CBSE Class 12 Chemistry Biomolecules NEET JEE Questions Set 01

Single Correct Answer

Question 1. Which of the following compounds can be detected by Molisch’s test? (AIEEE 2012)
(a) Sugars
(b) Amines
(c) Primary alcohols
(d) Nitro compounds
Answer:
Ans. (a)
Molisch’s test (α-naphthol + conc. H₂SO₄, violet ring) is the general test for carbohydrates, i.e., sugars.

Teacher's Note:
1. Molisch’s test is given by all carbohydrates (free or combined), so it is a group test, not a test to tell one sugar from another.

Question 2. A sugar ‘X’ dehydrates very slowly under acidic condition to give furfural. Which on further reaction with resorcinol gives the coloured product after some time. Sugar ‘X’ is (JEE Main 2022)
(a) Aldopentose
(b) Aldotetrose
(c) oxalic acid
(d) Ketotetrose
Answer:
Ans. (a)
This test is based upon Seliwanoff’s test which is used to distinguish between aldoses and ketoses. Ketoses give this test more rapidly than aldoses because they are more rapidly dehydrated than aldoses.
Aldopentose (X) \( \xrightarrow[-\mathrm{H_2O}]{\mathrm{H^+}} \) Furfural \( \xrightarrow{\text{Resorcinol}} \) Red coloured dye
Since the sugar ‘X’ dehydrates slowly to give furfural, therefore, sugar (X) is an aldopentose.

Teacher's Note:
1. Pentoses give furfural, hexoses give 5-hydroxymethylfurfural on dehydration; ketoses react faster than aldoses.

Question 3. The reaction of concentrated sulphuric acid with carbohydrates (C₁₂H₂₂O₁₁) is an example of (NEET 2020)
(a) sulphonation
(b) dehydration
(c) oxidation
(d) reduction
Answer:
Ans. (b)
Conc. H₂SO₄ removes the elements of water from sugar and leaves behind carbon (charring): C₁₂H₂₂O₁₁ \( \xrightarrow{\text{conc. H}_2\text{SO}_4} \) 12C + 11H₂O. Hence it is a dehydration reaction.

Teacher's Note:
1. Charring of sugar shows the dehydrating power of conc. H₂SO₄; no S atom enters the molecule, so it is not sulphonation.

Question 4. The structure of D-(+)-glucose isThe structure of L-(−)-glucose is (JEE Advanced 2015)
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (a)
L-(−)-Glucose is the mirror image (enantiomer) of D-(+)-glucose. So the configuration at every chiral carbon (C-2 to C-5) is reversed: the –OH groups are on the left, right, left, left from C-2 to C-5, as in structure (a).

Teacher's Note:
1. For an enantiomer, flip every stereocentre, not just the last one; changing only C-5 gives a diastereomer (L-idose), as in option (c).

Question 5. The correct corresponding order of names of four aldoses with configuration given belowrespectively, is (NEET Phase-II 2016)
(a) L-erythrose, L-threose, L-erythrose, D-threose
(b) D-threose, D-erythrose, L-threose, L-erythrose
(c) L-erythrose, L-threose, D-erythrose, D-threose
(d) D-erythrose, D-threose, L-erythrose, L-threose
Answer:
Ans. (d)
The configuration (D or L) is decided by the –OH on C-3 (the chiral carbon farthest from CHO): right = D, left = L.
When both –OH groups are on the same side, the sugar is erythrose; when they are on opposite sides, it is threose. Hence the four structures are D-erythrose, D-threose, L-erythrose and L-threose.

Teacher's Note:
1. Erythrose: both OH on the same side; threose: OH on opposite sides. Read D/L only from the lowest chiral carbon.

Question 6. L-isomer of a compound ‘A’ (C₄H₈O₄) gives a positive test with [Ag(NH₃)₂]⁺. Treatment of ‘A’ with acetic anhydride yields a triacetate derivative. Compound ‘A’ produced an optically active compound (B) and optically inactive compound (C) on treatment with bromine water and HNO₃ respectively. Compound (A) is (JEE Main 2022)
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (a)
L-sugars have OH group at the penultimate carbon towards left. Since sugar ‘A’ gives +ve test with Tollens’ reagent, it must be an aldose. Since the sugar A yields triacetate, it must contain three –OH groups.
L-sugar (both OH on left) \( \xrightarrow{\mathrm{Br_2/H_2O}} \) HOOC–CH(OH)–CH(OH)–CH₂OH, optically active (B); \( \xrightarrow{\mathrm{HNO_3}} \) HOOC–CH(OH)–CH(OH)–COOH with both OH on the left, which has a plane of symmetry and is optically inactive (C).
Thus, option (a) is correct.

