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ChemistryClass 123 markshard

Haloalkanes and Haloarenes

Distinguish between the SN1 and SN2 mechanisms of nucleophilic substitution.

Reveal model answer + marking points

In the SN2 (substitution nucleophilic bimolecular) mechanism, the reaction occurs in a single step; the nucleophile attacks the carbon from the side opposite the leaving group, and bond breaking and bond making occur simultaneously. Its rate depends on both the substrate and the nucleophile, and it gives inversion of configuration (Walden inversion); it is favoured by primary halides. In the SN1 (substitution nucleophilic unimolecular) mechanism, the reaction occurs in two steps through a carbocation intermediate; its rate depends only on the substrate, and it usually gives a racemic mixture. It is favoured by tertiary halides and polar protic solvents.

Marking-scheme points

  • SN2: one step, backside attack, inversion; rate depends on both reactants; favoured by primary halides
  • SN1: two steps via carbocation; rate depends only on substrate; racemisation
  • SN1 favoured by tertiary halides and polar protic solvents
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ChemistryClass 122 marksmedium

Haloalkanes and Haloarenes

Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Haloalkanes and Haloarenes

What is a dehydrohalogenation (elimination) reaction of a haloalkane? State Saytzeff's rule.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Haloalkanes and Haloarenes

What is a chiral molecule? What is meant by optical activity?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Haloalkanes and Haloarenes

What is a Grignard reagent? How is it prepared?

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Haloalkanes and Haloarenes

Why is chloroform stored in dark coloured bottles filled up to the brim?

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Alcohols, Phenols and Ethers

Classify alcohols as primary, secondary and tertiary with one example each.

Reveal model answer + marking points

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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ChemistryClass 123 marksmedium

Alcohols, Phenols and Ethers

How are alcohols prepared by the hydration of alkenes and by the reduction of aldehydes and ketones?

Reveal model answer + marking points

(1) Acid-catalysed hydration of alkenes: an alkene adds water in the presence of dilute sulphuric acid following Markovnikov's rule to give an alcohol, e.g. CH2=CH2 + H2O -> CH3CH2OH. (2) Reduction of carbonyl compounds: aldehydes on reduction (with H2/Ni or NaBH4 or LiAlH4) give primary alcohols (R-CHO -> R-CH2OH), and ketones on reduction give secondary alcohols (R-CO-R' -> R-CH(OH)-R'). Alcohols can also be prepared from Grignard reagents reacting with carbonyl compounds.

R-CHO + 2[H] -> R-CH2OH

Marking-scheme points

  • Hydration of alkenes (Markovnikov): CH2=CH2 + H2O -> CH3CH2OH
  • Reduction of aldehyde -> primary alcohol
  • Reduction of ketone -> secondary alcohol
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ChemistryClass 122 marksmedium

Alcohols, Phenols and Ethers

Why is phenol more acidic than ethanol?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Alcohols, Phenols and Ethers

Write the reaction of ethanol with sodium metal and its dehydration to ethene.

Reveal model answer + marking points

(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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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 123 marksmedium

Aldehydes, Ketones and Carboxylic Acids

Why do aldehydes and ketones undergo nucleophilic addition reactions? Give one example.

Reveal model answer + marking points

The carbonyl group (C=O) is polar because oxygen is more electronegative than carbon, so the carbon carries a partial positive charge and the oxygen a partial negative charge. This makes the carbonyl carbon electrophilic, so it is readily attacked by nucleophiles, leading to nucleophilic addition. For example, with hydrogen cyanide (HCN), the addition gives a cyanohydrin: R-CHO + HCN -> R-CH(OH)-CN. Aldehydes are more reactive than ketones towards nucleophilic addition due to less steric hindrance and a greater positive charge on the carbonyl carbon.

R-CHO + HCN -> R-CH(OH)-CN

Marking-scheme points

  • C=O is polar; carbonyl carbon is electrophilic (partial positive)
  • Nucleophile attacks the carbonyl carbon (nucleophilic addition)
  • Example: R-CHO + HCN -> R-CH(OH)-CN (cyanohydrin)
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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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