CARBOHYDRATES General introduction and importance of biomolecules. CARBOHYDRATES - Classification; aldoses and ketoses – monosaccharides (glucose and fructose) and constituent: NEET MCQs with Answers

11 free questions on CARBOHYDRATES General introduction and importance of biomolecules. CARBOHYDRATES - Classification; aldoses and ketoses – monosaccharides (glucose and fructose) and constituent for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: monosaccharides (glucose and fructose) and constituent monosaccharides of oligosaccharides (sucrose, lactose, and maltose).

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Question 1

Previous-year question
Sucrose on hydrolysis gives [NEET 2020]
  1. α-D-Glucose + β-D-Glucose
  2. α-D-Glucose + β-D-Fructose
  3. α-D-Fructose + β-D-Fructose
  4. α-D-Glucose + β-D-Fructose only in acid
Show answer
Answer: (B) α-D-Glucose + β-D-Fructose
Explanation: Hydrolysis breaks the glycosidic bond and liberates the two constituent units in the forms in which they were combined, namely α-D-glucose and β-D-fructose.

Question 2

Previous-year question
The difference between amylose and amylopectin is [NEET 2018]
  1. Amylopectin has 1 → 4 α-linkages and 1 → 6 α-linkages
  2. Amylose has 1 → 4 α-linkages and 1 → 6 β-linkages
  3. Amylose is made up of glucose and galactose
  4. Amylopectin has 1 → 4 α-linkages and 1 → 6 β-linkages
Show answer
Answer: (A) Amylopectin has 1 → 4 α-linkages and 1 → 6 α-linkages
Explanation: Amylose has only α-1,4 links and is therefore unbranched, whereas amylopectin has the same α-1,4 chains together with α-1,6 links at the branch points; both are built exclusively from α-D-glucose.

Question 3

Previous-year question
Which one given below is a non-reducing sugar? [NEET 2016]
  1. Maltose
  2. Lactose
  3. Glucose
  4. Sucrose
Show answer
Answer: (D) Sucrose
Explanation: In sucrose the anomeric carbons of both monosaccharide units are locked in the glycosidic bond, so it cannot open to give a free carbonyl group and does not reduce Tollens' or Fehling's reagent.

Question 4

How many stereoisomers are possible for the open-chain aldopentose CHO–CHOH–CHOH–CHOH–CH₂OH?
  1. 6
  2. 16
  3. 4
  4. 8
Show answer
Answer: (D) 8
Explanation: In the open-chain aldopentose, C-2, C-3 and C-4 each carry four different groups, giving three asymmetric carbons. The two ends (–CHO and –CH₂OH) differ, so no meso form is possible. The number of stereoisomers is therefore 2³ = 8.

Question 5

Glucose and fructose both have the molecular formula C₆H₁₂O₆. Which statement correctly distinguishes them?
  1. Both are aldohexoses, and they differ only in the direction in which they rotate plane polarised light
  2. Glucose is a ketohexose with its keto group at carbon 2, and fructose is an aldohexose known as dextrose
  3. Glucose is an aldopentose, and fructose is a ketohexose with its ketonic group at carbon 1
  4. Glucose is an aldohexose, also known as dextrose, and fructose is a ketohexose with its ketonic group at carbon 2
Show answer
Answer: (D) Glucose is an aldohexose, also known as dextrose, and fructose is a ketohexose with its ketonic group at carbon 2
Explanation: Glucose has an aldehyde group and six carbon atoms, so it is an aldohexose, and it is also called dextrose. Fructose has the same molecular formula but carries a ketonic group at carbon 2, so it is a ketohexose. The two are therefore functional isomers of each other in the open-chain form.

Question 6

D-(–)-Fructose, a ketohexose, exists in two cyclic forms. Which description of these forms is correct?
  1. The –OH at C-4 adds to the keto group at C-2, giving a four-membered ring, and the two anomers differ at C-4
  2. The –OH at C-5 adds to an aldehyde group at C-1, giving a six-membered pyranose ring, and the two anomers differ at C-1
  3. The –OH at C-6 adds to the keto group at C-2, giving a six-membered pyranose ring, and the two anomers differ at C-5
  4. The –OH at C-5 adds to the keto group at C-2, giving a five-membered furanose ring, and the two anomers differ at C-2
Show answer
Answer: (D) The –OH at C-5 adds to the keto group at C-2, giving a five-membered furanose ring, and the two anomers differ at C-2
Explanation: Fructose has its ketonic group at C-2, and the –OH at C-5 adds to this carbonyl group. The ring contains four carbon atoms and one oxygen atom, so it is five-membered and is called furanose by analogy with furan. C-2 is the former carbonyl carbon, so it is the anomeric carbon where the two forms differ.

