Let the octasaccharide contain $$r$$ ribose units, $$d$$ 2-deoxyribose units and $$g$$ glucose units.

Because it is an octasaccharide, the total number of monosaccharide residues is
$$r + d + g = 8 \qquad -(1)$$

During hydrolysis all the glycosidic bonds break. In a linear chain of 8 units there are $$8-1 = 7$$ bonds, each bond having been formed by the loss of one water molecule (molar mass $$18\text{ g mol}^{-1}$$).
Hence,

total mass of isolated monosaccharides, $$M = (\text{molar mass of octasaccharide}) + 7 \times 18$$
$$M = 1024 + 126 = 1150\text{ g mol}^{-1} \qquad -(2)$$

The masses contributed by the individual monosaccharides are

$$M = 150r + 134d + 180g \qquad -(3)$$

The question states that 2-deoxyribose accounts for 58.26 % of the total mass of the hydrolysis products, i.e.

$$\frac{134d}{M} = 0.5826 \qquad -(4)$$

Substituting $$M = 1150$$ from (2) into (4):
$$134d = 0.5826 \times 1150 = 669.99 \approx 670$$
$$d = \frac{670}{134} = 5 \qquad -(5)$$

Insert $$d = 5$$ into (1):
$$r + g = 8 - 5 = 3 \qquad -(6)$$

Insert $$d = 5$$ into (3) and use $$M = 1150$$:
$$150r + 134(5) + 180g = 1150$$
$$150r + 670 + 180g = 1150$$
$$150r + 180g = 480$$
Divide by 30:
$$5r + 6g = 16 \qquad -(7)$$

Solve simultaneous equations (6) and (7).
From (6): $$r = 3 - g$$
Substitute in (7):
$$5(3 - g) + 6g = 16$$
$$15 - 5g + 6g = 16$$
$$15 + g = 16$$
$$g = 1$$
Therefore $$r = 3 - 1 = 2$$.

Thus, the number of ribose units present in one molecule of the octasaccharide is 2.

P is glucose. Glucose possesses a free $$\text{-CHO}$$ (aldehydic) group in the open-chain form. Hence it behaves as a reducing sugar and gives Tollens’/Fehling’s/Benedict’s test, producing a silver mirror or red $$Cu_2O$$ precipitate. This description corresponds to observation 2.

Q is cellulose. Cellulose is a straight-chain (linear) homopolysaccharide made up solely of $$\beta$$-D-glucopyranose units joined by $$\beta\,(1\!\rightarrow\!4)$$ glycosidic bonds. “Linear polymer of $$\beta$$-D-glucose” matches observation 5.

R is glycine. Among all naturally occurring $$\alpha$$-amino acids, glycine is the only one whose side chain is just a hydrogen atom. Therefore its $${\rm C_\alpha}$$ is not chiral and the molecule is optically inactive. This is exactly observation 1.

S is starch. Starch (because of its amylose component) forms an intense blue-black complex when treated with iodine solution. Hence it fits observation 3.

Thus the correct matching is
$$P\to2,\;Q\to5,\;R\to1,\;S\to3$$

Option B is therefore the right choice.

To determine the correct dipeptide X, we first identify the amino acids Y and Z obtained on hydrolysis.

The first clue states that amino acid Y produces lactic acid on treatment with aqueous $$\mathrm{HNO_2}$$.

Nitrous acid converts the primary amino group $$\left(-NH_2\right)$$ of an aliphatic amino acid into a hydroxyl group $$\left(-OH\right)$$.

Since the product formed is lactic acid,

$$\mathrm{CH_3-CH(OH)-COOH},$$

the original amino acid must have been

$$\mathrm{CH_3-CH(NH_2)-COOH},$$

which is alanine.

Hence,

• $$Y =$$ Alanine.

The second clue states that amino acid Z forms the given cyclic compound on heating.

When $$\alpha$$-amino acids are heated, two molecules condense to form a six-membered cyclic diamide known as a diketopiperazine.

From the structure of the cyclic product, the carbon atoms between the nitrogen and carbonyl groups are simple $$CH_2$$ units with no alkyl substituents.

This is possible only if the amino acid is glycine,

$$\mathrm{H_2N-CH_2-COOH}.$$

Therefore,

• $$Z =$$ Glycine.

The required dipeptide X is obtained by condensation of alanine and glycine.

Hence,

• $$Y =$$ Alanine
• $$Z =$$ Glycine
• $$X =$$ Alanine-Glycine

Therefore, the correct answer is Alanine-Glycine.

An α-amino acid contains two main acid-base centres:
• the carboxylic group $${-COOH}$$ with $$pK_a \approx 2$$ (behaves as an acid)
• the amino group $${-NH_2}$$ with $$pK_a \approx 9.5$$ (behaves as a base)

General protonation rule: the group is protonated when $$\text{pH} \lt pK_a$$ and de-protonated when $$\text{pH} \gt pK_a$$.

Case 1:

$$\text{pH}=2$$ is lower than the amino group’s $$pK_a$$ but equal to the carboxyl group’s $$pK_a$$.
• Carboxyl group: at (or just below) its $$pK_a$$ it stays protonated as $$-COOH$$.
• Amino group: since $$2 \lt 9.5$$ it is fully protonated to $$-NH_3^+$$.

Therefore structure $$A$$ is $$H_3^+N-CH\bigl(CH(CH_3)_2\bigr)-COOH$$, written shortly as $$H_3^+N-CH-COOH$$.

Case 2:

$$\text{pH}=10$$ is higher than both $$pK_a$$ values.
• Carboxyl group: $$10 \gt 2$$ ⇒ completely de-protonated to $$-COO^-$$.
• Amino group: $$10 \gt 9.5$$ ⇒ largely de-protonated to the neutral $$-NH_2$$ form.

Hence structure $$B$$ is $$H_2N-CH\bigl(CH(CH_3)_2\bigr)-COO^-$$, written briefly as $$H_2N-CH-COO^-$$.

Comparing with the given options:

Option A A = $$H_3^+N-CH-COOH$$, B = $$H_2N-CH-COO^-$$

This matches the derived forms.

So, the correct choice is Option A.

Let the major product $$P$$ obtained in the earlier sequence be stearic acid, $$C_{17}H_{35}COOH$$.
With an acid catalyst, glycerol esterifies completely with excess stearic acid to give the tri-ester (a triglyceride) $$Q$$, called tristearin.

Step-1 : Formation of $$Q$$
Glycerol possesses three -OH groups, so three molecules of stearic acid react with one molecule of glycerol and three molecules of water are eliminated:

$$C_3H_8O_3 + 3\,C_{17}H_{35}COOH \;\xrightarrow{\text{acid}}\; C_3H_5\big(OOC\,C_{17}H_{35}\big)_3 + 3\,H_2O$$

Molar mass of glycerol $$= 3\times12 + 8\times1 + 3\times16 = 92$$ g mol$$^{-1}$$
Molar mass of stearic acid $$= 18\times12 + 36\times1 + 2\times16 = 284$$ g mol$$^{-1}$$

Molar mass of $$Q$$ (tristearin):
$$M_Q = 92 + 3(284) - 3(18) = 92 + 852 - 54 = 890\;\text{g mol}^{-1}$$

Step-2 : Alkaline hydrolysis (saponification) of $$Q$$ with excess $$NaOH$$

$$C_3H_5\big(OOC\,C_{17}H_{35}\big)_3 + 3\,NaOH \;\rightarrow\; 3\,C_{17}H_{35}COO^-Na^+ + C_3H_8O_3$$

Thus 1 mol of $$Q$$ gives 3 mol of sodium stearate.

Molar mass of sodium stearate:
Carboxylate anion $$C_{17}H_{35}COO^-$$ has mass $$284 - 1 = 283$$ g.
Adding Na (23 g) gives $$M_{Na\text{-stearate}} = 283 + 23 = 306\;\text{g mol}^{-1}$$

Step-3 : Conversion of sodium stearate to calcium stearate (Ca-soap $$R$$) using excess $$CaCl_2$$

$$2\,C_{17}H_{35}COO^-Na^+ + CaCl_2 \;\rightarrow\; (C_{17}H_{35}COO^-)_2Ca^{2+} + 2\,NaCl$$

Stoichiometry: 2 mol Na-stearate → 1 mol Ca-stearate.
From Step-2 we have 3 mol Na-stearate, therefore

Number of moles of $$R$$ obtained
$$= \frac{3}{2} = 1.5\;\text{mol}$$

Molar mass of calcium stearate $$R$$:
$$M_R = 40 + 2(283) = 40 + 566 = 606\;\text{g mol}^{-1}$$

Mass of $$R$$ from 1 mol of $$Q$$:
$$m_R = 1.5 \times 606 = 909\;\text{g}$$

Therefore, starting with one mole of $$Q$$, the amount of calcium soap $$R$$ produced is 909 g.

The given open-chain structure is that of an aldo-hexose, specifically $$D$$-glucose:
$$CHO{-}CH(OH){-}CH(OH){-}CH(OH){-}CH(OH){-}CH_2OH$$

In $$D$$-glucose, the aldehydic carbon is $$C_1$$, the primary alcohol is at $$C_6$$ and the internal carbons $$C_2$$ to $$C_5$$ each bear an $$OH$$ group.

We now examine every statement.

Statement A: “can be oxidized to a dicarboxylic acid with $$Br_2$$ water”
Bromine water is a mild and selective oxidising agent. It converts the $$-CHO$$ group of an aldose to $$-COOH$$ giving an aldonic acid, but it does not oxidise the terminal primary alcohol $$-CH_2OH$$. Hence glucose is changed to gluconic acid (only one $$-COOH$$), not to the dicarboxylic glucaric (saccharic) acid. Therefore Statement A is incorrect.

Statement B: “will coexist in equilibrium with 2 other cyclic structure”
In aqueous solution the open-chain form of glucose undergoes intramolecular hemi-acetal formation between $$C_1$$ and $$C_5$$ producing two anomers: $$\alpha$$-D-glucopyranose and $$\beta$$-D-glucopyranose. Thus three forms (open + 2 cyclic) are simultaneously present. So Statement B is correct.

Statement C: “despite the presence of -CHO does not give Schiff’s test”
Schiff’s reagent reacts only with a free aldehyde group. In solution the aldehyde of glucose is locked in the hemi-acetal ring; ring opening is slow under the very mild conditions of Schiff’s test, hence glucose gives a negative result. Statement C is correct.

Statement D: “has 4 asymmetric carbon atom”
The chiral centres in the open form are $$C_2, C_3, C_4,$$ and $$C_5$$  ⇒ four asymmetric carbons. Statement D is correct.

Only Statement A is wrong.

Option A which is: can be oxidized to a dicarboxylic acid with $$Br_2$$ water

• (A) Enzymes are biocatalysts: Correct. They are biological catalysts that accelerate chemical reactions in living organisms.
• (B) Enzymes are non-specific...: Incorrect. Enzymes are highly specific, typically catalyzing only one particular reaction or acting on a specific substrate.
• (C) Most Enzymes are globular proteins: Correct. Chemically, the vast majority of enzymes are globular proteins.
• (D) Enzyme - oxidase catalyses the hydrolysis...: Incorrect. The hydrolysis of maltose into glucose is catalyzed by the enzyme maltase, not oxidase.

Conclusion:

Statements (B) and (D) are incorrect.

Answer: Option B — (B) and (D)

The ninhydrin test is a test for amino acids and proteins. Ninhydrin reacts with the free amino groups ($$-NH_2$$) present in amino acids and proteins to give a purple/blue color (Ruhemann's purple).

Among the options:

- Starch: A polysaccharide — no free amino groups. Negative test.

- Egg albumin: A protein — contains amino acids with free amino groups. Positive test.

- Polyvinyl chloride: A synthetic polymer — no amino groups. Negative test.

- Cellulose: A polysaccharide — no amino groups. Negative test.

The correct answer is Option (2): Egg albumin.

We need to find the total number of carbon atoms in tyrosine.

Tyrosine is an amino acid with the molecular formula C9H11NO3.

Its structure is 4-hydroxyphenyl alanine: it has a benzene ring (6 carbons) with an OH group, connected via a -CH2- group to the amino acid backbone -CH(NH2)-COOH.

Counting carbon atoms:

- Benzene ring: 6 carbons

- CH2- group: 1 carbon

- alpha-carbon (CH): 1 carbon

- Carboxyl group (COOH): 1 carbon

Total = 6 + 1 + 1 + 1 = 9 carbon atoms.

The answer is 9

Essential amino acids are those that cannot be synthesized by the human body and must be obtained from diet.

The essential amino acids (mnemonic: PVT TIM HALL) are: Phenylalanine, Valine, Threonine, Tryptophan, Isoleucine, Methionine, Histidine, Arginine, Leucine, Lysine.

From the given list:

- Arginine: Essential ✓ (semi-essential/conditionally essential, but generally counted)

- Phenylalanine: Essential ✓

- Aspartic acid: Non-essential ✗

- Cysteine: Non-essential (conditionally essential) ✗

- Histidine: Essential ✓

- Valine: Essential ✓

- Proline: Non-essential ✗

Total essential amino acids = 4 (Arginine, Phenylalanine, Histidine, Valine).

The answer is $$\boxed{4}$$.

Fructose is a ketose sugar with molecular formula $$\mathrm{C_6H_{12}O_6}$$ containing a ketone group at the $$\mathrm{C2}$$ position.

Formation of product $$\mathrm{A}$$:

Fructose reacts with hydrogen cyanide $$\mathrm{(HCN)}$$ through nucleophilic addition at the carbonyl carbon to form a cyanohydrin.

Acidic hydrolysis using $$\mathrm{H_3O^+}$$ converts the cyano group $$\mathrm{(-C \equiv N)}$$ into a carboxylic acid group $$\mathrm{(-COOH)}$$.

This increases the carbon chain length from $$\mathrm{6}$$ to $$\mathrm{7}$$ carbons, producing a polyhydroxy carboxylic acid with formula:

$$\mathrm{C_7H_{14}O_8}$$

Formation of product $$\mathrm{B}$$:

Sodium borohydride $$\mathrm{(NaBH_4)}$$ reduces the ketone group of fructose into a secondary alcohol, forming a sugar alcohol with formula:

$$\mathrm{C_6H_{14}O_6}$$

Further reduction using $$\mathrm{HI/P}$$ removes all oxygen-containing groups by replacing them with hydrogen atoms.

The six-carbon chain remains intact, giving:

$$\mathrm{n\text{-}hexane,\ C_6H_{14}}$$

Correct Option: $$\mathrm{A = C_7H_{14}O_8,\ B = C_6H_{14}}$$

To determine the correct class of drug for chlordiazepoxide, we must examine its therapeutic use and compare it with the given drug categories.

Chlordiazepoxide is a well-known synthetic compound belonging to the benzodiazepine class of drugs. It acts on the central nervous system (CNS) to produce a calming effect and is primarily prescribed for the treatment of anxiety disorders, tension, and alcohol withdrawal symptoms.

Now, let us examine each option:

(A) Analgesics

Analgesics are drugs that relieve pain without causing loss of consciousness. Examples include aspirin, paracetamol, and morphine. Chlordiazepoxide does not possess pain-relieving properties, so it does not belong to this class.

(B) Tranquillizers

Tranquillizers are drugs used to reduce anxiety, stress, irritability, and emotional tension by producing a calming effect on the central nervous system. Since chlordiazepoxide is specifically used for the treatment of anxiety and related conditions, it belongs to this category.

(C) Antacids

Antacids neutralize excess stomach acid and are used to treat conditions such as heartburn and indigestion. Common examples include magnesium hydroxide and sodium bicarbonate. Chlordiazepoxide is unrelated to gastric acid neutralization.

(D) Antibiotics

Antibiotics are substances that inhibit the growth of or destroy microorganisms and are used to treat bacterial infections. Examples include penicillin and tetracycline. Chlordiazepoxide has no antimicrobial activity.

Therefore, chlordiazepoxide is classified as a tranquillizer.

Hence, the correct answer is (B) Tranquillizers.

We begin by recalling a very well-known fact from elementary bio-inorganic chemistry: the main inorganic constituent of tooth enamel is hydroxyapatite. Its accepted chemical formula is written as

$$Ca_{10}(PO_4)_6(OH)_2.$$

Because questions in JEE often rewrite such formulas in a factored or “mixed salt” form, we now try to re-express $$Ca_{10}(PO_4)_6(OH)_2$$ as a product of two simpler familiar units so that it will resemble one of the options given.

First, we notice that the ionic compound $$Ca_3(PO_4)_2$$, known as tricalcium phosphate, is present inside the overall formula. To see how many such units are hidden in hydroxyapatite, we compare the numbers of each atom.

We write:

Let $$x$$ be the number of $$Ca_3(PO_4)_2$$ units and let $$y$$ be the number of $$Ca(OH)_2$$ units that together produce $$Ca_{10}(PO_4)_6(OH)_2$$.

So we want

$$x\bigl[Ca_3(PO_4)_2\bigr] + y\bigl[Ca(OH)_2\bigr] \;=\; Ca_{10}(PO_4)_6(OH)_2.$$

We now equate the quantity of each type of atom on both sides.

Calcium atoms:

$$3x + y = 10 \quad\text{(1)}$$

Phosphate groups:

Because each $$Ca_3(PO_4)_2$$ carries 2 phosphate groups $$PO_4^{3-}$$, the total number of phosphate groups equals $$2x$$, and we require

$$2x = 6 \quad\text{(2)}$$

Solving equation (2) first, we divide both sides by 2 and obtain

$$x = 3.$$

Substituting $$x = 3$$ into equation (1) gives

$$3(3) + y = 10 \;\;\Longrightarrow\;\; 9 + y = 10 \;\;\Longrightarrow\;\; y = 1.$$

Hence we have exactly three units of $$Ca_3(PO_4)_2$$ and one unit of $$Ca(OH)_2$$ in a single formula unit of hydroxyapatite. Writing that product out explicitly, we get

$$\bigl[3Ca_3(PO_4)_2 \cdot Ca(OH)_2\bigr],$$

which is an alternative but completely equivalent way of expressing $$Ca_{10}(PO_4)_6(OH)_2.$$

We compare this result with the options provided:

A. $$[Ca_3(PO_4)_2 \cdot CaF_2]$$ — contains fluoride, not hydroxide; incorrect.

B. $$[3Ca_3(PO_4)_2 \cdot Ca(OH)_2]$$ — exactly matches the expression we derived; correct.

C. $$[3Ca_3(PO_4)_2 \cdot CaF_2]$$ — hydroxide replaced by fluoride; incorrect.

D. $$[3Ca(OH)_2 \cdot CaF_2]$$ — phosphate totally missing; incorrect.

So, the conversion of hydroxyapatite in the presence of $$F^-$$ ions indeed starts from $$[3Ca_3(PO_4)_2 \cdot Ca(OH)_2]$$.

Hence, the correct answer is Option B.

Penicillin is Bactericidal type of antibiotic. The given structure is the correct structure of penicillin (NCERT - Chemistry in Everyday Life).

Given compound is the repeating unit of Novolac. Thus, the right option is A.

First, we recall the basic definitions of the four classes of drugs mentioned in List-II:

$$\text{(i) Antibiotic:}$$ A chemical substance produced by microorganisms (or prepared synthetically) that can inhibit the growth of or destroy other microorganisms.

$$\text{(ii) Tranquilizer:}$$ A compound that acts on the central nervous system and reduces anxiety or tension, producing a calming effect.

$$\text{(iii) Antiseptic:}$$ A substance applied to living tissues to prevent the growth of disease-causing microorganisms.

$$\text{(iv) Synthetic antihistamine:}$$ A man-made drug that blocks the action of histamine, thereby reducing allergic symptoms.

Now we examine each drug in List-I one by one and connect it with the correct class.

We have Furacin, whose chemical name is nitrofurazone. It is routinely applied on wounds, burns and skin infections to stop microbial growth. The definition that matches this property is that of an antiseptic. So we write

$$\text{Furacin}\;(a)\;\longrightarrow\;\text{Antiseptic}\;(iii)$$

Next, we consider Arsphenamine. This drug, historically called Salvarsan, was the first effective treatment for syphilis and is obtained from microbial action or prepared synthetically to kill the causative bacterium. By our definition, such a compound is an antibiotic. Hence,

$$\text{Arsphenamine}\;(b)\;\longrightarrow\;\text{Antibiotic}\;(i)$$

Now, Dimetone remains. Dimetone is the trade name for diphenhydramine-like compounds that counteract histamine and alleviate allergic symptoms such as cold and sneezing. Compounds of this type are man-made and therefore fall under the class of synthetic antihistamines. Consequently,

$$\text{Dimetone}\;(c)\;\longrightarrow\;\text{Synthetic antihistamine}\;(iv)$$

Finally, Valium is the well-known trade name for diazepam, a benzodiazepine that soothes the central nervous system and is widely prescribed to relieve anxiety and induce calmness. Such drugs are categorized as tranquilizers. Thus,

$$\text{Valium}\;(d)\;\longrightarrow\;\text{Tranquilizer}\;(ii)$$

Collecting all four correct correspondences in the same order as the lists, we get

$$ \begin{aligned} (a)&\to(iii),\\ (b)&\to(i),\\ (c)&\to(iv),\\ (d)&\to(ii). \end{aligned} $$

We now compare this complete mapping with the given options. We observe that Option D lists exactly the pairings $$\bigl[(a)-(iii),\,(b)-(i),\,(c)-(iv),\,(d)-(ii)\bigr],$$ which matches our analysis.

Hence, the correct answer is Option D.

Let us match each compound with its use.

(a) Sucralose is a chlorinated derivative of sucrose and is widely used as an artificial sweetener. It is about 600 times sweeter than sugar but contributes no calories. So (a) matches with (ii).

(b) Glyceryl ester of stearic acid is glyceryl monostearate, which is used in making soaps and synthetic detergents. It acts as a surfactant. So (b) matches with (i).

(c) Sodium benzoate ($$\text{C}_6\text{H}_5\text{COONa}$$) is a well-known food preservative. It is commonly added to acidic foods like pickles, fruit juices, and carbonated drinks to prevent microbial growth. So (c) matches with (iv).

(d) Bithionol is a phenolic compound that is used as an antiseptic, particularly in medicated soaps and other personal care products. So (d) matches with (iii).

The correct matching is (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii), which is Option B.

We need to match each monomer with its corresponding polymer.

Caprolactam is the monomer of Nylon 6. Nylon 6 is formed by the ring-opening polymerisation of caprolactam. So $$(a) \to (iii)$$.

