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ChemistryClass 122 marksmedium

Alcohols, Phenols and Ethers

How can you distinguish between primary, secondary and tertiary alcohols using the Lucas test?

Reveal model answer + marking points

The Lucas test uses the Lucas reagent (a mixture of concentrated hydrochloric acid and anhydrous zinc chloride), which converts alcohols into alkyl chlorides that appear as an insoluble oily layer (turbidity). A tertiary alcohol reacts immediately and gives turbidity at once (because it forms the most stable carbocation). A secondary alcohol gives turbidity within about five minutes. A primary alcohol does not react at room temperature and gives no turbidity in the cold. Thus the rate of turbidity distinguishes the three classes.

Marking-scheme points

  • Lucas reagent = conc. HCl + anhydrous ZnCl2
  • Tertiary alcohol: immediate turbidity
  • Secondary: turbidity in about 5 min; primary: no turbidity in the cold
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ChemistryClass 122 marksmedium

Alcohols, Phenols and Ethers

What is Williamson's ether synthesis? Write the general reaction.

Reveal model answer + marking points

Williamson's synthesis is an important laboratory method for preparing symmetrical and unsymmetrical ethers. In it, a sodium alkoxide (or sodium phenoxide) reacts with a primary alkyl halide by nucleophilic substitution (SN2) to give an ether: R-O-Na + R'-X -> R-O-R' + NaX. A primary alkyl halide should be used (secondary and tertiary halides tend to undergo elimination). This method is especially useful for preparing mixed (unsymmetrical) ethers.

R-O-Na + R'-X -> R-O-R' + NaX

Marking-scheme points

  • Sodium alkoxide + alkyl halide -> ether (SN2)
  • R-O-Na + R'-X -> R-O-R' + NaX
  • Use a primary alkyl halide; good for mixed ethers
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ChemistryClass 122 marksmedium

Alcohols, Phenols and Ethers

What is the Reimer-Tiemann reaction?

Reveal model answer + marking points

The Reimer-Tiemann reaction is used to introduce an aldehyde group onto the benzene ring of phenol. When phenol is treated with chloroform (CHCl3) and aqueous sodium hydroxide (NaOH) and the product is then hydrolysed with acid, an -CHO group is introduced mainly at the ortho position, giving 2-hydroxybenzaldehyde (salicylaldehyde). The reactive intermediate is dichlorocarbene (:CCl2).

phenol + CHCl3 + NaOH -> salicylaldehyde

Marking-scheme points

  • Phenol + CHCl3 + NaOH, then acid hydrolysis
  • Introduces a -CHO group at the ortho position
  • Gives salicylaldehyde (2-hydroxybenzaldehyde); intermediate is dichlorocarbene
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ChemistryClass 122 markshard

Alcohols, Phenols and Ethers

Why is ortho-nitrophenol steam volatile while para-nitrophenol is not?

Reveal model answer + marking points

In ortho-nitrophenol, the -OH group and the -NO2 group are close together on adjacent carbons, so they form an intramolecular hydrogen bond (within the same molecule), a process called chelation. As a result the molecules do not associate with one another and it is volatile (steam volatile). In para-nitrophenol, the two groups are far apart, so they form intermolecular hydrogen bonds between different molecules, which associate them into larger units, making it less volatile and not steam volatile.

Marking-scheme points

  • ortho-nitrophenol forms intramolecular hydrogen bonding (chelation)
  • So its molecules do not associate -> volatile/steam volatile
  • para-nitrophenol forms intermolecular hydrogen bonds -> less volatile
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ChemistryClass 122 marksmedium

Aldehydes, Ketones and Carboxylic Acids

How are aldehydes and ketones prepared by the oxidation of alcohols?

Reveal model answer + marking points

Aldehydes and ketones can be prepared by the controlled oxidation of alcohols. Oxidation of a primary alcohol gives an aldehyde (which can be further oxidised to a carboxylic acid), R-CH2OH -> R-CHO. To stop at the aldehyde stage, a mild oxidising agent such as pyridinium chlorochromate (PCC) is used. Oxidation of a secondary alcohol gives a ketone, R-CH(OH)-R' -> R-CO-R'. Tertiary alcohols are not easily oxidised as they have no hydrogen on the carbon bearing the -OH group.

R-CH2OH -> R-CHO; R2CHOH -> R2CO

Marking-scheme points

  • Primary alcohol -> aldehyde (use mild oxidant like PCC to stop there)
  • Secondary alcohol -> ketone
  • Tertiary alcohols resist oxidation (no H on the -OH carbon)
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ChemistryClass 122 marksmedium

Aldehydes, Ketones and Carboxylic Acids

What is the aldol condensation reaction?

