A couple of reactions of glucose have been a source of great confusion to me. One of them is it's reaction with the Brady's reagent. It's pretty debatable whether the result is positive or negative. Could someone please shed some conclusive light onto this?

Furthermore, Glucose itself successfully produces an oxime with hydroxylamine. In contrast, it's pentaacetate fails to do so. Why does this occur? Why can't the pentaacetate, like glucose, "break out" of it's cyclic form to give a positive hydroxylamine test?

Edit: I found out about the oxime test, I had overlooked the fact that glucose pentaacetate doesn't have any hemiacetal. The Brady's test though, is still open for debate.

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    $\begingroup$ I have found that glucose fails to show the 2,4 DNP test and fails to show nucleophilic addition by sodium bisulphite. Since the cyclic form of glucose is in equilibrium with the open form, I cannot think of any reason why they do not, however. $\endgroup$
    – Charles
    Commented Aug 28, 2015 at 6:38

1 Answer 1


From Wikipedia:

The reaction of $\ce{NH2OH}$ with an aldehyde or ketone produces an oxime.
$$\ce {R2C=O + NH2OH·HCl + NaOH → R2C=NOH + NaCl + H2O}$$

For the penta-acetate ester of glucose not to react with hydroxylamine would indicate that the ester is more stable, or there is steric hindrance to the reaction. Since hydroxylamine is heat sensitive, I am guessing that you did not heat the reaction. Given time, you should see the oxime form.

From Wikipedia:

A positive test is signaled by a yellow, orange or red precipitate (known as a dinitrophenylhydrazone). $$\ce {RR'C=O + C6H3(NO2)2NHNH2 → C6H3(NO2)2NHNCRR' + H2O}$$

2,4-DNP also reacts with an aldehyde (or ketone), so both glucose and the penta-acetate ester of glucose should react. If you get a different color for the penta-acetate ester, it is due to the substituents on the glucose.


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