Teacher's Note:
1. An aldaric acid is optically inactive (meso) only when the two halves are mirror images, i.e., for erythrose (both OH on the same side).

Question 7. Glucose on prolonged heating with HI gives (JEE Main 2018)
(a) n-hexane
(b) 1-hexene
(c) hexanoic acid
(d) 6-iodohexanal
Answer:
Ans. (a)
On prolonged heating with HI, glucose is completely reduced to n-hexane, which shows that all its six carbon atoms are linked in a straight chain.

Teacher's Note:
1. This reaction is the evidence for the unbranched six-carbon chain of glucose.

Question 8. Glucose molecule reacts with X number of molecules of phenylhydrazine to yield osazone. The value of X is
(a) three
(b) two
(c) one
(d) four
Answer:
Ans. (a)
One molecule of phenylhydrazine forms the phenylhydrazone at C-1, the second oxidises C-2 (CHOH → C=O) and is itself reduced to aniline and NH₃, and the third condenses with the new C=O group to give the osazone. So X = 3.

Teacher's Note:
1. Glucose, fructose and mannose give the same osazone because only C-1 and C-2 are involved.

Question 9. D(+)-Glucose reacts with hydroxylamine and yields an oxime. The structure of the oxime would be (AIPMT 2014)
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (d)
Option (d) has the right orientation of OH groups at C₂, C₃, C₄ and C₅ (right, left, right, right, as in D-glucose).

Teacher's Note:
1. Only C-1 changes in oxime formation (CHO → CH=NOH); the configuration of all the chiral carbons stays exactly as in D-glucose.

Question 10. α-D(+)-glucose and β-D(+)-glucose are (AIEEE 2008; AIIMS 2010)
(a) enantiomers
(b) conformers
(c) epimers
(d) anomers
Answer:
Ans. (d)
α- and β-D-glucose differ only in the configuration at C-1, the anomeric (hemiacetal) carbon. Such isomers are called anomers.

Teacher's Note:
1. Anomers are a special kind of epimers: they differ only at the anomeric carbon (C-1 in aldoses).

Question 11. For the given below hemiacetal (X), the correct pyranose structure is (JEE Main 2022)
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (d)
Option (d) has all the groups on the left in Fischer projection (X) correctly projected up and all the groups on the right are projected below in the pyranose structure. Thus, option (d) is correct.

Teacher's Note:
1. In converting Fischer to Haworth, left goes up and right goes down; check every carbon, including the anomeric C-1.

Question 12. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution? (JEE Main 2017)
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer:
Ans. (c)
Ester (c) in presence of aqueous KOH solution undergoes hydrolysis. The –OCOCH₃ group at the anomeric carbon becomes –OH (hemiacetal), which opens to the α-hydroxy carbonyl compound.
Since hydrolysis gives an α-hydroxy carbonyl compound which gives +ve Tollens’ test, it behaves as a reducing sugar in aq. KOH solution.

Teacher's Note:
1. A glycoside (OCH₃ at the anomeric carbon) is stable to base, but an ester is hydrolysed by KOH and frees the anomeric OH.

Question 13. In the following structure, anomeric carbon is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
Ans. (a)
C₁ is the anomeric carbon. It is the carbon attached to the ring oxygen that carries the –OH group (hemiacetal carbon).

Teacher's Note:
1. In an aldopyranose, the anomeric carbon is the ring carbon bonded to two oxygens: the ring O and an OH.

Question 14. What is the relationship between the given structures (look at the arrows)? (J & K CET 2018)
(a) Enantiomers
(b) Anomers
(c) Diastereomers
(d) Metamers
Answer:
Ans. (b)
Stereoisomers differing only in configuration at C₁ are called anomers. Here only the –OH at C-1 (marked by the arrows) is below the ring in one and above the ring in the other.

Teacher's Note:
1. Look for the single changed centre: one carbon only = epimers; and if that carbon is the anomeric carbon, they are anomers.

Question 15. Which set of terms correctly identifies the carbohydrate shown?1. Pentose   2. Hexose   3. Aldose   4. Ketose   5. Pyranose   6. Furanose
(a) 1, 3 and 6
(b) 1, 3 and 5
(c) 2, 3 and 5
(d) 2, 3 and 6
Answer:
Ans. (a)
The ring has four carbons and one oxygen (furanose). Counting the CH₂OH carbon, the molecule has five carbon atoms (pentose), and the anomeric carbon carries an H and an OH, so it comes from an aldehyde group (aldose). Hence it is an aldopentose in the furanose form: 1, 3 and 6.