Question 7

Compare the acid hydrolysis of starch with that of sucrose as routes to glucose. Which statement is correct?
  1. Starch boiled with dilute H₂SO₄ at 393 K gives glucose and galactose in equal amounts, whereas sucrose gives only glucose
  2. Starch hydrolyses in water at 298 K without any acid to give glucose, whereas sucrose gives only fructose
  3. Starch boiled with concentrated HNO₃ gives glucose and fructose in equal amounts, whereas sucrose gives only glucose
  4. Starch boiled with dilute H₂SO₄ at 393 K under pressure gives only glucose, whereas sucrose gives glucose and fructose in equal amounts
Show answer
Answer: (D) Starch boiled with dilute H₂SO₄ at 393 K under pressure gives only glucose, whereas sucrose gives glucose and fructose in equal amounts
Explanation: Starch is a polymer of α-glucose, so its complete hydrolysis yields glucose alone; commercially this is done by boiling with dilute H₂SO₄ at 393 K under pressure. Sucrose is made of one glucose unit and one fructose unit, so boiling it with dilute acid gives the two sugars in equal amounts.
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Question 8

Glucose on oxidation with nitric acid gives saccharic acid, and gluconic acid on the same treatment gives the same saccharic acid. What does this common product show about the open-chain structure of glucose?
  1. A primary alcoholic (–OH) group is present
  2. All six carbon atoms lie in a straight chain
  3. The carbonyl group is a keto group at C-2
  4. Five –OH groups are attached to different carbon atoms
Show answer
Answer: (A) A primary alcoholic (–OH) group is present
Explanation: Gluconic acid already has –COOH at C-1 but keeps the –CH₂OH group at the other end of the chain. Nitric acid converts this end to a second –COOH, giving the same dicarboxylic acid from both compounds. This indicates a primary alcoholic group in glucose. The straight chain is shown by the reaction with HI, and the five –OH groups by acetylation.

Question 9

Natural starch is separated into its two components. Which description correctly identifies the component present in larger proportion by mass?
  1. Amylose, about 15–20% of starch, water-insoluble and branched through C1–C6 linkages
  2. Amylose, about 80–85% of starch, water-soluble and an unbranched chain of α-D-glucose
  3. Amylopectin, about 15–20% of starch, water-soluble and an unbranched chain of α-D-glucose
  4. Amylopectin, about 80–85% of starch, water-insoluble and branched through C1–C6 linkages
Show answer
Answer: (D) Amylopectin, about 80–85% of starch, water-insoluble and branched through C1–C6 linkages
Explanation: Starch is a mixture of amylose and amylopectin. Amylose is the water-soluble, unbranched component and forms only about 15–20% of starch. Amylopectin is the water-insoluble component and forms about 80–85%. Its chains use C1–C4 glycosidic linkages, and branches arise through C1–C6 linkages.

Question 10

When the open-chain formula of glucose is written with the –CHO group at the top, which carbon atom decides that (+)-glucose has the D-configuration, and how?
  1. C-5, the lowest asymmetric carbon, whose –OH lies on the left as in (–)-glyceraldehyde
  2. C-2, the asymmetric carbon next to the aldehyde group, whose –OH lies on the left as in (–)-glyceraldehyde
  3. C-5, the lowest asymmetric carbon, whose –OH lies on the right as in (+)-glyceraldehyde
  4. C-6, the last carbon of the chain, whose –OH lies on the right as in (+)-glyceraldehyde
Show answer
Answer: (C) C-5, the lowest asymmetric carbon, whose –OH lies on the right as in (+)-glyceraldehyde
Explanation: To assign D or L to a monosaccharide, the lowest asymmetric carbon in the chain is compared with glyceraldehyde. In glucose this is C-5, because C-6 carries two hydrogens and is not asymmetric. Its –OH lies on the right, matching (+)-glyceraldehyde, so the sugar belongs to the D-series.

Question 11

Cane sugar is boiled with dilute acid until hydrolysis is complete. The neutralised solution is then treated with bromine water. Which organic compounds are present at the end?
  1. Gluconic acid and unchanged fructose
  2. Saccharic acid and unchanged fructose
  3. Gluconic acid only, formed from both hexoses
  4. Unchanged glucose and an acid formed from fructose
Show answer
Answer: (A) Gluconic acid and unchanged fructose
Explanation: Hydrolysis of sucrose gives equal amounts of D-glucose and D-fructose. Bromine water is a mild oxidant that converts the aldehyde group of glucose to –COOH, which gives gluconic acid. It does not attack the keto group of fructose, so fructose remains unchanged. A stronger oxidant such as nitric acid is needed to form saccharic acid.

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