Chemistry · Biomolecules · NEET
They are two separate labels that just happen to both point 'right'. 'D' is about STRUCTURE: it means the -OH group on the lowest chiral carbon (C-5, the one just above the CH2OH end) is drawn on the RIGHT side in the Fischer projection. 'L' would mean that -OH is on the left. The '+' is about BEHAVIOUR: glucose is dextrorotatory, so it rotates plane-polarised light to the right (clockwise). Important for NEET: D does not force +. A 'D' sugar can be laevorotatory. In glucose both happen to be right, so we write D-(+)-glucose.
By its reactions in the open-chain form. Glucose reacts with hydroxylamine (NH2OH) to form an oxime and adds one molecule of HCN to form a cyanohydrin. Both of these reactions are given only by a carbonyl group (C=O). It also reduces Tollens' reagent (silver mirror) and Fehling's solution, which only aldehydes do among the two carbonyl types. So glucose must contain a -CHO (aldehyde) group, not a keto group.
When glucose is heated with concentrated hydroiodic acid (HI) for a long time, it gives n-hexane. Since HI removes all the oxygen atoms, the carbon skeleton left is a straight 6-carbon chain (n-hexane). This proves glucose has 6 carbon atoms joined in an unbranched chain.
Glucose reacts with acetic anhydride to form a pentaacetate (five -OH groups get acetylated). This shows there are 5 -OH groups. They must be on 5 different carbons, because two -OH groups on the same carbon (a gem-diol) would be unstable and lose water.
The LOWEST chiral (asymmetric) carbon, which is the carbon farthest from the -CHO group but still bonded to an -OH. In glucose this is C-5. If its -OH is on the right in the Fischer projection, the sugar is D; if on the left, it is L. You do NOT look at C-2, C-3 or C-4 for the D/L label. The 2016 NEET aldose question tests exactly this idea.
Because in water glucose exists mostly as a cyclic (ring) hemiacetal, so a truly free -CHO group is only present in a very tiny amount. Schiff's reagent and sodium bisulphite (NaHSO3) need a large amount of free aldehyde, so they give a negative result. Reactions like Tollens', HCN and NH2OH still work because the small open-chain amount is used up and more keeps forming. This exact point was asked in NEET 2024.
The reagents with which glucose does NOT react to give the corresponding test/product are: A. Tollen's reagent B. Schiff's reagent C. HCN D. NH2OH E. NaHSO3. Choose the correct option.
The correct corresponding order of names of the four aldoses with the given Fischer configurations (1)-(4), respectively, is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Glucose is C6H12O6. It is an aldohexose, meaning it has 6 carbons and one aldehyde (-CHO) group.
Glucose is an aldose because it has an aldehyde group (-CHO) at C-1. Fructose, in contrast, is a ketose with a keto group.
No. 'D' describes the position of the -OH on the lowest chiral carbon (C-5) in the Fischer projection. '+' or dextrorotatory describes rotation of polarised light. A D sugar can be laevorotatory; in glucose both are right, so we write D-(+)-glucose.
Prolonged heating with HI reduces glucose to n-hexane. Since all oxygen is removed and a straight 6-carbon chain remains, this proves glucose has 6 carbons in an unbranched chain.
Five -OH groups, shown by formation of glucose pentaacetate with acetic anhydride. They lie on five different carbons (C-2 to C-6).
It cannot explain why glucose fails Schiff's and NaHSO3 tests, and why it shows two forms (anomers) with mutarotation. These are explained by the cyclic (ring) structure, studied next.
