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800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.
ChemistryClass 123 marksmedium
Aldehydes, Ketones and Carboxylic Acids
Why are carboxylic acids more acidic than phenols? How do electron-withdrawing groups affect their acidity?
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Carboxylic acids are more acidic than phenols because the carboxylate ion formed after the loss of the proton is stabilised by resonance in which the negative charge is spread equally over two electronegative oxygen atoms, making it very stable. In the phenoxide ion the negative charge is mainly on one oxygen and partly on less electronegative ring carbons, so it is less stabilised. Electron-withdrawing groups (like -Cl or -NO2) increase acidity because they further disperse and stabilise the negative charge of the carboxylate ion (for example, chloroacetic acid is stronger than acetic acid), while electron-releasing groups decrease acidity.
Marking-scheme points
- ✓Carboxylate ion: charge spread equally over two oxygen atoms (very stable)
- ✓Phenoxide ion is less stabilised
- ✓Electron-withdrawing groups increase acidity; electron-releasing groups decrease it
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Aldehydes, Ketones and Carboxylic Acids
Write the esterification reaction of a carboxylic acid and the decarboxylation reaction.
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(1) Esterification: a carboxylic acid reacts with an alcohol in the presence of a little concentrated sulphuric acid (as catalyst) to form an ester and water, R-COOH + R'-OH -> R-COO-R' + H2O; this reaction is reversible and gives esters that have fruity smells. (2) Decarboxylation: the sodium salt of a carboxylic acid, when heated with soda lime (NaOH + CaO), loses carbon dioxide to give a hydrocarbon (alkane) with one carbon less, R-COONa + NaOH -> R-H + Na2CO3.
R-COOH + R'-OH -> R-COO-R' + H2O
Marking-scheme points
- ✓Esterification: R-COOH + R'-OH -> ester + water (conc. H2SO4 catalyst)
- ✓Decarboxylation: R-COONa + NaOH (soda lime) -> R-H + Na2CO3
- ✓Decarboxylation removes CO2 and gives an alkane with one less carbon
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Aldehydes, Ketones and Carboxylic Acids
What is the iodoform test? Which compounds give a positive result?
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The iodoform test is used to detect the presence of a methyl ketone group (CH3-CO-) or a group that can be oxidised to it, such as CH3-CH(OH)-. When such a compound is warmed with iodine and sodium hydroxide (or sodium hypoiodite), it gives a yellow precipitate of iodoform (CHI3) with a characteristic smell. Compounds like acetaldehyde, acetone, ethanol and isopropanol give a positive iodoform test.
Marking-scheme points
- ✓Detects a methyl ketone (CH3-CO-) or CH3-CH(OH)- group
- ✓Warm with I2 and NaOH -> yellow precipitate of iodoform (CHI3)
- ✓Positive for ethanol, acetaldehyde, acetone, isopropanol
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Amines
How are amines classified? Give one example of each.
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Amines are derivatives of ammonia in which one, two or three hydrogen atoms are replaced by alkyl or aryl groups. They are classified as: primary (1 degree) amine, in which one hydrogen of ammonia is replaced (e.g. CH3NH2, methylamine); secondary (2 degree) amine, in which two hydrogens are replaced (e.g. (CH3)2NH, dimethylamine); and tertiary (3 degree) amine, in which all three hydrogens are replaced (e.g. (CH3)3N, trimethylamine).
Marking-scheme points
- ✓Amines are derivatives of ammonia (H replaced by alkyl/aryl)
- ✓Primary: one H replaced (CH3NH2)
- ✓Secondary: two replaced ((CH3)2NH); Tertiary: three replaced ((CH3)3N)
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Amines
How are primary amines prepared by the reduction of nitro compounds and by Hofmann's bromamide reaction?
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(1) Reduction of nitro compounds: a nitro compound is reduced (with H2/Ni, or Sn/HCl, or Fe/HCl) to a primary amine; for example, nitrobenzene is reduced to aniline, C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O. (2) Hofmann's bromamide degradation: an amide is treated with bromine and a strong alkali (Br2 + NaOH/KOH) to give a primary amine with one carbon atom less than the amide, R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O.
