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The questions your board exam loves to ask
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 122 markseasy
Coordination Compounds
State any four applications of coordination compounds.
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(1) In biological systems: haemoglobin (a complex of iron) carries oxygen, and chlorophyll (a complex of magnesium) is essential for photosynthesis. (2) In metallurgy: metals like silver and gold are extracted and purified using complex formation (cyanide process). (3) In medicine: cisplatin is used in cancer treatment and EDTA complexes are used to treat lead poisoning. (4) In analytical chemistry and electroplating: complexes are used in the estimation of metal ions and in electroplating of metals.
Marking-scheme points
- ✓Biological: haemoglobin (Fe), chlorophyll (Mg)
- ✓Metallurgy: extraction of silver and gold (cyanide process)
- ✓Medicine (cisplatin), analysis and electroplating
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Haloalkanes and Haloarenes
Distinguish between haloalkanes and haloarenes with one example each.
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Haloalkanes (alkyl halides) are compounds in which one or more hydrogen atoms of an aliphatic hydrocarbon (alkane) are replaced by halogen atoms; the halogen is attached to an sp3 carbon (e.g. CH3Cl, chloromethane). Haloarenes (aryl halides) are compounds in which the halogen atom is directly attached to an sp2 carbon of an aromatic ring (e.g. C6H5Cl, chlorobenzene). Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution.
Marking-scheme points
- ✓Haloalkane: halogen on sp3 (aliphatic) carbon (CH3Cl)
- ✓Haloarene: halogen on sp2 carbon of an aromatic ring (C6H5Cl)
- ✓Haloarenes are less reactive to nucleophilic substitution
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Haloalkanes and Haloarenes
Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?
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Haloarenes are less reactive than haloalkanes towards nucleophilic substitution mainly because of: (1) resonance - the lone pair of the halogen delocalises into the ring, giving the carbon-halogen bond a partial double-bond character, so it is shorter and stronger and harder to break; (2) the halogen is attached to an sp2 hybridised carbon which is more electronegative and holds the shared electrons more tightly than the sp3 carbon in haloalkanes; and (3) repulsion between the electron-rich aromatic ring and the approaching nucleophile.
Marking-scheme points
- ✓Resonance gives the C-X bond partial double-bond character (stronger)
- ✓Halogen on sp2 carbon (more electronegative, holds electrons tightly)
- ✓Electron-rich ring repels the incoming nucleophile
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Haloalkanes and Haloarenes
What is a dehydrohalogenation (elimination) reaction of a haloalkane? State Saytzeff's rule.
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Dehydrohalogenation is a beta-elimination reaction in which a haloalkane loses a hydrogen halide (HX) when heated with an alcoholic solution of potassium hydroxide, forming an alkene. The hydrogen is removed from the beta-carbon (the carbon next to the one bearing the halogen). Saytzeff's rule states that in such an elimination the preferred (major) product is the more highly substituted (more stable) alkene, that is, the alkene formed by removal of the hydrogen from the beta-carbon having the fewer hydrogen atoms.
Marking-scheme points
- ✓Beta-elimination of HX with alcoholic KOH gives an alkene
- ✓Hydrogen removed from the beta-carbon
- ✓Saytzeff's rule: the more substituted (more stable) alkene is the major product
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Haloalkanes and Haloarenes
What is a chiral molecule? What is meant by optical activity?
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A chiral molecule is one that is non-superimposable on its mirror image, just as the left and right hands are not superimposable; it usually contains at least one carbon atom bonded to four different groups (an asymmetric or chiral carbon). Optical activity is the property of such a chiral substance to rotate the plane of plane-polarised light; the two non-superimposable mirror-image forms are called enantiomers, one rotating the light to the right (dextrorotatory) and the other to the left (laevorotatory).
Marking-scheme points
- ✓Chiral molecule: non-superimposable on its mirror image (chiral carbon)
- ✓Optical activity: rotates the plane of plane-polarised light
- ✓Mirror-image forms = enantiomers (dextro- and laevorotatory)
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Haloalkanes and Haloarenes
What is a Grignard reagent? How is it prepared?
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A Grignard reagent is an alkyl or aryl magnesium halide, R-Mg-X, which is a very important and reactive organometallic compound used in organic synthesis. It is prepared by the reaction of a haloalkane (or haloarene) with magnesium metal in the presence of dry ether: R-X + Mg -> R-Mg-X (in dry ether). Grignard reagents are highly reactive and must be prepared under anhydrous conditions, because even traces of water or moisture decompose them to alkanes.
