Q.Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol.

Showing the 12 most recent of 24 on this concept.

1. CBSE 2026Set 56/2/11 markMCQ
   Q.Identify the correct increasing order of boiling points of the given compounds : (A) Propan-1-ol < butan-1-ol < butan-2-ol < pentan-1-ol (B) Pentan-1-ol < butan-1-ol < butan-2-ol < Propan-1-ol (C) Propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol (D) Butan-1-ol < Butan-2-ol < Propan-1-ol < Pentan-1-ol

   ›Reveal solutionSolution

   Boiling points of alcohols depend on chain length (more carbons → higher bp) and branching (more branching → lower bp). The correct order is Propan-1‑ol < butan‑2‑ol < butan‑1‑ol < pentan‑1‑ol, which matches option (C).

   Why boiling points of alcohols behave this way

   Alcohols boil at much higher temperatures than hydrocarbons of similar mass because of hydrogen bonding between the –OH groups. Two factors control the boiling point within a family of alcohols:

   1. Chain length – A longer carbon chain means more surface area for van der Waals forces. These weak attractions add up, so a larger molecule needs more energy (higher temperature) to escape into the vapour phase. For straight‑chain alcohols, boiling point rises steadily as the number of carbons increases.

   2. Branching – When the –OH group is attached to a secondary or tertiary carbon (as in butan‑2‑ol), the molecule becomes more compact. A compact shape reduces the surface area available for van der Waals interactions, so the boiling point drops compared to its straight‑chain isomer. The hydrogen‑bonding ability is roughly the same for all isomers (one –OH per molecule), so the difference comes from the weaker London forces in the branched form.

   Watch out

   A common mistake is to think that branching increases boiling point because the molecule looks “more crowded”. In reality, branching decreases the surface area and therefore weakens the intermolecular forces. Always compare chain length first, then branching.

   Step‑by‑step reasoning

   1. Identify the compounds and their carbon counts

      • Propan‑1‑ol: 3 carbons, straight chain.
      • Butan‑1‑ol: 4 carbons, straight chain.
      • Butan‑2‑ol: 4 carbons, branched (the –OH is on carbon 2).
      • Pentan‑1‑ol: 5 carbons, straight chain.

   2. Order by chain length

      Longer chain → higher boiling point. So the 5‑carbon alcohol (pentan‑1‑ol) should have the highest bp, and the 3‑carbon alcohol (propan‑1‑ol) the lowest. The two 4‑carbon alcohols will sit in between.

   3. Compare the two C₄ isomers

      Butan‑1‑ol is a straight‑chain primary alcohol. Butan‑2‑ol is a secondary alcohol with a branched shape. Because branching reduces surface area, butan‑2‑ol has weaker van der Waals forces and therefore a lower boiling point than butan‑1‑ol.

   4. Assemble the full order

      From lowest to highest boiling point:

      • Propan‑1‑ol (3C, straight)
      • Butan‑2‑ol (4C, branched)
      • Butan‑1‑ol (4C, straight)
      • Pentan‑1‑ol (5C, straight)

      This matches option (C).

   Tip

   You can remember the pattern as: more carbons → higher bp; more branching → lower bp (for same number of carbons). For alcohols, the hydrogen‑bonding contribution is roughly constant per molecule, so the van der Waals forces decide the order.

   ✓Final answer

   The correct increasing order of boiling points is Propan‑1‑ol < butan‑2‑ol < butan‑1‑ol < pentan‑1‑ol, which corresponds to option (C).

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): Boiling point of alkanes decreases with increase in molecular mass. Reason (R): Intermolecular Vander Waals forces increase with increase in molecular size or surface area of the molecules.

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

   (c) Assertion (A) is true, but Reason (R) is false.

   (d) Assertion (A) is false, but Reason (R) is true.

   ›Reveal solutionSolution

   The Assertion has the trend backwards — alkane boiling points INCREASE with molecular mass — while the Reason correctly describes why (stronger van der Waals forces with larger surface area).

   Assertion: 'Boiling point of alkanes decreases with increase in molecular mass' — this is false. In reality, as molecular mass (chain length) increases, boiling point increases (e.g. methane $-162°C$ → butane $-0.5°C$ → octane $126°C$).

