Question

Brady’s reagent is,
A)$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$
B)$KMn{O_4}/NaI{O_4}$
C) 

D) 

Answer 1

Hint: We know that Brady's reagent and also called Borche's reagent. It is a red to orange solid and substituted hydrazine and is most commonly used to qualitatively test for carbonyl groups linked with aldehydes and ketones.

Complete step by step answer:
First, we see the structure of Brady’s reagent,

We know that an aqueous solution of \[2,{\text{ }}4 - dinitrophenyl{\text{ }}hydrazine\] is called Brady’s reagent. It reacts with carbonyl compounds to yield a colored precipitate. These precipitates have a sharp melting point. The melting points of the precipitates confirm the carbonyl compounds.
Therefore, the option C is correct.

Additional Information:
Detection of carbonyl compound using Brady’s reagent:
The carbonyl compound is just mixed with an acid solution of Brady’s reagent in methanol. The derivatives are orange colored crystalline solids called \[\;2,4 - dinitrophenylhydrazones\]. These crystals are filtered off and purified by recrystallization. Their melting temperatures are measured. Brady’s reagent also can be used as a test for the presence of a carbonyl compound because orange crystals appear when it's added to either an Aldehyde or a Ketone.
The difference between the Aldehyde and Ketone is the presence of a hydrogen atom. The presence of the hydrogen atom makes it very easy to oxidize aldehydes, they are fast reduction agents.

Note:
We must remember that Brady's reagent is used for the qualitative identification of Ketone or Aldehyde functional group carbonyl functionality. Aldehydes and ketones react to yellow, orange, or reddish-orange precipitates while alcohols do not react. This is a successful method for checking for the existence of a drug or representing its absence.