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 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Match the following List – I (Compound) / List – II (b.p / K) A. $n-C_4H_9OH$ / I. 310.5 B. $(C_2H_5)_2NH$ / II. 350.8 C. $n-C_4H_9NH_2$ / III. 390.3 D. $C_2H_5N(CH_3)_2$ / IV. 329.3 The correct answer is (A) A-IV, B-II, C-I, D-III (B) A-III, B-IV, C-I, D-II (C) A-III, B-IV, C-II, D-I (D) A-II, B-III, C-IV, D-I

   ›Reveal solutionSolution

   This tests the classic boiling-point trend among an alcohol, a 1° amine, a 2° amine, and a 3° amine of comparable molecular weight, driven by hydrogen-bonding capacity. The match is A-III, B-IV, C-II, D-I.

   Concept and Intuition

   For molecules of similar size, boiling point tracks how strongly molecules can hydrogen-bond to each other:

   • Alcohols (O–H) hydrogen-bond most strongly (O is more electronegative than N, and the O–H bond is highly polarized), so they have the highest boiling points among comparably-sized compounds.
   • Primary amines have two N–H bonds per molecule available for intermolecular hydrogen bonding — next highest.
   • Secondary amines have only one N–H bond — weaker hydrogen bonding, lower boiling point than primary amines.
   • Tertiary amines have no N–H bond at all (nitrogen's lone pair can still accept a hydrogen bond from something else, but the molecule itself cannot donate one), so they rely mainly on weaker dipole–dipole and dispersion forces — lowest boiling point of the four.

   Step-by-Step Solution

   1. A. $n$-$C_4H_9OH$ (n-butanol): a primary alcohol — strongest H-bonding → highest boiling point among the four, 390.3 K → list item III. A-III.
   2. C. $n$-$C_4H_9NH_2$ (n-butylamine): a primary amine, two N–H bonds → next highest, 350.8 K → list item II. C-II.
   3. B. $(C_2H_5)_2NH$ (diethylamine): a secondary amine, one N–H bond → lower still, 329.3 K → list item IV. B-IV.
   4. D. $C_2H_5N(CH_3)_2$ (N,N-dimethylethylamine): a tertiary amine, no N–H bond → lowest boiling point, 310.5 K → list item I. D-I.
   5. Ranking confirms: $390.3 > 350.8 > 329.3 > 310.5$, i.e. alcohol > 1° amine > 2° amine > 3° amine, exactly as the hydrogen-bonding argument predicts.

   Common Mistakes

   • Ranking amines purely by molecular weight/size rather than by how many N–H bonds are available for hydrogen bonding.
   • Assuming a secondary amine boils higher than a primary amine of similar formula weight — it's actually lower, because a 2° amine has only one N–H donor versus two for a 1° amine.

   ✓Final answer

   The correct option is (C) — A-III, B-IV, C-II, D-I.

   ANSWER: C

2. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Among the hydrides of group 15 elements, the hydride with highest boiling point is $\underline{A}$ and the hydride with lowest boiling point is $\underline{B}$. What are A and B respectively? (A) $BiH_3,\ NH_3$ (B) $BiH_3,\ PH_3$ (C) $NH_3,\ PH_3$ (D) $NH_3,\ SbH_3$

   ›Reveal solutionSolution

   Boiling points of group 15 hydrides dip after ammonia (loss of H-bonding) then rise with increasing molar mass; highest is BiH3, lowest is PH3.

   Concept and Intuition

   $NH_3$ has strong intermolecular hydrogen bonding (N is small and highly electronegative), giving it an unusually high boiling point for its size. Once H-bonding is lost going to $PH_3$, boiling point drops sharply because only weak van der Waals (London dispersion) forces operate. As you continue down the group ($AsH_3\to SbH_3\to BiH_3$), molecular size and mass increase steadily, so van der Waals forces strengthen again and boiling point rises — eventually exceeding even $NH_3$.

   Step-by-Step Solution

   1. Approximate boiling points: $NH_3\approx -33°C$, $PH_3\approx -87.7°C$, $AsH_3\approx -55°C$, $SbH_3\approx -17°C$, $BiH_3\approx +17°C$.
   2. Lowest of these is $PH_3$ (the H-bonding of $NH_3$ is gone, and molecular mass is still small).
   3. Highest of these is $BiH_3$ (largest, heaviest molecule with strongest dispersion forces).
   4. So A (highest) = $BiH_3$, B (lowest) = $PH_3$.

   Common Mistakes

   • Assuming $NH_3$ must be the highest-bp hydride of the whole series just because of H-bonding — $BiH_3$ actually surpasses it due to its much greater size.
   • Picking $NH_3$ as the lowest by mistakenly thinking H-bonding lowers boiling point.

   ✓Final answer

   The correct option is (B) — $BiH_3, PH_3$.

   ANSWER: B

3. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The correct order of boiling points of the compounds given below is A) Methoxy ethane B) Propan-1-ol C) Propanal D) Propanone (A) C > B > A > D (B) B > D > C > A (C) B > C > D > A (D) C > A > B > D

   ›Reveal solutionSolution

   Tests ranking boiling points by intermolecular forces: H-bonding alcohol > dipolar ketone > dipolar aldehyde > weakly-polar ether.

