Q.(A) Define the following term :

Showing the 12 most recent of 36 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The number of secondary alcoholic groups present in the end product 'Y' of the given reaction sequence is $C_6H_{12}O_6 \xrightarrow{HCN} X \xrightarrow{H_2O\ |\ H^+} Y$ (A) 4 (B) 3 (C) 5 (D) 6

   ›Reveal solutionSolution

   This tests the Kiliani–Fischer chain-extension sequence (aldose + HCN → cyanohydrin → hydrolysis → one-carbon-longer aldonic acid) and careful counting of secondary alcohol groups in the product. The answer is 5.

   Concept and Intuition

   Glucose's open-chain form has an aldehyde at C1. Adding HCN to an aldehyde forms a cyanohydrin: the carbonyl carbon becomes a new stereocentre bearing both $-OH$ and $-CN$, and crucially the $-CN$ carbon becomes a new carbon added to the chain. When that nitrile is hydrolysed ($H_2O/H^+$) it becomes a carboxylic acid carbon. So overall, a 6-carbon aldose becomes a 7-carbon acid with one extra $-CH(OH)-$ unit inserted next to the original carbonyl position — this is exactly how sugar chains are lengthened in the Kiliani–Fischer synthesis.

   Step-by-Step Solution

   1. Write glucose's open-chain structure: $C1(CHO)-C2(CHOH)-C3(CHOH)-C4(CHOH)-C5(CHOH)-C6(CH_2OH)$.
   2. HCN addition at the carbonyl (C1): the carbonyl carbon becomes $-CH(OH)(CN)-$, i.e. a new nitrile carbon is attached to old C1. So $X$: $NC-CH(OH)[old\,C1]-CH(OH)[C2]-CH(OH)[C3]-CH(OH)[C4]-CH(OH)[C5]-CH_2OH[C6]$.
   3. Hydrolysis ($H_2O/H^+$) converts the terminal $-CN$ to $-COOH$: $Y$: $HOOC[\text{new carbon}]-CH(OH)[\text{old }C1]-CH(OH)[C2]-CH(OH)[C3]-CH(OH)[C4]-CH(OH)[C5]-CH_2OH[C6]$ — a 7-carbon chain. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Observe the following reactions I. $\mathrm{D\text{-}Glucose \xrightarrow{NH_2OH}}$ II. $\mathrm{D\text{-}Glucose \xrightarrow[(ii)\ NH_2OH]{(i)\ (CH_3CO)_2O}}$ Correct statement regarding the reactions I and II is (A) Oxime is formed in both the reactions I, II (B) Oxime is not formed in both the reactions I, II (C) Oxime is formed in reaction I but oxime is not formed in reaction II (D) Oxime is not formed in reaction I but oxime is formed in reaction II

   ›Reveal solutionSolution

   This is the classic experiment proving glucose's cyclic hemiacetal structure: plain

   glucose reacts with hydroxylamine (free $-CHO$), but glucose pentaacetate does

   not (no free $-CHO$ left to react). Answer: (C).

   Concept and Intuition

   D-Glucose is not fixed as a straight chain — it exists mainly as a cyclic hemiacetal

   (pyranose) in equilibrium with a small amount of the open-chain aldehyde form. Reagents

   that test for a free carbonyl (like $NH_2OH$, Schiff's reagent, or $NaHSO_3$) react via

   this open-chain minority form, so plain glucose behaves as if it has a free $-CHO$.

   If, however, glucose is first acetylated with excess acetic anhydride, every $-OH$ group gets esterified — including the anomeric $-OH$ that would otherwise open up to

   expose the aldehyde. The resulting glucose pentaacetate is locked in the cyclic

   form with no accessible carbonyl, so it fails to react with hydroxylamine or Schiff's

   reagent. This experiment is one of the classical pieces of evidence for the cyclic

   (not open-chain) structure of glucose.

   Step-by-Step Solution

   1. Reaction I: $\text{D-Glucose} \xrightarrow{NH_2OH} \text{oxime}$ — plain glucose's open-chain aldehyde tautomer reacts normally, forming the oxime.
   2. Reaction II: $\text{D-Glucose} \xrightarrow{(i)\ (CH_3CO)_2O} \text{glucose pentaacetate}$ — this step acetylates all $-OH$ groups, including the anomeric one, locking the ring closed. …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Choose the correct statements A) The first amino acid in a protein chain is in the left end and termed as C-terminal amino acids. B) Long protein chain folded upon itself like a hollow woolen ball is called tertiary structure. C) In general, only right handed helices are observed in protein. D) In general, polysaccharides will have glycosidic bond formed by dehydration. (A) A, B and C (B) B, C and D (C) A, C and D (D) A, B and D

   ›Reveal solutionSolution

   Of four claims about protein/polysaccharide structure, only the "first amino acid = C-terminal" claim is wrong (it should be N-terminal); tertiary structure, right-handed helices, and dehydration-formed glycosidic bonds are all correct.

