Biomolecules — NEET Chemistry MCQs with answers

Biomolecules is a Class 12 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 450 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Carbohydrates · Amino acids and proteins · Nucleic acids · Vitamins, enzymes and hormones · Proteins and amino acids · Biomolecules & Polymers · Vitamins and Enzymes · Hormones · Amino acids · Polymers · Lipids · Biomolecules · Enzymes · Proteins · Vitamins · Biomolecular linkages

8 free Biomolecules practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Carbohydrates

   Carbohydrates that cannot be hydrolysed further into simpler polyhydroxy aldehyde or ketone units are classified as:

   • A.Disaccharides
   • B.Monosaccharides
   • C.Polysaccharides
   • D.Oligosaccharides
   Show answer and explanation

   Answer: B. Monosaccharides

   Monosaccharides are the fundamental carbohydrate units; they cannot be split into smaller polyhydroxy aldehyde/ketone molecules by hydrolysis, unlike oligo- and polysaccharides which are built from them.

   Carbohydrates are classified by hydrolysis behaviour. Monosaccharides (glucose, fructose, ribose) are single polyhydroxy aldehyde or ketone units and give no smaller carbohydrate on hydrolysis. Oligosaccharides yield 2-10 monosaccharide units, and polysaccharides yield many monosaccharide units on hydrolysis. Hence the defining, non-hydrolysable class is monosaccharides.

   Common mistake: Mixing up which class is the 'building block' versus which classes are built from it

2. Question 2 · easy · Carbohydrates

   A sugar is termed a 'reducing sugar' if, in aqueous solution, it possesses:

   • A.A ring oxygen atom
   • B.A glycosidic linkage to another sugar unit
   • C.At least one -OH group on an asymmetric carbon
   • D.A free aldehyde or a free/potentially free ketone group (via a free hemiacetal OH)
   Show answer and explanation

   Answer: D. A free aldehyde or a free/potentially free ketone group (via a free hemiacetal OH)

   A sugar reduces Fehling's/Tollens' reagent only if it has a free anomeric carbon, i.e. an open-chain -CHO or an -OH at the hemiacetal (anomeric) carbon that can equilibrate to the open-chain carbonyl form.

   Reducing character in carbohydrates arises from the ability of the sugar to exist, even partly, in an open-chain form with a free -CHO (aldose) or from a free hemiacetal -OH at C1/C2 that can tautomerise to expose a reactive carbonyl. This free carbonyl is oxidised by mild oxidants such as Tollens' or Fehling's reagent, giving the positive test. If the anomeric carbon is locked in a glycosidic bond (as in sucrose), no free carbonyl can form and the sugar is non-reducing.

   Common mistake: Attributing reducing power to chirality or ring oxygen rather than the free anomeric carbon

3. Question 3 · hard · Carbohydrates

   D-Glucose is heated with excess HI and red phosphorus (prolonged reaction). This reaction is chemically significant because it establishes which structural feature of glucose?

   • A.That glucose contains a five-membered oxygen-containing ring
   • B.That glucose contains an aldehyde group at C1
   • C.That all six carbons of glucose form a straight (unbranched) chain
   • D.That glucose has the D-configuration at C5
   Show answer and explanation

   Answer: C. That all six carbons of glucose form a straight (unbranched) chain

   Prolonged reaction of glucose with HI/red phosphorus reduces every C-OH and the C=O to C-H, converting glucose entirely into n-hexane. Since a single unbranched product (n-hexane) is obtained, all six carbon atoms of glucose must be joined in a straight chain.

   HI with red phosphorus is a powerful, exhaustive reducing system that removes all oxygen functions, converting -OH and >C=O groups into -H. When glucose (C6H12O6) is subjected to this treatment for a long time, the sole organic product is n-hexane (CH3-(CH2)4-CH3). Since n-hexane has an unbranched six-carbon skeleton, and no rearrangement of the carbon skeleton occurs under these conditions, the six carbons of glucose must already have been linked in a straight chain before reduction. This is one of the classical degradation experiments used to deduce glucose's open-chain skeletal structure, distinct from experiments (like penta-acetate/pentamethyl ether formation) used to count -OH groups or establish ring size.

   Common mistake: Attributing this classical reaction's significance to ring size or functional group identity instead of chain linearity

4. Question 4 · medium · Carbohydrates

   D-Glucose is treated with bromine water at room temperature. What is the major product?

