The product of the following reaction will be: CHO-(CHOH)4-CH2OH xrig…

The correct answer is Option (4) → A: COOH–(CHOH)₄–CH₂OH    B: COOH–(CHOH)₄–COOH

The starting compound, \( \ce{CHO-(CHOH)4-CH2OH} \), is glucose. It has an aldehyde group at one end and a primary alcohol group at the other, and its oxidation product depends on the strength of the oxidising agent:

• Bromine water is a mild oxidant. It oxidises only the –CHO group to –COOH, giving gluconic acid (A).

      \( \displaystyle \ce{CHO-(CHOH)4-CH2OH + [O]} \xrightarrow{\ce{Br2}\text{ water}} \ce{COOH-(CHOH)4-CH2OH} \)

• Nitric acid is a strong oxidant. It also oxidises the –\( \ce{CH2OH} \) group to –COOH, giving the dicarboxylic acid saccharic acid (B).

      \( \displaystyle \ce{COOH-(CHOH)4-CH2OH + 2[O]} \xrightarrow{\ce{HNO3}} \ce{COOH-(CHOH)4-COOH + H2O} \)

So A is gluconic acid, COOH–(CHOH)₄–CH₂OH, and B is saccharic acid, COOH–(CHOH)₄–COOH.

Why the other options are incorrect

• Option (1) an unbranched alkane (A) and a nitroalkane (B) — Incorrect. An unbranched hydrocarbon (n-hexane) is obtained only by heating glucose with HI; neither bromine water nor nitric acid removes the oxygen atoms.

• Option (2) A: COOH–(CHOH)₄–COOH, B: COOH–(CHNO₂)₄–COOH — Incorrect. Bromine water is too mild to oxidise the –\( \ce{CH2OH} \) group, and nitric acid acts as an oxidant, not a nitrating agent, here.

• Option (3) A: CHO–(CHOH)₄–COOH, B: COOH–(CHOH)₄–COOH — Incorrect. Bromine water oxidises the aldehyde end, not the alcohol end, so A cannot still contain –CHO.

NCERT (2026-27 edition):
Glucose – “… 4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.”

“… 6. On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose.”