   Concept and Intuition

   A polypeptide has directionality: one end carries a free –NH₂ group (the N-terminus, conventionally drawn/numbered first, "left" end) and the other a free –COOH group (the C-terminus, "right" end). Beyond primary structure (sequence) and secondary structure (regular local folding like the α-helix, predominantly right-handed in natural proteins), further compact 3-D folding into a globular shape is tertiary structure. Separately, polysaccharides are built by joining monosaccharide units via glycosidic bonds, formed through dehydration (loss of a water molecule per bond), just like peptide bonds in proteins.

   Step-by-Step Solution

   1. A: the residue with the free amino group is by convention the FIRST/N-terminal amino acid (drawn on the left) — calling it "C-terminal" is a direct mix-up of the two termini. FALSE.
   2. B: further folding of an already-folded (secondary structure) polypeptide chain into a compact, roughly spherical, hollow-ball-like shape is precisely the description used for tertiary structure. TRUE. …

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4. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.D-Glucose does not react with which of the following reagents? I. $NaHSO_3$ II. $NH_2OH$ III. $(CH_3CO)_2O$ IV. Schiff's reagent (A) II, III only (B) I, II, III only (C) I, II only (D) I, IV only

   ›Reveal solutionSolution

   Glucose fails to react with sodium bisulphite and Schiff's reagent — the classic evidence that its carbonyl exists mainly as a cyclic hemiacetal, not a free aldehyde.

   Concept and Intuition

   Glucose behaves like a typical aldehyde in many respects (oxidation to gluconic acid, cyanohydrin formation, oxime formation, acetylation of its five -OH groups), which is used to establish its structure. But it notably FAILS the two classic 'free aldehyde' spot tests — the bisulphite addition reaction and the Schiff's test — because in solution glucose exists overwhelmingly in the cyclic hemiacetal (pyranose) form, and the small equilibrium amount of open-chain aldehyde isn't sufficient/fast enough for these particular fast, stoichiometric addition reactions, even though it's enough to drive the slower oxidation and condensation reactions via Le Chatelier's principle.

   Step-by-Step Solution

   1. I. $NaHSO_3$: Glucose does NOT form the expected bisulphite addition compound — a key piece of evidence against a simple free-aldehyde structure.
   2. II. $NH_2OH$: Glucose DOES react, forming an oxime, confirming the presence of a carbonyl group (via the open-chain tautomer).
   3. III. $(CH_3CO)_2O$: Glucose DOES react, undergoing acetylation to give glucose pentaacetate (confirms 5 -OH groups). …

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5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A carbonyl compound X($\mathrm{C_8H_8O}$) gives yellow precipitate with NaOI. Hemiacetal of X with methanol/dry HCl is (A) $\mathrm{C_6H_5-CH_2-CH(OH)-OCH_3}$ (drawn as a benzene ring with a $-CH_2CH(OH)OCH_3$ substituent) (B) $\mathrm{C_6H_5-CH_2-CH(OCH_3)_2}$ (drawn as a benzene ring with a $-CH_2CH(OCH_3)_2$ substituent) (C) $\mathrm{C_6H_5-C(OH)(CH_3)(OCH_3)}$ (drawn as a benzene ring whose ring carbon bears a central carbon substituted with -OH, -CH_3 and -OCH_3) (D) $\mathrm{C_6H_5-C(OH)(OCH_3)_2}$ (drawn as a benzene ring whose ring carbon bears a central carbon substituted with -OH and two -OCH_3 groups)

   ›Reveal solutionSolution

   X is acetophenone ($\mathrm{C_6H_5COCH_3}$, iodoform-positive, $\mathrm{C_8H_8O}$); its hemiacetal with methanol is $\mathrm{C_6H_5-C(OH)(CH_3)(OCH_3)}$.