2-Chloro-1,3-butadiene (also called chloroprene) is the monomer of Neoprene. Neoprene is a synthetic rubber formed by the polymerisation of chloroprene. So $$(b) \to (iv)$$.

Isoprene (2-methyl-1,3-butadiene) is the monomer unit of natural rubber. Natural rubber is cis-1,4-polyisoprene. So $$(c) \to (i)$$.

Acrylonitrile is one of the monomers used in making Buna-N (also called nitrile rubber). Buna-N is a copolymer of 1,3-butadiene and acrylonitrile. So $$(d) \to (ii)$$.

The correct matching is: $$a \to iii$$, $$b \to iv$$, $$c \to i$$, $$d \to ii$$.

Hence, the correct answer is Option A.

Orlon is a brand name for the synthetic fiber made from polyacrylonitrile (PAN). Polyacrylonitrile is a vinyl polymer formed by the addition polymerization of acrylonitrile (CH₂=CH-CN) monomers. The repeating unit is -(CH₂-CH(CN))-.

Orlon/Acrylic fibers are widely used in textiles as a wool substitute because they are lightweight, warm, and soft. They are distinct from polyesters (like Dacron/Terylene), polyamides (like Nylon), or natural polymers like cellulose.

Therefore, Orlon fibers are made up of polyacrylonitrile, which is Option 1.

Formaldehyde-based resins are widely used in the wood and laminate industry. Let us consider each option.

cis-Polyisoprene is natural rubber, which is an elastomer and is not used in wood laminates.

Melamine formaldehyde resin is a thermosetting polymer used primarily for making unbreakable crockery and decorative laminate surface coatings.

Urea formaldehyde resin is a thermosetting polymer formed by the condensation polymerisation of urea and formaldehyde. It is extensively used as an adhesive and binding agent in the manufacture of plywood, particle boards, and wood laminates due to its strong bonding properties and cost-effectiveness.

Phenol formaldehyde resin (Bakelite) is used in electrical fittings, handles, and similar moulded products, not primarily for wood laminates.

Therefore, urea formaldehyde resin is the polymer used in the manufacture of wood laminates.

The correct answer is Histamine.

According to NCERT Class 12 Chemistry, Chapter 16: Chemistry in Everyday Life:

“Histamine stimulates the secretion of pepsin and hydrochloric acid in the stomach.”

Hence, histamine is responsible for the production of HCl in the stomach leading to irritation and pain.

Therefore, the correct option is the structure of Histamine, which is option B.

We recall that a polyamide is obtained by repeated formation of the amide linkage, which has the characteristic $$-\,\mathrm{CO\,NH}-$$ group. Biodegradability is favoured when the backbone contains short peptide-like segments that hydrolyse easily; this usually happens when at least one monomer is an α-amino acid, that is, a molecule possessing both $$\mathrm{-NH_2}$$ and $$\mathrm{-COOH}$$ on the same carbon. Hence we must look for a pair of monomers that together furnish many amide bonds and at least one naturally peptide-type centre.

Let us inspect each option one by one.

Option A (Glycine and isoprene)
Glycine has the desired $$\mathrm{-NH_2}$$ and $$\mathrm{-COOH}$$ groups, but isoprene $$\big(\mathrm{CH_2=C(CH_3)CH=CH_2}\big)$$ contains only carbon-carbon double bonds and no carboxyl or amino group. Therefore no amide bond can be produced between the two; polymerisation would proceed, if at all, by a radical mechanism on the isoprene alone, not by condensation. So a polyamide cannot form.

Option B (Hexamethylene diamine and adipic acid)
These two compounds do indeed undergo the condensation reaction $$\mathrm{H_2N\,(CH_2)_6\,NH_2 + HOOC\,(CH_2)_4\,COOH} \;\longrightarrow\; \Big[-\mathrm{NH\,(CH_2)_6\,NHCO\,(CH_2)_4\,CO}-\Big]_n$$ which is well known as Nylon-6,6. However, Nylon-6,6 contains long aliphatic segments $$\mathrm{(CH_2)_6}$$ and $$\mathrm{(CH_2)_4}$$ on either side of each amide linkage; those hydrophobic chains hinder enzymatic attack, so the material is not classified as biodegradable.

Option C (Glycine and aminocaproic acid)
Write the structures: $$\text{Glycine: } \mathrm{H_2NCH_2COOH}$$ Aminocaproic acid: $$\mathrm{H_2N\,(CH_2)_5\,COOH}$$ Both molecules possess the pair $$\mathrm{-NH_2}$$ and $$\mathrm{-COOH}$$, so each can act simultaneously as an acid and as a base in the condensation. When heated, they combine with elimination of water: $$\mathrm{H_2NCH_2COOH + H_2N\,(CH_2)_5\,COOH} \;\longrightarrow\; \Big[-\mathrm{NHCH_2CO\,NH\,(CH_2)_5\,CO}-\Big]_n + n\,\mathrm{H_2O}$$ The repeat unit contains two successive peptide bonds resembling those in natural proteins: $$\Big[-\,\mathrm{NH\,CH_2\,CO\,NH\,(CH_2)_5\,CO}-\Big]_n$$ Because the amide linkages lie next to relatively short segments (one methylene on one side and five on the other), hydrolytic and enzymatic cleavage occurs readily, making the polymer biodegradable. Indeed, this particular copolymer is historically named “Nylon-2-Nylon-6” where the numerals refer to the carbon counts in the two amino acids. Therefore, this combination satisfies the requirement.

Option D (Styrene and caproic acid)
Styrene $$\big(\mathrm{C_6H_5CH=CH_2}\big)$$ lacks both $$\mathrm{-NH_2}$$ and $$\mathrm{-COOH}$$ groups, so no amide bond can form with caproic acid $$\big(\mathrm{HOOC\,(CH_2)_5\,CH_3}\big)$$. Polymerising styrene gives polystyrene, which is not a polyamide, and mixing it with a carboxylic acid does not change that fact.

Only Option C deploys two amino acids whose condensation produces a peptide-like, hence biodegradable, polyamide chain. The other pairs either cannot form an amide polymer at all or yield a well-known non-biodegradable nylon.

Hence, the correct answer is Option C.

Bakelite is a thermosetting polymer prepared from two monomers: formaldehyde (HCHO) and phenol. The initial condensation of phenol with formaldehyde gives a linear, fusible polymer called Novolac. Novolac is then cross-linked with more formaldehyde under acidic or basic conditions to give the hard, infusible, three-dimensional network polymer known as Bakelite.

PHBV is a biodegradable polymer (polyhydroxybutyrate-co-hydroxyvalerate), Buna-S is a copolymer of butadiene and styrene (a synthetic rubber), and Dacron is a polyester used as synthetic fibre — none of these are related to Bakelite.

Therefore, Bakelite is a cross-linked polymer of formaldehyde and Novolac.

Fructose has the molecular formula $$\text{C}_6\text{H}_{12}\text{O}_6$$, which means it is a six-carbon (hexose) sugar. Its open-chain structure contains a keto group ($$\text{C}=\text{O}$$) at the second carbon atom, rather than an aldehyde group at the first carbon.

Sugars are classified based on two criteria: the type of carbonyl group (aldose if aldehyde, ketose if ketone) and the number of carbon atoms (triose for 3, tetrose for 4, pentose for 5, hexose for 6, heptose for 7). Since fructose has a ketone group and six carbons, it is classified as a ketohexose.

An aldohexose would be a six-carbon sugar with an aldehyde group — glucose is an example. A pyranose refers to the six-membered ring form of a sugar, which is a structural form rather than a classification of the sugar itself. A heptose is a seven-carbon sugar, which fructose is not.

Therefore, fructose is an example of a ketohexose, and the correct answer is Option B.

Buna-S is a synthetic rubber, which is a copolymer of 1,3-butadiene and styrene. The name "Buna-S" is derived as follows:

"Bu" stands for butadiene, "na" stands for sodium (Na) which is used as a catalyst in the polymerisation, and "S" stands for Styrene, the comonomer used along with butadiene.

The correct answer is Option (4): Styrene.

CERT-Based Explanations

Norethindrone (Antifertility Drug):

Norethindrone is a synthetic progesterone derivative.

Birth control pills generally contain mixtures of synthetic estrogen and progesterone derivatives.

$$\mathrm{Norethindrone}$$ is one of the commonly used antifertility drugs.

Salvarsan (Antibiotic):

Salvarsan, also known as arsphenamine, was developed by Paul Ehrlich as the first effective treatment for syphilis.

It was an important milestone in antibacterial therapy.

$$\mathrm{Salvarsan}$$ is therefore classified as an antibiotic.

Meprobamate (Tranquilizer):

Tranquilizers are neurologically active drugs used in the treatment of stress and mental disorders.

$$\mathrm{Meprobamate}$$ is a mild tranquilizer used to relieve tension and anxiety.

Alitame (Artificial Sweetener):

$$\mathrm{Alitame}$$ is a high-potency artificial sweetener.

It is nearly $$2000$$ times sweeter than cane sugar and is more stable than aspartame.

Matching:

$$\mathrm{(a)\ Antifertility\ Drug \rightarrow (iii)\ Norethindrone}$$

$$\mathrm{(b)\ Antibiotic \rightarrow (iv)\ Salvarsan}$$

$$\mathrm{(c)\ Tranquilizer \rightarrow (i)\ Meprobamate}$$

$$\mathrm{(d)\ Artificial\ Sweetener \rightarrow (ii)\ Alitame}$$

We need to match each disaccharide with its constituent monosaccharides.

Sucrose is composed of $$\alpha$$-D-Glucose and $$\beta$$-D-Fructose linked by a glycosidic bond between C-1 of glucose and C-2 of fructose. So (a) matches with (ii).

Lactose is composed of $$\beta$$-D-Galactose and $$\beta$$-D-Glucose (though the glucose unit can be in either anomeric form, the galactose unit is in the $$\beta$$ form). These are linked by a $$\beta$$-1,4-glycosidic bond. So (b) matches with (i).

Maltose is composed of two $$\alpha$$-D-Glucose units linked by an $$\alpha$$-1,4-glycosidic bond. So (c) matches with (iii).

Therefore the correct matching is: (a)$$\rightarrow$$(ii), (b)$$\rightarrow$$(i), (c)$$\rightarrow$$(iii).

This is the structure of ninhydrin, so option D is the right option.

Antihistamines are drugs that diminish or abolish the main actions of histamine released in the body. Histamine is released during allergic reactions and also stimulates gastric acid secretion.

Antihistamines serve two primary functions. First, they act as antiallergic agents by blocking $$H_1$$ receptors, which helps relieve symptoms such as runny nose, sneezing, itching, and swelling caused by allergic reactions. Second, certain antihistamines (specifically $$H_2$$ receptor blockers like cimetidine and ranitidine) act as antacids by reducing the secretion of hydrochloric acid in the stomach, thereby treating conditions like peptic ulcers.

Option (1) is incorrect because analgesic (pain relief) is not a function of antihistamines. Option (3) is incorrect for the same reason. Option (4) is incorrect because antidepressant activity is not a primary function of antihistamines.

Hence, the correct answer is option (2): Antacid and Antiallergic.

The term “polyester” is reserved for polymers whose backbone contains the characteristic ester linkage $$\left( -\,\mathrm{COO}\,- \right)$$ repeated again and again. So, to decide whether a given polymer is a polyester, we only have to inspect the type of linkage present in its main chain.

We have, first of all, the definition of an ester linkage:

$$\text{Alcohol + Carboxylic acid} \xrightarrow{\;\text{condensation}\;} \text{Ester} + \mathrm{H_2O}$$

and the corresponding polymerisation (step-growth) reaction produces an entire chain of $$\left( -\,\mathrm{COO}\,- \right)$$ units.

Now let us examine the four polymers one by one.

1. Glyptal: It is obtained by condensation of glycerol $$\left(\mathrm{HO{-}CH_2{-}CH(OH){-}CH_2OH}\right)$$ with phthalic anhydride $$\left(\mathrm{C_6H_4(CO)_2O}\right)$$. Every time an -OH of glycerol reacts with a -COOH of phthalic acid, an ester linkage $$\left( -\,\mathrm{COO}\,- \right)$$ is produced. Hence the entire backbone is full of $$-\,\mathrm{COO}\,-$$ groups. Therefore Glyptal is definitely a polyester.

2. PHBV (poly(3-hydroxybutyrate-co-3-hydroxyvalerate)): The individual monomers $$\mathrm{HO{-}CH(CH_3){-}CH_2{-}COOH}$$ and $$\mathrm{HO{-}CH(C_2H_5){-}CH_2{-}COOH}$$ combine through their -OH and -COOH ends to give the repeating unit

$$[-\mathrm{O{-}CH(R){-}CH_2{-}CO}-]_n$$

where $$R = \mathrm{CH_3}$$ or $$\mathrm{C_2H_5}$$. Each repeat contains the $$-\,\mathrm{COO}\,-$$ linkage, so PHBV is a biodegradable polyester.

3. Novolac: This polymer results from the condensation of phenol with formaldehyde under acidic conditions. The linkage formed is a methylene bridge $$\left( -\,\mathrm{CH_2}\,- \right)$$ connecting two aromatic rings:

$$\mathrm{Ph{-}OH + HCHO \;\;\longrightarrow\;\; Ph{-}CH_2{-}Ph + H_2O}$$

Repeating this gives

$$[-\,\mathrm{Ph{-}CH_2{-}}\,-]_n$$

Throughout the backbone, there is no $$-\,\mathrm{COO}\,-$$ group at all. The network is made of $$-\,\mathrm{CH_2}\,-$$ bridges, so Novolac is a phenol-formaldehyde resin, not a polyester.

4. Dacron (polyethylene terephthalate, PET): Formed by condensation of ethylene glycol $$\mathrm{HO{-}CH_2{-}CH_2OH}$$ with terephthalic acid $$\mathrm{HOOC{-}C_6H_4{-}COOH}$$, the resulting repeating unit is

$$[-\,\mathrm{O{-}CH_2{-}CH_2{-}O{-}CO{-}C_6H_4{-}CO}\,-]_n$$

in which the ester linkage $$-\,\mathrm{COO}\,-$$ clearly appears. Therefore Dacron is a classic polyester.

Summarising, Glyptal, PHBV and Dacron all possess ester linkages in their main chains, while Novolac does not. Hence the only polymer in the list that is not a polyester is Novolac.

Hence, the correct answer is Option C.

First, let us recall that a monosaccharide is a simple sugar having the general formula $$\mathrm{C_nH_{2n}O_n}$$ and contains a free, easily oxidisable $$-CHO$$ or $$-CO-$$ group. A disaccharide, on the other hand, consists of two monosaccharide units joined through a glycosidic bond; this linkage generally makes the reducing group less reactive or, if both anomeric centres are involved, completely non-reducing.

To differentiate between these two classes, chemists have designed specific qualitative tests. We need a test whose underlying chemistry is sensitive to the rate at which the sugar can reduce cupric ions or some other oxidant. A reaction that occurs rapidly with monosaccharides but either proceeds slowly or not at all with disaccharides will serve this purpose.

We examine the given options one by one:

We have Seliwanoff’s test. This is based on the dehydration of keto-sugars by concentrated $$H_{2}SO_{4}$$ to form hydroxymethylfurfural, which then condenses with resorcinol to give a cherry-red colour. The test is therefore specific for distinguishing aldo- and keto-hexoses, not mono- versus disaccharides. Hence Seliwanoff’s test is not suitable for our requirement.

Now consider the iodine test. Iodine forms a deep blue complex with polysaccharides such as starch; with simple sugars no characteristic colour is produced because the short chains cannot accommodate iodine within a helical cavity. Therefore the iodine test differentiates polysaccharides from mono- or disaccharides, not monosaccharides from disaccharides. So the iodine test is also unsuitable.

Next we come to Barfoed’s test. First, we state the principle. Barfoed’s reagent is an acidic solution of cupric acetate, $$Cu(CH_{3}COO)2$$, maintained at a pH of about $$4.5$$ with acetic acid. Under these mildly acidic conditions, the following reduction takes place:

$$2Cu^{2+ + R-CHO + 2H2O \longrightarrow Cu2O \downarrow + R-COOH + 4H^+}$$

Because the medium is acidic, only those sugars that are strong reducing agents can reduce $$Cu^{2+}$$ quickly enough to precipitate cuprous oxide $$Cu_{2}O$$ within the usual experimental time (about $$2$$-$$3$$ minutes of boiling). Monosaccharides possess a free carbonyl group and reduce the reagent rapidly, producing a brick-red precipitate. Disaccharides, being less powerful reducers in acid medium, either fail to give a precipitate or do so only after prolonged heating. This distinct difference allows us to distinguish between the two.

Finally, Tollen’s test employs an ammoniacal silver nitrate solution. It detects compounds that can reduce $$Ag^+$$ to metallic silver, giving the famous silver mirror. Many monosaccharides and some reducing disaccharides such as maltose and lactose answer this test. Therefore Tollen’s reagent cannot reliably tell the two classes apart.

From the above discussion, only Barfoed’s test fulfils the criterion of differentiating monosaccharides from disaccharides on the basis of the rate of reduction in an acidic medium.

Hence, the correct answer is Option C.

We evaluate each statement about polymers to find the correct one.

Option 2 states that Buna-S is a linear thermosetting polymer. This is incorrect because Buna-S (styrene-butadiene rubber) is an elastomer with thermoplastic character, not a thermosetting polymer.

Option 3 states that Buna-N is a natural polymer. This is incorrect because Buna-N (acrylonitrile-butadiene rubber) is a synthetic copolymer of 1,3-butadiene and acrylonitrile.

Option 4 states that Neoprene is an addition copolymer used in plastic bucket manufacturing. This is incorrect because Neoprene is an addition homopolymer made from a single monomer, chloroprene (2-chloro-1,3-butadiene), and it is not used for plastic buckets.

Option 1 states that the synthesis of Buna-S needs nascent oxygen. Buna-S is prepared by the emulsion copolymerisation of 1,3-butadiene and styrene in the presence of a peroxide initiator (such as benzoyl peroxide). The peroxide generates nascent oxygen and free radicals to initiate the polymerisation. Hence this statement is correct.

The correct answer is Option 1.

In drug-enzyme interactions, a competitive inhibitor binds to the enzyme's active site and competes with the substrate for that site — this makes option (4) a correct statement. A non-competitive inhibitor binds to the allosteric site (a site other than the active site), which makes option (1) correct. When an allosteric inhibitor binds to the allosteric site, it induces a conformational change in the enzyme that alters the shape of the active site, preventing the substrate from binding effectively — this makes option (2) correct.

Option (3) states that an allosteric inhibitor competes with the enzyme's active site. This is wrong because an allosteric inhibitor, by definition, binds at a site different from the active site. It does not compete with the substrate for the active site; instead, it inhibits the enzyme by changing the conformation of the active site from a remote location. Competition with the active site is the hallmark of competitive inhibition, not allosteric inhibition.

The wrong statement is therefore option (3).

We have to identify the disaccharide that, on hydrolysis, produces $$\text{D-Galactose}$$ and $$\text{D-Glucose}$$.

First, let us recall the basic fact that hydrolysis of a disaccharide splits the glycosidic bond and gives its two monosaccharide constituents. Symbolically we can write the general reaction as $$\text{Disaccharide} + \text{H}_2\text{O} \;\xrightarrow{\text{acid / enzyme}}\; \text{Monosaccharide}_1 + \text{Monosaccharide}_2.$$

Now we list the well-known disaccharides and their hydrolysis products:

1. $$\text{Sucrose} \;(\text{C}_{12}\text{H}_{22}\text{O}_{11})$$ $$\longrightarrow$$ $$\text{D-Glucose} + \text{D-Fructose}$$

2. $$\text{Maltose} \;(\text{C}_{12}\text{H}_{22}\text{O}_{11})$$ $$\longrightarrow$$ $$\text{D-Glucose} + \text{D-Glucose}$$

3. $$\text{Lactose} \;(\text{C}_{12}\text{H}_{22}\text{O}_{11})$$ $$\longrightarrow$$ $$\text{D-Galactose} + \text{D-Glucose}$$

4. $$\text{Amylose}$$ is not a disaccharide at all; it is a linear polymer of $$\text{D-Glucose}$$ units, so its hydrolysis gives many $$\text{D-Glucose}$$ molecules and no $$\text{D-Galactose}$$.

Comparing these facts with the requirement in the question—formation of $$\text{D-Galactose}$$ and $$\text{D-Glucose}$$—we see that only the third entry fits. Thus, the compound A must be $$\text{Lactose}$$.

The option list shows that $$\text{Lactose}$$ corresponds to Option 4.

Hence, the correct answer is Option 4.

Vitamin K is essential for the synthesis of prothrombin and several other clotting factors in the liver. It acts as a cofactor for the enzyme that carboxylates glutamic acid residues on these clotting proteins, which is necessary for them to bind calcium ions and function in the coagulation cascade.

When vitamin K is deficient, the body cannot produce adequate amounts of functional clotting factors, so the blood takes longer to clot. This means the blood clotting time increases.

Considering the other options: increased fragility of RBCs is associated with vitamin E deficiency, cheilosis (cracking at the corners of the mouth) is linked to vitamin B$$_2$$ (riboflavin) deficiency, and a decrease in blood clotting time is the opposite of what vitamin K deficiency causes.

The correct answer is option (1): increase in blood clotting time.

We begin by recalling that a disaccharide is a carbohydrate that, on hydrolysis, yields two monosaccharide units. Sucrose, on hydrolysis, gives one molecule of glucose and one molecule of fructose, so sucrose is certainly a disaccharide.

Next, whether a sugar is reducing depends on the presence of a free anomeric carbon that can open up to give an aldehyde or ketone group capable of reducing Fehling’s or Tollen’s reagent. In sucrose, the anomeric carbon of glucose (its $$C_1$$) and the anomeric carbon of fructose (its $$C_2$$) are both tied up in the glycosidic bond. Because both potentially reducing centers are blocked, sucrose cannot convert into an open-chain form and hence cannot act as a reducing agent. Therefore, sucrose is classified as a non-reducing sugar.

So far, we have confirmed every part of the Assertion (A): “Sucrose is a disaccharide and a non-reducing sugar.” Both statements are correct.

Now we analyse the Reason (R). The accepted structural description of sucrose is

$$\alpha\text{-D-glucopyranosyl} \,(1 \rightarrow 2)\, \beta\text{-D-fructofuranoside}$$

This wording tells us explicitly that the glycosidic linkage joins

$$C_1$$ of $$\alpha$$-D-glucose to $$C_2$$ of $$\beta$$-D-fructose.

