Carbohydrates

Stachyose is a tetrasaccharide consisting of two -D-galactose units, one -D-glucose unit and one -D-fructose unit
sequentially linked as gaI gal glc fru. It is naturally found in numerous vegetables and plants.

Step 1: Nature of lactose.
Lactose is a disaccharide composed of one glucose unit and one galactose unit.

Step 2: Hydrolysis of lactose.
On hydrolysis, the glycosidic bond in lactose breaks, releasing its constituent monosaccharides.

Step 3: Products formed.
The hydrolysis of one mole of lactose produces one mole of glucose and one mole of galactose.

Step 4: Conclusion.
Hence, the correct answer is mol Galactose mol Glucose.

Step 1: Understanding gluconic acid formation.
Gluconic acid is produced from glucose by the oxidation of glucose in the presence of bromine water, which oxidizes the aldehyde group to a carboxyl group.

Step 2: Analyzing the options.
(A) dil. Nitric acid: This reagent is a strong oxidizing agent, but it is not typically used to convert glucose to gluconic acid.
(B) Acetyl chloride: Acetyl chloride is used in esterification reactions, not for converting glucose to gluconic acid.
(C) Bromine water: Correct — Bromine water is used to oxidize glucose to gluconic acid by oxidizing the aldehyde group.
(D) Acetic anhydride: This is used in acetylation reactions, not for oxidation of glucose.

Step 3: Conclusion.
The correct answer is (C) Bromine water, as it is the reagent used to oxidize glucose into gluconic acid.

Step 1: Understand raffinose structure.
Raffinose is a trisaccharide composed of three monosaccharide units linked together.

Step 2: Components of raffinose.
Raffinose consists of one molecule each of glucose, fructose, and galactose.

Step 3: Hydrolysis reaction.
On hydrolysis, raffinose breaks into its constituent monosaccharides in equimolar amounts.

Step 4: Conclusion.
Hydrolysis of one mole of raffinose produces one mole each of glucose, fructose, and galactose.

Step 1: Understanding the Reaction.
Maltose, when hydrolyzed, breaks down into two glucose molecules. Upon reaction with excess dilute nitric acid (HNO ), it undergoes oxidation to form saccharic acid (also known as glucaric acid), which is a dicarboxylic acid. This oxidation involves the conversion of the aldehyde group in glucose to a carboxylic acid group.

Step 2: Analyzing the options.
(A) Gluconic acid: Incorrect. Gluconic acid is obtained by the oxidation of glucose at the aldehyde group, but here the reaction leads to the formation of saccharic acid.
(B) Saccharic acid: Correct. Saccharic acid (glucaric acid) is formed by the oxidation of glucose, and this is the product formed in the given reaction.
(C) n-Hexane: Incorrect. This is an alkane, which is not related to the reaction described here.
(D) Glucoxime: Incorrect. Glucoxime is a derivative of glucose formed by reacting glucose with hydroxylamine, not nitric acid.

Step 3: Conclusion.
The correct product formed is Saccharic acid, corresponding to option (B).

Step 1: Identifying glyceraldehyde.
Glyceraldehyde is an aldose, a type of monosaccharide with the molecular formula C O H . It is a simple sugar with a three-carbon backbone.

Step 2: Analyzing the options.
(A) C O H : This is incorrect as it does not match the molecular formula of glyceraldehyde.
(B) C O H : This is incorrect as glyceraldehyde has three carbon atoms, not four.
(C) C O H : Correct — This is the molecular formula of glyceraldehyde.
(D) C O H : This is incorrect as it does not correspond to the structure of glyceraldehyde.

Step 3: Conclusion.
The correct answer is (C) C O H , which is the molecular formula of glyceraldehyde.

Glucose is a reducing sugar because it contains an aldehyde group in its open-chain form. The presence of the aldehyde group allows glucose to donate electrons and reduce other compounds, which is the defining characteristic of reducing sugars. In its cyclic form, glucose forms a hemiacetal, but when in an open-chain form, it exposes the aldehyde group. Reducing sugars can reduce compounds such as Benedict’s solution or Fehling's solution, which is a common test for the presence of reducing sugars.

Glucose is a reducing sugar because it contains an aldehyde group in its open-chain form. The presence of the aldehyde group allows glucose to donate electrons and reduce other compounds, which is the defining characteristic of reducing sugars. In its cyclic form, glucose forms a hemiacetal, but when in an open-chain form, it exposes the aldehyde group.