   • A.Glucaric acid (a dicarboxylic acid formed by oxidation of both C1 and C6)
   • B.Glucose pentaacetate
   • C.Gluconic acid (monocarboxylic acid from oxidation of only C1)
   • D.Sorbitol (hexahydric alcohol from reduction of C1)
   Show answer and explanation

   Answer: C. Gluconic acid (monocarboxylic acid from oxidation of only C1)

   Br2 water is a mild, selective oxidising agent that oxidises only the aldehyde group (C1) of glucose to a carboxylic acid, giving gluconic acid; it does not touch the C6 primary alcohol.

   Glucose's open-chain form has an aldehyde at C1 and a primary alcohol at C6. Mild oxidants such as bromine water (or dilute HOCl) selectively oxidise only the more easily oxidised aldehyde group, converting -CHO to -COOH and leaving the -CH2OH at C6 untouched. This gives gluconic acid, a monocarboxylic acid (pentahydroxy hexanoic acid). Stronger oxidants like dilute HNO3 oxidise both ends to give the dicarboxylic saccharic (glucaric) acid.

   Common mistake: Assuming any oxidant oxidises glucose fully to the dicarboxylic acid

5. Question 5 · medium · Carbohydrates

   Which observation about D-glucose could NOT be explained by its open-chain (acyclic) structure alone, and led chemists to propose a cyclic hemiacetal form?

   • A.Glucose gives a positive Tollens' test
   • B.Glucose exists in two crystalline forms with different specific rotations (+112° and +19°) that both converge to +52.7° in solution (mutarotation), and glucose does not give the Schiff's test
   • C.Glucose reacts with hydroxylamine to form an oxime
   • D.Glucose forms a cyanohydrin with HCN
   Show answer and explanation

   Answer: B. Glucose exists in two crystalline forms with different specific rotations (+112° and +19°) that both converge to +52.7° in solution (mutarotation), and glucose does not give the Schiff's test

   The open-chain structure predicts a single compound with one specific rotation and a positive Schiff's test (since Schiff's reagent should detect a free aldehyde), but real glucose shows two distinct crystalline forms (mutarotation) and does not restore colour to Schiff's reagent — this anomalous behaviour is explained only by a cyclic hemiacetal structure with an anomeric centre.

   Straight-chain structures with a genuine free aldehyde should give a negative-to-positive colour with Schiff's reagent and should exist as a single compound with one fixed optical rotation. But glucose is isolated in two crystalline forms (α, [α]D = +112°, and β, [α]D = +19°) whose solutions slowly change rotation to a common equilibrium value of +52.7° — mutarotation — and glucose fails Schiff's test, unlike a typical free aldehyde. These facts are explained by glucose existing predominantly as a six-membered cyclic hemiacetal (pyranose form): ring closure between the C5-OH and the C1 aldehyde creates a new stereocentre at C1 (the anomeric carbon), giving two diastereomeric anomers (α and β) that interconvert in solution via the small equilibrium amount of open-chain form, causing mutarotation. The hemiacetal -OH at C1, though giving positive Tollens'/Fehling's tests (via the open-chain equilibrium), reacts too slowly/differently with Schiff's reagent to give the classic aldehyde colour change.

   Common mistake: Thinking all listed carbonyl-type reactions of glucose are equally explained (or equally anomalous) without distinguishing mutarotation/Schiff's test as the key anomaly

6. Question 6 · medium · Carbohydrates

   A food chemist wants to test whether a sugar sample is sucrose or a mixture containing invert sugar. Which observation would confirm the presence of invert sugar rather than pure sucrose?

   • A.The sample gives a positive Fehling's/Tollens' test, consistent with a mixture of free glucose and fructose formed by hydrolysis of the glycosidic bond
   • B.The sample gives a negative Fehling's test
   • C.The sample shows no glycosidic bond hydrolysis products on TLC
   • D.The sample rotates plane-polarised light strongly dextrorotatory and does not change on acid hydrolysis
   Show answer and explanation

   Answer: A. The sample gives a positive Fehling's/Tollens' test, consistent with a mixture of free glucose and fructose formed by hydrolysis of the glycosidic bond

   Invert sugar is the equimolar mixture of free glucose and fructose obtained on hydrolysis of sucrose; since both monosaccharides have free anomeric carbons, invert sugar (unlike sucrose) gives a positive Fehling's/Tollens' test.