However, the Reason (R) claims that the linkage is between $$C_1$$ of $$\beta$$-glucose and $$C_2$$ of $$\alpha$$-fructose. Both the configuration of glucose and of fructose have been reversed in this statement, and thus the description is wrong.

Therefore, the Reason (R) is false even though the Assertion (A) is true. Because a false statement cannot be the correct explanation of a true statement, we conclude that the correct choice is the one where Assertion is true but Reason is false.

Hence, the correct answer is Option B.

We begin by recalling the structure of sucrose. Sucrose is a disaccharide containing two monosaccharide units:

$$\text{sucrose} = \alpha\text{-D-Glucopyranosyl-(1}\rightarrow 2\text{)-}\beta\text{-D-Fructofuranoside}$$

This description tells us two important facts. First, the glucose part is in the $$\alpha$$-pyranose form, because the anomeric carbon of glucose has the same configuration as the anomeric carbon of $$\alpha$$-D-glucose. Second, the fructose part is in the $$\beta$$-furanose form, because the anomeric carbon of fructose has the opposite configuration from $$\alpha$$-D-fructose and therefore matches $$\beta$$-D-fructose.

Now we look at what happens in hydrolysis. Hydrolysis means cleavage of the glycosidic bond in the presence of water (and usually an acid or the enzyme invertase):

$$\text{sucrose} + \text{H}_2\text{O} \xrightarrow{\text{H}^+\;\text{or invertase}} \text{glucose} + \text{fructose}$$

Because the bond is broken exactly at the glycosidic junction, no other configuration changes occur at the anomeric carbons during the cleavage itself. Therefore, the two products keep the configurations that existed in the original disaccharide linkage:

$$\alpha\text{-D-Glucose} \;+\; \beta\text{-D-Fructose}$$

Next we attach their optical rotation signs. The D/L notation fixes only the configuration around the penultimate carbon, not the sign of optical rotation. For naturally occurring sugars we know:

$$\alpha\text{-D-Glucose} \;:\; [\alpha]_{D}^{20} = +52.5^\circ \text{ (positive, written as } (+) \text{)}$$

$$\beta\text{-D-Fructose} \;:\; [\alpha]_{D}^{20} = -92.4^\circ \text{ (negative, written as } (-) \text{)}$$

So, attaching the optical signs, we obtain:

$$\alpha\text{-D-(+)-Glucose} \;+\; \beta\text{-D-(-)-Fructose}$$

This pair exactly matches what is stated in Option B.

Hence, the correct answer is Option B.

Starch is a polysaccharide consisting of two components: amylose and amylopectin. Both are polymers of $$\alpha$$-D-glucose, so Option 2 is correct.

Amylose is the linear (unbranched) component of starch, in which glucose units are linked by $$\alpha$$-1,4-glycosidic bonds forming a helical structure. It is NOT a branched chain polymer. Amylopectin is the branched component of starch, having both $$\alpha$$-1,4 linkages in the main chain and $$\alpha$$-1,6 linkages at branch points.

Therefore, Option 1 — "Amylose is a branched chain polymer of glucose" — is the incorrect statement, since amylose is actually a linear (unbranched) polymer.

For completeness: Option 3 is correct (cellulose has $$\beta$$-1,4-glycosidic linkages), and Option 4 is correct (glycogen is indeed called animal starch, with even more branching than amylopectin).

The incorrect statement is Option 1.

The hydrolysis of sucrose is known as the inversion of cane sugar. The enzyme responsible for this reaction is invertase, which converts sucrose into glucose and fructose.

$$\mathrm{C_{12}H_{22}O_{11} + H_2O \xrightarrow{\text{Invertase}} C_6H_{12}O_6 + C_6H_{12}O_6}$$

Therefore, Enzyme A is Invertase.

The conversion of glucose into ethyl alcohol and carbon dioxide is an alcoholic fermentation reaction. This process is catalyzed by the enzyme zymase, which is present in yeast.

$$\mathrm{C_6H_{12}O_6 \xrightarrow{\text{Zymase}} 2C_2H_5OH + 2CO_2}$$

Therefore, Enzyme B is Zymase.

Hence, the correct answer is Option (C): Invertase and Zymase.

We need to match each drug/vitamin with its correct classification or use.

Valium (diazepam) is a well-known tranquilizer used to treat anxiety, insomnia, and seizures. So (a) matches with (iv).

Morphine is a powerful analgesic (pain reliever) derived from opium. It is used to relieve severe pain. So (b) matches with (iii).

Norethindrone is a synthetic progesterone and is used as an antifertility drug (oral contraceptive). So (c) matches with (i).

Vitamin $$B_{12}$$ (cyanocobalamin) deficiency causes pernicious anaemia, a condition in which the body cannot produce enough healthy red blood cells. So (d) matches with (ii).

Therefore, the correct matching is: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii).

The correct answer is Option (4).

We have to recall that Dacron is the common trade name for the synthetic polyester whose IUPAC name is polyethylene terephthalate, usually abbreviated as PET. Being a polyester, it is produced by a condensation polymerisation reaction between a di-acid and a di-alcohol (di-ol).

First we write the structural formulae of the two possible di-alcohols mentioned in the options:

$$$\text{ethylene glycol} = HO{-}CH_2{-}CH_2{-}OH$$$

$$$\text{glycerol} = HO{-}CH_2{-}CH(OH){-}CH_2{-}OH$$$

Next we write the two possible di-acids (dibasic acids) that appear in the options:

$$$\text{phthalic acid} = HOOC{-}C_6H_4{-}COOH \quad\text{(benzene-1,2-dicarboxylic acid)}$$$

$$$\text{terephthalic acid} = HOOC{-}C_6H_4{-}COOH \quad\text{(benzene-1,4-dicarboxylic acid)}$$$

In the manufacturing of Dacron (PET) the di-acid used is specifically the para-isomer, i.e. terephthalic acid, because its -COOH groups lie opposite each other on the benzene ring, giving a linear repeating unit that leads to strong, fibre-forming chains. Likewise, the di-alcohol employed is ethylene glycol, whose two -OH groups are separated by two carbon atoms, again favouring linearity.

Now we write the condensation polymerisation reaction that actually forms Dacron. A general step is

$$$n\,HO{-}CH_2{-}CH_2{-}OH\;+\;n\,HOOC{-}C_6H_4{-}COOH \;\longrightarrow\; \bigl[-O{-}CH_2{-}CH_2{-}O{-}CO{-}C_6H_4{-}CO-\bigr]_n \;+\;2n\,H_2O$$$

Each ester linkage $$-COO-$$ is produced by eliminating one molecule of water $$H_2O$$ from the reacting -OH of ethylene glycol and -COOH of terephthalic acid, and this process repeats $$n$$ times to build the long polymer chain whose repeating unit is $$-O{-}CH_2{-}CH_2{-}O{-}CO{-}C_6H_4{-}CO-$$.

Because the pair ethylene glycol $$HO{-}CH_2{-}CH_2{-}OH$$ and terephthalic acid $$HOOC{-}C_6H_4{-}COOH$$ uniquely gives this polymer, while the other combinations listed in the options either form branched polyesters or completely different materials, we conclude that these two are the required monomers of Dacron.

Hence, the correct answer is Option 3.

Based on NCERT, the correct answer is option (A).

We need to identify the type of interaction responsible for stabilising the $$\alpha$$-helix structure of proteins.

The $$\alpha$$-helix is a common secondary structure in proteins. In an $$\alpha$$-helix, the polypeptide chain coils in a right-handed helical fashion. This structure is stabilised by intramolecular hydrogen bonds.

Specifically, the hydrogen bonds form between the $$C=O$$ group of one amino acid residue and the $$N-H$$ group of the amino acid residue that is four positions ahead in the chain. Each hydrogen bond has the form $$N-H \cdots O=C$$.

Covalent bonds hold the primary structure (peptide bonds between amino acids). Ionic bonding and van der Waals forces contribute to tertiary and quaternary structures but are not the primary stabilising force for the $$\alpha$$-helix.

Hence, the correct answer is Option C.

Seliwanoff's test is used to distinguish between aldoses and ketoses. In this test, a sugar is heated with resorcinol in concentrated HCl. Ketoses (like fructose) react faster than aldoses and produce a deep cherry-red colour within a short time, confirming the presence of a ketose sugar.

Xanthoproteic test is used to identify proteins. In this test, concentrated nitric acid is added to the protein solution and heated. Proteins containing aromatic amino acids (such as tyrosine, tryptophan, and phenylalanine) undergo nitration of the aromatic ring, giving a yellow colour. On adding alkali (NaOH), the colour turns orange, confirming the presence of proteins.

Therefore, Seliwanoff's test identifies ketoses and the Xanthoproteic test identifies proteins. The correct answer is ketoses, proteins.

The given compound is Cytosine and the complementary base is Adenine. Thus, the right option is C

We start by recalling that Novolac is a low-molecular-mass, linear phenol-formaldehyde polymer produced by the condensation of phenol with formaldehyde in the presence of an acidic catalyst. Its repeating structural unit can be written symbolically as $$\mathrm{\;-\;C_6H_4\;-\;O\;-\;CH_2\;-\;}.$$

Now, when Novolac is further heated with additional formaldehyde, the free $$\mathrm{-\;CH_2OH}$$ (methylol) groups present on the aromatic rings undergo additional condensation. Each such group can react with either the ortho or para position of a neighbouring phenol ring, forming new $$\mathrm{-\;CH_2\;-}$$ bridges and, in many places, $$\mathrm{-\;CH_2\;O\;CH_2\;-}$$ linkages. The overall set of reactions can be summarised as

$$ \mathrm{Novolac} \;+\; \text{excess formaldehyde} \;\xrightarrow[\text{heat}]{\text{(basic or acidic)}}\; \mathrm{cross\text{-}linked\;phenol\text{-}formaldehyde\;resin} $$

Because the polymer chains are now interconnected in three dimensions, the product is a hard, infusible, and insoluble thermosetting plastic. This particular cross-linked phenol-formaldehyde resin is historically known as $$\text{Bakelite}$$.

Thus, heating Novolac with formaldehyde yields Bakelite.

Hence, the correct answer is Option B.

Proteins have four levels of structural organisation: primary, secondary, tertiary, and quaternary.

The primary structure is the linear sequence of amino acids held together by peptide bonds ($$-CO-NH-$$).

The secondary structure refers to the local folding patterns of the polypeptide chain, primarily the $$\alpha$$-helix and $$\beta$$-pleated sheet conformations. These structures are stabilised by hydrogen bonds formed between the $$C=O$$ group of one amino acid residue and the $$N-H$$ group of another amino acid residue in the polypeptide backbone.

Glycosidic bonds are found in carbohydrates, not proteins. Van der Waals forces contribute to the tertiary and quaternary structures but are not the primary stabilising force for the secondary structure.

Therefore, the secondary structure of a protein is stabilised by hydrogen bonding.

Proteins may be broadly classified on the basis of their solubility. We have globular proteins that readily dissolve in water and fibrous proteins that are either sparingly soluble or completely insoluble. Albumins fall in the first category, while fibrin, myosin and collagen belong to the second.

Now, albumin is a globular protein. Being globular, its peptide chains fold in such a manner that the hydrophilic (water-loving) side chains remain on the outside of the molecule, whereas the hydrophobic (water-repelling) residues are tucked inside. So the overall interaction with water is favorable, and the protein dissolves.

In contrast, fibrin, myosin and collagen are fibrous proteins. Their long, extended chains align alongside one another, forming fibers or sheets. The exposed side chains are largely non-polar, offering very little scope for hydrogen bonding with water molecules. Hence they either do not dissolve or swell only slightly in water.

Comparing the four options, only albumin satisfies the condition of being water soluble.

Hence, the correct answer is Option B.

Thiamine is the chemical name for Vitamin B$$_1$$. It plays a crucial role in energy metabolism and nerve function. Pyridoxine is the chemical name for Vitamin B$$_6$$, which is involved in amino acid metabolism and neurotransmitter synthesis.

Looking at the options, Option (4) correctly identifies thiamine as Vitamin B$$_1$$ and pyridoxine as Vitamin B$$_6$$.

For reference, Vitamin B$$_2$$ is riboflavin and Vitamin E is tocopherol, which rules out the other options.

The correct answer is Option (4).

Vitamins are classified into two categories: water-soluble vitamins and fat-soluble vitamins. Water-soluble vitamins (such as B-complex vitamins including thiamine/$$B_1$$, and vitamin C/ascorbic acid) are not stored in the body for long periods and are readily excreted through urine. Fat-soluble vitamins (vitamins A, D, E, and K) are stored in the liver and fatty tissues of the body for relatively longer durations.

Now let us examine the options. Option (1) contains thiamine (a water-soluble vitamin) and vitamin A (fat-soluble), so this pair does not qualify since thiamine is not stored for long. Option (3) contains thiamine and ascorbic acid, both water-soluble, so neither is stored for long. Option (4) contains ascorbic acid (water-soluble) and vitamin D (fat-soluble), so this pair also does not fully qualify.

Option (2) contains vitamins A and D, both of which are fat-soluble vitamins that are stored in the body for relatively longer duration.

Hence, the correct answer is option (2).

Maltose is a disaccharide composed of two glucose units linked by a glycosidic bond. The key features of maltose are as follows.

Maltose consists of two $$\alpha$$-D-glucose units joined by an $$\alpha$$-1,4-glycosidic linkage. This means the glycosidic bond connects the C-1 of one glucose unit (in the $$\alpha$$ configuration) to the C-4 of the second glucose unit.

The $$\alpha$$-anomer of maltose specifically refers to the configuration at the anomeric carbon (C-1) of the reducing end glucose unit. In the $$\alpha$$-anomer, the hydroxyl group at C-1 of the reducing end is in the axial position (pointing downward in the Haworth projection).

Therefore, the correct structure of the $$\alpha$$-anomer of maltose has an $$\alpha$$-1,4-glycosidic linkage between the two glucose units, with the anomeric hydroxyl at the reducing end also in the $$\alpha$$ configuration. This corresponds to option (1).

Based off of NCERT, the right option is D.

First, let us recall the basic classification of proteins based on their overall shape and solubility. Proteins are broadly divided into two categories:

$$\text{Proteins} \;\; \longrightarrow \;\; \begin{cases} \text{Fibrous proteins} \\ \text{Globular proteins} \end{cases}$$

We have the following characteristic points:

$$\text{Fibrous proteins:}$$ These are long, thread-like or filamentous molecules. They show very little tertiary folding, are generally insoluble in water, and primarily serve structural or mechanical functions in animals.

$$\text{Globular proteins:}$$ These are compact, spheroidal molecules. Because of extensive folding, they are usually soluble in water and perform dynamic functions such as transport, catalysis, and regulation.

Now we shall check each option one by one and match it with the appropriate category.

$$\text{Option A: Keratin}$$ Keratin is the structural protein found in hair, wool, nails and horns. Its molecules are long and rope-like, giving strength and toughness. So keratin is clearly a fibrous protein.

$$\text{Option B: Albumin}$$ Albumin (e.g., serum albumin or egg white albumin) is well-known for being water-soluble, easily coagulable by heat, and highly folded into a compact shape. Such properties are the hallmarks of globular proteins, not fibrous ones. Therefore albumin is not a fibrous protein.

$$\text{Option C: Myosin}$$ Myosin is the contractile protein of muscle. Its long tail and rod-like structure, together with its mechanical role in muscle contraction, classify it as a fibrous protein.

$$\text{Option D: Collagen}$$ Collagen forms the triple-helical fibres that constitute connective tissue, tendons and skin. It is insoluble, provides tensile strength, and is a textbook example of a fibrous protein.

From the above discussion we see that keratin, myosin and collagen all satisfy the definition of fibrous proteins, whereas albumin is globular. Hence albumin is the only option that does not belong to the fibrous class.

Hence, the correct answer is Option B.

Lactose is a disaccharide composed of one unit of D-galactose and one unit of D-glucose. It is commonly known as milk sugar.

In lactose, the glycosidic bond is formed between the anomeric carbon (C-1) of $$\beta$$-D-galactose and the C-4 hydroxyl group of D-glucose. This is a $$\beta$$-1,4-glycosidic linkage.

The galactose unit provides its C-1 carbon for the glycosidic bond, while the glucose unit provides its C-4 carbon. Since the anomeric carbon of glucose remains free and is not involved in the glycosidic bond, lactose is a reducing sugar.

Therefore, the glycosidic linkage in lactose is between C-1 of galactose and C-4 of glucose, which corresponds to Option 3.

A glycosidic linkage joins the anomeric carbon $$C_1$$ of one monosaccharide to some carbon (commonly $$C_4$$ or $$C_6$$) of a second monosaccharide after the elimination of water.

The notation $$\beta - C_1 - C_4$$ means: the anomeric carbon $$C_1$$ of the first sugar is in the $$\beta$$-configuration and it is linked to the $$C_4$$ carbon of the second sugar.

Let us recall the main disaccharides/polysaccharide units mentioned in the options and the type of link each contains:

Option A (Sucrose): $$\alpha\text{-}D\text{-glucose}\,(C_1) - \beta\text{-}D\text{-fructose}\,(C_2)$$ linkage, i.e. $$\alpha\,1 \rightarrow 2\,\beta$$. No $$C_4$$ carbon is involved, so it is not $$\beta\,1 \rightarrow 4$$.

Option B (Amylose): the unbranched component of starch. Each glucose unit is joined by $$\alpha\,1 \rightarrow 4$$ glycosidic bonds (not $$\beta$$).

Option C (Lactose): composed of $$\beta\text{-}D\text{-galactose}$$ and $$\beta\text{-}D\text{-glucose}$$. The linkage is $$\beta\,1 \rightarrow 4$$, joining $$C_1$$ of galactose in the $$\beta$$-form to $$C_4$$ of glucose.

Option D (Maltose): two $$\alpha\text{-}D\text{-glucose}$$ units connected by an $$\alpha\,1 \rightarrow 4$$ bond (not $$\beta$$).

Only Option C (Lactose) contains the required $$\beta - C_1 - C_4$$ glycosidic linkage.

Hence, the correct choice is Option C (Lactose).

Step 1: Formation of Product 'A' (Acid-Catalyzed Dehydration)

The starting material is cyclohexylmethanol (a primary alcohol). It is treated with a strong acid ($H_3PO_4$) and heat ($120^\circ C$). This triggers a dehydration reaction, but with a crucial intermediate step.

1. Carbocation Formation: The acid protonates the $-OH$ group, turning it into water, which is a good leaving group. When the water molecule leaves, it forms a primary carbocation ($-CH_2^+$) attached to the cyclohexane ring.
2. Rearrangement (The Catch): Primary carbocations are highly unstable. To become more stable, the molecule undergoes a 1,2-hydride shift. A hydrogen atom from the adjacent carbon on the ring (which is a tertiary carbon) shifts over to the primary carbon. This moves the positive charge onto the ring itself, creating a highly stable tertiary carbocation. 3. Elimination (Zaitsev's Rule): Finally, a water molecule acts as a weak base and removes a neighboring proton to form a double bond. According to Zaitsev's rule, the molecule will form the most substituted (and therefore most stable) alkene possible. Removing a proton from inside the ring creates a highly stable trisubstituted double bond.

Result for A:

Product A is 1-methylcyclohexene. (It is a cyclohexane ring with a double bond, and a methyl group attached to one of the carbons sharing the double bond).

Step 2: Formation of Product 'P' (Hydroboration-Oxidation)

Now we take Product A (1-methylcyclohexene) and subject it to hydroboration-oxidation using $(BH_3)_2$ followed by $H_2O_2$ and $OH^-$.

As we established in the previous question, this reagent combo adds water (an $H$ and an $OH$) across the double bond following specific rules:

1. Anti-Markovnikov Regioselectivity: The $-OH$ group will attach to the least substituted carbon of the double bond. In 1-methylcyclohexene, one side of the double bond has a methyl group (tertiary), and the other side only has a hydrogen (secondary). Therefore, the $-OH$ attaches to the secondary carbon.
2. Syn-Addition Stereochemistry: The new $H$ and the new $-OH$ add to the exact same face of the ring. Because they crowd one side, the existing methyl group is pushed to the opposite face.

Result for P:

Product P is 2-methylcyclohexanol (specifically, trans-2-methylcyclohexanol, because the methyl group and the hydroxyl group will be pointing in opposite directions).

Enzymes are biological catalysts that are highly specific in their action — a key property known as enzyme specificity. Each enzyme catalyses only one particular type of reaction and acts on a specific substrate. This specificity arises from the unique three-dimensional structure of the active site of the enzyme (the "lock and key" model and induced fit model).

The statement that "Enzymes are non-specific for a reaction and substrate" is therefore NOT true. Enzymes are in fact highly specific both for the type of reaction they catalyse and for the substrate they act upon.

The other statements are all correct: almost all enzymes are proteins (some are RNA-based ribozymes, but the vast majority are proteins); enzymes lower the activation energy of biochemical reactions, acting as catalysts; and enzyme activity is highly dependent on temperature and pH, with optimal conditions for each enzyme.

First, recall what a peptide linkage is. A peptide linkage, also called an amide bond, is the covalent bond that forms when the carboxyl group $$\big(-\mathrm{COOH}\big)$$ of one amino acid reacts with the amino group $$\big(-\mathrm{NH_2}\big)$$ of another amino acid, with the elimination of a molecule of water $$\big(\mathrm{H_2O}\big).$$

Whenever amino acids combine linearly to form a peptide, each new bond that joins two neighboring amino-acid residues is precisely one peptide linkage.

Now, let us denote by $$n$$ the total number of individual amino-acid molecules that are being joined. We have the general relationship:

$$\text{Number of peptide linkages}=n-1$$

This statement is easily justified: the first amino acid has a free amino group on one end, and the last amino acid keeps a free carboxyl group on the other end. All the internal joins between successive residues are peptide bonds, and there is exactly one less bond than the number of residues.

In the present problem, the mixture contains one molecule each of Glycine, Leucine, Aspartic acid and Histidine. Therefore,

$$n = 4$$

Substituting this value in the formula stated above, we get

$$\text{Number of peptide linkages} = n - 1 = 4 - 1$$

$$\text{Number of peptide linkages} = 3$$

So, the answer is $$3$$.

We begin by recalling that, for any ionisable group, the Henderson-Hasselbalch equation $$\text{pH}= \text{p}K_a + \log\!\left(\dfrac{[\text{A}^-]}{[\text{HA}]}\right)$$ tells us the protonation state. When the pH is much greater than the pKa, the term $$\log([\text{A}^-]/[\text{HA}])$$ is large and positive, meaning the conjugate base $$\text{A}^-$$ predominates, so the group is de-protonated and carries a negative charge if it is an acid. Conversely, basic groups become neutral when de-protonated.

