Q.12 What is the major product formed in the given reaction?

In carbohydrate chemistry, bromine water at mildly acidic pH is a classic reagent used to distinguish aldoses from ketoses and to convert aldose sugars into aldonic acids. In competitive exams, questions often test whether you know that Br₂/H₂O (pH ≈ 5–6) oxidizes only the aldehyde group of an aldose without affecting the primary alcohol at C‑6 or the secondary alcohols.

Major product of Br₂/H₂O oxidation (MCQ answer)

The major product is the corresponding aldonic acid, formed by selective oxidation of the anomeric aldehyde of the sugar with Br₂/H₂O at pH 6, so the correct option is the structure in which only C‑1 is oxidized to COOH and the terminal CH₂OH remains unchanged (the option matching a monocarboxylic acid).

Reaction concept: Br₂/H₂O oxidation of aldoses

• Br₂ in water is a mild oxidizing agent that selectively oxidizes the aldehyde group (‑CHO) of an aldose to a carboxylic acid (‑COOH), giving an aldonic acid.
• Under these mildly acidic conditions (pH 5–6), the other hydroxyl groups and the terminal CH₂OH group are not oxidized, so no dicarboxylic (aldaric) acids are formed.
• Stronger oxidants such as dilute nitric acid convert both the aldehyde and terminal CH₂OH groups into carboxylic acids to form aldaric acids, which is a different reaction and is not occurring here.

Step‑by‑step solution of the MCQ

The cyclic sugar shown in the question is an aldose (a hemiacetal form with an anomeric carbon derived from an aldehyde). In aqueous solution, this cyclic form is in equilibrium with a small amount of open‑chain aldehyde, which is what actually undergoes oxidation with bromine water.

Ring–chain equilibrium

• The pyranose (or furanose) ring opens to generate the corresponding open‑chain aldose with one terminal ‑CHO group and one terminal CH₂OH group.
• All internal carbons carry secondary alcohol groups (‑CHOH‑), which remain intact during this oxidation.

Selective oxidation by Br₂/H₂O at pH 6

• Br₂/H₂O oxidizes the aldehyde carbon (C‑1) to a carboxylic acid without oxidizing the terminal CH₂OH at the other end of the chain.
• The product is therefore an aldonic acid of the original sugar: HO‑(CHOH)₄‑CH₂OH becomes HO‑(CHOH)₄‑COOH only at the aldehyde end.

Matching with the given options

• The correct Fischer projection must show COOH at one end and CH₂OH at the other, with exactly the same pattern of OH/H configuration on the internal chiral carbons as the parent aldose.
• Any option that shows COOH at both ends corresponds to an aldaric acid, which would require a stronger oxidizing agent such as HNO₃, not bromine water.

Thus, the major product is the monocarboxylic aldonic acid structure (COOH at C‑1 and CH₂OH at C‑6, with unchanged stereochemistry at the middle carbons), and that option should be selected as the answer.

Detailed analysis of each option

Option (A): Dicarboxylic acid

• This structure has COOH groups at both ends of the chain, meaning both the aldehyde and the terminal CH₂OH group have been oxidized.
• Such a product is an aldaric acid, which is produced when an aldose is oxidized by stronger oxidizing agents like dilute nitric acid, not by mild Br₂/H₂O.
• Therefore, option (A) is incorrect for oxidation of an aldose sugar with bromine water at pH 6.

Option (B): Aldehydo‑acid (CHO and COOH together)

• Option (B) shows an aldehyde (CHO) on one end and a carboxylic acid (COOH) on the other, which cannot result from simple, selective oxidation of an aldose aldehyde.
• Br₂/H₂O converts the existing aldehyde fully into COOH; it does not create a new aldehyde at the opposite end of the molecule, so a CHO–(CHOH)ₙ–COOH pattern is not expected.
• Hence, option (B) is also incorrect.

Option (C): Monocarboxylic aldonic acid (correct)

• Option (C) displays COOH at one end and CH₂OH at the other, with the internal stereocenters retaining the same OH/H configuration as the starting aldose.
• This matches the definition of an aldonic acid, formed when the aldehyde group of an aldose is selectively oxidized by bromine water while the primary alcohol at the other end remains unoxidized.
• Therefore, option (C) correctly represents the major product of the reaction and is the correct answer.

Option (D): Dialdehyde

• Option (D) shows CHO at both ends of the carbon chain, which would correspond to a dialdehyde, a structure not obtained by this oxidation.
• Oxidation with Br₂/H₂O cannot introduce an additional aldehyde; instead, it destroys the aldehyde by converting it into a carboxylic acid, so this option contradicts the reaction type.
• Thus, option (D) is also incorrect.

Key exam tips for Br₂/H₂O and sugars

• Remember that bromine water at pH 5–6 selectively oxidizes the aldehyde of an aldose to a single COOH, giving an aldonic acid, while leaving CH₂OH and secondary OH groups unchanged.
• If you see a sugar product with two COOH groups (aldaric acid), think of nitric acid or other strong oxidants, not bromine water.
• This reaction is widely used as a test for aldoses vs ketoses, because ketoses do not react under these mild conditions and do not decolorize bromine water.