R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O
Marking-scheme points
- ✓Reduction of nitro compound -> primary amine (nitrobenzene -> aniline)
- ✓Hofmann bromamide: amide + Br2 + NaOH -> amine with one carbon less
- ✓R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O
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Amines
Why are amines basic in nature? Compare the basic strength of amines in the gaseous phase.
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Amines are basic because the nitrogen atom has a lone pair of electrons which it can donate to a proton (or to a Lewis acid), forming a bond; thus amines can accept a proton. In the gaseous phase (or a non-aqueous medium), only the inductive effect of the alkyl groups operates: more alkyl groups increase the electron density on nitrogen, so the basic strength follows the order tertiary > secondary > primary > ammonia. In aqueous solution the order changes because of the combined effect of inductive effect, solvation (hydrogen bonding of the cation) and steric hindrance.
Marking-scheme points
- ✓Amines are basic: nitrogen lone pair accepts a proton
- ✓Gas phase (inductive effect only): tertiary > secondary > primary > ammonia
- ✓Aqueous order differs due to solvation and steric effects
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Amines
How does the Hinsberg test distinguish between primary, secondary and tertiary amines?
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The Hinsberg test uses Hinsberg's reagent (benzenesulphonyl chloride, C6H5SO2Cl). A primary amine reacts to form a sulphonamide that is soluble in alkali (because the N-H hydrogen is acidic). A secondary amine forms a sulphonamide that is insoluble in alkali (it has no N-H hydrogen left to ionise). A tertiary amine does not react at all with the reagent (it has no replaceable hydrogen on nitrogen). Thus the three classes of amine are distinguished by their behaviour with Hinsberg's reagent.
Marking-scheme points
- ✓Reagent: benzenesulphonyl chloride (Hinsberg's reagent)
- ✓Primary amine: product soluble in alkali; secondary: product insoluble
- ✓Tertiary amine: does not react
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Amines
What is the carbylamine reaction? What is it used for?
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The carbylamine reaction (isocyanide test) is a reaction in which a primary amine (aliphatic or aromatic) is heated with chloroform and alcoholic potassium hydroxide to form an isocyanide (carbylamine), which has an extremely unpleasant (foul) smell: R-NH2 + CHCl3 + 3KOH -> R-NC + 3KCl + 3H2O. Secondary and tertiary amines do not give this reaction. Because only primary amines respond, the carbylamine reaction is used as a test to detect primary amines.
R-NH2 + CHCl3 + 3KOH -> R-NC + 3KCl + 3H2O
Marking-scheme points
- ✓Primary amine + CHCl3 + alcoholic KOH -> isocyanide (foul smell)
- ✓R-NH2 + CHCl3 + 3KOH -> R-NC + 3KCl + 3H2O
- ✓Given only by primary amines -> test for primary amines
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Amines
How is benzenediazonium chloride prepared? Write one of its reactions (Sandmeyer reaction).
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Benzenediazonium chloride is prepared by diazotisation: aniline is treated with nitrous acid (formed from sodium nitrite and hydrochloric acid) at a low temperature of 0 to 5 degrees C, C6H5NH2 + NaNO2 + 2HCl -> C6H5N2Cl + NaCl + 2H2O. In the Sandmeyer reaction, the diazonium group is replaced by a halogen or cyanide using the corresponding copper(I) salt; for example, C6H5N2Cl + CuCl -> C6H5Cl + N2. Diazonium salts are very useful for introducing many groups into the benzene ring.
C6H5NH2 + NaNO2 + 2HCl -> C6H5N2Cl + NaCl + 2H2O
Marking-scheme points
- ✓Diazotisation: aniline + NaNO2 + HCl at 0-5 degrees C -> C6H5N2Cl
- ✓Sandmeyer reaction: replace N2+ with Cl/Br/CN using Cu(I) salts
- ✓C6H5N2Cl + CuCl -> C6H5Cl + N2
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Amines
Why is aniline less basic than ethylamine (or ammonia)?
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Aniline is less basic than ethylamine (and than ammonia) because in aniline the lone pair of electrons on the nitrogen atom is delocalised (drawn) into the benzene ring by resonance. As a result the lone pair is less available for donation to a proton, so aniline accepts a proton less readily and is a weaker base. In ethylamine, the alkyl group has an electron-releasing (+I) effect that increases the electron density on nitrogen and makes the lone pair more available, so it is more basic.