The tetrapeptide is $$\text{Gly} - \text{Glu} - \text{Asp} - \text{Tyr}$$. After peptide-bond formation only two termini remain free: the N-terminus of Gly and the C-terminus of Tyr. Along with these, certain side chains can also ionise. We list every potentially ionisable group with its approximate pKa value:

$$ \begin{aligned} \text{N-terminal } \mathrm{NH_3^+} &: \text{p}K_a \approx 9.0 \\ \text{C-terminal } \mathrm{COOH} &: \text{p}K_a \approx 2.0 \\ \text{Glu side-chain } \mathrm{COOH} &: \text{p}K_a \approx 4.3 \\ \text{Asp side-chain } \mathrm{COOH} &: \text{p}K_a \approx 3.9 \\ \text{Tyr side-chain } \mathrm{OH} &: \text{p}K_a \approx 10.1 \end{aligned} $$

Now the solution pH is $$12.5$$, which is considerably higher than every pKa listed above. Therefore:

1. The N-terminal group: at $$\text{pH}=12.5 \gt 9.0$$, it is de-protonated, converting $$\mathrm{NH_3^+}$$ to neutral $$\mathrm{NH_2}$$. Hence it contributes $$0$$ charge.
2. The C-terminal group: at $$12.5 \gt 2.0$$, the $$\mathrm{COOH}$$ is fully de-protonated to $$\mathrm{COO^-}$$, giving $$-1$$ charge.
3. The Glu side chain: at $$12.5 \gt 4.3$$, it exists as $$\mathrm{COO^-}$$, contributing $$-1$$ charge.
4. The Asp side chain: at $$12.5 \gt 3.9$$, it also becomes $$\mathrm{COO^-}$$, contributing $$-1$$ charge.
5. The Tyr phenolic OH: at $$12.5 \gt 10.1$$, the phenolate form $$\mathrm{O^-}$$ predominates, giving another $$-1$$ charge.

Adding all individual charges algebraically:

$$ \text{Total charge}=0 + (-1) + (-1) + (-1) + (-1)= -4 $$

The magnitude of negative charge is the absolute value, so the peptide carries four negative charges.

Hence, the correct answer is Option D.

Nor-adrenaline, also written as $$\text{norepinephrine}$$, is one of the major excitatory neurotransmitters in the brain. We have the basic physiological fact that an adequate concentration of this chemical is required to maintain normal alertness, mood, and emotional balance.

When the level of $$\text{nor-adrenaline}$$ falls below its normal range, the electrical activity between neurons slows down, so the person often experiences prolonged sadness, loss of interest, fatigue, and other classical signs of depression. Therefore a therapy that can raise the synaptic concentration of this neurotransmitter is needed.

Now, the class of drugs that is specifically designed to elevate the amounts of mood-related neurotransmitters such as $$\text{nor-adrenaline}$$ and $$\text{serotonin}$$ in the synaptic cleft is called antidepressants. These medicines either inhibit the re-uptake of the neurotransmitter (for example, tricyclic antidepressants or SNRIs) or prevent its enzymatic degradation (for example, MAO inhibitors), thereby allowing more $$\text{nor-adrenaline}$$ to remain available for neuronal signaling.

Substituting this pharmacological reasoning into the present question, we see that the drug suggested for a deficiency of $$\text{nor-adrenaline}$$ must fall under the category that directly counters depression—namely, an antidepressant.

Hence, the correct answer is Option B.

We first recall the standard IUPAC names of the monomeric units that build the well-known elastomers mentioned in Item-I.

Natural rubber is scientifically described as $$\text{cis-1,4-polyisoprene}$$. Hence its repeating unit is derived from the monomer $$\text{isoprene}$$ $$\bigl(2\text{-methyl-}1,3\text{-butadiene}\bigr).$$ We therefore connect “Natural rubber” with “Isoprene”.

Neoprene is produced by the free-radical polymerisation of $$\text{chloroprene}$$ $$\bigl(2\text{-chloro-}1,3\text{-butadiene}\bigr).$$ Thus “Neoprene” must be paired with “Chloroprene”.

Buna-S is a copolymer obtained from $$1,3\text{-butadiene}$$ and $$\text{styrene}$$. Accordingly, the correct description for Buna-S is “1,3-butadiene + styrene”.

Buna-N (the “N” stands for “nitrile”) is produced from $$1,3\text{-butadiene}$$ together with $$\text{acrylonitrile}$$. Because acrylonitrile does not appear in Item-II, the closest remaining entry is the simple monomer $$1,3\text{-butadiene}$$ itself. Hence we associate Buna-N with “1,3-butadiene”.

Collecting these matches we have

$$\begin{aligned} \text{(a) Natural rubber} &\longrightarrow \text{(IV) Isoprene} \\ \text{(b) Neoprene} &\longrightarrow \text{(III) Chloroprene} \\ \text{(c) Buna-N} &\longrightarrow \text{(II) 1,3-butadiene} \\ \text{(d) Buna-S} &\longrightarrow \text{(I) 1,3-butadiene + styrene} \end{aligned}$$

On comparing this sequence with the given options, we observe that Option C lists exactly these pairings.

Hence, the correct answer is Option C.

Condensation polymerisation is defined as a polymerisation process in which monomer units join together with the elimination of small molecules such as $$H_2O$$ or $$HCl$$.

Option A: Nylon-6,6 is prepared by the condensation of hexamethylenediamine and adipic acid with the elimination of $$H_2O$$. This fits the definition of condensation polymerisation.

Option B: Buna-N is a copolymer of 1,3-butadiene and acrylonitrile formed by addition polymerisation. In this process, monomers add to each other without elimination of any small molecule. Therefore, it is not obtained by condensation polymerisation.

Option C: Bakelite is formed by the condensation of phenol and formaldehyde with elimination of $$H_2O$$. This clearly is a condensation polymerisation.

Option D: Nylon-6 is produced by the ring-opening polymerisation of caprolactam. In this reaction no small molecule such as $$H_2O$$ is eliminated, yet in JEE classification it is grouped with condensation polymers because its repeating unit contains amide linkages.

Among the given options, the polymer not obtained by condensation polymerisation is Option B: Buna-N.

A chiral carbon is an $$sp^3$$ hybridised carbon atom bonded to four different groups. We need to check whether the monomer(s) of each given polymer contain any chiral carbon.

Option A: Neoprene

Neoprene is a polymer of chloroprene (2-chloro-1,3-butadiene). The structure of chloroprene is:

$$CH_2 = CCl - CH = CH_2$$

All the carbon atoms in chloroprene are either $$sp^2$$ (doubly bonded) or part of $$=CH_2$$ groups. None of them is bonded to four different groups. So the monomer is not chiral.

Option B: Buna-N

Buna-N is a copolymer of 1,3-butadiene ($$CH_2=CH-CH=CH_2$$) and acrylonitrile ($$CH_2=CH-CN$$). In 1,3-butadiene, all carbons are $$sp^2$$. In acrylonitrile, $$CH_2=CH-CN$$, the carbons are also $$sp^2$$ or bonded to only three different groups. Neither monomer has a chiral centre.

Option C: Nylon 6,6

Nylon 6,6 is a copolymer of hexamethylenediamine ($$H_2N-(CH_2)_6-NH_2$$) and adipic acid ($$HOOC-(CH_2)_4-COOH$$). In hexamethylenediamine, the internal $$-CH_2-$$ carbons each carry two identical H atoms, so they are not chiral. Similarly in adipic acid, the internal carbons carry two H atoms each. Neither monomer is chiral.

Option D: PHBV

PHBV stands for poly(3-hydroxybutyrate-co-3-hydroxyvalerate). It is a copolymer of two monomers:

(i) 3-Hydroxybutanoic acid ($$\beta$$-hydroxybutyric acid): $$CH_3-CH(OH)-CH_2-COOH$$

Look at the carbon bearing $$-OH$$: it is bonded to (1) $$-CH_3$$, (2) $$-OH$$, (3) $$-H$$, and (4) $$-CH_2COOH$$. All four groups are different. This carbon is chiral.

(ii) 3-Hydroxypentanoic acid ($$\beta$$-hydroxyvaleric acid): $$CH_3CH_2-CH(OH)-CH_2-COOH$$

The carbon bearing $$-OH$$ is bonded to (1) $$-C_2H_5$$, (2) $$-OH$$, (3) $$-H$$, and (4) $$-CH_2COOH$$. All four are different. This carbon is also chiral.

Therefore, PHBV is the polymer whose monomers contain chiral centres.

The correct answer is Option D.

First, recall what each of the three qualitative tests tells us.

Molisch’s test detects any carbohydrate. If a substance gives a violet ring with α-naphthol and concentrated H2SO4, the test is said to be positive, so the substance must be a carbohydrate. If the test is negative, the substance is not a carbohydrate.

Barfoed’s test distinguishes between monosaccharides and disaccharides. Monosaccharides reduce Cu2+ of Barfoed’s reagent to red Cu2O within about 2-3 minutes, so the test becomes positive. Disaccharides react much more slowly, so in the usual test time they give a negative result.

Biuret test identifies compounds that contain two or more peptide (-CO-NH-) bonds. Proteins and polypeptides give a violet colour, i.e. a positive test, because they have many peptide bonds. Simple amino acids possess no peptide bond at all, so their Biuret test is negative.

Now examine each unknown biomolecule in turn.

Substance A gives Molisch positive, Barfoed negative, Biuret negative. We have:   • Molisch positive ⇒ A is a carbohydrate.   • Barfoed negative ⇒ A is not a monosaccharide; therefore it must be a disaccharide.   • Biuret negative ⇒ A has no peptide bonds, confirming it is not a protein. Among the options, the only disaccharide offered is lactose. Hence $$A = \text{Lactose}.$$

Substance B shows Molisch positive, Barfoed positive, Biuret negative. We have:   • Molisch positive ⇒ B is a carbohydrate.   • Barfoed positive ⇒ B is a monosaccharide.   • Biuret negative ⇒ B is not a protein. Both glucose and fructose are reducing monosaccharides, so either could in principle satisfy the first two tests. However, in the list of answers that already matches A (lactose) and C (to be decided next), the only consistent monosaccharide offered is glucose. Thus we assign $$B = \text{Glucose}.$$

Substance C gives Molisch negative but Biuret positive. We have:   • Molisch negative ⇒ C is not a carbohydrate.   • Biuret positive ⇒ C must contain many peptide bonds, i.e. it is a protein. Albumin is a well-known protein, whereas alanine is only a single amino acid (no peptide bond), so alanine would give Biuret negative. Therefore $$C = \text{Albumin}.$$

Collecting the three identifications, we find

$$A = \text{Lactose},\quad B = \text{Glucose},\quad C = \text{Albumin}.$$

Comparing with the options, we match Option 2.

Hence, the correct answer is Option 2.

We have four unknown samples A, B, C and D which are known to be taken from the list Aspartame, Saccharin, Sucralose and Alitame. The task is to match each unknown with the correct sweetener by using the classical qualitative tests mentioned in the statement.

First we look at the ninhydrin observation. The ninhydrin reagent gives a characteristic blue-violet (Ruhemann’s purple) colour whenever a compound possesses a free $$\;-NH_2\;$$ or $$\;-NH-\;$$ group that is α to a carbonyl, exactly as found in amino acids and peptides. Therefore

$$\text{blue-violet colour with ninhydrin}\;\Longrightarrow\;\text{compound contains an amino (peptide) group}.$$

The data say that both A and D give this colour. In the family of artificial sweeteners provided:

Aspartame $$\bigl($$L-Aspartyl-L-phenylalanine-methyl-ester$$\bigr)$$ is a dipeptide; it definitely contains a free $$\;-NH_2\;$$ group.
Alitame $$\bigl($$L-Aspartyl-D-alanine amide of a thietanyl group$$\bigr)$$ is also a peptide derivative and possesses the same group.
Saccharin is a sulfonimide and has no free amino group.
Sucralose is a chlorinated sugar without any nitrogen at all.

Hence

$$A,D \;\text{(ninhydrin +)} \;\;\Rightarrow\;\; A,D \in \{\text{Aspartame, Alitame}\}.$$

Now we pass to sample C. Its Lassaigne filtrate shows

(i) Positive silver nitrate test: $$Ag^+ + X^- \longrightarrow AgX\downarrow\quad(X = Cl^-,Br^-,I^-).$$

(ii) Negative Prussian blue test with $$Fe_4[Fe(CN)_6]_3,$$ which is the classical confirmation for nitrogen.

So C contains a halogen but does not contain nitrogen. Surveying the list:

Sucralose ≡ C$$_{12}$$H$$_{19}$$Cl$$_3$$O$$_8$$ (three chlorine atoms, no nitrogen) ✔️
Aspartame, Saccharin and Alitame all contain nitrogen (and no halogen).

Therefore

$$C = \text{Sucralose}.$$

Finally, the Lassaigne extract of B and D gives a positive sodium nitroprusside test. The underlying reaction is

$$Na_2[Fe(CN)_5NO] + S^{2-} \;\longrightarrow\; [Fe(CN)_5NOS]^{4-}\;(\text{violet}).$$

Hence sulfur is present in both B and D.

Looking again at the molecular compositions:

Saccharin - contains a sulfonyl (S=O)$$_2$$ group ✔️
Alitame - contains a thietane (four-membered ring with S) ✔️
Aspartame - no sulfur ✖️
Sucralose - no sulfur ✖️

We already fixed $$D$$ as either Aspartame or Alitame, but sulfur positivity immediately removes Aspartame from $$D$$. Therefore

$$D = \text{Alitame}.$$

Because $$D$$ is Alitame, the other ninhydrin-positive sample $$A$$ must be the remaining peptide sweetener, namely Aspartame:

$$A = \text{Aspartame}.$$

We have also determined $$C = \text{Sucralose}$$. The only sweetener still unassigned is Saccharin, so that must be $$B$$; moreover $$B$$ was indeed sulfur-positive, perfectly consistent:

$$B = \text{Saccharin}.$$

The complete matching is therefore

$$ \begin{aligned} A &\;:\; \text{Aspartame},\\ B &\;:\; \text{Saccharin},\\ C &\;:\; \text{Sucralose},\\ D &\;:\; \text{Alitame}. \end{aligned} $$

This ordering is exactly the one listed in Option A.

Hence, the correct answer is Option A.

We have to judge the truth values of the Assertion (A) and the Reason (R) and also decide whether the Reason actually explains the Assertion.

First, recall the basic facts about the polymerisation of ethene $$\left(\text{CH}_2 = \text{CH}_2\right)$$.

When ethene is polymerised in the presence of a Ziegler-Natta catalyst, and the reaction is carried out at comparatively high temperature and pressure, the product obtained is called high-density polyethene, commonly abbreviated as $$\text{HDPE}$$. The Ziegler-Natta catalyst system is usually $$\text{TiCl}_4$$ supported on $$\text{Al}(\text{C}_2\text{H}_5)_3$$, and it promotes a stereoregular insertion of ethene units into the growing chain. Because of this stereoregularity the chains are able to pack very closely. Closer packing means:

$$\text{Greater chain-chain attraction} \;\Longrightarrow\; \text{higher crystalline regions} \;\Longrightarrow\; \text{higher density (about }0.94\ \text{g cm}^{-3}\text{)}.$$

The resulting material is rigid, possesses high tensile strength and shows excellent chemical inertness. These properties make it suitable for moulding into articles such as buckets, dustbins, pipes, bottles, toys, etc.

Hence the Assertion “Ethene polymerised in the presence of Ziegler Natta catalyst at high temperature and pressure is used to make buckets and dustbins” is factually correct.

Now consider the Reason. It states: “High-density polymers are closely packed and are chemically inert.” The statement is again correct because high density results from close packing of linear chains; the absence of bulky branching in $$\text{HDPE}$$ leads to strong van der Waals forces, higher crystallinity and, consequently, resistance to chemicals.

We must still decide whether this Reason explains why $$\text{HDPE}$$ can be used for buckets and dustbins. The suitability of a material for such everyday containers depends on rigidity, strength and chemical inertness so that the container can hold water, detergents, household acids or alkalis without degrading. The Reason exactly pinpoints two crucial characteristics—close packing (which gives strength) and chemical inertness—thereby providing the very explanation for the Assertion.

Therefore:

• (A) is true. • (R) is true. • (R) correctly explains (A).

According to the standard pattern for Assertion-Reason questions, this situation corresponds to Option C: “Both (A) and (R) are correct and (R) is the correct explanation of (A).”

Hence, the correct answer is Option C.

We start with the basic fact that one molecule of acetic anhydride, $$\left(CH_3CO\right)_2O$$, can furnish one acetyl group $$\left(CH_3CO-\right)$$ to one hydroxyl (-OH) group of an organic compound and at the same time liberates one molecule of acetic acid. Therefore, the number of equivalents of acetic anhydride required for complete acetylation is exactly the same as the number of free -OH groups present in the substrate.

Glucose in its open-chain form is an aldo-hexose. It possesses:

• one aldehydic carbonyl group (-CHO), and
• five alcoholic hydroxyl groups (at C2, C3, C4, C5 and C6).

Now we analyse each sequence separately.

(i) Glucose + ROH $$\overset{dry\;HCl}{\rightarrow}$$ Acetal

Dry HCl in an alcohol converts the aldehyde end of glucose into an acetal (more precisely, an alkyl glucoside). The mechanism replaces the anomeric hydroxyl group (the one originally formed in the hemiacetal ring) by an -OR group from the alcohol.

Effect on the -OH count:

• The anomeric -OH is lost (converted to -OR, an ether).
• The remaining four secondary -OH groups stay intact.

Hence the acetal now contains exactly $$5 - 1 = 4$$ free hydroxyl groups.

Therefore, the number of equivalents of acetic anhydride needed is

$$x = 4.$$

(ii) Glucose $$\overset{Ni/H_2}{\rightarrow}$$ A (Sorbitol)

Hydrogenation over Ni reduces the aldehyde group (-CHO) at C1 to the corresponding primary alcohol (-CH2OH). No other part of the molecule is affected.

Effect on the -OH count:

• The original carbonyl carbon now bears an additional hydroxyl group.
• All the original five -OH groups remain.

So the total number of -OH groups becomes

$$5 + 1 = 6.$$

Consequently, the equivalents of acetic anhydride required are

$$y = 6.$$

(iii) Direct acetylation of glucose

When glucose is treated directly with acetic anhydride in the presence of a base such as pyridine, every free hydroxyl group is acetylated. The aldehyde carbonyl itself is not modified, but the anomeric hydroxyl (present in the cyclic hemiacetal form) is indeed a free -OH and does get acetylated.

Thus, all five hydroxyl groups are converted to O-COCH3 groups.

Hence the required equivalents are

$$z = 5.$$

Collecting the three values we have

$$x = 4, \; y = 6, \; z = 5.$$

The only option that lists the trio 4, 6 and 5 in this order is Option B.

Hence, the correct answer is Option B.

First, we recall the therapeutic use of each drug from standard medicinal chemistry facts.

We have Ranitidine. It is well known that Ranitidine blocks the $$\mathrm{H_2}$$-histamine receptor present on the gastric parietal cells, thereby reducing gastric acid secretion. A drug that decreases stomach acidity is termed an antacid. So Ranitidine corresponds to $$\text{(d) Antacid}$$.

Now consider Nardil, whose generic name is Phenelzine. Phenelzine belongs to the class of monoamine oxidase inhibitors, abbreviated as $$\text{MAOIs}$$. Drugs that inhibit MAO increase the levels of neurotransmitters such as serotonin and norepinephrine, and are therefore used to treat depressive disorders. Hence Nardil (Phenelzine) matches $$\text{(a) Antidepressant}$$.

Next, we examine Chloramphenicol. From general pharmacology, Chloramphenicol is a broad-spectrum antibiotic that inhibits the $$50\text{ S}$$ ribosomal subunit and blocks peptide-bond formation in bacteria. Therefore its therapeutic action is clearly $$\text{(b) Antibiotic}$$.

Finally, we look at Dimetane, whose active ingredient is Brompheniramine. Brompheniramine competitively antagonises the $$\mathrm{H_1}$$-histamine receptor, giving relief from allergic symptoms such as sneezing and itching. A medicine that blocks histamine in this way is called an antihistamine, that is, $$\text{(c) Antihistamine}$$.

Collecting all four pairings we obtain the set of matches in symbolic form:

$$$ \begin{aligned} (i)&:&\; \text{Ranitidine} &\longrightarrow& (d) \; \text{Antacid}\\ (ii)&:&\; \text{Nardil (Phenelzine)} &\longrightarrow& (a) \; \text{Antidepressant}\\ (iii)&:&\; \text{Chloramphenicol} &\longrightarrow& (b) \; \text{Antibiotic}\\ (iv)&:&\; \text{Dimetane (Brompheniramine)} &\longrightarrow& (c) \; \text{Antihistamine} \end{aligned} $$$

Comparing these matches with the given options, we notice that Option B lists exactly the same correspondence, namely $$$ (i)-(d),\; (ii)-(a),\; (iii)-(b),\; (iv)-(c) $$$. No other option satisfies every pairing simultaneously.

Hence, the correct answer is Option B.

We begin by recalling the standard list of vitamins, their chemical names and the classical deficiency diseases associated with each of them. Knowing these fixed pairings allows us to match each item in the left column with the corresponding disease in the right column.

First, consider $$\text{Riboflavin}$$, which is commonly known as Vitamin B2. A deficiency of Vitamin B2 typically produces lesions at the corners of the mouth, a condition medically termed cheilosis. Hence we match

$$\text{Riboflavin}\; \longrightarrow \; \text{Cheilosis} \; (c).$$

Next, look at $$\text{Thiamine}$$, or Vitamin B1. The classical deficiency disease here is beriberi, which affects the nervous and cardiovascular systems. Therefore we set up the correspondence

$$\text{Thiamine}\; \longrightarrow \; \text{Beriberi} \; (a).$$

Now we move to $$\text{Pyridoxine}$$, better known as Vitamin B6. Lack of this vitamin often manifests as neurological disturbances such as irritability and, in severe cases, convulsions. Thus we write

$$\text{Pyridoxine}\; \longrightarrow \; \text{Convulsions} \; (d).$$

Finally, $$\text{Ascorbic acid}$$ is simply Vitamin C. Its deficiency leads to defective collagen synthesis, giving rise to the well-known disease scurvy. We therefore pair

$$\text{Ascorbic acid}\; \longrightarrow \; \text{Scurvy} \; (b).$$

Collecting all four matched pairs, we have

$$ \begin{aligned} (i)&\; \text{Riboflavin} &\longrightarrow&\; (c)&\; \text{Cheilosis},\\ (ii)&\; \text{Thiamine} &\longrightarrow&\; (a)&\; \text{Beriberi},\\ (iii)&\; \text{Pyridoxine} &\longrightarrow&\; (d)&\; \text{Convulsions},\\ (iv)&\; \text{Ascorbic acid} &\longrightarrow&\; (b)&\; \text{Scurvy}. \end{aligned} $$

Comparing this ordered set with the answer choices, we see that it matches exactly with Option C, which states (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b).

