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


> 
> 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](http://www.meritnation.com/ask-answer/question/why-glucose-doesnot-react-with-schiff-s-reagent-and-2-4-dn/biomolecules/1595234), [this](https://en.wikipedia.org/wiki/Osazone) and this Stack Exchange one too [Does glucose react with Brady's reagent?](https://chemistry.stackexchange.com/questions/22269/does-glucose-react-with-bradys-reagent)


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




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.  

[![enter image description here](https://i.stack.imgur.com/TTMBl.jpg)](https://i.stack.imgur.com/TTMBl.jpg)


[1](https://i.stack.imgur.com/TTMBl.jpg): [Lloyd, E. A. and Doherty, D. G., *J. Am. Chem. Soc*., **1952**, *74*, 4214.](https://pubs.acs.org/doi/10.1021/ja01136a601)  

[2](https://pubs.acs.org/doi/10.1021/ja01136a601): [Shemyakin, M. M., Maimind, V. I., Ermolaev, K. M. and Bamdas, E. M., *Tetrahedron*, **1965**, *21*, 2771.](https://www.sciencedirect.com/science/article/pii/S0040402001983620)  



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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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