   Reason: 'Intermolecular van der Waals forces increase with increase in molecular size or surface area' — this is true, and it is the correct underlying physical reason boiling points rise with chain length (larger molecules have greater surface contact area, so stronger van der Waals/London dispersion forces, requiring more energy to separate molecules into the gas phase).

   Since the Reason is a true, well-established fact but the Assertion states the opposite trend, the Assertion is false while the Reason is true.

   ✓Final answer

   (d) — Assertion is false, Reason is true.

3. CBSE 2025Set X11 markMCQ
   Q.Two compounds 'A' and 'B' were being tested for their boiling points. It was observed that 'A' started boiling after 'B', when both were subjected to same conditions. If the compound 'B' is acetone, which of the following can be compound 'A'?

   (a) Propanal

   (b) Propan-1-ol

   (c) Methoxyethane

   (d) n-Butane

   ›Reveal solutionSolution

   "A boils after B" → A has the higher boiling point; among the options only propan-1-ol (H-bonding) boils higher than acetone, so A = propan-1-ol.

   B is acetone (propanone), b.p. $\approx 56^\circ\text{C}$. "A started boiling after B" means A needs a higher temperature, i.e. A has a higher boiling point than acetone. Comparing approximate boiling points of the options (all $\text{C}_3$/$\text{C}_4$ molecules of similar mass):

   Compound	Approx. b.p.	Reason
   (a) Propanal	$\approx 49^\circ\text{C}$	dipole–dipole only
   (b) Propan-1-ol	$\approx 97^\circ\text{C}$	strong intermolecular H-bonding
   (c) Methoxyethane	$\approx 8^\circ\text{C}$	weak dipole, no H-bond
   (d) n-Butane	$\approx 0^\circ\text{C}$	only London forces

   Only propan-1-ol boils higher than acetone, because its –OH group allows extensive intermolecular hydrogen bonding.

   ✓Final answer

   (b) Propan-1-ol

4. CBSE 2025Set X11 markMCQ
   Q.Select the correct order of melting points of isomeric dichlorobenzenes.

   (a) o-dichlorobenzene > m-dichlorobenzene > p-dichlorobenzene

   (b) p-dichlorobenzene > m-dichlorobenzene > o-dichlorobenzene

   (c) p-dichlorobenzene > o-dichlorobenzene > m-dichlorobenzene

   (d) m-dichlorobenzene > o-dichlorobenzene > p-dichlorobenzene

   ›Reveal solutionSolution

   Melting point depends on how well molecules pack in the crystal; the symmetrical para isomer packs best (highest m.p.), giving the order para > ortho > meta.

   For isomeric dichlorobenzenes, melting point is governed mainly by crystal packing / molecular symmetry rather than by intermolecular force magnitude:

   • p-dichlorobenzene is the most symmetrical, so it packs most efficiently into the crystal lattice and has the highest melting point ($\approx 53^\circ\text{C}$).
   • o-dichlorobenzene ($\approx -17^\circ\text{C}$) packs better than the meta isomer.
   • m-dichlorobenzene ($\approx -25^\circ\text{C}$) is the least symmetrically packed and has the lowest melting point.

   Hence the correct order is: para > ortho > meta.

   ✓Final answer

   (c) p-dichlorobenzene > o-dichlorobenzene > m-dichlorobenzene

5. CBSE 2025Set X11 markMCQ
   Q.Sufficient amount of 2-methylpropan-2-ol heated with 20% phosphoric acid at 358 K gives main product 'X' with the elimination of water and tert-butyl alcohol undergoes dehydration when it is passed over heated copper at 573 K gives 'Y' Pick the correct statement regarding X and Y.

   (a) The boiling points of 'X' and 'Y' are equal

   (b) The boiling point of 'X' is greater than the boiling point of 'Y'

   (c) The boiling point of 'X' is lesser than the boiling point of 'Y'

   (d) At room temperature both 'X' and 'Y' exists as a solids

   ›Reveal solutionSolution

   Acid dehydration and passing over hot copper both convert 2-methylpropan-2-ol to the same alkene (2-methylpropene), so X = Y and their boiling points are equal — option (a).