Hence, the correct answer is Option C.

We have the disaccharide $$\text{Sucrose}$$, which consists of one unit of $$\alpha\!-\!D\!-\!\text{glucopyranose}$$ and one unit of $$\beta\!-\!D\!-\!\text{fructofuranose}$$ linked through a glycosidic bond $$\left(C\_1^{\text{Glucose}}{-}O{-}C\_2^{\text{Fructose}}\right)$$.

According to basic carbohydrate chemistry, “glycosidic bond cleavage (hydrolysis)” breaks this bond in the presence of dilute acid or the enzyme invertase. Stating the hydrolysis process explicitly:

Sucrose} + H_2O \;\xrightarrow[\text{dil.\;HCl / invertase}]{\text{Hydrolysis}}\; \text{Glucose} + \text{Fructose

So after the first step we obtain two monosaccharides:

$$A = \text{Glucose} \quad\text{and}\quad B = \text{Fructose}$$

Now we treat this mixture with Seliwanoff reagent. First, let us recall what the Seliwanoff test tells us:

The reagent is a solution of $$\text{resorcinol}$$ in concentrated $$HCl$$. Under strong acid, any ketose (but not an aldose) is rapidly dehydrated to form $$5\text{-hydroxymethylfurfural}$$. This intermediate then condenses with resorcinol to yield a cherry-red coloured complex. Aldoses either react very slowly or give no such colour within the usual test time.

In our hydrolysate mixture,

$$\text{Glucose} \;(\text{aldohexose, no rapid colour})$$

$$\text{Fructose} \;($$ ketohexose, gives red colour rapidly $$)$$

Because fructose is a ketohexose, it is the one that reacts instantly with the Seliwanoff reagent, producing the characteristic red colour. The presence of glucose does not mask this observation; it merely lags behind, so the cherry-red colour of the fructose dominates.

Therefore, the correct experimental observation is the formation of a red colour.

Hence, the correct answer is Option C.

According to the NCERT Chemistry Class 12 textbook (Chapter: Chemistry in Everyday Life), the given structure corresponds to Brompheniramine.

Brompheniramine belongs to the class of drugs known as antihistamines. Antihistamines act by competing with histamine for its receptor sites, thereby preventing histamine from producing its physiological effects. As a result, they are commonly used to relieve symptoms associated with allergic reactions, such as sneezing, itching, runny nose, and watery eyes.

Hence, the given compound is classified as an anti-histamine.

Correct Answer: Option D - Anti-histamine.

First, let us recall a basic biological fact: histamine exerts many of its physiological effects by binding to specialized cell-surface proteins called histamine receptors, mainly of the $$H_1$$ type in the context of allergy.

Now, the class of drugs to which Terfenadine (brand name “Seldane”) belongs is known as “second-generation antihistamines.” The word “antihistamine” itself literally means “against histamine,” so, from the very name, we already suspect that the drug must somehow oppose the action of histamine rather than enhance it.

We have two major ways in which a substance could oppose histamine:

(i) It might stop histamine from being released in the first place, i.e. inhibit its secretion from mast cells.

(ii) It might allow histamine to be released but then block, or antagonize, the receptors so that histamine cannot bind and trigger its usual effects.

Terfenadine belongs to category (ii). Pharmacologically, it is described as an $$H_1$$-receptor antagonist, or more informally, an “$$H_1$$ blocker.” That is, it fits into the $$H_1$$ receptor’s binding site without activating the receptor, thus preventing endogenous histamine from attaching and initiating the cascade that produces itching, swelling, sneezing, and other allergic symptoms.

So, rather than “activating” the histamine receptor (which would mimic histamine) or “helping in secretion” or “inhibiting secretion,” Terfenadine actually inhibits the action of the histamine receptor itself. It neutralizes the receptor’s response, even though histamine molecules might still be floating around in the extracellular fluid.

Comparing this mechanistic description with the options provided, we see that option D—“Inhibits the action of histamine receptor”—matches perfectly.

Hence, the correct answer is Option 4.

First, let us recall that $$\text{maltose}$$ is a disaccharide obtained by the condensation of two $$\alpha\text{-D-glucopyranose}$$ units. These two monosaccharide rings are linked through their anomeric carbon of the first glucose (C-1) and the hydroxyl group attached to C-4 of the second glucose. This linkage is called an $$\alpha(1\rightarrow 4)$$ glycosidic bond.

An anomeric carbon that is part of a free (unbound) ring possesses the $$-OR$$ and $$-OH$$ pattern that characterises a hemiacetal. In contrast, when that anomeric carbon loses its $$-OH$$ during glycoside formation and now bears two $$-OR$$ groups, it is converted into an acetal.

We have one glucose residue acting as the glycosyl donor; its anomeric $$-OH$$ participates in the condensation and is replaced by the $$-O$$ bridge to the second glucose. Therefore the anomeric carbon of this first glucose now has two $$-OR$$ substituents and is an acetal.

The second glucose residue is the glycosyl acceptor; its own anomeric carbon (C-1) does not participate in the bond formation, so it retains the original $$-OH$$ group on the anomeric carbon along with one $$-OR$$ group already present inside the ring. Hence this carbon still displays the $$-OH$$ + $$-OR$$ pattern of a hemiacetal.

Summarising, in maltose we find:

Thus, the molecule contains one acetal group and one hemiacetal group.

Hence, the correct answer is Option C.

First, let us recall the biological definition of an essential amino acid. An essential amino acid is one that the human body cannot synthesize in sufficient quantity, so it must be obtained directly from the diet. Conversely, a non-essential amino acid can be synthesized within the body from other metabolic intermediates.

The universally accepted list of essential amino acids for humans contains nine members, which we may write symbolically inside math delimiters as $$\text{His, Ile, Leu, Lys, Met, Phe, Thr, Trp, Val}.$$ Translating these abbreviations, the list is:

Histidine, Isoleucine, Leucine, Lysine, Methionine, Phenylalanine, Threonine, Tryptophan, and Valine.

Now we examine each option given in the problem and compare it with this list.

Option B gives $$\text{Leucine},$$ which we immediately see in the essential list. So Leucine is an essential amino acid.

Option C gives $$\text{Valine},$$ which also appears in the essential list. Therefore Valine is likewise essential.

Option D gives $$\text{Lysine},$$ another name present in the essential list. Hence Lysine is essential as well.

Option A gives $$\text{Tyrosine}.$$ Tyrosine does not appear in the nine-member essential list. The human body can synthesize Tyrosine from the essential amino acid Phenylalanine through the following metabolic transformation:

$$\text{Phenylalanine} \;\xrightarrow[\text{phenylalanine hydroxylase}]{\;\text{O}_2,\;\text{BH}_4\;} \;\text{Tyrosine}.$$

Because this conversion is readily carried out in normal metabolism, Tyrosine does not need to be supplied by the diet and therefore is classified as non-essential.

From the above comparison, only Tyrosine fails the criterion of being essential.

Hence, the correct answer is Option A.

We begin by examining each of the four statements about glucose one by one, keeping in mind the structural features of glucose in aqueous solution and in its derivatives.

We know that glucose is an aldohexose. In dilute aqueous solution it exists mainly in its two cyclic (hemi-acetal) forms, denoted $$\alpha$$-D-glucopyranose and $$\beta$$-D-glucopyranose, which interconvert through the open-chain aldehydic form. Let us now analyse the assertions.

Statement A. “Glucose exists in two crystalline forms $$\alpha$$ and $$\beta$$.”

The two cyclic stereoisomers just mentioned crystallise separately. The $$\alpha$$-form melts at $$\;146^{\circ}\text{C}$$, whereas the $$\beta$$-form melts at $$\;150^{\circ}\text{C}$$. They are obtained by crystallising glucose from water at different temperatures. Hence Statement A is true.

Statement C. “Glucose reacts with hydroxylamine to form oxime.”

The aldehydic carbonyl group of the small fraction of open-chain glucose is capable of the usual nucleophilic addition reactions. Hydroxylamine ($$\mathrm{NH_2OH}$$) adds to an aldehyde or ketone to yield an oxime according to the general equation

$$\mathrm{R{-}CHO + NH_2OH \;\longrightarrow\; R{-}CH{=}NOH + H_2O}$$

Substituting $$\mathrm{R = (CHOH)_4CH_2OH}$$ for glucose, the same transformation occurs. Therefore Statement C is true.

Statement D. “The pentaacetate of glucose does not react with hydroxylamine to give oxime.”

In glucose pentaacetate, all five hydroxyl groups are converted to acetate esters. The anomeric hydroxyl at $$\mathrm{C\!-\!1}$$ is also acetylated, so the molecule is locked in the cyclic acetal form and cannot open to reveal the $$\mathrm{-CHO}$$ group. With no free carbonyl available, hydroxylamine cannot form an oxime. Consequently Statement D is true.

Statement B. “Glucose gives Schiff’s test for aldehyde.”

Schiff’s reagent detects free, un-hindered aldehyde groups; a positive result is the immediate appearance of a magenta colour. In solution the aldehyde functionality of glucose is effectively masked because the molecule overwhelmingly prefers its intramolecular hemi-acetal forms. Although the open-chain form exists in equilibrium, its concentration is too low and its carbonyl is too hindered to decolourise Schiff’s reagent rapidly. Experimentally, glucose gives a negative (or at best very faint and slow) Schiff’s test. Therefore Statement B is not true.

We have now classified all four statements. Only Statement B fails, while A, C and D are correct.

Hence, the correct answer is Option B.

The sugar $$\text{glucose}$$ has the molecular formula $$\mathrm{C_6H_{12}O_6}$$ and in its open-chain (acyclic) form it possesses

$$\mathrm{\;CHO\!-\![CH(OH)]_4\!-\!CH_2OH}$$

i.e. an aldehydic group $$-CHO$$ at carbon 1 and a primary alcoholic group $$-CH_2OH$$ at carbon 6, the remaining four carbons each bearing a secondary $$-OH$$ group.

When only the aldehydic carbon is oxidised, we replace $$-CHO$$ by the carboxylic group $$-COOH$$ while leaving the alcohol at carbon 6 untouched. The reaction therefore becomes

$$$\mathrm{CHO\!-\![CH(OH)]_4\!-\!CH_2OH \;\xrightarrow[\text{mild}]{[O]} \;COOH\!-\![CH(OH)]_4\!-\!CH_2OH}$$$

The product so obtained is called gluconic acid. Because only one of the two possible oxidisable groups is converted into $$-COOH$$, the process is termed partial oxidation. Gluconic acid is therefore a monocarboxylic acid, not a dicarboxylic one.

Let us now examine each given statement.

Statement A: "Gluconic acid can form cyclic (acetal/hemiacetal) structure." A cyclic acetal (or hemiacetal) in carbohydrates originates from an intramolecular reaction between the aldehydic carbon 1 and an $$-OH$$ group on carbon 4 or carbon 5. Gluconic acid no longer contains an aldehyde; carbon 1 is already oxidised to the acid group $$-COOH$$. Hence such an intramolecular acetal formation is impossible. So statement A is false.

Statement B: "Gluconic acid is a dicarboxylic acid." As written above, gluconic acid possesses only one $$-COOH$$ group (at carbon 1). Carbon 6 still retains $$-CH_2OH$$. Therefore gluconic acid is monocarboxylic, not di-carboxylic. Statement B is false.

Statement C: "Gluconic acid is a partial oxidation product of glucose." We have just shown that mild oxidising agents such as $$\mathrm{Br_2/H_2O}$$ convert only the aldehyde of glucose to the acid without affecting the primary alcohol. Because the molecule is oxidised only in part, the term "partial oxidation product" is fully appropriate. Statement C is therefore true.

Statement D: "Gluconic acid is obtained by oxidation of glucose with $$\mathrm{HNO_3}$$." Nitric acid is a strong oxidising agent; it converts both the aldehyde at carbon 1 and the primary alcohol at carbon 6 into carboxylic groups, giving the dicarboxylic acid $$\mathrm{HOOC\!-\![CH(OH)]_4\!-\!COOH}$$ (called glucaric or saccharic acid). Hence oxidation with $$\mathrm{HNO_3}$$ does not yield gluconic acid. Statement D is false.

Among the four alternatives, only statement C satisfies the chemical facts.

Hence, the correct answer is Option C.

We have to find the statement that does not match the real structure and properties of lactose. Let us collect the accepted facts about the sugar first.

Lactose is made when one molecule of $$\beta\text{-D-galactose}$$ joins with one molecule of $$\text{D-glucose}$$ by removing a molecule of water. The removal of water creates the glycosidic bridge. Symbolically one writes

$$\text{C}_6\text{H}_{12}\text{O}_6+\text{C}_6\text{H}_{12}\text{O}_6-\text{H}_2\text{O}= \text{C}_{12}\text{H}_{22}\text{O}_{11}.$$

So lactose really has the molecular formula $$\text{C}_{12}\text{H}_{22}\text{O}_{11}.$$ Now we test the four statements one after another.

Testing option A. In lactose the anomeric carbon $$C_1$$ of galactose keeps the $$\beta$$-configuration and forms a bridge to the $$C_4$$ of the glucose unit. Chemists therefore write a $$\beta(1\!\rightarrow\!4)$$-glycosidic linkage. Option A, however, talks about an $$\alpha$$-glycosidic linkage. The given linkage description is therefore wrong.

Testing option B. After the bridge is made the anomeric carbon of the glucose molecule is still a free hemi-acetal. A free hemi-acetal can open to the aldehyde form and reduce Fehling’s reagent. Hence lactose is a reducing sugar and it does give Fehling’s test. Thus option B is a true statement.

Testing option C. We have already verified the formula $$\text{C}_{12}\text{H}_{22}\text{O}_{11}.$$ Next we count hydroxyl groups. Before condensation each monosaccharide possesses five -OH groups.

Number of -OH groups before condensation $$5+5=10.$$ During bridge formation one -OH from the anomeric carbon of galactose is lost completely and the -OH on $$C_4$$ of glucose loses only its hydrogen, so the total count of complete hydroxyl groups decreases by one: $$10-1=9.$$

However, the oxygen that remained as the bridge is no longer part of a hydroxyl; hence two complete hydroxyl groups disappear. A quick recount on the actual structure indeed shows eight -OH groups. Option C therefore matches the accepted structure and is true.

Testing option D. The glycosidic bond in lactose is cleaved by dilute acid or by the enzyme lactase. One then simply obtains the two original monosaccharides in the same configuration:

$$\text{lactose}+ \text{H}_2\text{O}\;\xrightarrow[\text{H}^+]{\text{hydrolysis}}\; \text{D(+)-glucose}+ \text{D(+)-galactose}.$$

Hence option D is also true.

Only option A contradicts the real $$\beta(1\!\rightarrow\!4)$$ linkage of lactose, while the other three statements agree with accepted carbohydrate chemistry.

Hence, the correct answer is Option A.

We start by recalling the molecular formula of histamine. Histamine is known to be $$\mathrm{C_5H_9N_3}$$, that is, each molecule contains 5 carbon atoms, 9 hydrogen atoms and 3 nitrogen atoms.

Now, to find the mass percentage of nitrogen, we first need the molar mass of histamine. We list the atomic masses (in g mol−1) that will be used:

$$\text{Atomic mass of C}=12.01,\qquad\text{Atomic mass of H}=1.008,\qquad\text{Atomic mass of N}=14.01$$

We next calculate the total mass contributed by each element in one mole of histamine.

For carbon we have 5 atoms, so

$$m_{\text{C}} = 5 \times 12.01 = 60.05$$

For hydrogen we have 9 atoms, so

$$m_{\text{H}} = 9 \times 1.008 = 9.072$$

For nitrogen we have 3 atoms, so

$$m_{\text{N}} = 3 \times 14.01 = 42.03$$

The molar mass of histamine is obtained by adding these three contributions:

$$M_{\mathrm{C_5H_9N_3}} = 60.05 + 9.072 + 42.03 = 111.152\;\text{g mol}^{-1}$$

We now apply the general formula for percentage composition of an element in a compound:

$$\text{Mass percentage of an element} = \frac{\text{Total mass of that element in one mole of compound}}{\text{Molar mass of the compound}}\times 100$$

Substituting the values for nitrogen we get

$$\%\ \text{N} = \frac{42.03}{111.152}\times 100$$

Carrying out the division first,

$$\frac{42.03}{111.152} \approx 0.378$$

and then multiplying by 100,

$$0.378 \times 100 = 37.8$$

So, the answer is $$37.8\%$$.

First, we recall the definition: a carbon atom is called chiral when it is attached to four different substituents, so that its mirror image is not super-imposable on the original molecule.

Next, we note that in any naturally occurring L-α-amino acid (except glycine) the α-carbon (the carbon directly attached to the -NH2 group, the -COOH group, the side chain R, and a hydrogen atom) satisfies this condition and is therefore chiral. Hence, each residue of Isoleucine (Ile), Arginine (Arg) and Proline (Pro) automatically supplies one chiral α-carbon.

Now, we must examine whether the side chains contribute any additional chiral centres.

(i) For Isoleucine, the side chain is $$-CH(CH_3)CH_2CH_3$$. The first carbon after the α-carbon (called the β-carbon) is attached to four different groups: $$-CH_3,\ -H,\ -CH_2CH_3,$$ and the rest of the backbone. Thus this β-carbon is also chiral. Therefore, Isoleucine contributes $$1\ (\text{α}) + 1\ (\text{β}) = 2$$ chiral carbons.

(ii) For Arginine, the side chain $$-(CH_2)_3-NH-C(=NH)NH_2$$ contains no carbon that is connected to four different groups; all of them are either methylene (-CH2-) or guanidinium carbons with multiple bonds. Hence Arginine contributes only its single α-chiral carbon, giving $$1$$ chiral centre from Arginine.

(iii) For Proline, although its side chain forms a five-membered ring with the α-nitrogen, the α-carbon still bears four different groups (-NH- in the ring, -CO-, -H, and the rest of the ring), so Proline supplies exactly $$1$$ chiral carbon. There are no other stereogenic carbons in the ring.

Adding all the contributions, we have

$$ \text{Total chiral carbons}=2\ (\text{Ile}) + 1\ (\text{Arg}) + 1\ (\text{Pro}) = 4. $$

So, the answer is $$4$$.

We first recall that sucrose is a disaccharide obtained by joining an $$\alpha$$-D-glucopyranose unit with a $$\beta$$-D-fructofuranose unit through an $$\alpha(1\rightarrow2)\beta$$ glycosidic linkage. In other words, the anomeric carbon $$C_1$$ of glucose is linked to the anomeric carbon $$C_2$$ of fructose. Our task is to count how many carbon atoms in the whole sucrose molecule are chiral, i.e. attached to four different substituents.

Let us examine the glucose part first. In the six-membered ring (pyranose form) of D-glucose we number the ring atoms clockwise starting from the anomeric centre. The atoms and their substituents are:

$$ \begin{aligned} C_1 &:~\text{attached to } H,\;O\!-\!\text{fructose},\;O(\text{ring}),\;C_2 \\ C_2 &:~\text{attached to } OH,\;H,\;C_1,\;C_3 \\ C_3 &:~\text{attached to } OH,\;H,\;C_2,\;C_4 \\ C_4 &:~\text{attached to } OH,\;H,\;C_3,\;C_5 \\ C_5 &:~\text{attached to } OH,\;CH_2OH,\;C_4,\;O(\text{ring}) \end{aligned} $$

At every one of $$C_1,\,C_2,\,C_3,\,C_4,$$ and $$C_5$$ the four groups listed are all different, so each of these five carbons is chiral. Thus the glucose moiety contributes

$$ 5 \text{ chiral carbons.} $$

Now we consider the fructose part. In the five-membered furanose ring of D-fructose we number the ring atoms starting at the anomeric carbon $$C_2$$ (because $$C_1$$ is the exocyclic $$CH_2OH$$ group). The relevant atoms are:

$$ \begin{aligned} C_2 &:~\text{attached to } H,\;O\!-\!\text{glucose},\;O(\text{ring}),\;C_3 \\ C_3 &:~\text{attached to } OH,\;H,\;C_2,\;C_4 \\ C_4 &:~\text{attached to } OH,\;H,\;C_3,\;C_5 \\ C_5 &:~\text{attached to } CH_2OH,\;OH,\;C_4,\;O(\text{ring}) \end{aligned} $$

Again, each of $$C_2,\,C_3,\,C_4,$$ and $$C_5$$ is bonded to four distinct groups, and each is therefore chiral. Hence the fructose moiety contributes

$$ 4 \text{ chiral carbons.} $$

Adding the contributions from both halves of the molecule, we obtain

$$ \text{Total chiral carbons} = 5 \;(\text{glucose}) + 4 \;(\text{fructose}) = 9. $$

So, the answer is $$9$$.

First, we recall that a chiral centre (also called an asymmetric carbon) is defined as a carbon atom that is attached to four different groups. Because the four substituents are all different, the carbon becomes a centre of optical activity and gives rise to non-superimposable mirror images (enantiomers).

Now, let us write the condensed structural formula of the amino acid threonine. In its open-chain form it is

$$\mathrm{HO{-}CH{-}(CH_3)\,{-}\,CH(NH_2)\,{-}\,COOH}$$

We can see three carbon atoms in the main chain, which we label for clarity as follows:

$$\mathrm{C_1\;-\;C_2\;-\;C_3}$$

where

• $$C_1$$ is the carbon of the carboxylic acid group $$\mathrm{COOH},$$
• $$C_2$$ is the carbon that bears the $$\mathrm{NH_2}$$ group,
• $$C_3$$ is the carbon that is bonded to the $$\mathrm{OH}$$ group and the $$\mathrm{CH_3}$$ group.

We examine each carbon one by one to test whether it is attached to four different groups.

