Identify the incorrect statement about glucose

Glucose exists as a 5-membered ring

The incorrect statement is (1) Glucose exists as a 5-membered ring.

Let us look at each of the statements :

1. Glucose exists as a 5-membered ring: Glucose exists as a 6-membered ring structure. 

The limitations shown by the open chain structure of glucose can be explained by its cyclic structure. It was proposed that glucose can form a six-membered ring in which \(−OH\) at \(C-5\) can add to the \(−CHO\) group and can form a cyclic hemiacetal structure. This explains the absence of \(−CHO\) group and also the existence of glucose in \(α, β\) forms.

The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at \(C-1\), called anomeric carbon (the aldehyde carbon before cyclization), and the corresponding \(α\) and \(β\) forms are called anomers. It should be noted that \(α\) and \(β\)-forms of glucose are not mirror images of each other, hence are not enantiomers. The six-membered cyclic structure of glucose is called pyranose structure (\(α\) or \(β\)), in analogy with pyran.

Pyran is a six-membered ring with one oxygen and five carbon atoms in the ring. The cyclic structure of glucose can be more accurately shown by Haworth structure as given below:

2. It forms n-hexane on prolonged heating with \(HI\):

Glucose on reduction with HI and red P at 373 K gives a mixture of n-hexane and 2-iodohexane.

3. Acetylation of glucose takes place with acetic anhydride:

On acetylation with acetic anhydride, glucose gives a pentaacetate. This confirms that glucose contains five –OH groups.

4. It does not give 2,4-DNP test:

Glucose does not give the 2,4-DNP test because it exists in a cyclic hemiacetal form. The 2,4-DNP test is used to identify aldehydes and ketones, and it is based on the reaction of the carbonyl group (C=O) with 2,4-dinitrophenylhydrazine (2,4-DNP) to form a hydrazone.

Glucose exists primarily in the cyclic hemiacetal form, in which the carbonyl group is involved in the formation of the hemiacetal ring. This ring structure prevents the carbonyl group from reacting with 2,4-DNP, and therefore glucose does not give a positive test result.

The cyclic hemiacetal form of glucose is more stable than the open-chain aldehyde form, which is why it is the predominant form in solution. The cyclic form also has a lower energy barrier for mutarotation, which is the process of interconversion between the α- and β-anomers of glucose.

Therefore, the cyclic hemiacetal form of glucose is responsible for its inability to give a positive 2,4-DNP test. This test is only useful for identifying aldehydes and ketones that exist in the open-chain form with a free carbonyl group.