R-X + Mg -> R-Mg-X (dry ether)
Marking-scheme points
- ✓Grignard reagent = alkyl/aryl magnesium halide (R-Mg-X)
- ✓Prepared: R-X + Mg -> R-Mg-X in dry ether
- ✓Very reactive; must be kept anhydrous (water decomposes it)
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Haloalkanes and Haloarenes
Why is chloroform stored in dark coloured bottles filled up to the brim?
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Chloroform (CHCl3) is stored in dark coloured bottles filled completely up to the brim because in the presence of air (oxygen) and sunlight it undergoes slow oxidation to form a highly poisonous gas, phosgene (carbonyl chloride, COCl2): 2CHCl3 + O2 -> 2COCl2 + 2HCl. Filling the bottle to the brim leaves no air space, and the dark bottle keeps out light; together these prevent the oxidation of chloroform to phosgene.
2CHCl3 + O2 -> 2COCl2 + 2HCl
Marking-scheme points
- ✓Chloroform is oxidised in air and light to poisonous phosgene (COCl2)
- ✓2CHCl3 + O2 -> 2COCl2 + 2HCl
- ✓Dark bottle filled to the brim excludes light and air
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Alcohols, Phenols and Ethers
Classify alcohols as primary, secondary and tertiary with one example each.
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Alcohols are classified according to the type of carbon atom to which the -OH group is attached. In a primary (1 degree) alcohol the -OH is on a carbon attached to only one other carbon, e.g. ethanol (CH3CH2OH). In a secondary (2 degree) alcohol the -OH is on a carbon attached to two other carbons, e.g. propan-2-ol ((CH3)2CHOH). In a tertiary (3 degree) alcohol the -OH is on a carbon attached to three other carbons, e.g. 2-methylpropan-2-ol ((CH3)3COH).
Marking-scheme points
- ✓Primary: -OH carbon attached to one carbon (ethanol)
- ✓Secondary: -OH carbon attached to two carbons (propan-2-ol)
- ✓Tertiary: -OH carbon attached to three carbons (2-methylpropan-2-ol)
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Alcohols, Phenols and Ethers
Why is phenol more acidic than ethanol?
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Phenol is more acidic than ethanol because the phenoxide ion formed after the loss of the proton is stabilised by resonance (the negative charge is delocalised into the benzene ring), which makes phenol lose its proton more easily. In contrast, the ethoxide ion from ethanol is not resonance-stabilised, and the alkyl group has an electron-releasing (+I) effect that further destabilises the ethoxide ion. Hence phenol ionises more readily and is a stronger acid than ethanol.
Marking-scheme points
- ✓Phenoxide ion is stabilised by resonance (charge delocalised into ring)
- ✓Ethoxide ion is not resonance-stabilised
- ✓Alkyl (+I) effect destabilises ethoxide, so ethanol is weaker acid
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Alcohols, Phenols and Ethers
Write the reaction of ethanol with sodium metal and its dehydration to ethene.
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(1) Reaction with sodium: alcohols react with active metals like sodium to liberate hydrogen gas and form sodium alkoxide, 2C2H5OH + 2Na -> 2C2H5ONa + H2. This shows the acidic nature of the -OH group. (2) Dehydration: when ethanol is heated with concentrated sulphuric acid at about 443 K (170 degrees C), it loses a molecule of water to form ethene, C2H5OH -> CH2=CH2 + H2O. Concentrated H2SO4 acts as a dehydrating agent.
C2H5OH -> CH2=CH2 + H2O
Marking-scheme points
- ✓With sodium: 2C2H5OH + 2Na -> 2C2H5ONa + H2 (shows acidic -OH)
- ✓Dehydration with conc. H2SO4 at 443 K gives ethene
- ✓C2H5OH -> CH2=CH2 + H2O
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Alcohols, Phenols and Ethers
How can you distinguish between primary, secondary and tertiary alcohols using the Lucas test?
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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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Alcohols, Phenols and Ethers
What is Williamson's ether synthesis? Write the general reaction.
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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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Alcohols, Phenols and Ethers
What is the Reimer-Tiemann reaction?
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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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Alcohols, Phenols and Ethers
Why is ortho-nitrophenol steam volatile while para-nitrophenol is not?
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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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Aldehydes, Ketones and Carboxylic Acids
How are aldehydes and ketones prepared by the oxidation of alcohols?
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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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Aldehydes, Ketones and Carboxylic Acids
What is the aldol condensation reaction?
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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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Aldehydes, Ketones and Carboxylic Acids
What is the Cannizzaro reaction?
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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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Aldehydes, Ketones and Carboxylic Acids
How can you distinguish between an aldehyde and a ketone using chemical tests?
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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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Aldehydes, Ketones and Carboxylic Acids
How are carboxylic acids prepared from primary alcohols and from Grignard reagents?
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(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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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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