For $$C_1$$ (the carboxyl carbon): the two atoms of the double bond in $$\mathrm{C=O}$$ make the carbon trigonal planar; moreover, $$C_1$$ is part of a double bond and is not attached to four separate substituents. Hence $$C_1$$ cannot be chiral.

For $$C_2$$ (the α-carbon): the four groups attached are

$$\begin{aligned} &1.\; \mathrm{NH_2} \\ &2.\; \mathrm{H} \\ &3.\; \mathrm{COOH} \\ &4.\; \mathrm{CH(OH)CH_3} \end{aligned}$$

All four groups are distinct, so $$C_2$$ is a chiral centre.

For $$C_3$$ (the β-carbon bearing the hydroxyl group): the four groups attached are

$$\begin{aligned} &1.\; \mathrm{OH} \\ &2.\; \mathrm{H} \\ &3.\; \mathrm{CH_3} \\ &4.\; \mathrm{CH(NH_2)COOH} \end{aligned}$$

Again, each of these four substituents is different from the others, so $$C_3$$ is also a chiral centre.

We have now checked every carbon atom in the molecule. Only $$C_2$$ and $$C_3$$ satisfy the condition of having four distinct groups. Therefore,

$$\text{Number of chiral centres} = 2$$

So, the answer is $$2$$.

Each amino acid residue in a peptide may contribute ionisable (titratable) groups that are NOT involved in peptide-bond formation.

Peptide given: Asp - Glu - Lys

Step 1 Identify the terminal (α) groups that remain free
• N-terminus (α-NH3+) comes from the first residue (Asp). ⇒ 1 group
• C-terminus (α-COO-) belongs to the last residue (Lys). ⇒ 1 group

Step 2 Check the side chains of each residue
• Aspartic acid (Asp) side chain: one extra -COOH group. ⇒ 1 group
• Glutamic acid (Glu) side chain: one extra -COOH group. ⇒ 1 group
• Lysine (Lys) side chain: one extra -NH3+ (ε-amino) group. ⇒ 1 group

Step 3 Add all ionisable groups
Total groups = 1 (N-terminal α-NH3+) + 1 (C-terminal α-COO-) + 1 (Asp side -COOH) + 1 (Glu side -COOH) + 1 (Lys side -NH3+) = 5

Hence, the tripeptide Asp - Glu - Lys possesses 5 ionisable groups.

Answer: 5

1. Step 1: Diazotization and Substitution

   Treatment of the primary aliphatic amine ($$-\text{NH}_2$$) with nitrous acid ($$\text{NaNO}_2 / \text{H}_3\text{O}^\oplus$$) converts it into a highly unstable diazonium salt, which decomposes immediately in the aqueous medium to form a hydroxyl ($$-\text{OH}$$) group:

   $$\text{HOOC--}(\text{CH}_2)_4\text{--NH}_2 \xrightarrow{\text{NaNO}_2/\text{H}_3\text{O}^\oplus} \text{HOOC--}(\text{CH}_2)_4\text{--OH}$$



2. Step 2: Condensation Polymerization

   Heating this bifunctional $\omega$-hydroxy acid monomer initiates self-esterification (condensation polymerization) accompanied by the continuous loss of water ($$\text{H}_2\text{O}$$) molecules, yielding a linear polyester chain:

   $$n\,\text{HOOC--}(\text{CH}_2)_4\text{--OH} \longrightarrow \left[\text{--O--}(\text{CH}_2)_4\text{--C}(=\text{O})\text{--}\right]_n + n\,\text{H}_2\text{O}$$

Conclusion:

The conversion of the amine into an alcohol yields a hydroxy acid monomer that polymerizes into a polyester structure rather than a polyamide.

Answer: Option showing the polyester repeating unit: $$\left[\text{--O--}(\text{CH}_2)_4\text{--C}(=\text{O})\text{--}\right]_n$$

First, we recall the basic definition of a thermosetting polymer. A thermosetting polymer is a polymer that, once molded or set by heating, becomes infusible and insoluble. This irreversible hardening occurs because extensive cross-linking forms a three-dimensional network. After this curing, reheating cannot soften or melt the material; instead, it simply chars or decomposes.

Now, we consider each option one by one and decide whether it exhibits such cross-linked, permanently hard characteristics.

We have Option A, Bakelite. Bakelite is the trade name for phenol-formaldehyde resin. During its synthesis, phenol reacts with formaldehyde through a condensation reaction, initially producing a linear Novolac. Upon heating in the presence of a curing agent such as hexamethylenetetramine, Novolac undergoes further condensation to create extensive $$−CH_{2}−$$ cross-links connecting the aromatic rings. This yields a rigid, three-dimensional network. Because of these cross-links, Bakelite cannot be remelted; it is a classic thermosetting polymer used for electrical switches, handles of utensils, and radio and television casings.

Next, we look at Option B, PVC, or polyvinyl chloride. PVC is obtained by addition polymerisation of vinyl chloride $$\left(CH_{2}=CHCl\right)$$. The structure is essentially linear with slight branching; there is no extensive cross-linking. When we heat PVC, it can soften and then be moulded, and upon cooling it solidifies again. This reversibility identifies PVC as a typical thermoplastic, not a thermoset.

Then, Option C is Nylon 6. Nylon 6 is a polyamide formed by ring-opening polymerisation of caprolactam $$\left((CH_{2})5C(O)NH\right)$$. The polymer chain contains repeating $$−NH−(CH_{2})5−CO−$$ units. Although hydrogen bonding imparts high tensile strength and crystallinity, the polymer chains are still essentially linear. Heating Nylon 6 allows it to melt and be reshaped, which is the hallmark of a thermoplastic, not a thermosetting polymer.

Finally, Option D is Buna-N, also called NBR (nitrile-butadiene rubber). It is a copolymer of acrylonitrile and butadiene obtained through free-radical addition polymerisation. Buna-N is an elastomer, meaning it exhibits elasticity; it can be vulcanised to introduce some sulphur cross-links, but even vulcanised rubber does not possess the infusible, insoluble three-dimensional network characteristic of true thermosets like Bakelite.

So, comparing all four substances, only Bakelite meets the strict criterion of a thermosetting polymer with irreversible cross-linking formed upon curing.

Hence, the correct answer is Option A.

• (A) Norethindrone: A synthetic progesterone derivative widely used as an antifertility drug.
  $$\implies \text{(A)} \rightarrow \text{(Q)}$$
• (B) Ofloxacin: A fluoroquinolone derivative used as a broad-spectrum antibiotic.
  $$\implies \text{(B)} \rightarrow \text{(P)}$$
• (C) Equanil: A medication used to manage anxiety and hypertension, classified as a tranquilizer (not an analgesic, but typically pairs with the remaining relevant therapeutic class or specific target line in textbook questions).
  $$\implies \text{(C)} \rightarrow \text{(R)}$$

Conclusion:

Aligning the medicinal categories yields the correct sequence match.

Answer: (A) → (Q); (B) → (P); (C) → (R)

The given polymer structure is a Urea-Formaldehyde resin, a cross-linked copolymer synthesized via a condensation polymerization reaction between Urea and Formaldehyde. During the initial stage of the reaction, the nucleophilic nitrogen atoms of urea attack the electrophilic carbonyl carbon of formaldehyde to form an intermediate known as mono=methylol urea. As polymerization progresses, this intermediate undergoes a condensation reaction with neighboring urea molecules, eliminating a molecule of water for every link formed. Because the oxygen atom from the formaldehyde is entirely removed as part of the leaving water molecule, only the stable methylene group remains in the final polymer backbone to bridge the symmetrical units provided by the urea. Therefore, among the choices provided, both Urea and Formaldehyde are the correct constitutional monomers of this polymer.

We first recall the definition of an epimer. Two monosaccharides that differ only in the configuration around a single chiral carbon atom (other than the anomeric carbon) are called epimers. Mathematically, if a molecule has several chiral centres $$C\_1, C\_2, C\_3, \ldots , C\_n$$ and two structures are identical at all centres except one $$C\_k$$, then they are epimers at $$C\_k$$.

Now, D-Glucose and D-Galactose are both aldohexoses, so each contains six carbon atoms arranged linearly in the Fischer projection as:

$$\text{CHO - C}_2\text{H(OH) - C}_3\text{H(OH) - C}_4\text{H(OH) - C}_5\text{H(OH) - CH}_2\text{OH}$$

In the Fischer projection for D-Glucose, the orientations of the OH groups (from $$C\_2$$ to $$C\_5$$) are:

$$C_2$$: Right, $$C_3$$: Left, $$C_4$$: Right, $$C_5$$: Right

For D-Galactose, the corresponding orientations are:

$$C_2$$: Right, $$C_3$$: Left, $$C_4$$: Left, $$C_5$$: Right

We carefully compare these two sets.

At $$C\_2$$ both molecules have the OH group on the right. At $$C\_3$$ both have the OH on the left. At $$C\_5$$ both have the OH on the right.

The only discrepancy appears at $$C\_4$$:

• In D-Glucose, the OH at $$C\_4$$ is on the right.
• In D-Galactose, the OH at $$C\_4$$ is on the left.

Since every other chiral centre matches, D-Glucose and D-Galactose differ only in the configuration at carbon number 4. Therefore they are C-4 epimers.

Looking back at the options, Option B states “C - 4”. This matches our conclusion.

Hence, the correct answer is Option B.

We have the substance “noradrenaline,” which is also called “norepinephrine.” In human physiology, noradrenaline is released from the termini of sympathetic nerve fibers and from the adrenal medulla. Its primary role is to convey chemical messages from one neuron to the next or from a neuron to an effector organ. Any chemical that carries such signals across a synapse is defined as a neurotransmitter.

Now, let us compare this role with the meanings of the four given options. An antihistamine counteracts the effects of histamine during allergic reactions, an antacid neutralizes excess gastric acid, and an antidepressant is a drug used to treat clinical depression by altering neurotransmitter levels but is itself not necessarily a neurotransmitter. Only the term neurotransmitter exactly matches the intrinsic biological function of noradrenaline.

So, after reviewing the definitions, the only accurate classification for noradrenaline is “neurotransmitter.”

Hence, the correct answer is Option B.

First, we remind ourselves of what the word “copolymer” means. A copolymer is a polymer that is formed when two different monomer molecules link together in the same long chain. Thus, to identify the monomers of PHBV we must recall the two specific β-hydroxy carboxylic acids that join alternately to build up this polymer.

The name given in the question is “poly-β-hydroxybutyrate-co-β-hydroxyvalerate.” Let us break this down:

• “β-hydroxybutyrate” is the conjugate base of β-hydroxybutyric acid.
• “β-hydroxyvalerate” is the conjugate base of β-hydroxyvaleric acid.

We must therefore translate these common names into proper IUPAC names so that we can match them with the options.

Step 1 - Identifying β-hydroxybutyric acid.
The parent chain “butyric” tells us there are four carbon atoms (butanoic acid). The “β-hydroxy” prefix tells us that the hydroxyl group lies on the β-carbon, which is the third carbon from the carboxyl carbon when we start numbering at the -COOH group. Hence, its IUPAC name is $$3\text{-hydroxybutanoic\ acid}.$$ The structure is $$CH_3-CH(OH)-CH_2-COOH.$$

Step 2 - Identifying β-hydroxyvaleric acid.
“Valeric” corresponds to a five-carbon straight chain (pentanoic acid). Once again “β-hydroxy” places the -OH on the β-carbon, i.e. the third carbon from -COOH. Therefore its IUPAC name is $$3\text{-hydroxypentanoic\ acid},$$ with structure $$CH_3-CH_2-CH(OH)-CH_2-COOH.$$

Now, let us compare these IUPAC names with the options given:

A. 3-hydroxybutanoic acid and 4-hydroxypentanoic acid ⟹ wrong second position
B. 3-hydroxybutanoic acid and 2-hydroxypentanoic acid ⟹ wrong second position
C. 3-hydroxybutanoic acid and 3-hydroxypentanoic acid ⟹ both positions correctly at 3 (β-position)
D. 2-hydroxybutanoic acid and 3-hydroxypentanoic acid ⟹ wrong first position

Only Option C contains both monomers in which the hydroxyl group is indeed on the β-carbon (carbon-3) of each respective acid, matching the common names β-hydroxybutyric acid and β-hydroxyvaleric acid that build PHBV.

Hence, the correct answer is Option C.

We start by recalling how each of the given polymers is manufactured and which catalyst or reaction condition is characteristically employed.

High density polythene, commonly abbreviated as HDPE, is obtained by polymerising ethene in the presence of the well-known Ziegler-Natta catalyst, a mixture of $$\text{TiCl}_4$$ and $$\text{(C}_2\text{H}_5)_3\text{Al}$$. So we can write the correspondence as

High density polythene}\; \longrightarrow \; \text{Ziegler-Natta catalyst (III)

Polyacrylonitrile, abbreviated as PAN, is produced by the free-radical polymerisation of acrylonitrile. A peroxide such as benzoyl peroxide or hydrogen peroxide supplies these free radicals. Therefore, the link is

Polyacrylonitrile}\; \longrightarrow \; \text{Peroxide catalyst (I)

Novolac is a phenol-formaldehyde resin formed by the condensation of phenol with formaldehyde when an acid or a base acts as the catalyst. Thus, we have

$$\text{Novolac} \; \longrightarrow \;$$ Acid or base catalyst (IV)

Nylon 6 is synthesised by the ring-opening polymerisation (a step-growth process) of $$\epsilon$$-caprolactam. This polymerisation is carried out at a high temperature of about $$533 \text{ K}$$ and a pressure of roughly $$10^4 \text{ Pa}$$, so it is categorised as condensation at high temperature and pressure. Hence,

Nylon 6}\; \longrightarrow \; \text{Condensation at high temperature \& pressure (II)

Collecting all four pairings, we obtain

$$\begin{aligned} (a)&: \text{HDPE} &\rightarrow&\; (III) \\ (b)&: \text{Polyacrylonitrile} &\rightarrow&\; (I) \\ (c)&: \text{Novolac} &\rightarrow&\; (IV) \\ (d)&: \text{Nylon 6} &\rightarrow&\; (II) \end{aligned}$$

Looking at the given options, this exact sequence corresponds to Option B.

Hence, the correct answer is Option B.

NCERT Facts

• Repeating Unit: The backbone consists of a repeating chain unit featuring a quaternary carbon bonded to two methyl groups ($$-\text{CH}_2-\text{C}(\text{CH}_3)_2-$$).
• Monomer Identification: This polymer is formed via the addition polymerization of isobutylene (2-methylpropene, $$\text{CH}_2=\text{C}(\text{CH}_3)_2$$).

Conclusion:

Polymerizing isobutylene monomers results directly in polyisobutylene.

Answer: Option A — Polyisobutylene

We need to determine which of the three statements about potassium ions ($$K^+$$) are correct.

Statement I: "They activate many enzymes."

This is correct. Potassium ions are known to act as enzyme activators. They are required for the activity of several enzymes, including pyruvate kinase, which is involved in glycolysis. $$K^+$$ ions help in maintaining the proper conformation of enzymes for catalytic activity.

Statement II: "They participate in the oxidation of glucose to produce ATP."

This is correct. Potassium ions play a role in the oxidation of glucose to produce ATP (adenosine triphosphate). The $$K^+$$ ions are essential cofactors in the glycolytic pathway, where they activate pyruvate kinase, a key enzyme in the final step of glycolysis that produces ATP.

Statement III: "Along with sodium ions, they are responsible for the transmission of nerve signals."

This is correct. The transmission of nerve impulses (nerve signals) depends on the movement of $$Na^+$$ and $$K^+$$ ions across the nerve cell membrane. The sodium-potassium pump ($$Na^+/K^+$$-ATPase) maintains the concentration gradient of these ions across the cell membrane, which is essential for generating action potentials and transmitting nerve signals.

Since all three statements (I, II, and III) are correct, the answer is Option A: I, II and III.

There are six carbons between two nitrogens in Nylon-6 hence the name. The structure is to be remembered.

We begin by recalling the definition of a condensation polymer. A condensation polymer is produced when monomer molecules join together with the simultaneous elimination of small molecules such as water, $$\mathrm{HCl}$$, methanol, etc. This is in contrast to an addition polymer, where the monomers add together without the loss of any small molecule.

Now we look at each option in turn and relate it to the definition.

First, consider Option A, Nylon 6,6. Nylon 6,6 is prepared from two different monomers: hexamethylenediamine, $$\mathrm{H_2N{-}(CH_2)_6{-}NH_2}$$, and adipic acid, $$\mathrm{HOOC{-}(CH_2)_4{-}COOH}$$. When these react, each -COOH group of adipic acid combines with an -NH2 group of the diamine, eliminating a molecule of water. Stating the general reaction,

$$\mathrm{H_2N{-}(CH_2)_6{-}NH_2 \;+\; HOOC{-}(CH_2)_4{-}COOH \;\longrightarrow\; -NH{-}(CH_2)_6{-}NHCO{-}(CH_2)_4{-}CO- \;+\; H_2O}$$

Because water is eliminated at every step while the chain grows, the process clearly fits the definition of condensation polymerisation. So Nylon 6,6 is a condensation polymer.

Next, look at Option B, Neoprene. Neoprene is obtained by the free-radical polymerisation of chloroprene, $$\mathrm{CH_2=CCl{-}CH=CH_2}$$. No small molecule is released during this process, so it is an addition polymer, not a condensation polymer.

Option C, Buna-S, is produced by copolymerising 1,3-butadiene, $$\mathrm{CH_2=CH{-}CH=CH_2}$$, with styrene, $$\mathrm{C_6H_5CH=CH_2}$$. Again, no small molecule is eliminated; therefore Buna-S is an addition polymer.

Finally, Option D, Teflon (polytetrafluoroethylene), is formed by the polymerisation of tetrafluoroethene, $$\mathrm{CF_2=CF_2}$$, without the loss of any by-product. Hence, Teflon is also an addition polymer.

Summarising, only Nylon 6,6 among the given choices satisfies the criterion for a condensation polymer because its synthesis involves the stepwise elimination of water molecules.

Hence, the correct answer is Option A.

The problem asks us to decide which one of the four given qualitative tests is not suitable for detecting or identifying amino acids. To reach the conclusion, we shall recall, one by one, what chemical group each test responds to, and then match that information with the functional groups present in amino acids.

First, let us note that a typical α-amino acid contains two principal functional groups: an amino group $$\left( -\mathrm{NH_2} \right)$$ and a carboxyl group $$\left( -\mathrm{COOH} \right)$$. Depending on the side chain, extra groups such as phenyl, hydroxyl, or sulfur may also be present, but the defining feature is the $$- \mathrm{NH_2}$$ group on the carbon that also bears the $$- \mathrm{COOH}$$ group.

Now we analyse each test:

Biuret test: The Biuret test is based on the complexation of $$\mathrm{Cu^{2+}}$$ ions in alkaline medium with two or more peptide (-CONH-) linkages. Even a single amino acid does not contain a peptide bond by itself, yet when several amino acids are linked (as in di-, tri-, or polypeptides), the test becomes positive and yields a violet or purple colour. In many practical laboratory manuals, the Biuret test is nevertheless mentioned in the context of proteins and peptides—derivatives of amino acids. Hence, although a free amino acid gives a very faint or negative response, the test is broadly connected with amino‐acid chemistry.

Xanthoproteic test: This test relies on the nitration of activated aromatic rings (such as those present in the side chains of tyrosine, tryptophan, and phenylalanine) by concentrated $$\mathrm{HNO_3}$$, producing a yellow nitro derivative. On making the solution alkaline, the colour deepens to orange. Because many amino acids possess aromatic rings, the Xanthoproteic test is employed specifically to identify such amino acids.

Ninhydrin test: The Ninhydrin reaction is a classic general test for free amino acids. Ninhydrin oxidatively deaminates $$\alpha$$-amino acids, giving $$\mathrm{CO_2}$$, an aldehyde with one fewer carbon, and ammonia. Simultaneously, the reduced ninhydrin condenses with the liberated ammonia to form Ruhemann’s purple, a deep blue-violet colour. Therefore, the Ninhydrin test is explicitly designed for detecting amino acids.

Barfoed test: Barfoed’s reagent is an acidic solution of $$\mathrm{Cu^{2+}}$$ ions (commonly $$\mathrm{Cu(CH_3COO)_2}$$ in dilute acetic acid). The reagent is reduced by monosaccharides—that is, simple reducing sugars—to metallic copper or cuprous oxide $$\left(\mathrm{Cu_2O}\right)$$, giving a brick-red precipitate upon heating. The operationally important point is that the reagent is kept acidic so that only monosaccharides, and not disaccharides, reduce it at an appreciable rate. The Barfoed test has nothing to do with amino groups or peptide bonds; it is designed solely to distinguish monosaccharides from higher carbohydrates.

We now compare: Biuret, Xanthoproteic, and Ninhydrin have direct or at least partial applicability to amino acids or their polypeptide derivatives, whereas Barfoed’s test concerns reducing sugars and does not respond to amino or peptide functional groups at all.

Therefore, the test that cannot be used for identifying amino acids is Barfoed’s test.

Hence, the correct answer is Option C.

We begin by recalling the general idea of basicity in amino acids. The basic strength of an amino acid depends upon the availability of a lone pair of electrons on a nitrogen atom that can readily accept a proton, $$\text{H}^+.$$ The more easily this lone pair can be donated, the higher will be the basicity of the molecule.

Now, let us examine each of the four given amino acids in turn, always keeping in mind the presence (or absence) of additional nitrogen-containing groups that can provide extra basic sites beyond the normal $$-NH_{2}$$ of the amino acid backbone.

For clarity, we write the structural fragments of the side chains:

• Lysine has the side chain $$-CH_{2}-CH_{2}-CH_{2}-CH_{2}-NH_{2}.$$ Thus, in addition to the backbone $$-NH_{2}$$ group, lysine possesses another terminal $$-NH_{2}$$ group on the side chain, giving it two basic nitrogens.

• Asparagine has the side chain $$-CH_{2}-CONH_{2}.$$ The amide nitrogen in $$-CONH_{2}$$ is involved in resonance with the adjacent carbonyl group $$C=O,$$ which withdraws electron density and reduces the availability of its lone pair. Hence, this nitrogen is a poor base.

• Serine has the side chain $$-CH_{2}-OH.$$ There is no additional nitrogen present; the -OH group is neutral and actually exhibits weak acidity rather than basicity. Therefore, serine has only the single backbone $$-NH_{2}$$ and is not strongly basic.

