Multiple options may be correct (+4 / partial / −1). Four statements, four benches. Predict first, then run the experiment — the verdicts below are computed from what you observe, not looked up.
Nitric acid is a strong oxidising agent: it attacks an aldehyde and a primary alcohol alike. Put something in the flask, look at which ends are oxidisable, and heat it.
Fehling's reagent is alkaline, and that one word is the whole story. Run the test on a ketose that has no aldehyde group of its own and see what the copper does.
Sucrose is one glucose and one fructose held together by a glycosidic bond. Break that bond and count what comes out — then watch the polarimeter needle change sides, which is where the name comes from.
The paper gives you the L sugars. Invert sugar is made of the D ones. Sort that out first, then mix and measure — and watch what happens if you forget one step of the arithmetic.
D/L is configuration. (+)/(−) is which way the light turns. They are independent — that is the whole trap in this statement.
Tap a tile, then tap a slot. Get it right and the polarimeter locks in.
Every verdict below came out of a bench you ran. The traps are the wrong turns this question is built to catch — open one only after you have formed your own view.
Written as you work. Only what you actually demonstrated appears here.
Test all four statements on the bench above and the derivation opens by itself.
Statements tested: 0 of 4 — oxidise both compounds, run Fehling's on fructose, hydrolyse sucrose and finish the average
Nitric acid is a strong oxidising agent. Open-chain glucose offers it two oxidisable positions — the C1 aldehyde and the C6 primary alcohol — and it takes both.
Gluconic acid is simply glucose whose C1 has already been oxidised — by a mild oxidant such as bromine water, which stops there. Its C6 −CH2OH is still untouched, so HNO3 takes it to the same dicarboxylic acid.
Both routes land on saccharic acid. The statement's first half is fine; the clause "whereas glucose is not oxidized to saccharic acid" is what breaks it. (A) is incorrect.
Fructose has no aldehyde of its own. But Fehling's solution is strongly basic, and base removes the acidic α-hydrogen at C1 to give a 1,2-enediol. That enediol can collapse back in three directions.
This is the Lobry de Bruyn–van Ekenstein rearrangement. It produces glucose and mannose — two aldohexoses differing only at C2, which is exactly the "another aldohexose" the statement names. Aldoses reduce Cu(II) to Cu2O, the red precipitate.
So a ketose does give a positive test, and for exactly the reason stated. (B) is correct.
One glycosidic bond, one water, two monosaccharides — one D-glucose and one D-fructose, in strictly equal numbers. That equimolar mixture is what "invert sugar" names. (C) is correct.
Sucrose is dextrorotatory, [α] = +66.5°. The mixture it hydrolyses into is laevorotatory. The sign of the rotation flips during the reaction — the rotation inverts. It has nothing to do with any configuration inverting.
Start from the printed data, which is for the L sugars. Enantiomers rotate equally and oppositely, so the D sugars follow by a sign flip:
Invert sugar is equimolar, and glucose and fructose have the same molar mass (180.16), so equal moles means equal mass and the specific rotation of the mixture is the plain average:
−40° is what you get by adding the two and never dividing. It is the right pair of numbers with the last step missing. (D) is incorrect.
Marking: +4 for both; +2 for either one alone; −1 for including (A) or (D).
| Simulation ID | ADV-2026-P2-CHE-Q09 |
| Exam / Year / Paper | JEE Advanced · 2026 · Paper 2 |
| Subject | Chemistry (Organic) |
| Chapter | Biomolecules |
| Topic | Monosaccharides, reducing sugars and optical rotation |
| Subtopics | Mild versus strong oxidation of glucose · Lobry de Bruyn–van Ekenstein rearrangement · Fehling's test and reducing sugars · Hydrolysis of sucrose and invert sugar · Specific rotation of a mixture · D/L configuration versus (+)/(−) rotation |
| Question type | Multiple correct MCQ (+4 / partial / −1) |
| Difficulty | Moderate–Hard |
| Answer | — run the experiment |