   Concept and Intuition

   The iodoform (yellow precipitate with $\mathrm{NaOI}$, i.e. $\mathrm{I_2/NaOH}$) test is positive for any compound with a $\mathrm{CH_3-CO-}$ group (methyl ketones) or that can be oxidised to one (like ethanol/secondary alcohols bearing a $\mathrm{CH_3CH(OH)-}$ group). Given the molecular formula $\mathrm{C_8H_8O}$ and an aromatic-sized carbon count, the methyl ketone that fits is acetophenone, $\mathrm{C_6H_5-CO-CH_3}$. A hemiacetal forms when one equivalent of alcohol adds across a carbonyl (catalysed by dry HCl): the carbonyl oxygen picks up the proton to become $-\mathrm{OH}$, while the alcohol's oxygen bonds to the former carbonyl carbon as $-\mathrm{OR}$ — all substituents originally on that carbon (here, the phenyl and the methyl) remain attached to the same, now sp$^3$, carbon.

   Step-by-Step Solution

   1. Identify X: $\mathrm{C_8H_8O}$, gives iodoform test $\Rightarrow$ a methyl ketone. Acetophenone $\mathrm{C_6H_5COCH_3}$ fits ($6\ (\text{ring}) + 1\ (\text{C=O}) + 1\ (\text{CH}_3) = 8$ carbons; one O).
   2. Hemiacetal formation mechanism: methanol's OH oxygen attacks the electrophilic carbonyl carbon of $\mathrm{C_6H_5-C(=O)-CH_3}$, protonated by dry HCl. …

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6. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Which of the following contain $\alpha$-D-glucose units ?

   a) cane sugar b) milk sugar c) cellulose d) amylose (A) a, d (B) a, b (C) b, c (D) c, d

   ›Reveal solutionSolution

   Tests recognising which common carbohydrates are actually built from $\alpha$-D-glucose units, as opposed to $\beta$-D-glucose or a non-glucose sugar.

   Concept and Intuition

   A saccharide "contains $\alpha$-D-glucose units" only when a glucose unit's anomeric carbon is tied down in the $\alpha$ configuration by a glycosidic bond (as in a polysaccharide chain, or in sucrose where glucose's anomeric carbon links to fructose), or the free sugar is drawn/crystallised as its $\alpha$ anomer feeding a fixed linkage.

   Step-by-Step Solution

   1. (a) Cane sugar = sucrose = $\alpha$-D-glucose $(1\rightarrow2)$ $\beta$-D-fructose. The glucose unit's anomeric carbon (C1) is locked in the $\alpha$ configuration by this glycosidic bond — contains $\alpha$-D-glucose. ✓
   2. (b) Milk sugar = lactose = $\beta$-D-galactose $(1\rightarrow4)$ D-glucose. Here it is galactose, not glucose, that supplies its anomeric carbon to the bond; the glucose unit's own anomeric carbon is FREE (this is why lactose is a reducing sugar and shows mutarotation) — not fixed as $\alpha$-D-glucose. ✗
   3. (c) Cellulose is a linear polymer of $\beta$-D-glucose units joined by $\beta(1\rightarrow4)$ linkages — glucose, but the WRONG anomer ($\beta$, not $\alpha$). ✗ …

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7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.Consider the following. Statement-I : Primary structure of protein represents its constitution Statement-II : $\alpha$-Helix and $\beta$-pleated sheet structure of protein represent tertiary structure of it Correct answer is (A) Both statement-I and statement-II are correct (B) Both statement-I and statement-II are not correct (C) Statement-I is correct, but statement-II is not correct (D) Statement-I is not correct, but statement-II is correct

   ›Reveal solutionSolution

   This tests the protein structure hierarchy: primary = sequence (constitution), secondary = local folding patterns like $\alpha$-helix/$\beta$-sheet, tertiary = overall 3-D folding. Statement-I is right; Statement-II wrongly calls a secondary-structure feature 'tertiary'.

   Concept and Intuition

   Proteins are described at four structural levels: primary (linear amino-acid sequence — the actual covalent 'constitution' of the molecule), secondary (regular, repeating local folding patterns held by hydrogen bonds along the peptide backbone — the $\alpha$-helix and $\beta$-pleated sheet are the two classic examples), tertiary (the overall three-dimensional folding of the whole chain, stabilised by disulfide bonds, hydrogen bonds, and hydrophobic interactions among side chains), and quaternary (arrangement of multiple polypeptide subunits). Recognising which level a named feature belongs to is the entire test here.