Teacher's Note:
1. Count all carbons, not just ring carbons; an H on the anomeric carbon means aldose, a CH₂OH there means ketose.

Question 16. The following carbohydrate is (IIT Paper II, 2011)
(a) a ketohexose
(b) an aldohexose
(c) an α-furanose
(d) an α-pyranose
Answer:
Ans. (b)
Since the given carbohydrate has five carbon and one oxygen atom in the ring, therefore, it has pyranose ring. Further, since C₁–OH is β-oriented, therefore, it is a β-pyranose. Further, the carbohydrate has in all six carbon atoms and also has one H and one OH group at C₁, therefore, it is an aldohexose. Thus, option (b) is correct.

Teacher's Note:
1. A ring with five C and one O is a pyranose; the H on C-1 shows it came from an aldehyde, so it is an aldose.

Question 17. Which of the following statements is correct? (JEE Main 2020)
(a) Gluconic acid is obtained by oxidation of glucose with HNO₃.
(b) Gluconic acid is a dicarboxylic acid.
(c) Gluconic acid can form cyclic (acetal/hemiacetal) structure.
(d) Gluconic acid is a partial oxidation product of glucose.
Answer:
Ans. (d)
Glucose on partial oxidation with Br₂–H₂O gives gluconic acid (only the –CHO is oxidised to –COOH). HNO₃ gives the dicarboxylic saccharic acid.

Teacher's Note:
1. Mild oxidant (Br₂ water) gives the monocarboxylic gluconic acid; strong oxidant (HNO₃) gives the dicarboxylic saccharic acid.

Question 18. Which of the following statement is not true for glucose? (JEE Main 2020)
(a) Glucose reacts with hydroxylamine to form oxime
(b) The pentaacetate of glucose does not react with hydroxylamine to give oxime
(c) Glucose exists in two crystalline forms α and β
(d) Glucose gives Schiff’s test for aldehyde
Answer:
Ans. (d)
Glucose does not give Schiff’s reagent test characteristic of aldehyde, because in solution it exists almost entirely in the cyclic hemiacetal form.

Teacher's Note:
1. Failure of Schiff’s test, NaHSO₃ addition and pentaacetate–oxime reaction are the evidence for the cyclic structure of glucose.

Question 19. (i) Glucose + ROH \( \xrightarrow{\text{dry HCl}} \) Acetal \( \xrightarrow[(\mathrm{CH_3CO})_2\mathrm{O}]{x\ \text{eq. of}} \) acetyl derivative
(ii) Glucose \( \xrightarrow{\mathrm{Ni/H_2}} \) A \( \xrightarrow[(\mathrm{CH_3CO})_2\mathrm{O}]{y\ \text{eq. of}} \) acetyl derivative
(iii) Glucose \( \xrightarrow[(\mathrm{CH_3CO})_2\mathrm{O}]{z\ \text{eq. of}} \) acetyl derivative
‘x’, ‘y’ and ‘z’ in these reactions are respectively (JEE Main 2020)
(a) 5, 4 and 5
(b) 4, 6 and 5
(c) 4, 5 and 5
(d) 5, 6 and 5
Answer:
Ans. (b)
(i) CH₂OH–(CHOH)₄–CHO + ROH \( \xrightarrow{\text{Dry HCl}} \) cyclic acetal (methyl glucoside) with four free –OH groups \( \xrightarrow{(\mathrm{CH_3CO})_2\mathrm{O}} \) tetraacetate. Thus, x is 4.
(ii) CH₂OH–(CHOH)₄–CHO \( \xrightarrow{\mathrm{Ni/H_2}} \) CH₂OH–(CHOH)₄–CH₂OH \( \xrightarrow{(\mathrm{CH_3CO})_2\mathrm{O}} \) CH₂OAc–(CHOAc)₄–CH₂OAc (hexaacetate). Thus, y is 6.
(iii) CH₂OH–(CHOH)₄–CHO \( \xrightarrow{(\mathrm{CH_3CO})_2\mathrm{O}} \) CH₂OAc–(CHOAc)₄–CHO (pentaacetate). Thus, z is 5.
∴ Option (b) is correct.

Teacher's Note:
1. Count the free OH groups each time: the glycoside uses one OH, while reduction to sorbitol creates a sixth OH.