• Histidine contains an imidazole ring with two nitrogens. One of these nitrogens has a lone pair that can accept a proton, making histidine noticeably basic. However, quantitative pKa data show that the side-chain imidazole nitrogen has a pKa of about $$6.0,$$ whereas the terminal $$-NH_{2}$$ of lysine’s side chain has a pKa of about $$10.5.$$ A larger pKa for the conjugate acid corresponds to a stronger base.

We can summarise the essential trend by writing:

$$\text{Basicity} \propto \text{p}K_a(\text{conjugate acid})$$

and then substituting the approximate numerical values:

$$\begin{aligned} \text{Lysine : } & pK_a \approx 10.5 \\ \text{Histidine : } & pK_a \approx 6.0 \\ \text{Asparagine : } & pK_a \text{ very low (amide; non-basic)} \\ \text{Serine : } & \text{no extra basic site} \end{aligned}$$

Since $$10.5 > 6.0,$$ the side-chain amine of lysine is able to accept a proton much more readily than the imidazole nitrogen of histidine. Asparagine and serine lack strongly basic nitrogens altogether.

Therefore, lysine exhibits the greatest overall basicity among the four options.

Hence, the correct answer is Option A.

Key Differences Between DNA and RNA Bases:

• Adenine (A), Cytosine (C), and Guanine (G): Found in both DNA and RNA.
• Thymine (T): Found exclusively in DNA.
• Uracil (U): Serves as the structural replacement for thymine and is found only in RNA.

Conclusion:

Uracil is the unique pyrimidine base characteristic of RNA molecules.

We begin by recalling the basic facts about starch. Starch, which is the principal storage polysaccharide of plants, is made of two structurally different components - amylose and amylopectin. Amylose is an unbranched polymer, whereas amylopectin is a highly branched polymer. Our task is to identify the nature of the monosaccharide units and the type of glycosidic linkages present in amylopectin.

First, we note the type of monosaccharide. Both amylose and amylopectin are built only from glucose units, not from any other hexose. Moreover, the ring form involved in starch is the α-anomer of D-glucose. That is, every glucose residue appears in the $$\alpha$$-D-glucopyranose configuration rather than the $$\beta$$-form.

Now, let us remind ourselves of the linkages. In an unbranched chain (the straight portion of the polymer), each glucose joins to the next through a glycosidic bond formed between the anomeric carbon $$C_1$$ of one residue and the $$C_4$$ hydroxyl of the next. This is written symbolically as a $$C_1 - C_4$$ or $$1 \rightarrow 4$$ linkage, and because the anomeric carbon is in the $$\alpha$$-orientation, we call it an $$\alpha(1 \rightarrow 4)$$ glycosidic bond.

Because amylopectin is branched, there must be additional linkages at branch points. Experimentally it is known that the branch arises when the anomeric carbon of one glucose connects to the $$C_6$$ hydroxyl of another glucose unit. This second type of bond is therefore an $$\alpha(1 \rightarrow 6)$$ linkage, often stated as a $$C_1 - C_6$$ linkage.

Putting these facts together we can summarise: amylopectin consists exclusively of $$\alpha$$-D-glucose units, with straight-chain segments held by $$\alpha(1 \rightarrow 4)$$ or $$C_1 - C_4$$ linkages, and branch points introduced via $$\alpha(1 \rightarrow 6)$$ or $$C_1 - C_6$$ linkages.

We now compare this conclusion with the four options given in the question. Option C reads: “$$\alpha$$-D-glucose, $$C_1 - C_4$$ and $$C_1 - C_6$$ linkages.” That matches exactly with the description we have just derived.

Hence, the correct answer is Option C.

We are asked to decide which one of the four classical qualitative tests can tell fructose apart from glucose. Let us first recall the structural difference between the two sugars. Glucose is an aldohexose; that is, the carbonyl group is an aldehyde, written schematically as $$\mathrm{-CHO}$$ at the C-1 position. Fructose, on the other hand, is a ketohexose; its carbonyl group is a ketone, written as $$\mathrm{>C=O}$$ at the C-2 position. Therefore any test that responds differently to an aldose versus a ketose will serve to distinguish the two.

We now examine each test:

  • Fehling’s test and Benedict’s test are both based on the same principle: an alkaline cupric ion complex $$\mathrm{Cu^{2+}}$$ is reduced by any reducing sugar to give a red precipitate of cuprous oxide $$\mathrm{Cu_2O}$$. Both glucose and fructose possess free or enolisable carbonyl groups, so both reduce the reagent: $$\text{Reducing sugar} + 2\;\mathrm{Cu^{2+}} + 5\;\mathrm{OH^-} \;\longrightarrow\; \text{oxidized sugar} + \mathrm{Cu_2O}\downarrow + 3\;\mathrm{H_2O}.$$ Since the outcome is the same for the two sugars, neither Fehling’s nor Benedict’s test can differentiate them.

  • Barfoed’s test employs a weakly acidic solution of cupric acetate to distinguish monosaccharides from disaccharides because monosaccharides reduce $$\mathrm{Cu^{2+}}$$ more rapidly under mildly acidic conditions. Both glucose and fructose are monosaccharides, so both will give the same positive result; again, no distinction is possible.

  • Seliwanoff’s test is specifically designed to distinguish ketoses from aldoses. The reagent is a solution of resorcinol in concentrated hydrochloric acid. Under these strongly acidic conditions, a ketose such as fructose undergoes dehydration much faster than an aldose like glucose, producing hydroxymethylfurfural, which then condenses with resorcinol to yield a characteristic cherry-red complex: $$\text{Ketose} \;\xrightarrow[\text{conc. } \mathrm{HCl}]{\text{dehydration}} \text{hydroxymethylfurfural}$$ $$\text{hydroxymethylfurfural} + \text{resorcinol} \;\longrightarrow\; \text{red condensation product}.$$ An aldose forms the same furfural derivative far more slowly, so glucose either remains colorless or develops only a faint pink colour after a prolonged period. Thus, by observing the rapid appearance of an intense red colour with fructose and its absence (or delayed appearance) with glucose, the two sugars can be clearly distinguished.

Because only Seliwanoff’s test reacts differently with fructose and glucose, it is the unique choice that fulfills the requirement.

Hence, the correct answer is Option D.

We first recall what maltose is. Maltose is a disaccharide whose molecular formula is $$\mathrm{C_{12}H_{22}O_{11}}$$ and whose structure consists of two units of $$\alpha\text{-D-glucopyranose}$$ (that is, two $$\alpha$$-D-glucose molecules) joined together through an $$\alpha(1 \rightarrow 4)$$ glycosidic linkage.

Whenever a disaccharide is treated with dilute mineral acid such as $$\text{dil.\,HCl}$$ in the presence of water and heat, the process that occurs is called acid-catalysed hydrolysis. The general chemical fact we use is:

“A glycosidic bond is cleaved by water in the presence of dilute acid, giving back its constituent monosaccharide units.”

Writing the reaction in symbolic form, we have

$$\text{Disaccharide} + \mathrm{H_2O} \xrightarrow[\text{heat}]{\text{dil.\,HCl}} \text{Monosaccharide}_1 + \text{Monosaccharide}_2.$$

Substituting maltose specifically, the equation becomes

$$\mathrm{C_{12}H_{22}O_{11}} + \mathrm{H_2O} \;\xrightarrow[\text{heat}]{\text{dil.\,HCl}}\; \mathrm{C_6H_{12}O_6} + \mathrm{C_6H_{12}O_6}.$$

Each $$\mathrm{C_6H_{12}O_6}$$ appearing on the right-hand side represents a molecule of D-glucose, because both halves of maltose are $$\alpha$$-D-glucose units to begin with. No other monosaccharide (such as fructose or galactose) is present in maltose, so none can appear after hydrolysis. Thus the only product is D-glucose.

So, upon treatment of maltose with dilute $$\mathrm{HCl}$$, we obtain exclusively D-glucose.

Hence, the correct answer is Option A.

We start by recalling the definition of a stereocentre (also called a chiral centre). A stereocentre is a carbon atom that is attached to four different groups, so that interchanging any two of those groups produces a non-superimposable mirror image.

Now we draw in mind the Fischer projection of D-glucose in its open-chain (linear) form:

$$$\mathrm{HO{-}CH_2{-}C(=O)-HC(OH)-HC(OH)-HC(OH)-HC(OH)-CH_2OH}$$$

Written carbon by carbon, the skeleton is

$$\begin{aligned} \text{C}_1 &: \; \mathrm{CHO} \\ \text{C}_2 &: \; \mathrm{H{-}C(OH)-} \\ \text{C}_3 &: \; \mathrm{H{-}C(OH)-} \\ \text{C}_4 &: \; \mathrm{H{-}C(OH)-} \\ \text{C}_5 &: \; \mathrm{H{-}C(OH)-} \\ \text{C}_6 &: \; \mathrm{CH_2OH} \end{aligned}$$

We examine every carbon one by one:

• $$\text{C}_1$$ is part of the aldehyde group $$\mathrm{-CHO}$$ and is double-bonded to oxygen; therefore it is not bonded to four different substituents. It is achiral.

• $$\text{C}_2$$ has the four different groups $$\mathrm{-CHO,\; OH,\; H,\; CH(OH)-}$$. This satisfies the chiral condition, so $$\text{C}_2$$ is a stereocentre.

• $$\text{C}_3$$ is attached to $$\mathrm{-C(=O)H,\; OH,\; H,\; CH(OH)-}$$ (all four groups different). Thus $$\text{C}_3$$ is also a stereocentre.

• $$\text{C}_4$$ in the same way has four different groups, so $$\text{C}_4$$ is a stereocentre.

• $$\text{C}_5$$ is bonded to $$\mathrm{-C(OH)-,\; OH,\; H,\; CH_2OH}$$, again four distinct groups; hence $$\text{C}_5$$ is a stereocentre.

• $$\text{C}_6$$ has two identical hydrogens (it is $$\mathrm{CH_2OH}$$), so it is not chiral.

Counting, the linear form possesses

$$4 \text{ stereocentres: } C_2,\, C_3,\, C_4,\, C_5.$$

Next we convert glucose into its more stable cyclic (pyranose) form. The aldehyde carbon $$\text{C}_1$$ reacts intramolecularly with the hydroxyl on $$\text{C}_5$$, producing a hemiacetal. In the Haworth representation this looks like:

$$$\text{(Fructose open-chain structure: HO-CH}_2\text{-C(=O)-HC(OH)-HC(OH)-HC(OH)-CH}_2\text{OH)}$$$

The crucial observation is that during ring formation $$\text{C}_1$$—formerly achiral—now becomes bonded to four different groups: $$\mathrm{O\,(ring)},\; H,\; OH\,(anomeric),\; \text{and } C_2$$. Therefore $$\text{C}_1$$ turns into a new stereocentre, called the anomeric carbon.

The pre-existing centres $$\text{C}_2,\, C_3,\, C_4,\, C_5$$ remain chiral as before. Hence, in the cyclic form we have

$$5 \text{ stereocentres: } C_1,\, C_2,\, C_3,\, C_4,\, C_5.$$

Summarising the count:

$$\text{Linear (open)} = 4,\quad \text{Cyclic (pyranose)} = 5.$$

Comparing with the given options, the pair (4 and 5) corresponds to Option C.

Hence, the correct answer is Option C.

The starting material is a dipeptide composed of two amino acids: Asparagine (Asn) at the N-terminus and Serine (Ser) at the C-terminus. Combined, they contain several nucleophilic functional groups capable of reacting:

• N-terminal alpha-amine group: Found on the Asparagine residue (-NH2).
• Side-chain amide group: Found on the Asparagine side chain (-CONH2).
• Side-chain hydroxyl group: Found on the Serine side chain (-OH).
• C-terminal carboxylic acid group: Found on the Serine residue (-COOH).
• Peptide bond amide group: The linkage holding the two amino acids together (-CONH-).

2. The Reagents and Reaction Type

The reagents are acetic anhydride (CH3CO)2O in excess and triethylamine (NEt3) as a base.

This is an acetylation reaction. Acetic anhydride introduces acetyl groups (CH3CO-) to nucleophilic atoms like nitrogen and oxygen. Because it is explicitly added in excess, every nucleophilic group that is strong enough to react will become fully acetylated.

3. Comparing Nucleophilicity: What Reacts and What Doesn't?

Not all nitrogen and oxygen atoms in a peptide react equally with acetic anhydride. Here is how they behave:

• The N-terminal Amine (-NH2) Reacts: Primary aliphatic amines are highly nucleophilic. It readily attacks acetic anhydride to form a stable amide group (-NH-COCH3).
• The Serine Hydroxyl (-OH) Reacts: The aliphatic alcohol group on Serine is nucleophilic enough to attack acetic anhydride, converting the alcohol into an ester group (-O-COCH3).
• The Amide Groups Do NOT React: Both the side-chain amide (-CONH2) of Asparagine and the internal peptide backbone amide (-CONH-) are very poor nucleophiles. This is because the lone pairs on these nitrogen atoms are heavily delocalized into their adjacent carbonyl groups through resonance. Therefore, they remain unchanged.
• The Carboxylic Acid (-COOH) Does NOT get permanently acetylated: While it can form a temporary mixed anhydride, under these standard conditions it reverts back or remains as the free carboxylic acid (-COOH) in the final isolated peptide product.

First, recall the definition of $$pK_a$$. By the relation $$pK_a = -\log_{10}K_a,$$ a lower $$pK_a$$ value corresponds to a larger acid-dissociation constant $$K_a,$$ hence to a stronger acid. Consequently, when we arrange substances in increasing order of $$pK_a,$$ we are really writing them from the strongest acid (smallest $$pK_a$$) to the weakest acid (largest $$pK_a$$).

Every α-amino acid possesses at least two titratable groups: an α-carboxyl group (-COOH) with $$pK_{a1}\,{\approx}\,2$$ and an α-ammonium group (-NH3+) with $$pK_{a2}\,{\approx}\,9$$. Amino acids that contain an additional ionisable group in the side chain (R-group) have a third $$pK_a,$$ usually written $$pK_{aR}$$. For such amino acids, it is the $$pK_{aR}$$ that is characteristic of the nature of the side chain and varies widely from one amino acid to another, while the α-carboxyl and α-ammonium values stay in their usual narrow ranges.

We now list the pertinent numeric values taken from standard aqueous titration tables:

• Glycine (no ionisable R-group):
   $$pK_{a1}(\text{-COOH}) \approx 2.34$$,    $$pK_{a2}(\text{-NH}_3^+) \approx 9.60$$.

• Aspartate (extra -COOH in the side chain):
   $$pK_{a1}(\text{α-COOH}) \approx 2.09,$$    $$pK_{aR}(\text{β-COOH}) \approx 3.86,$$    $$pK_{a2}(\text{-NH}_3^+) \approx 9.82.$$
Among these, the R-group value $$pK_{aR} = 3.86$$ is the one that distinguishes Aspartate from other amino acids.

• Lysine (extra ε-NH3+ group):
   $$pK_{a1}(\text{α-COOH}) \approx 2.18,$$    $$pK_{a2}(\text{-NH}_3^+\text{ on α-C}) \approx 8.95,$$    $$pK_{aR}(\text{ε-NH}_3^+) \approx 10.53.$$
Here the characteristic value is the basic side-chain $$pK_{aR}=10.53.$$

• Arginine (guanidinium side chain):
   $$pK_{a1}(\text{α-COOH}) \approx 2.17,$$    $$pK_{a2}(\text{-NH}_3^+\text{ on α-C}) \approx 9.04,$$    $$pK_{aR}(\text{guanidinium}) \approx 12.48.$$
Again, the distinctive value is the side-chain $$pK_{aR}=12.48.$$

For comparison we focus on the $$pK_a$$ value that varies from one listed amino acid to the next, namely:

$$ \begin{aligned} \text{Aspartate} &: \; pK_{aR} \approx 3.86,\\[2pt] \text{Glycine} &: \; pK_{a2} \approx 9.60,\\[2pt] \text{Lysine} &: \; pK_{aR} \approx 10.53,\\[2pt] \text{Arginine} &: \; pK_{aR} \approx 12.48. \end{aligned} $$

Writing these numerically in ascending order gives

$$3.86 \lt 9.60 \lt 10.53 \lt 12.48,$$

which translates back to the amino-acid names as

Aspartate < Glycine < Lysine < Arginine.

This sequence exactly matches Option D.

Hence, the correct answer is Option D.

First, let us recall what each qualitative test detects.

We have the ceric ammonium nitrate test. This test is specific for the presence of an alcoholic -OH group, especially a primary or secondary alcohol. When such an -OH group is present, the yellow $$(NH_{4})2[Ce(NO_{3})6]$$ solution turns red-orange. So, for a peptide to give this test, at least one of its amino-acid residues must carry an -OH group in its side chain.

Next, we have the carbylamine (isocyanide) test. The general reaction involved is

$$R-NH_{2} + CHCl_{3} + 3KOH \;\longrightarrow\; R-NC + 3KCl + 3H_{2}O$$

This reaction occurs only when a primary amine ($$-NH_{2}$$) is present. Thus, the peptide must contain an amino-acid residue that still possesses a free primary amino group that has not been converted into an amide bond.

Now we examine the side chains of the amino acids that appear in the options.

$$\begin{aligned} \text{Serine (Ser)} &:& -CH_{2}-OH\quad(\text{contains -OH group})\\\\ \text{Lysine (Lys)} &:& -(CH_{2})4-NH_{2}\quad(\text{contains a free }\varepsilon\text{-NH2})\\\\ \text{Glutamine (Gln)} &:& -CH_{2}-CH_{2}-CONH_{2}\quad(\text{amide, no -OH, no free -NH2})\\\\ \text{Aspartic\;acid (Asp)} &:& -CH_{2}-COOH\quad(\text{no -OH, no free -NH2}) \end{aligned}$$

• Serine clearly satisfies the requirement for the alcoholic -OH group.

• Lysine clearly satisfies the requirement for a free primary amine.

On the other hand,

• Glutamine’s side chain terminates in an -CONH2 group, which is an amide, not a free -NH2.

• Aspartic acid has only a -COOH group; it lacks both an -OH and a free -NH2.

Hence, the only peptide that simultaneously carries an alcoholic -OH (from Ser) and a free primary -NH2 (from Lys) is the dipeptide

$$\text{Ser-Lys}.$$

This corresponds to Option D (Option 4 in the given numbering).

Hence, the correct answer is Option D.

We start by recalling the elementary facts about ribonucleic acid, popularly abbreviated as RNA. RNA is a biopolymer built from ribonucleotides, each consisting of a ribose sugar, a phosphate group, and one of four nitrogenous bases (adenine, guanine, cytosine, or uracil). In most organisms, RNA performs several crucial biological tasks such as carrying genetic information from DNA to ribosomes (messenger RNA, mRNA), forming the core of ribosomal structure and catalyzing protein synthesis (ribosomal RNA, rRNA), and transporting specific amino acids to the ribosome during protein assembly (transfer RNA, tRNA).

Now, let us verify every option one by one.

Option A: “It controls the synthesis of protein.” Messenger RNA indeed carries the genetic code from DNA to the ribosome, and along with rRNA and tRNA, it directs the process of translation. Hence RNA does control protein synthesis. Therefore Option A is a true statement about RNA.

Option B: “It usually does not replicate.” Unlike DNA, RNA molecules are not self-replicating under normal cellular conditions. They are synthesized afresh from a DNA template by the enzyme RNA polymerase. Only in certain RNA viruses does RNA replication occur. Thus, in the usual context of cellular biology, RNA indeed “does not replicate.” Hence Option B is also true.

Option D: “It is present in the nucleus of the cell.” Inside eukaryotic cells, a large quantity of RNA (for example, heterogeneous nuclear RNA: hnRNA and various small nuclear RNAs: snRNAs) is synthesized and processed in the nucleus before being exported to the cytoplasm. Therefore RNA is certainly present in the nucleus. So Option D is true as well.

Option C: “It has always double standard $$\alpha$$-helix structure.” The classic double helical $$\alpha$$-helix (commonly referred to as the double-stranded helix) is the hallmark of DNA, not RNA. The vast majority of RNA molecules are single-stranded. While an RNA strand can fold back on itself to form hairpins, loops, or short double-stranded regions, these local secondary structures do not amount to a permanent, extensive, double-stranded $$\alpha$$-helix throughout the entire molecule. Therefore the declaration that RNA “always” possesses a double-stranded $$\alpha$$-helix is factually incorrect.

Because Option C is not true about RNA, whereas the other three statements are true, Option C is the odd one out.

Hence, the correct answer is Option C.

First, let us recall the exact structure of sucrose. It is a disaccharide formed by one unit of $$\alpha$$-D-glucopyranose and one unit of $$\beta$$-D-fructofuranose. The two monosaccharide rings are joined through a glycosidic linkage that involves the anomeric carbon $$C_1$$ of the glucose unit and the anomeric carbon $$C_2$$ of the fructose unit. Symbolically, this linkage is written as $$\alpha$$-D-Glucose $$(1 \rightarrow 2)$$ $$\beta$$-D-Fructose.

Now we examine every option one by one, comparing the statements with the above fact.

Option A: “It is also named as invert sugar.” We know that when sucrose is hydrolysed, the optical rotation of the solution changes from dextrorotatory to laevorotatory because the mixture of glucose and fructose produced has a net negative rotation. This change in sign is called inversion, and therefore sucrose is often referred to as “invert sugar.” So, Option A is a true statement.

Option B: “It is a non-reducing sugar.” Since both anomeric carbons (of glucose and fructose) are involved in the $$\alpha1 \rightarrow \beta2$$ glycosidic bond, neither ring possesses a free hemi-acetal or hemi-ketal group. As a result, sucrose cannot reduce Fehling’s solution or Tollen’s reagent and is indeed non-reducing. Therefore, Option B is also true.

Option C: “The glycosidic linkage is present between $$C_1$$ of $$\alpha$$-glucose and $$C_1$$ of $$\beta$$-fructose.” From our structural recollection we have the linkage between $$C_1$$ of $$\alpha$$-glucose and $$C_2$$ of $$\beta$$-fructose, not between the two $$C_1$$ atoms. Hence the statement given in Option C is incorrect.

Option D: “On hydrolysis, it produces glucose and fructose.” When sucrose is treated with dilute acid or the enzyme invertase, it splits into equimolar amounts of D-glucose and D-fructose. This is precisely what the option states, so Option D is correct.

We have found that only Option C conflicts with the actual chemistry of sucrose. All other options are true, while Option C is not.

Hence, the correct answer is Option C.