   2-Methylpropan-2-ol is a tertiary alcohol (tert-butyl alcohol), $\text{(CH}_3)_3\text{C–OH}$.

   • With 20% phosphoric acid at 358 K it undergoes acid-catalysed dehydration (elimination of water) to give the alkene: $\text{(CH}_3)_3\text{C–OH} \rightarrow \text{(CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O}$, so X = 2-methylpropene (isobutylene).
   • When passed over heated copper at 573 K, primary alcohols give aldehydes and secondary alcohols give ketones by dehydrogenation, but a tertiary alcohol has no $\alpha$-H on the carbinol carbon, so instead it undergoes dehydration to the alkene. Hence Y = 2-methylpropene as well.

   Since X and Y are the same compound, they have identical boiling points, and 2-methylpropene is a gas (b.p. $\approx 266$ K), not a solid.

   ✓Final answer

   (a) The boiling points of 'X' and 'Y' are equal (both X and Y are the same alkene, 2-methylpropene)

6. CBSE 2025Set D1 markMCQ
   Q.At room temperature, formaldehyde is

   (a) gas

   (b) liquid

   (c) solid

   (d) none of these

   ›Reveal solutionSolution

   Formaldehyde, the first member of the aldehyde series, is a gas at ordinary temperature (b.p. about −19 °C).

   Formaldehyde (methanal, HCHO) is the lowest aldehyde. It has a very low boiling point (about −19 °C), so at room temperature it exists as a colourless, pungent-smelling gas. Its 40% aqueous solution is called formalin. The higher aldehydes (e.g. acetaldehyde onwards) are liquids.

   ✓Final answer

   (A) gas — formaldehyde is a gas at room temperature.

7. CBSE 2025Set ANNUAL1 mark
   Q.Fill in the blank: The boiling point of methanol is ________ K.

   ›Reveal solutionSolution

   Methanol (CH3OH), the smallest alcohol, boils at about 338 K (64.7 degrees C) at atmospheric pressure.

   Methanol's boiling point of ~338 K is relatively low among common alcohols because of its small molecular size (weaker van der Waals/London forces), even though, like other alcohols, it is capable of intermolecular hydrogen bonding, which raises its boiling point well above hydrocarbons of comparable molar mass.

   ✓Final answer

   338 K (~65 degrees C).

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following compounds has highest melting point?

   (a) 1,2-dichlorobenzene (ortho-dichlorobenzene, structure drawn)

   (b) 1,3-dichlorobenzene (meta-dichlorobenzene, structure drawn)

   (c) 1,4-dichlorobenzene (para-dichlorobenzene, structure drawn)

   (d) All have same melting point.

   ›Reveal solutionSolution

   Melting point depends on how efficiently molecules pack into a crystal lattice, not just molecular weight — the highly symmetric para isomer packs far better than the less symmetric ortho and meta isomers, giving it a much higher melting point.

   All three dichlorobenzenes have the same molecular formula and molecular weight, so their melting-point difference comes purely from crystal packing efficiency. p-Dichlorobenzene is linear and symmetric, so molecules stack very closely and regularly in the solid lattice, maximising van der Waals contact — this raises its melting point sharply (~53°C). The ortho and meta isomers are less symmetric, pack less efficiently, and have much lower melting points (both well below 0°C).

   ✓Final answer

   (c) 1,4-dichlorobenzene (para-dichlorobenzene) — its high molecular symmetry lets it pack most efficiently in the solid lattice.

9. CBSE 2025Set ANNUAL1 mark
   Q.Arrange the following compounds in increasing order of their boiling points: $CH_3CHO$, $CH_3CH_2OH$, $CH_3OCH_3$, $CH_3CH_2CH_3$

   ›Reveal solutionSolution

   Boiling point here tracks the strength of intermolecular forces: propane (only weak van der Waals forces) boils lowest, dimethyl ether (weak dipole-dipole, no H-bonding) next, acetaldehyde (stronger dipole-dipole from the polar C=O) next, and ethanol (hydrogen-bonded) boils highest.

   All four compounds have comparable molar mass (propane 44, dimethyl ether 46, acetaldehyde 44, ethanol 46 g mol$^{-1}$), so the boiling-point order is decided almost entirely by the type of intermolecular attraction available, not by size:

   • $CH_3CH_2CH_3$ (propane): a non-polar hydrocarbon; molecules are held together only by weak instantaneous dipole–induced dipole (London/van der Waals) forces. Lowest boiling point (real value ≈ −42 °C).