Question 20. Which of the following is not true about glucose? (NEET 2020)
(a) It contains five hydroxyl groups
(b) It is a reducing sugar
(c) It is an aldopentose
(d) It is an aldohexose
Answer:
Ans. (c)
Option (c) is wrong since glucose is not an aldopentose but is actually an aldohexose.

Teacher's Note:
1. Glucose: six carbons (hexose), an aldehyde group (aldo-) and five OH groups.

Question 21. Sucrose on hydrolysis gives (NEET 2020)
(a) α-D-glucose + β-D-glucose
(b) α-D-glucose + β-D-fructose
(c) α-D-fructose + β-D-fructose
(d) β-D-glucose + α-D-fructose
Answer:
Ans. (b)
Sucrose on hydrolysis gives a mixture of α-D-glucose and β-D-fructose.

Teacher's Note:
1. In sucrose, C-1 of α-D-glucose is joined to C-2 of β-D-fructose, so hydrolysis releases exactly these two anomers.

Question 22. Which one of the following is a non-reducing sugar? (NEET Phase-I 2016)
(a) Glucose
(b) Sucrose
(c) Maltose
(d) Lactose
Answer:
Ans. (b)
Sucrose is a non-reducing sugar while all others are reducing sugars.

Teacher's Note:
1. Sucrose is non-reducing because both anomeric carbons (C-1 of glucose and C-2 of fructose) are used in the glycosidic bond.

Question 23. Which one of the following does not exhibit the phenomenon of mutarotation? (AIPMT Prelims 2010)
(a) (−) Fructose
(b) (+) Sucrose
(c) (+) Lactose
(d) (+) Maltose
Answer:
Ans. (b)
Only monosaccharides, i.e., (−)-fructose, and reducing disaccharides, i.e., (+)-lactose and (+)-maltose, show mutarotation. Since (+)-sucrose is a non-reducing sugar, it does not show mutarotation. Thus, option (b) is correct.

Teacher's Note:
1. Mutarotation needs a free hemiacetal (anomeric) OH; reducing sugars have it, sucrose does not.

Question 24. The beta and alpha glucose have different specific rotations. When either is dissolved in water, their rotation changes until the same fixed value results. This is called (AIIMS 2008)
(a) epimerization
(b) racemisation
(c) anomerisation
(d) mutarotation
Answer:
Ans. (d)
The spontaneous change in specific rotation of an optically active compound with time, to reach an equilibrium value (+52.5° for glucose), is called mutarotation.

Teacher's Note:
1. α-D-Glucose (+111°) and β-D-glucose (+19.2°) both reach +52.5° through the open-chain form.

Question 25. When sugar ‘X’ is boiled with dilute H₂SO₄ in alcoholic solution, two isomers ‘A’ and ‘B’ are obtained. ‘A’ on oxidation with HNO₃ yields saccharic acid whereas B shows laevorotation. The compound ‘X’ is (JEE Main 2022)
(a) Maltose
(b) Sucrose
(c) Lactose
(d) Starch
Answer:
Ans. (b)
Sucrose on hydrolysis gives two isomers (C₆H₁₂O₆): glucose (A), which on oxidation with HNO₃ gives saccharic acid, and fructose (B), which is laevorotatory.

Teacher's Note:
1. Two different isomeric hexoses on hydrolysis point to sucrose (or lactose); the laevorotatory one is fructose.

Question 26. The formation of ‘A’ and ‘B’ for the following reaction sequence are:
Fructose \( \xrightarrow[\mathrm{H_3O^+}]{\mathrm{HCN}} \) A
Fructose \( \xrightarrow[(ii)\ \mathrm{HI,\ P}]{(i)\ \mathrm{NaBH_4}} \) B (JEE Main 2022)
(a) A = C₇H₁₄O₈, B = C₆H₁₄
(b) A = C₇H₁₂O₇, B = C₇H₁₄O
(c) A = C₇H₁₂O₈, B = C₆H₁₄
(d) A = C₇H₁₄O₈, B = C₆H₁₄O₆
Answer:
Ans. (a)
Fructose (C₆H₁₂O₆), CH₂OH–CO–(CHOH)₃–CH₂OH \( \xrightarrow[(ii)\ \mathrm{H_3O^+}]{(i)\ \mathrm{HCN}} \) CH₂OH–C(OH)(COOH)–(CHOH)₃–CH₂OH, A (C₇H₁₄O₈)
Fructose \( \xrightarrow{\mathrm{NaBH_4}} \) CH₂OH–(CHOH)₄–CH₂OH \( \xrightarrow[\text{Reduction}]{\mathrm{HI,\ P}} \) CH₃–(CH₂)₄–CH₃, B (C₆H₁₄, n-hexane)
Thus, option (a) is correct.

