Glucose on oxidation with nitric acid results in formation of one of…

The correct answer is Option (4) → Saccharic acid

The open-chain structure of glucose has an aldehyde group (–CHO) at C1, four secondary –OH groups, and one primary alcoholic group (–CH₂OH) at C6. Each reaction of glucose reveals one feature of this structure:

• Bromine water (mild oxidant) oxidises only –CHO to –COOH, giving gluconic acid: this shows the aldehyde group.

• Nitric acid (strong oxidant) oxidises both ends: –CHO and the primary –\( \ce{CH2OH} \) both become –COOH, giving the dicarboxylic acid saccharic acid (glucaric acid).

    \( \displaystyle \underset{\text{glucose}}{\ce{OHC-(CHOH)4-CH2OH}} + \ce{3[O]} \xrightarrow{\ce{HNO3}} \underset{\text{saccharic acid}}{\ce{HOOC-(CHOH)4-COOH}} + \ce{H2O} \)

Only a primary alcohol group can be oxidised to –COOH without losing a carbon, so the formation of saccharic acid shows that glucose contains a primary alcoholic group.

Why the other options are incorrect

• Option (1) Hexane — Incorrect. It is formed on prolonged heating with HI; it shows that the six carbons are in a straight chain.

      \( \displaystyle \ce{OHC-(CHOH)4-CH2OH} \xrightarrow[\Delta]{\ce{HI}} \underset{\text{n-hexane}}{\ce{CH3-(CH2)4-CH3}} \)

• Option (2) Cyanohydrin — Incorrect. It is formed with HCN; it shows the presence of a carbonyl group.

      \( \displaystyle \ce{OHC-(CHOH)4-CH2OH + HCN -> NC-CH(OH)-(CHOH)4-CH2OH} \)

• Option (3) Gluconic acid — Incorrect. It is formed with bromine water; it shows the aldehyde group, since only C1 is oxidised.

      \( \displaystyle \ce{OHC-(CHOH)4-CH2OH} \xrightarrow{\ce{Br2}\text{ water}} \underset{\text{gluconic acid}}{\ce{HOOC-(CHOH)4-CH2OH}} \)

NCERT (2026-27 edition):
Glucose – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: