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In its open-chain form, glucose possesses an aldehyde group; however, glucose does not test positive with 2,4-dinitrophenylhydrazine (which typically forms a yellow/orange/red precipitate with carbonyl compounds).

Why is this so? Is this a reflection of the cyclic hemiacetal form (where the aldehyde group is not present)?

Both forms are in equilibrium so I'd expect it to give a positive test.

glucose forms

Consider also that glucose gives a positive test in both the Tollens' and Benedict's test. While it is true that less than one in hundred exists in the aqueous solution as the cyclic form, this is not sufficient to explain why it does not give a positive test with 2,4-dinitrophenylhydrazine but yet gives a positive test in Tollens' test and also in Benedict's test.

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  • $\begingroup$ Interestingly, aqueous glucose reacts to give a positive test in both the Tollens' test and Benedict's test. It is indeed peculiar why it gives a negative test with 2,4-DNPH. $\endgroup$ – Tan Yong Boon Oct 16 at 6:58
  • $\begingroup$ It was proposed by my classmate that it is perhaps due to the differences in reaction conditions. The conditions for the Tollens' test and Benedict's test are alkaline. However, in the reaction with 2,4-DNPH, the reaction medium is supposedly neutral. This would mean that the rate of conversion between the cyclic and linear forms is rather low in the neutral medium (i.e. the conversion reaction is uncatalysed). In the alkaline medium, the $\ce {OH^-}$ acts as a catalyst. This can allow for the rapid conversion from the linear form to the cyclic form when the linear form reacts with DNPH. $\endgroup$ – Tan Yong Boon Oct 16 at 7:03
  • $\begingroup$ What I can say about this is that the isomerisation of the cyclic hemiacetal form of Glucose in the open chain aldehyde may explain the positive Tollens' and Benedict's test, but may not be the correct reason, because if it has to be the isomerization only, hemiacetals will break more easily in acidic medium than basic conditions and therefore glucose should have been more prone towards positive 2,4-DNPH test. I believe that there should be some other explanation for this kind of observation, other than isomerisation to aldehyde. $\endgroup$ – Soumik Das Oct 19 at 12:33
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The test is based on imine formation. The formation of imine from carbonyl compound occurs in equilibrium ergo the principle of le Chatelier is applicable to the test. According to this principle the increase of the concentration of at least one of the products results in the decrease of degree of conversion of the reactant.

I guess that you are trying to perform the test in aqueous solution, because glucose has poor solubility in ethanol. The second product of the conversion of carbonyl to imine is water. So, performing the reaction in the tremendous excess of water, you are in fact push the degree of conversion toward zero in accordance with the principle of le Chatelier.

That's why the expected precipitate is not formed.

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We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.

  • $\begingroup$ I upvoted this answer because the link ("occurs in equilibrium") is such a complete description of the mechanism and the equilibrium. The link also mentions the catalytic effect of acidity. And it mentions the use of a dehydrating agent to form the products. $\endgroup$ – James Gaidis Oct 16 at 13:53
  • $\begingroup$ I have down-voted this answer because it does only tangentially address the question. The link might contain more useful information, however, it would be preferable to quote the relevant portions, and leave the link as reference. In its current state, this is rather a comment than an actual answer. $\endgroup$ – Martin - マーチン Oct 16 at 14:05
  • $\begingroup$ Imine and eanmine formation is reversible but the formation of DNP's, phenyl hydrazones, osazones, oximes and semicarbazides are not reversible under the mild conditions applied to imines and enamines. The latter group is used as derivatizing agents and are stable. These derivatizing agents all have N-N or N-O bonds. Was the DNP derivatization done under the same conditions applied to phenylhydrazone formation of glucose? $\endgroup$ – user55119 Oct 22 at 18:51

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