Teacher's Note:
1. HCN adds one carbon (C₆ → C₇) and hydrolysis turns CN into COOH; HI/P removes all oxygen to give the alkane.

One or More Correct Answers

Question 27. Among the following statements about the molecules X and Y, the one(s) which is (are) correct is (are) (West Bengal JEE 2014)
(a) X and Y are diastereomers
(b) X and Y are enantiomers
(c) X and Y both are aldohexoses
(d) X is a D-sugar and Y is an L-sugar
Answer:
Ans. (b, c, d)
X and Y are actually enantiomers and not diastereomers. Although both X and Y are aldohexoses but X is a D-sugar and Y is a L-sugar.

Teacher's Note:
1. Every stereocentre of Y is the reverse of X, so they are mirror images (enantiomers), not diastereomers.

Question 28. Which of the following statements are correct?
(a) α- and β-D-glucopyranose are anomers
(b) β-D-glucopyranose is more stable than α-D-glucopyranose
(c) galactose is a C₄-epimer of glucose
(d) invert sugar is laevorotatory
Answer:
Ans. (a, b, c, d)
(a) They differ only at C-1, so they are anomers. (b) In β-D-glucopyranose all the bulky groups, including the C-1 OH, are equatorial, so it is more stable. (c) Galactose differs from glucose only at C-4. (d) Invert sugar has [α] = −20°, i.e., it is laevorotatory.

Teacher's Note:
1. At equilibrium about 64% of glucose is the β-anomer because all its substituents are equatorial.

Matrix Match Questions

Teacher's Note:
1. Nucleoside = base + sugar; adding phosphate gives the nucleotide, the monomer of DNA/RNA.

Integer and Numerical Answer Questions

Question 30. Total number of possible aldohexoses belonging to D-series is.
Answer:
Ans. 8
An aldohexose has 4 chiral carbons, so in all 2⁴ = 16 aldohexoses are possible. Out of these 8 belong to D-series and 8 to the L-series.

Teacher's Note:
1. Fixing the configuration at C-5 (D-series) leaves three free chiral centres: 2³ = 8.

Question 31. When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form is
OHC–CH₂–CHOH–CHOH–CHOH–CH₂OH (IIT 2012)
Answer:
Ans. 8
In the pyranose structure, the chiral carbons are C-1 (anomeric), C-3, C-4 and C-5 (C-2 is CH₂).
Since configuration at C₅ is fixed for D-series, therefore, there are three other chiral centres and hence there are 2³ = 8 stereoisomers.

Teacher's Note:
1. Ring formation makes C-1 a new chiral centre; do not forget it when counting stereoisomers of the cyclic form.

Assertion–Reason Questions

Question 32. Statement-1: Glucose gives a reddish brown precipitate with Fehling’s solution.
Statement-2: Reaction of glucose with Fehling’s solution gives CuO and gluconic acid. (IIT Paper II, 2007)
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct statement-2. The reaction of glucose with Fehling’s solution gives Cu₂O and gluconic acid.

Teacher's Note:
1. The red-brown precipitate is copper(I) oxide, Cu₂O, not black CuO.

Question 33. Statement-1: Fructose reduces Fehling’s solution and Tollens’ reagent.
Statement-2: Fructose does not contain any aldehyde group.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (b)
Correct explanation. Fructose contains the α-ketol group, i.e., –COCH₂OH which gets converted into the α-aldol group, i.e., –CHOHCHO group under basic condition of the reaction (Lobry de Bruyn van Ekenstein rearrangement) and hence reduces Fehling’s solution and Tollens’ reagent.

Teacher's Note:
1. Statement-2 is true but cannot explain why fructose reduces; the real reason is base-catalysed rearrangement of the α-hydroxy ketone.

Question 34. Statement-1: Keratin is a globular protein.
Statement-2: Enzymes are globular proteins.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (d)
Correct statement-1. Keratin is a fibrous protein. Statement-2 is true.

Teacher's Note:
1. Keratin (hair, nails, wool) is fibrous; enzymes, insulin and albumins are globular.

Question 35. Statement-1: The two strands of DNA are complementary.
Statement-2: Cytosine always pairs with guanine and thymine pairs with adenine.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Because of the specific base pairing (A with T and G with C), the sequence of one strand fixes the sequence of the other; hence the two strands are complementary and Statement-2 explains Statement-1.

Teacher's Note:
1. Complementary does not mean identical: one strand is the base-paired partner of the other.