   Step-by-Step Solution

   1. Statement-I: the primary structure is defined as the sequence in which amino acids are linked via peptide bonds — this is indeed the molecule's basic constitution (which amino acid, in which order). True. …

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8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.During the denaturation of proteins, which of the following structures will remain intact? (A) Secondary & tertiary ($2^\circ$, $3^\circ$) (B) Tertiary only ($3^\circ$) (C) Primary only ($1^\circ$) (D) Primary and secondary ($1^\circ$, $2^\circ$)

   ›Reveal solutionSolution

   Denaturation breaks the weak interactions maintaining secondary/tertiary folding, but the covalent peptide-bond backbone (primary structure) survives untouched.

   Concept and Intuition

   Protein structure is organised in levels: primary (amino-acid sequence, held by strong covalent peptide bonds), secondary (α-helices/β-sheets, held by hydrogen bonds), and tertiary (overall 3-D fold, held by a mix of hydrogen bonds, hydrophobic effects, ionic interactions, and disulfide bridges). Denaturing agents (heat, extreme pH, chaotropic agents) disrupt these weak non-covalent interactions but do not have enough energy to break the strong covalent peptide bonds of the backbone.

   Step-by-Step Solution

   1. Denaturation unfolds the protein by breaking the hydrogen bonds/hydrophobic interactions responsible for secondary structure.
   2. It also destroys tertiary structure (loss of the specific 3-D globular shape), since tertiary structure depends on similarly weak interactions (plus occasionally disulfide bonds, which some strong denaturants can also break). …

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9. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.Match the following List-I A) Glycosidic bond B) Hollow woollen ball folding C) Enzyme D) Esterbond List-II I) Trypsin II) Bond between phosphate and 5th carbon of sugar III) tertiary structure of protein IV) formed by dehydration (A) A-II, B-I, C-IV, D-III (B) A-IV, B-III, C-I, D-II (C) A-III, B-II, C-I, D-IV (D) A-IV, B-I, C-II, D-III

   ›Reveal solutionSolution

   Matching each biomolecule term to its correct description gives A-IV, B-III, C-I, D-II — answer (B).

   Concept and Intuition

   • Glycosidic bond: forms between two monosaccharides via a dehydration (condensation) reaction, releasing a water molecule.
   • The image of a protein folding "like a hollow woollen ball" is a common descriptive analogy for a protein's tertiary structure — the overall 3D folding of a single polypeptide chain.
   • Trypsin is a well-known digestive protease and is the standard textbook example cited for "enzyme."
   • The ester bond joining a phosphate group to the 5′-carbon hydroxyl of a sugar (as in nucleotides) is a phosphoester linkage.

   Step-by-Step Solution

   1. A) Glycosidic bond → IV (formed by dehydration).
   2. B) "Hollow woollen ball" folding → III (tertiary structure of protein).
   3. C) Enzyme → I (Trypsin, a textbook enzyme example). …

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10. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
    Q.Choose the correct statements among the following I. In exothermic reactions 'P' is at a lower level than 'S'. II. Enzyme activity can be affected by temperature etc. with alteration of tertiary structure of protein. III. Low temperature denature the protein. IV. Ligases can remove the groups from substrate (A) I, II (B) III, IV (C) II, IV (D) I, III

    ›Reveal solutionSolution

    Exothermic reactions do place products at lower energy than substrates, and temperature does affect activity via tertiary-structure changes — both true. Low temperature inactivates (not denatures) enzymes, and ligases join (not remove) — both false. Correct set: I, II — option (A).

    Concept and Intuition

    Enzyme-catalysed reaction energetics: in an exothermic reaction, the products (P) are at a lower free-energy level than the substrates (S), with the difference released as heat — this is basic thermodynamics of catalysed reactions and is independent of the enzyme itself (the enzyme only lowers the activation energy, not the net energy change). Enzyme activity is highly temperature-sensitive because temperature affects the enzyme's three-dimensional (tertiary) structure, which determines the shape of its active site; this is why activity rises with temperature up to an optimum and then falls off as heat begins to disrupt this structure (denaturation) at high temperatures. Crucially, LOW temperature does not denature an enzyme — it merely reduces the kinetic energy of molecules, slowing the reaction reversibly (activity resumes when temperature rises again); denaturation (irreversible loss of structure) is a HIGH-temperature (or extreme pH) phenomenon. Ligases are enzymes that join two molecules together (often using ATP, e.g., DNA ligase sealing nicks) — they do not remove groups from a substrate; that role belongs to hydrolases/lyases.