   • $CH_3OCH_3$ (dimethyl ether): the C–O–C linkage gives the molecule a small permanent dipole, so molecules attract each other by dipole–dipole forces, stronger than propane's dispersion forces alone but the ether oxygen has no O–H bond, so no hydrogen bonding is possible. Boils higher than propane (real value ≈ −24 °C).

   • $CH_3CHO$ (acetaldehyde): the C=O group is strongly polar (larger bond dipole than the C–O–C of an ether), giving stronger dipole–dipole attractions than the ether, though again no O–H/N–H bond means no hydrogen bonding. Boils higher than the ether (real value ≈ 20 °C).

   • $CH_3CH_2OH$ (ethanol): possesses an O–H bond, so molecules are linked by intermolecular hydrogen bonding, by far the strongest of the forces here, requiring the most energy to separate the molecules into vapour. Boils highest (real value ≈ 78 °C).

   ✓Final answer

   Increasing boiling point: $CH_3CH_2CH_3 < CH_3OCH_3 < CH_3CHO < CH_3CH_2OH$

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The correct order of boiling points of alcohols having the same number of Carbon atoms is ...................... .

    (a) 2° > 1° > 3°

    (b) 1° > 2° > 3°

    (c) 3° > 1° > 2°

    (d) 3° > 2° > 1°

    ›Reveal solutionSolution

    Among isomeric alcohols, boiling point falls as branching increases, because branching reduces the effective surface area available for intermolecular hydrogen bonding and van der Waals interactions.

    All isomeric alcohols with the same molecular formula can hydrogen-bond through their –OH group, but a straight-chain (primary) alcohol packs more efficiently and has a larger surface area for van der Waals contact between molecules than a branched (tertiary) one. Branching (as in tertiary alcohols) shields the –OH group and reduces intermolecular association, lowering the boiling point.

    ✓Final answer

    The correct order is 1° > 2° > 3° (option b).

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.Which of the following has the highest melting point?

    (a) o-xylene

    (b) m-xylene

    (c) p-xylene

    (d) Toluene

    ›Reveal solutionSolution

    p-Xylene has the highest melting point because its high molecular symmetry allows the most efficient crystal packing.

    Among the three xylene isomers (o-, m-, p-) and toluene, melting point depends heavily on how symmetrically the molecules can pack into a solid lattice (not just on molecular weight or boiling point). p-Xylene, with its methyl groups symmetrically placed at opposite (1,4) positions on the ring, packs most efficiently into a crystal lattice, giving it a distinctly higher melting point than o-xylene, m-xylene (the least symmetric, hence lowest melting point of the three), or toluene (which has no second substituent to create this comparison).

    ✓Final answer

    (C) p-xylene.

12. CBSE 2024Set ANNUAL1 mark
    Q.Why has propanol higher boiling point than propane?

    ›Reveal solutionSolution

    Boiling point depends on the strength of intermolecular forces that must be overcome; propanol's -OH group enables hydrogen bonding between molecules, a much stronger force than the weak van der Waals (London dispersion) forces that are all propane has.

    Propane (CH3-CH2-CH3) is a non-polar hydrocarbon with no functional group capable of hydrogen bonding. Its molecules are held together only by weak van der Waals (induced-dipole) forces, so relatively little energy is needed to separate them - it boils at a very low temperature (-42 degree C) and is a gas at room temperature.

    Propanol (CH3-CH2-CH2-OH) has a polar -OH group. The highly electronegative oxygen, bonded to a hydrogen, allows propanol molecules to form hydrogen bonds with each other (O-H...O). Hydrogen bonding is a much stronger intermolecular force than van der Waals forces, so significantly more thermal energy is required to break these bonds and vaporise the liquid - hence propanol boils at a much higher temperature (~97 degree C) and is a liquid at room temperature.

    ✓Final answer

    Because propanol molecules are held together by (much stronger) intermolecular hydrogen bonding via the -OH group, while propane has only weak van der Waals forces - so propanol needs far more energy to boil.