
Glucose does not react with which of the following?
A) $2,4 - DNP$
B) Schiff’s reagent
C) $NaHS{O_3}$
D) All of these
Answer
569.4k+ views
Hint:As we know that $2,4 - DNP$, $NaHS{O_3}$ and Schiff’s reagent are characteristic reagents involved in the chemical reactions of the aldehydic group and we also know that glucose also contains an aldehydic group in its open structure.
Complete answer:
As we know that $2,4 - DNP$, $NaHS{O_3}$ and Schiff’s reagent are characteristic reagents involved in the chemical reactions of the aldehydic group and are generally used to distinguish between an aldehyde and ketone group but still glucose does not undergo any reaction with these reagents as glucose actually exists in the cyclic hemiacetal form with only a small amount of the open chain form in equilibrium.
We know that aldehydic group reacts with bisulphates of $NaHS{O_3}$ to form a new product but Glucose does not react with this reagent as its aldehydic group is not free because the fifth and first carbon of glucose forms an oxy linkage. $NaHS{O_3}$ is rarely used to differentiate between aldehyde and ketones as it does not easily react with ketones and the reason being their less reactivity than aldehydes.
Similarly, glucose does give a positive test with schiff’s base as it does not possess any free aldehydic group.
Therefore from the above explanation the correct answer is (D).
Note:Remember that all the three given reagents are used to differentiate between the aldehyde and ketones, $2,4 - DNP$ when reacts with any aldehyde or ketone it turns the solution into a yellow colour, resulting in a positive test, similarly when Schiff’s base is added to any aldehyde or ketone it turns into reddish violet colour indicating the positive test and $NaHS{O_3}$ does not give a good result with ketonic group because they are less reactive than aldehyde groups.
Complete answer:
As we know that $2,4 - DNP$, $NaHS{O_3}$ and Schiff’s reagent are characteristic reagents involved in the chemical reactions of the aldehydic group and are generally used to distinguish between an aldehyde and ketone group but still glucose does not undergo any reaction with these reagents as glucose actually exists in the cyclic hemiacetal form with only a small amount of the open chain form in equilibrium.
We know that aldehydic group reacts with bisulphates of $NaHS{O_3}$ to form a new product but Glucose does not react with this reagent as its aldehydic group is not free because the fifth and first carbon of glucose forms an oxy linkage. $NaHS{O_3}$ is rarely used to differentiate between aldehyde and ketones as it does not easily react with ketones and the reason being their less reactivity than aldehydes.
Similarly, glucose does give a positive test with schiff’s base as it does not possess any free aldehydic group.
Therefore from the above explanation the correct answer is (D).
Note:Remember that all the three given reagents are used to differentiate between the aldehyde and ketones, $2,4 - DNP$ when reacts with any aldehyde or ketone it turns the solution into a yellow colour, resulting in a positive test, similarly when Schiff’s base is added to any aldehyde or ketone it turns into reddish violet colour indicating the positive test and $NaHS{O_3}$ does not give a good result with ketonic group because they are less reactive than aldehyde groups.
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