Marking-scheme points
- ✓In aniline the N lone pair is delocalised into the ring (resonance)
- ✓Lone pair is less available for protonation -> weaker base
- ✓In ethylamine, +I effect of the alkyl group makes N more basic
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Amines
What is the coupling reaction of diazonium salts? Give one example.
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The coupling reaction is a reaction in which a diazonium salt reacts with an electron-rich aromatic compound such as phenol or an aromatic amine to form a brightly coloured azo compound (containing the -N=N- linkage). For example, benzenediazonium chloride reacts with phenol in a mildly alkaline medium to give p-hydroxyazobenzene (an orange dye). These coupling reactions are used to prepare a large number of azo dyes.
Marking-scheme points
- ✓Diazonium salt + phenol/aromatic amine -> coloured azo compound (-N=N-)
- ✓Example: benzenediazonium chloride + phenol -> p-hydroxyazobenzene
- ✓Used to make azo dyes
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Biomolecules
Classify carbohydrates into three types based on hydrolysis, with one example of each.
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Based on their behaviour on hydrolysis, carbohydrates are classified as: (1) monosaccharides - simple sugars that cannot be hydrolysed further into smaller units (e.g. glucose, fructose); (2) oligosaccharides - which give 2 to 10 monosaccharide units on hydrolysis, the most common being disaccharides (e.g. sucrose, which gives glucose and fructose, and maltose); and (3) polysaccharides - which give a large number of monosaccharide units on hydrolysis (e.g. starch, cellulose and glycogen).
Marking-scheme points
- ✓Monosaccharides: cannot be hydrolysed further (glucose, fructose)
- ✓Oligosaccharides (disaccharides): give 2-10 units (sucrose, maltose)
- ✓Polysaccharides: give many units (starch, cellulose)
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Biomolecules
Distinguish between reducing and non-reducing sugars with examples.
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A reducing sugar is a carbohydrate that has a free aldehyde or ketone group (a free hemiacetal), so it can reduce Tollens' reagent and Fehling's solution; all monosaccharides (glucose, fructose) and many disaccharides (maltose, lactose) are reducing sugars. A non-reducing sugar has no free aldehyde or ketone group (the reducing groups of both units are involved in the glycosidic bond), so it does not reduce Tollens' or Fehling's reagents; sucrose is the common example of a non-reducing sugar.
Marking-scheme points
- ✓Reducing sugar: has a free aldehyde/ketone group; reduces Tollens'/Fehling's
- ✓Examples: glucose, fructose, maltose, lactose
- ✓Non-reducing sugar: no free reducing group (e.g. sucrose)
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Biomolecules
State two evidences that show the presence of an aldehyde group and an alcohol group in glucose.
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Evidence for an aldehyde (-CHO) group in glucose: (1) it reduces Tollens' reagent to give a silver mirror and Fehling's solution to give a red precipitate; (2) it reacts with hydroxylamine to form an oxime and adds one molecule of HCN to form a cyanohydrin, confirming a carbonyl group. Evidence for alcohol (-OH) groups: glucose reacts with acetic anhydride to form a penta-acetate, showing the presence of five -OH groups. Thus glucose is a polyhydroxy aldehyde.
Marking-scheme points
- ✓Aldehyde group: reduces Tollens' and Fehling's; forms oxime and cyanohydrin
- ✓Alcohol groups: forms penta-acetate with acetic anhydride (five -OH groups)
- ✓Glucose is a polyhydroxy aldehyde
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Biomolecules
State two differences between starch and cellulose.
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(1) Starch is a polymer of alpha-glucose units and consists of two components, amylose (a linear chain) and amylopectin (a branched chain), joined by alpha-glycosidic linkages; cellulose is a straight-chain polymer of beta-glucose units joined by beta-glycosidic linkages. (2) Starch is the main storage carbohydrate (food reserve) in plants and can be digested by humans, whereas cellulose is a structural material (the main component of plant cell walls) and cannot be digested by humans (we lack the enzyme cellulase).