    Step-by-Step Solution

    1. Statement I: In exothermic reactions, product (P) is at a lower energy level than substrate (S) — TRUE, by definition of an exothermic process (energy released, product lower in energy). …

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11. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.Sucrose when boiled with dilute $\mathrm{HCl}$ gives two functional isomers X and Y. X gives monocarboxylic acid with bromine water but not Y. The number of –OH groups in cyclic structure of X is (A) 6 (B) 4 (C) 5 (D) 3

    ›Reveal solutionSolution

    X (from sucrose hydrolysis) is glucose, identified because bromine water oxidises only aldoses; its cyclic pyranose form carries 5 free $-\mathrm{OH}$ groups.

    Concept and Intuition

    Sucrose is a non-reducing disaccharide of glucose and fructose joined through their anomeric carbons. Acid hydrolysis (dil. HCl, heat) breaks the glycosidic bond, releasing free glucose and fructose — this equimolar mixture is called invert sugar. Bromine water is a mild, selective oxidant: it oxidises the free aldehyde group of an aldose (like glucose) to a carboxylic acid, but it does not oxidise ketoses (like fructose), because ketones are far more resistant to this kind of mild oxidation than aldehydes.

    Step-by-Step Solution

    1. Sucrose $\xrightarrow{\text{dil. HCl}, \Delta}$ glucose + fructose (invert sugar) — these are the two isomeric monosaccharides X and Y ($\mathrm{C_6H_{12}O_6}$ each).
    2. Bromine water oxidises the aldehyde of glucose to gluconic acid (a monocarboxylic acid), but does not react with fructose's ketone group. So X = glucose, Y = fructose.
    3. Draw the cyclic (pyranose, 6-membered) ring of glucose: ring atoms are the ring-O, C1, C2, C3, C4, C5, with C6 ($\mathrm{CH_2OH}$) exocyclic on C5. …

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12. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.Which of the following represents the correct structure of $\beta-D-(-)$-Fructofuranose ? (A) [FIGURE] (Haworth furanose ring with O at the top vertex; upper-left ring carbon carries an exocyclic $\mathrm{HOH_2C}-$ group, with H outside and H just inside it; upper-right ring carbon carries an exocyclic $-CH_2OH$ group, with HO inside and OH outside; bottom-left ring carbon carries OH outside; bottom-right ring carbon carries H outside) (B) [FIGURE] (Haworth furanose ring with O at the top vertex; upper-left ring carbon carries an exocyclic $\mathrm{HOH_2C}-$ group, with H outside and H inside; upper-right ring carbon carries OH directly, with H inside and $CH_2OH$ outside; bottom-left ring carbon carries OH outside; bottom-right ring carbon carries OH outside) (C) [FIGURE] (Haworth furanose ring with O at the top vertex; upper-left ring carbon carries an exocyclic $\mathrm{HOH_2C}-$ group, with H outside and H inside; upper-right ring carbon carries OH directly, with HO inside and $CH_2OH$ outside; bottom-left ring carbon carries OH outside; bottom-right ring carbon carries H outside) (D) [FIGURE] (Haworth furanose ring with O at the top vertex; upper-left ring carbon carries H directly, with H inside; upper-right ring carbon carries OH directly, with HO inside; the exocyclic $\mathrm{HOH_2C}-$ and $-CH_2OH$ groups instead sit on the bottom-left and bottom-right ring carbons respectively, which also carry OH outside and H outside)

    ›Reveal solutionSolution

    This tests the Haworth structure of β-D-fructofuranose — the 5-membered ring form of D-fructose found in sucrose. Matching each option's substituent placement to the known structure gives option (C).

    Concept and Intuition

    D-Fructose is a ketohexose; its furanose (5-membered) ring forms between the carbonyl carbon C2 and the C5-oxygen, releasing C1 ($CH_2OH$) and C6 ($CH_2OH$) as exocyclic groups on C2 and C5 respectively. The anomeric carbon here is C2 (not C1, since fructose's carbonyl is a ketone at C2). The α/β distinction refers to the orientation of the new OH generated at C2 upon ring closure, relative to the reference (C5) configuration — exactly analogous to the α/β distinction at C1 in glucopyranose.

    Step-by-Step Solution

    1. Identify the ring: 5-membered furanose ring = O, C2, C3, C4, C5 (C1 and C6 are exocyclic $CH_2OH$ groups).
    2. C5 (derived from a D-sugar's terminal configuration) carries its $CH_2OH$ (C6) group "up" in the standard Haworth convention for D-sugars — matching the exocyclic $CH_2OH$ shown at the upper-right ring carbon in the options.
    3. C2 (anomeric) carries the exocyclic $HOH_2C-$ (C1) group at the upper-left ring carbon, plus its anomeric OH; for the β-anomer this OH sits on the same face as the C5→C6 reference substituent. …

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