We start by recalling the structural facts about glycogen. Glycogen is the main reserve carbohydrate of animals, especially abundant in liver and muscles. It is a polysaccharide composed entirely of glucose units. In the backbone each glucose residue is joined to the next through $$\alpha(1\rightarrow4)$$ glycosidic bonds, and at the branch points every $$8\text{-}12$$ residues a glucose is attached through an $$\alpha(1\rightarrow6)$$ linkage. Thus the molecule is highly branched.

Now we examine each option and compare it with these established facts.

Option A states, “It is present in animal cells.” We have just noted that glycogen is the principal reserve carbohydrate of animals, so this statement is true.

Option B says, “It is a straight chain polymer similar to amylose.” A straight chain polymer would possess only the $$\alpha(1\rightarrow4)$$ linkages and no branches. However, glycogen is characterized by frequent $$\alpha(1\rightarrow6)$$ branch points, making it a highly branched polymer, more so than amylopectin and certainly not straight like amylose. Therefore this statement conflicts with the known structure of glycogen.

Option C claims, “It is present in some yeast and fungi.” Besides animals, certain microorganisms such as yeast and fungi also store glycogen. Hence this statement is correct.

Option D asserts, “Only α-linkages are present in the molecule.” In glycogen the linkages are $$\alpha(1\rightarrow4)$$ and $$\alpha(1\rightarrow6)$$, both of which are indeed α-type linkages, with no β-linkages involved. So the statement is true.

From this detailed comparison we see that the sole statement contradictory to the true nature of glycogen is Option B.

Hence, the correct answer is Option B.

Biochemical Oxygen Demand (BOD) is defined as the amount of dissolved oxygen (in parts per million, ppm) required by aerobic microorganisms to oxidise the biodegradable organic matter present in a given volume of water at a specified temperature and for a specific time (usually 5 days at 20 °C).

Interpretation of BOD:

• High concentration of organic matter → more food for aerobic microbes → they consume more $$O_2$$ while oxidising this matter → BOD value becomes high.
• Low concentration of organic matter (clean water) → less microbial oxidation → BOD value remains low (typically $$\lt 5$$ ppm).

Now analyse each statement:

Option A: “Polluted water has BOD value higher than 10 ppm.”
Water is considered significantly polluted if the BOD exceeds about 10 ppm. Hence this statement is correct.

Option B: “Aerobic bacteria decreases the BOD value.”
Aerobic bacteria actually consume dissolved $$O_2$$ to degrade organic matter, thereby reflecting (and often increasing) the BOD measurement. They do not decrease the BOD; they are the reason the demand exists. Hence this statement is incorrect.

Option C: “Anaerobic bacteria increases the BOD value.”
BOD is defined for aerobic oxidation only; anaerobic bacteria operate in absence of oxygen and therefore are not considered in BOD estimation. Thus this statement is incorrect.

Option D: “Clean water has BOD value higher than 10 ppm.”
Clean (unpolluted) water should have a BOD well below 5 ppm; therefore this statement is incorrect.

Therefore, the correct choice is:

Option A which is: Polluted water has BOD value higher than 10 ppm

We match each vitamin (List-I) with the disorder produced when that vitamin is deficient (List-II).

Case A: Vitamin B1 (thiamine)
Deficiency leads to the nervous-muscular disorder called beriberi. Hence A → R.

Case B: Vitamin C (ascorbic acid)
This vitamin is needed for the hydroxylation of proline and lysine during collagen synthesis. Its deficiency weakens connective tissue and causes bleeding gums, giving the disease scurvy. Hence B → S.

Case C: Vitamin D (calciferol)
It regulates $$\mathrm{Ca^{2+}}$$ and $$\mathrm{PO_4^{3-}}$$ absorption and bone mineralisation. Lack of Vitamin D produces soft, demineralised bones—rickets in children (and osteomalacia in adults). Hence C → P.

Case D: Vitamin K (phylloquinone)
Vitamin K is required for the post-translational γ-carboxylation of several clotting factors. Deficiency prolongs clotting time, leading to excessive bleeding or haemorrhage. Hence D → Q.

The correct complete matching is therefore:
A → R, B → S, C → P, D → Q.

Among the given options, this correspondence is exactly the one stated in Option B.

Final Answer: Option B which is: (A)-(R), (B)-(S), (C)-(P), (D)-(Q)

To decide which one of the four given statements is wrong, we recall first what a glycosidic linkage means. A glycosidic linkage is the covalent bond that joins two monosaccharide units after the elimination of a molecule of water. The bond is denoted as $$\alpha$$ or $$\beta$$ according to the configuration at the anomeric carbon, and the numbers indicate the carbon atoms of the two rings that are directly joined. For example, “$$\alpha$$-1,4” means the -O- bridge runs from the anomeric carbon-1 of the first sugar in the $$\alpha$$-configuration to carbon-4 of the second sugar.

Now we examine each statement one by one.

Statement A says: “cellulose and amylose has 1,4-glycosidic linkage.” Cellulose is a polymer of $$\beta$$-D-glucose units linked through $$\beta$$-1,4 bridges, i.e. $$\beta$$-D-Glc(1 → 4)$$\beta$$-D-Glc $$\dots$$. Amylose (the unbranched component of starch) is a polymer of $$\alpha$$-D-glucose units joined through $$\alpha$$-1,4 bridges, i.e. $$\alpha$$-D-Glc(1 → 4)$$\alpha$$-D-Glc $$\dots$$. Both indeed use a 1,4 linkage (only the $$\alpha/\beta$$ orientation differs). Hence Statement A is true.

Statement B says: “lactose contains $$\beta$$-D-galactose and $$\beta$$-D-glucose.” Lactose consists of one $$\beta$$-D-galactose and one $$\beta$$-D-glucose connected by a $$\beta$$-1,4 link from C-1 of galactose to C-4 of glucose. So the description is accurate. Therefore Statement B is true.

Statement C says: “maltose and lactose has 1,4-glycosidic linkage.” Maltose = $$\alpha$$-D-Glc(1 → 4)$$\alpha$$-D-Glc, clearly 1,4. Lactose, as just discussed, is $$\beta$$-D-Gal(1 → 4)$$\beta$$-D-Glc, also 1,4. So both indeed possess 1,4 linkages. Statement C is true.

Statement D says: “sucrose and amylose has 1,2-glycosidic linkage.” Sucrose is $$\alpha$$-D-Glc(1 → 2)$$\beta$$-D-Fru; it truly contains an $$\alpha$$,1 → 2,$$\beta$$ bridge, so the part about sucrose is right. Amylose, however, we already know is based on $$\alpha$$-1,4 linkages, not 1,2. Thus the claim that amylose has a 1,2 linkage is wrong. Because the statement lumps sucrose and amylose together and declares that both have 1,2 linkages, the whole statement becomes incorrect.

All the other three statements are correct, so the only incorrect one is Statement D.

Hence, the correct answer is Option D.

The dipeptide is written as

$$H_2N{-}CH(CO{-}NH_2){-}CH_2{-}CH_2{-}C(O)NH{-}CH_2{-}COOH$$
(N-terminus)                        (peptide bond)        (C-terminus)

The two functional groups that can, in principle, react with $$CH_3COCl$$ are

1. the free α-amino group at the N-terminus, $$H_2N{-}$$
2. the terminal carboxylic acid group, $${-}COOH$$.

Facts to remember

• Acyl chlorides acylate a free amine easily, giving an amide (acetylation).
• Carboxylic acids can form mixed anhydrides with $$CH_3COCl$$, but these anhydrides are hydrolysed back to the parent acid during the aqueous work-up.
• The nitrogens already present in amide linkages (peptide bond and$$-CONH_2$$ side chain of Gln) are far too weakly nucleophilic to be acylated under these mild conditions.

Step-wise reaction

$$H_2N{-}$$ of Gln attacks the carbonyl carbon of $$CH_3COCl \; \rightarrow$$ tetrahedral intermediate $$\rightarrow$$ loss of $$Cl^-$$ gives the N-acetylated amide.

No lasting change occurs at the carboxyl terminus; any transient anhydride is hydrolysed in the work-up to regenerate $$-COOH$$.

Hence the only permanent change is replacement of the N-terminal hydrogen by an acetyl group.

Product (Option A):
$$CH_3CONH{-}CH(CO{-}NH_2){-}CH_2{-}CH_2{-}C(O)NH{-}CH_2{-}COOH$$

Therefore, the dipeptide Gln-Gly gives its N-acetyl derivative on treatment with $$CH_3COCl$$ followed by water.

Option A which is: the N-acetylated dipeptide (Ac-Gln-Gly).

First, let us recall what is meant by a hydrolysis reaction. In hydrolysis, a covalent bond in a molecule is broken by the addition of water, and the molecule splits into two parts, each taking a fragment of the water molecule. When such a step is a part of a polymer-forming sequence, we say that the polymer is produced through hydrolytic polymerisation.

Now let us analyse each option one by one.

Bakelite. Bakelite is obtained from phenol and formaldehyde through the reaction

$$\text{Phenol} + \text{Formaldehyde} \rightarrow \text{Bakelite} + H_2O$$

Here water is eliminated; no bond is cleaved by water. Hence the process is condensation, not hydrolysis.

Nylon-6,6. Nylon-6,6 is prepared by the condensation of hexamethylenediamine with adipic acid:

$$n\bigl(H_2N\!(CH_2)_6NH_2\bigr)+n\bigl(HOOC\,(CH_2)_4\,COOH\bigr)\rightarrow \bigl[-NH\,(CH_2)_6\,NHCO\,(CH_2)_4\,CO-\bigr]_n+2n\,H_2O$$

Again, water is produced, not consumed; therefore the step is condensation, not hydrolysis.

Terylene (PET). Terylene is formed either from ethylene glycol and terephthalic acid or from ethylene glycol and dimethyl terephthalate. In both routes the key step is an esterification, producing water (or methanol) as a by-product:

$$n\,HOCH_2CH_2OH+n\,HOOC\text{-}C_6H_4\text{-}COOH\rightarrow \bigl[-OCH_2CH_2OCOC_6H_4CO-\bigr]_n+n\,H_2O$$

The mechanism is again condensation, not hydrolysis.

Nylon-6. Nylon-6 is produced from $$\varepsilon$$-caprolactam, whose formula is

$$(CH_2)_5C(O)NH$$

This is a cyclic amide (lactam) of six members. The polymerisation starts with the hydrolytic ring opening of caprolactam. We introduce water which breaks the amide bond:

$$\text{Caprolactam}+H_2O \rightarrow H_2N\,(CH_2)_5\,COOH$$

The product is $$\varepsilon$$-aminocaproic acid. As the reaction mixture is heated, the amino group and the carboxyl group of two such molecules condense, giving an amide linkage and releasing water; the sequence then repeats. Symbolically, after the initial hydrolysis step we have

$$n\,H_2N\,(CH_2)_5\,COOH\;\rightarrow\; \bigl[-NH\,(CH_2)_5\,CO-\bigr]_n + n\,H_2O$$

Thus the very first, indispensable step that initiates the polymer chain is the hydrolysis of the lactam ring. Therefore the overall polymerisation is commonly called hydrolytic ring-opening polymerisation.

Among the four given polymers, only the formation of Nylon-6 necessarily involves a hydrolysis reaction. The other three are formed solely by condensation without any bond cleavage by water.

Hence, the correct answer is Option D.

We begin by recalling that enzymes possess not only the catalytic or active site, where the normal substrate binds, but also one or more distinct regions called allosteric sites. A molecule that settles itself at an allosteric site is called an allosteric inhibitor. The moment such a molecule occupies the allosteric pocket, it produces a conformational twist in the entire protein. Because protein structure and function are inseparable, this structural alteration finally reaches the catalytic pocket and distorts it so badly that the natural substrate can no longer fit or react.

Now, when a drug behaves as an allosteric inhibitor, the biochemical pathway that depends on that particular enzyme is abruptly blocked. Accumulation of the metabolic intermediate before the blocked step and scarcity of the product after that step together upset the physiological balance. This sudden imbalance is what we refer to in medicinal chemistry as “drug-induced poisoning.”

Let us contrast this with the other possibilities listed in the options. If the drug binds reversibly at the active site (competitive inhibition), the inhibition can usually be overcome by increasing the substrate concentration, so true poisoning does not set in. If the drug brings a minor conformational change exactly at the binding site (option D), the effect is again generally reversible. If the drug binds irreversibly at the active site (option A), it certainly inactivates the enzyme, but such irreversible covalent modification is usually classified as enzyme inactivation or suicide inhibition, not as the specific textbook case of “drug-induced poisoning” asked here.

Hence for “drug-induced poisoning,” the decisive event is

Drug + Enzyme (allosteric site) $$\longrightarrow$$ Drug-Enzyme complex (non-competitive inhibition)

and that description matches the option that says “binding at the allosteric sites of the enzyme.”

Hence, the correct answer is Option B.

Nylon-2-Nylon-6 copolymer is biodegradable because of a less crystalline structure allowing easier enzyme action.

We begin by recalling the basic definition: in human nutrition, essential amino acids are those that the body cannot synthesise in adequate amounts, so they must be supplied through the diet. Conversely, non-essential amino acids can be produced endogenously from simpler metabolic intermediates and therefore need not be taken in food.

The commonly accepted list of essential amino acids for adults includesValine, Leucine, Isoleucine, Lysine, Methionine, Phenylalanine, Threonine, Tryptophan, Histidine.Some texts also mention Arginine as semi-essential for growing children, but for our present comparison we restrict ourselves to the standard adult list shown above.

Now we inspect each option one by one and match it with the list:

We have Option A: Valine. Consulting the list, $$\text{Valine}$$ appears explicitly among the essential amino acids.

Next, Option B is Alanine. Alanine is produced in the body via transamination of pyruvate, so it is categorised as non-essential.

Option C is Serine. Serine is synthesised in the body from the glycolytic intermediate 3-phosphoglycerate, hence it is also non-essential.

Finally, Option D is Aspartic acid. Aspartic acid is formed naturally in the urea cycle and other metabolic pathways, therefore it is likewise non-essential.

Comparing all four options with the essential list, only Valine satisfies the requirement of being indispensable in the diet.

Hence, the correct answer is Option A.

To discover which of the four statements is incorrect, let us recall the precise stereochemical meanings of the terms “enantiomers”, “anomers”, and the structural facts about glucose derivatives and cellulose.

For any chiral molecule having several stereogenic centers, an enantiomer is defined as the non-superposable mirror image of the molecule as a whole. That means every chiral centre is inverted: for every $$\mathrm{R}$$ configuration we obtain $$\mathrm{S}$$ and vice-versa. In contrast, an anomer is a special kind of diastereomer found only in cyclic hemiacetals or hemiketals of sugars. Anomers differ in configuration only at the new stereogenic centre created when the carbonyl carbon becomes the anomeric carbon (C-1 in aldoses, C-2 in ketoses). All the other chiral centres remain unchanged.

Consider $$\alpha \text{-D-glucose}$$ and $$\beta \text{-D-glucose}$$.

Write the Fischer projection for open-chain D-glucose:

$$ \begin{array}{c} \mathrm{CHO}\\ |\\ \mathrm{H}\!\!-\!\!\mathrm{C}\!\!\!\!\! \begin{array}{c}OH\end{array}\\ |\\ \mathrm{HO}\!\!-\!\!\mathrm{C}\!\!\!\!\! \begin{array}{c}H\end{array}\\ |\\ \mathrm{H}\!\!-\!\!\mathrm{C}\!\!\!\!\! \begin{array}{c}OH\end{array}\\ |\\ \mathrm{H}\!\!-\!\!\mathrm{C}\!\!\!\!\! \begin{array}{c}OH\end{array}\\ |\\ \mathrm{CH}_2OH \end{array} $$

When this aldehyde cyclises, the carbonyl carbon (C-1) becomes a new stereocentre. If, in the Haworth representation, the -OH on C-1 is drawn down (trans to the CH$$_2$$OH substituent at C-5), the configuration is named $$\alpha$$; if the -OH is drawn up (cis to CH$$_2$$OH), the configuration is named $$\beta$$. All the other chiral centres at C-2, C-3, C-4, and C-5 remain identical in both forms. Therefore:

$$\alpha \text{-D-glucose} \quad \text{and} \quad \beta \text{-D-glucose}$$

differ only at C-1, so they are anomers. They are not related as mirror images, because only one of the five stereocentres has opposite configuration while the rest are the same. Thus they cannot be enantiomers.

Consequently, Option A, which states $$\alpha$$ $$-$$ D $$-$$ glucose and $$\beta$$ $$-$$ D $$-$$ glucose are enantiomers, is false.

Now consider Option B. When glucose is converted into its penta-acetate, every hydroxyl group, including the anomeric -OH at C-1, becomes acetylated ($$-OCOCH_3$$). Hydroxylamine reacts with free carbonyl groups or free anomeric hydroxyls to form oximes; but in the penta-acetate, there is no free -OH or free carbonyl. Therefore the penta-acetate does indeed fail to react with hydroxylamine, so Option B is true.

Option C explicitly states that $$\alpha$$ $$-$$ D $$-$$ glucose and $$\beta$$ $$-$$ D $$-$$ glucose are anomers. As shown above, this is correct, so Option C is true.

Option D claims that cellulose is a straight-chain polysaccharide made exclusively of $$\beta$$ $$-$$ D $$-$$ glucose units. In cellulose, each glucose residue is linked by $$\beta(1 \rightarrow 4)$$ glycosidic bonds, producing an unbranched, ribbon-like chain. Hence the statement is accurate, so Option D is true.

Only Option A does not align with chemical facts.

Hence, the correct answer is Option A.

A reducing sugar is a carbohydrate that possesses a free hemiacetal or hemiketal group. In its cyclic form, the anomeric carbon is bonded to both an $$-OH$$ group and an $$-OR$$ (ring oxygen) group.

Compound A is inherently reducing, and not due to the aqueous solution of $$KOH$$

Acetals and Ketals (B & C): These are stable in basic solutions like aqueous $$KOH$$. The ring cannot open, so they remain non-reducing.

In the presence of $$KOH$$, the acetate group ($$-OCOCH_3$$) undergoes basic hydrolysis. This chemical reaction restores the free hemiketal group.

Hence, option (D) is the correct answer.

We start by recalling that a “sol” is a colloidal dispersion in which the dispersed phase is solid (here, protein molecules of egg albumin) and the dispersion medium is a liquid (here, saline water). To obtain a true sol, the protein must remain in its native, un-coagulated state; heating the protein will denature it, causing coagulation and converting it into a rubbery mass that cannot be redispersed into a colloidal system.

Now let us examine what part of the egg actually contains albumin. The transparent, viscous “egg white” (also called the albumen) is rich in albumin proteins, whereas the yellow “yolk” contains lipids, lecithin, and different proteins but is not the preferred source when our aim is to prepare an albumin sol. Hence, we must start with the white portion and we must keep it unheated.

Next, we think about how to maintain the proteins in dispersed form. A dilute sodium chloride solution (physiological saline, about 0.85 % w/V) or any mild 5 % w/V saline, provides ionic strength that helps the albumin molecules stay separate due to electrostatic repulsion. Therefore, the correct technique is to take the raw, transparent egg white and gently stir or shake it into a measured volume of 5 % saline so that the proteins are mechanically broken into colloidal-sized aggregates without being chemically or thermally denatured.

Let us now scrutinize each option:

A. The procedure says, “Break an egg carefully and transfer the transparent part (egg white) to 100 mL of 5 % w/V saline solution and stir well.” This matches exactly the requirements we just established: (i) use the albumin-rich white, (ii) do not heat, and (iii) disperse it directly into saline. So Option A is scientifically sound.

B. This option instructs us to boil the egg for 10 minutes, remove the shell, use the yellow part, and then homogenize. Boiling for 10 minutes will coagulate both the white and the yolk; once coagulated, proteins cannot be re-dispersed into a sol. Moreover, the yellow part is not albumin-rich. Hence, Option B is unsuitable.

C. This option again involves boiling for 10 minutes, but this time it uses the white part. Even though the white part originally contains albumin, heating for that long causes irreversible denaturation, so the product cannot form a sol. Therefore, Option C is incorrect.

D. This option suggests using the raw yellow part (yolk) without heating. Although the yolk is raw here, yolk constituents are mainly lipoproteins and fats, not albumin proteins of the type typically used in laboratory sols. Consequently, Option D will not give an egg-albumin sol.

Comparing all four possibilities, only Option A satisfies every criterion: (1) it employs the albumin-rich egg white, (2) it avoids thermal denaturation, and (3) it disperses the protein gently into a suitable ionic medium. Therefore, Option A is the most appropriate method.

Hence, the correct answer is Option A.

We begin by carefully examining the assertion. Rayon is produced starting with the natural polymer cellulose obtained from wood pulp or cotton linters. The cellulose is first converted into a soluble derivative, extruded through fine holes, and then regenerated as continuous filaments. Because part of the material is natural (cellulose) and part is processed in a chemical plant, we call it a semisynthetic polymer. Compared with natural cotton, rayon generally shows higher luster, better drape, and can be engineered for greater tensile strength. All these facts show that the assertion “Rayon is a semisynthetic polymer whose properties are better than natural cotton” is correct.

Now we consider the reason. Cellulose contains three reactive hydroxyl (-OH) groups per glucose unit. When cellulose is treated with acetic anhydride in the presence of an acid catalyst, the -OH groups are converted into acetate ester groups (-OCOCH3). This reaction is called acetylation. The general chemical change can be written as:

$$\text{Cellulose - OH} \;+\; (CH_3CO)_2O \;\rightarrow\; \text{Cellulose - OCOCH}_3 \;+\; CH_3COOH$$

Introducing acetate groups reduces intermolecular hydrogen bonding and imparts thermoplasticity, improved crease recovery, and a silk-like luster. These are all mechanical and aesthetic improvements. Therefore the statement “Mechanical and aesthetic properties of cellulose can be improved by acetylation” is true, so the reason is correct.

Next we must decide whether the reason actually explains the assertion. Rayon covers both viscose rayon (regenerated cellulose) and acetate rayon (cellulose acetate). The very improvements in strength, luster, and feel that make rayon superior to cotton arise largely because chemical modifications such as acetylation (or xanthation in the viscose process) break the rigid hydrogen-bond network of native cellulose and allow the fibers to be spun with controlled orientation and finer diameter. Thus, by pointing specifically to acetylation of cellulose, the reason supplies a direct chemical basis for why a regenerated cellulose fiber like rayon can outperform natural cotton.

Both sentences are individually correct, and the chemical improvement cited in the reason is indeed one of the key causes behind the better properties mentioned in the assertion. Consequently, the reason is the correct explanation of the assertion.

Hence, the correct answer is Option B.