Reveal model answer + marking points

The aldol condensation is a reaction of aldehydes or ketones that have at least one alpha-hydrogen atom, in the presence of a dilute base (such as dilute NaOH). Two molecules combine: the alpha-carbon of one adds to the carbonyl carbon of the other to give a beta-hydroxy aldehyde or ketone (an aldol). For example, two molecules of acetaldehyde give 3-hydroxybutanal (CH3CHO + CH3CHO -> CH3CH(OH)CH2CHO). On heating, the aldol loses water to give an alpha, beta-unsaturated carbonyl compound.

2CH3CHO -> CH3CH(OH)CH2CHO

Marking-scheme points

  • Needs an alpha-hydrogen and a dilute base
  • Product is a beta-hydroxy aldehyde/ketone (aldol)
  • Two acetaldehyde molecules -> 3-hydroxybutanal; loses water on heating
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ChemistryClass 122 marksmedium

Aldehydes, Ketones and Carboxylic Acids

What is the Cannizzaro reaction?

Reveal model answer + marking points

The Cannizzaro reaction is a disproportionation (self-oxidation-reduction) reaction shown by aldehydes that do not have an alpha-hydrogen atom, when they are treated with a concentrated alkali (such as concentrated NaOH). One molecule of the aldehyde is oxidised to a carboxylate salt and another is reduced to an alcohol. For example, two molecules of formaldehyde give methanol and sodium formate: 2HCHO + NaOH -> CH3OH + HCOONa. Benzaldehyde behaves similarly.

2HCHO + NaOH -> CH3OH + HCOONa

Marking-scheme points

  • Shown by aldehydes without an alpha-hydrogen, with conc. alkali
  • Disproportionation: one molecule oxidised, another reduced
  • 2HCHO + NaOH -> CH3OH + HCOONa
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ChemistryClass 122 marksmedium

Aldehydes, Ketones and Carboxylic Acids

How can you distinguish between an aldehyde and a ketone using chemical tests?

Reveal model answer + marking points

Aldehydes are easily oxidised and give positive results with mild oxidising agents, whereas ketones do not. (1) Tollens' test: on warming with Tollens' reagent (ammoniacal silver nitrate), an aldehyde gives a bright silver mirror on the walls of the test tube, while a ketone gives no reaction. (2) Fehling's test: on warming with Fehling's solution, an aliphatic aldehyde gives a brick-red precipitate of cuprous oxide (Cu2O), while a ketone does not react. Thus these tests distinguish aldehydes from ketones.

Marking-scheme points

  • Aldehydes are oxidised by mild oxidants; ketones are not
  • Tollens' test: aldehyde gives a silver mirror
  • Fehling's test: aliphatic aldehyde gives a brick-red precipitate (Cu2O)
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ChemistryClass 122 marksmedium

Aldehydes, Ketones and Carboxylic Acids

How are carboxylic acids prepared from primary alcohols and from Grignard reagents?

Reveal model answer + marking points

(1) From primary alcohols (or aldehydes) by oxidation: a primary alcohol is oxidised by a strong oxidising agent such as acidified potassium permanganate or potassium dichromate to give a carboxylic acid, R-CH2OH -> R-CHO -> R-COOH. (2) From Grignard reagents: a Grignard reagent reacts with carbon dioxide (dry ice) and the product is hydrolysed with acid to give a carboxylic acid with one more carbon atom, R-MgX + CO2 -> R-COOMgX, then hydrolysis gives R-COOH.

R-MgX + CO2 -> R-COOMgX -> R-COOH

Marking-scheme points

  • Oxidation of primary alcohol/aldehyde: R-CH2OH -> R-COOH
  • Grignard reagent + CO2 then hydrolysis -> carboxylic acid
  • Grignard method adds one carbon atom
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ChemistryClass 122 marksmedium

Aldehydes, Ketones and Carboxylic Acids

Write the esterification reaction of a carboxylic acid and the decarboxylation reaction.

Reveal model answer + marking points

(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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ChemistryClass 122 marksmedium

Aldehydes, Ketones and Carboxylic Acids

What is the iodoform test? Which compounds give a positive result?

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Amines

How are amines classified? Give one example of each.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Amines

How does the Hinsberg test distinguish between primary, secondary and tertiary amines?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Amines

What is the carbylamine reaction? What is it used for?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Amines

Why is aniline less basic than ethylamine (or ammonia)?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Amines

What is the coupling reaction of diazonium salts? Give one example.

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Biomolecules

Classify carbohydrates into three types based on hydrolysis, with one example of each.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Biomolecules

Distinguish between reducing and non-reducing sugars with examples.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Biomolecules

State two evidences that show the presence of an aldehyde group and an alcohol group in glucose.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Biomolecules

State two differences between starch and cellulose.

Reveal model answer + marking points

(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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