Marking-scheme points
- ✓Starch: alpha-glucose units (amylose + amylopectin), alpha-linkages
- ✓Cellulose: beta-glucose units, straight chain, beta-linkages
- ✓Starch is a food reserve (digestible); cellulose is structural (indigestible by humans)
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Biomolecules
What are proteins? What is meant by denaturation of a protein?
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Proteins are naturally occurring biomolecules that are polymers of alpha-amino acids joined by peptide bonds (-CO-NH-); they are essential for growth and maintenance of the body and act as enzymes, hormones and structural materials. Denaturation of a protein is the process in which a protein loses its natural three-dimensional shape (its secondary and tertiary structure) due to heat, change in pH, or addition of chemicals, while the primary structure (sequence of amino acids) remains intact. On denaturation the protein loses its biological activity; for example, the coagulation of egg white on boiling.
Marking-scheme points
- ✓Proteins are polymers of alpha-amino acids linked by peptide bonds
- ✓Denaturation: loss of secondary and tertiary structure (shape)
- ✓Caused by heat/pH/chemicals; loses biological activity (e.g. boiling egg white)
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Biomolecules
Name and briefly describe the four levels of protein structure.
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(1) Primary structure - the specific sequence in which the amino acids are linked in the polypeptide chain. (2) Secondary structure - the local folding of the chain into shapes such as the alpha-helix or beta-pleated sheet, held together by hydrogen bonds. (3) Tertiary structure - the overall three-dimensional folding of the whole polypeptide chain, giving a globular or fibrous shape. (4) Quaternary structure - the arrangement and association of two or more polypeptide chains (subunits), as in haemoglobin.
Marking-scheme points
- ✓Primary: sequence of amino acids
- ✓Secondary: alpha-helix or beta-pleated sheet (hydrogen bonds)
- ✓Tertiary: overall 3D fold; Quaternary: association of subunits (haemoglobin)
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Biomolecules
What are enzymes? State two of their characteristics.
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Enzymes are biological catalysts that speed up the biochemical reactions occurring in living organisms; chemically almost all enzymes are proteins (globular proteins). Characteristics: (1) they are highly specific, each enzyme usually catalysing only one particular reaction or type of reaction (lock-and-key specificity); and (2) they are highly efficient and work best under mild conditions of an optimum temperature (around body temperature) and an optimum pH; they are denatured and lose activity outside these conditions.
Marking-scheme points
- ✓Enzymes = biological catalysts (mostly proteins)
- ✓Highly specific (one enzyme, one reaction)
- ✓Work best at an optimum temperature and pH
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Biomolecules
How are vitamins classified? Name one deficiency disease caused by a lack of vitamin C and vitamin D.
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Vitamins are organic compounds required in small amounts in the diet for normal growth and health. They are classified into two groups: (1) fat-soluble vitamins (A, D, E and K), which are soluble in fats and oils and are stored in the liver and fatty tissues; and (2) water-soluble vitamins (the B group and C), which are soluble in water and must be supplied regularly in the diet as they are not stored. Deficiency of vitamin C causes scurvy, and deficiency of vitamin D causes rickets in children.
Marking-scheme points
- ✓Fat-soluble vitamins: A, D, E, K (stored in the body)
- ✓Water-soluble vitamins: B group and C (not stored)
- ✓Vitamin C deficiency: scurvy; vitamin D deficiency: rickets
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Biomolecules
State three differences between DNA and RNA.
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(1) DNA (deoxyribonucleic acid) contains the sugar deoxyribose, whereas RNA (ribonucleic acid) contains the sugar ribose. (2) DNA contains the nitrogenous bases adenine, guanine, cytosine and thymine, whereas RNA contains adenine, guanine, cytosine and uracil (uracil replaces thymine). (3) DNA is usually a double-stranded helix and stores the genetic (hereditary) information, whereas RNA is usually single-stranded and mainly takes part in protein synthesis.
Marking-scheme points
- ✓Sugar: deoxyribose in DNA, ribose in RNA
- ✓Bases: DNA has thymine, RNA has uracil (instead of thymine)
- ✓DNA double-stranded (stores genes); RNA single-stranded (protein synthesis)
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