   Sucrose is non-reducing because its glycosidic bond joins the anomeric carbons of both glucose (C1) and fructose (C2), leaving no free carbonyl/hemiacetal centre. Acid or enzymatic (invertase) hydrolysis breaks this glycosidic bond, releasing free D-glucose and D-fructose — this equimolar mixture is called invert sugar (named for the inversion of optical rotation from dextrorotatory sucrose to the net laevorotatory mixture, since fructose's strong (-) rotation outweighs glucose's (+) rotation). Because both glucose and fructose now have free anomeric centres, invert sugar reduces Fehling's and Tollens' reagents, unlike intact sucrose.

   Common mistake: Forgetting that hydrolysis, not the intact disaccharide, is what converts sucrose into a reducing mixture

7. Question 7 · medium · Carbohydrates

   Consider the following statements about D-glucose and D-fructose: (I) Glucose forms an oxime with hydroxylamine, confirming a carbonyl group. (II) Both glucose and fructose reduce Tollens' reagent in alkaline solution. (III) Fructose, despite having a ketone (not aldehyde) group, is a reducing sugar because base-catalysed tautomerisation converts it to an aldose form under the alkaline test conditions. (IV) Glucose and fructose are functional isomers with the molecular formula C6H12O6. Which of these statements are correct?

   • A.II and III only
   • B.I, II, III and IV
   • C.I and IV only
   • D.I, III and IV only
   Show answer and explanation

   Answer: B. I, II, III and IV

   All four statements are chemically accurate: glucose's carbonyl reacts with NH2OH (I); both sugars reduce Tollens' reagent under alkaline conditions (II) because fructose isomerises to an aldose via enediol tautomerisation under base (III); and glucose/fructose share the formula C6H12O6 while differing in functional group, making them functional isomers (IV).

   (I) Glucose's free -CHO reacts with NH2OH to give glucose oxime, standard carbonyl chemistry. (II) Tollens' test is performed in mildly alkaline (ammoniacal) medium; under these basic conditions, fructose undergoes base-catalysed enolisation (via a 1,2-enediol intermediate) that interconverts it with glucose and mannose, generating an aldose form that reduces Ag+ to metallic silver — hence fructose, though a ketose, is experimentally a reducing sugar under Tollens'/Fehling's conditions. (III) directly states this mechanism and is correct. (IV) Glucose (aldohexose) and fructose (ketohexose) share the molecular formula C6H12O6 but differ in the position/type of carbonyl group, making them functional group isomers. All four statements hold.

   Common mistake: Assuming ketoses can never reduce Tollens'/Fehling's reagent, ignoring the alkaline tautomerisation pathway

8. Question 8 · medium · Carbohydrates

   D-Glucose is treated with excess acetic anhydride in the presence of pyridine. What is the major product, and what does it establish about glucose's structure?

   • A.Glucose diacetate; it establishes that glucose has only two -OH groups
   • B.Glucose pentaacetate; it establishes that glucose contains a free aldehyde group reacting with acetic anhydride
   • C.Glucose pentaacetate; it establishes that glucose has five -OH groups (four alcoholic plus one hemiacetal/anomeric -OH)
   • D.Sodium gluconate; it establishes the presence of a carboxylic acid group
   Show answer and explanation

   Answer: C. Glucose pentaacetate; it establishes that glucose has five -OH groups (four alcoholic plus one hemiacetal/anomeric -OH)

   Acetic anhydride acetylates all free hydroxyl groups of glucose (esterification), including the anomeric -OH; the product, glucose pentaacetate, shows glucose has five acetylatable -OH groups, consistent with its cyclic hemiacetal structure (four ring -OH plus the anomeric -OH).

   Acetic anhydride/pyridine is a mild acetylating agent for hydroxyl groups (esterification: -OH + (CH3CO)2O -> -OCOCH3 + CH3COOH), and does not react with a carbonyl carbon under these conditions. When D-glucose (cyclic pyranose form, bearing four secondary/primary alcoholic -OH groups on C2, C3, C4, C6 plus one anomeric hemiacetal -OH on C1) is fully acetylated, all five -OH groups are converted to acetate esters, giving glucose pentaacetate. The fact that exactly five acetate groups are incorporated (verified by quantitative acetylation/analysis) confirms glucose has five -OH-bearing/acetylatable positions — consistent with the cyclic hemiacetal structure rather than a hypothetical structure with a different number of oxygen functions.

   Common mistake: Miscounting the acetylatable -OH groups or confusing acetylation with carbonyl-addition chemistry

Questions about Biomolecules for NEET

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