   Concept and Intuition

   For molecules of similar molar mass, boiling point is set by the strength of intermolecular forces. An –OH group enables strong hydrogen bonding (raising b.p. sharply above similarly-sized non-alcohols). A C=O group gives a fairly strong permanent dipole (ketones/aldehydes), but weaker than H-bonding. An ether has a weaker net dipole (bond dipoles partly oppose) and no H-bond donor, so it boils at the lowest temperature of the four functional classes here.

   Step-by-Step Solution

   1. B) Propan-1-ol, $\mathrm{CH_3CH_2CH_2OH}$: extensive intermolecular H-bonding via $-\mathrm{OH}$ gives it the highest boiling point of the four.
   2. D) Propanone (acetone), $\mathrm{CH_3COCH_3}$: a symmetric ketone with a strong dipole from C=O but no H-bond donor — boils next highest.
   3. C) Propanal, $\mathrm{CH_3CH_2CHO}$: also has a polar C=O, but the aldehyde's dipole/packing gives it a slightly lower boiling point than the ketone of the same carbon count.
   4. A) Methoxyethane, $\mathrm{CH_3-O-C_2H_5}$: an ether — only weak dipole-dipole/van der Waals forces, no H-bonding — has by far the lowest boiling point.
   5. Order (highest to lowest): B > D > C > A.

   Common Mistakes

   • Assuming aldehydes always boil higher than ketones of the same size — in this size range the ketone (acetone) actually boils a little higher than the aldehyde (propanal).
   • Forgetting that ethers, despite having an oxygen atom, cannot hydrogen-bond with each other and so boil much lower than alcohols of similar mass.

   ✓Final answer

   The correct option is (B) — B > D > C > A.

   ANSWER: B

4. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.Observe the following substances. Ethanol, acetic acid, ethylamine, trimethylamine, salicylic acid, ethanal. In the above list, the number of substances with H-bonding is (A) 4 (B) 3 (C) 5 (D) 2

   ›Reveal solutionSolution

   Tests recognizing which functional groups (O–H, N–H) enable hydrogen bonding; 4 of the 6 substances qualify.

   Concept and Intuition

   Hydrogen bonding needs a hydrogen atom covalently bonded to a small, highly electronegative atom — O, N, or F — so that the H carries a strong partial positive charge able to interact with a lone pair on a neighbouring electronegative atom. A carbonyl oxygen (as in an aldehyde) or a nitrogen with no attached H (as in a fully substituted tertiary amine) cannot act as an H-bond donor themselves.

   Step-by-Step Solution

   1. Ethanol ($\mathrm{C_2H_5OH}$): has an O–H group → capable of H-bonding.
   2. Acetic acid ($\mathrm{CH_3COOH}$): has a carboxylic O–H group → capable of H-bonding.
   3. Ethylamine ($\mathrm{C_2H_5NH_2}$): a primary amine with N–H bonds → capable of H-bonding.
   4. Trimethylamine ($\mathrm{N(CH_3)_3}$): a tertiary amine — nitrogen has no attached H, so it cannot donate a hydrogen bond → excluded.
   5. Salicylic acid: has both a carboxylic O–H and a phenolic O–H → capable of H-bonding.
   6. Ethanal ($\mathrm{CH_3CHO}$): an aldehyde with no O–H or N–H bond → excluded.
   7. Total qualifying substances: ethanol, acetic acid, ethylamine, salicylic acid = 4.

   Common Mistakes

   • Assuming any amine can hydrogen-bond, without checking whether it actually has an N–H bond (tertiary amines don't).
   • Assuming a carbonyl compound like an aldehyde can hydrogen-bond just because it contains oxygen.

   ✓Final answer

   The correct option is (A) — 4.

   ANSWER: A

5. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.What is the correct boiling point order of the following haloalkanes? i) 2-chloro 2-methylpropane ii) 1-Cholobutane iii) 2-Chlorobutane (A) i > ii > iii (B) ii > iii > i (C) i < ii < iii (D) i > iii > ii

   ›Reveal solutionSolution

   Among isomeric C₄H₉Cl haloalkanes, the straight-chain isomer boils highest and the most branched (tertiary) isomer boils lowest, giving the order ii > iii > i.

   Concept and Intuition

   For a set of structural isomers with the same molecular formula, boiling point is governed mainly by the strength of intermolecular van der Waals (London dispersion) forces, which depend on the surface area available for molecules to contact each other. A straight (unbranched) chain packs closely and has more surface contact, giving stronger dispersion forces and a higher boiling point. Branching makes the molecule more compact/spherical, reducing surface area and intermolecular contact, and hence lowering the boiling point.

