehyde group, it becomes chain. $\endgroup$ – Quark Dec 25 '15 at 4:48
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Although glucose 1a exists principally in the hemiacetal pyranose forms (α and β), nonetheless it reacts with one equivalent of phenylhydrazine to form the hydrazone 1b via the aldehydic open form. Two additional equivalents serve to form the osazone and the products of reduction of phenylhydrazine, ammonia and aniline.
The glucose 2,4-dinitrophenylhydrazone 1c has been reported1 but not the corresponding osazone. In their studies on the mechanism of osazone formation, Shemyakin and coworkers2 utilized p-nitrophenylhydrazine labeled with nitrogen-15 at the terminal nitrogen (the red N in the structures). Although glucose was not examined, the p-nitrophenylhydrazones of fructose (forms the same osazone as glucose), benzoin and 2-hydroxycylohexanone were studied. The labelled nitrogen was located in the ammonia that was liberated during osazone formation.
Structures 2-7 represent the essentials of the mechanism. Enolization of the hydrazone 2 produces the enol 3, which, via the zwitterion 4, undergoes oxidation of the carbon framework and reduction of the phenylhydrazine moiety. The transformation 3 --> 4 is essential. The basicity of the electron pair on the nitrogen in the 2,4-dinitrophenyl case is sufficiently reduced by delocalization into the aromatic ring, thereby repressing the proton transfer. This step must be completed before osazone formation can be accomplished.
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1: Lloyd, E. A. and Doherty, D. G., J. Am. Chem. Soc., 1952, 74, 4214.
2: Shemyakin, M. M., Maimind, V. I., Ermolaev, K. M. and Bamdas, E. M., Tetrahedron, 1965, 21, 2771.

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