   Step-by-Step Solution

   1. Identify the three isomers, all of formula $C_4H_9Cl$: (i) 2-chloro-2-methylpropane (tert-butyl chloride) — most branched, chlorine on a tertiary carbon; (ii) 1-chlorobutane — straight (unbranched) chain, chlorine on a primary carbon; (iii) 2-chlorobutane — chlorine on a secondary carbon, slightly branched.
   2. Rank by branching (least to most): (ii) unbranched < (iii) one branch point < (i) most branched (quaternary-like carbon skeleton around the C–Cl carbon).
   3. Since boiling point decreases with increasing branching for isomers, the order (highest bp to lowest) is: (ii) > (iii) > (i).
   4. This matches known experimental values: 1-chlorobutane (~78.5°C) > 2-chlorobutane (~68°C) > tert-butyl chloride (~51°C).

   Common Mistakes

   • Assuming a more "substituted" (tertiary) halide would have a higher boiling point by analogy with stability of carbocations — boiling point trends for branching go the opposite way from carbocation stability trends.

   ✓Final answer

   The correct option is (B) — ii > iii > i.

   ANSWER: B

6. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.Para-nitro phenol has higher boiling point than ortho-nitrophenol. This is due to (A) The presence of intermolecular hydrogen bonding between Para-nitro phenol molecules (B) The presence of intramolecular hydrogen bonding in Para-nitro phenol molecules (C) The presence of intermolecular hydrogen bonding between ortho-nitro phenol molecules (D) The absence of intramolecular hydrogen bonding between ortho-nitro phenol molecules

   ›Reveal solutionSolution

   Para-nitrophenol boils higher than ortho because ortho forms intramolecular H-bonding (chelation) while para is forced into intermolecular H-bonding, which needs more energy to break.

   Concept and Intuition

   Boiling point depends on the strength of the forces holding molecules together in the liquid. Ortho-nitrophenol's $\mathrm{-OH}$ and $\mathrm{-NO_2}$ are adjacent, so they hydrogen-bond to each other within the same molecule (a six-membered ring "chelate"). This uses up the $\mathrm{-OH}$'s hydrogen-bonding capacity internally, so ortho-nitrophenol molecules interact with each other only weakly (via van der Waals forces) — it boils low and is even steam-volatile. In para-nitrophenol the groups are on opposite ends of the ring and cannot reach each other, so the $\mathrm{-OH}$ of one molecule instead hydrogen-bonds to the $\mathrm{-NO_2}$/$\mathrm{-OH}$ of a neighbouring molecule — building an extended, harder-to-break intermolecular network, hence a higher boiling point.

   Step-by-Step Solution

   1. Identify the substitution pattern: ortho places $\mathrm{-OH}$ and $\mathrm{-NO_2}$ next to each other; para places them across the ring.
   2. Ortho: intramolecular H-bond forms a stable ring — no need for the molecule to H-bond with neighbours.
   3. Para: no intramolecular H-bond is geometrically possible, so $\mathrm{-OH}$ groups H-bond between different molecules.
   4. Intermolecular H-bonds must all be broken simultaneously to vaporise the liquid → higher boiling point for para.
   5. Hence para-nitrophenol has the higher boiling point due to intermolecular H-bonding among its own molecules.

   Common Mistakes

   • Assuming intramolecular H-bonding always raises boiling point — it's the opposite; it lowers it by removing the drive to associate with other molecules.
   • Mixing up which isomer has which type of bonding.

   ✓Final answer

   The correct option is (A) — The presence of intermolecular hydrogen bonding between para-nitrophenol molecules.

   ANSWER: A

7. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.Arrange the following in increasing order of their boiling points N-Ethylethanamine - I Butanamine - II N,N-dimethylethanamine - III (A) III > II > I (B) III > I > II (C) II > III > I (D) II > I > III

   ›Reveal solutionSolution

   Boiling point of amines of the same formula falls as 1° > 2° > 3°, because more N–H bonds mean stronger intermolecular hydrogen bonding.

   Concept and Intuition

   All three compounds share the molecular formula $C_4H_{11}N$, so molecular weight/dispersion forces are essentially comparable; the boiling-point differences are governed by hydrogen bonding capacity. A primary amine ($-NH_2$) has two N–H bonds and can form the most extensive intermolecular hydrogen-bond network, giving it the highest boiling point among the three classes for a given carbon count. A secondary amine ($-NH-$) has only one N–H bond, so it hydrogen-bonds less extensively (lower bp than the primary isomer). A tertiary amine has no N–H bond at all, so it cannot hydrogen-bond with itself, relying only on weaker dipole–dipole and dispersion forces, giving it the lowest boiling point.

   Step-by-Step Solution

   1. Classify each compound: Butanamine (II) = $\mathrm{CH_3CH_2CH_2CH_2NH_2}$, a primary amine (2 N–H bonds).
   2. N-Ethylethanamine (I) = diethylamine, $\mathrm{(C_2H_5)_2NH}$, a secondary amine (1 N–H bond).
   3. N,N-Dimethylethanamine (III) = $\mathrm{CH_3CH_2N(CH_3)_2}$, a tertiary amine (0 N–H bonds).
   4. Ranking by H-bonding strength (and hence boiling point): primary > secondary > tertiary, i.e. II > I > III.
   5. Checking each option against this true relative ranking (II's bp highest, then I, then III lowest), only option (D) has both pairwise relations (II > I and I > III) correctly stated.

   Common Mistakes

   • Assuming more substitution (like more alkyl branching) always raises boiling point — for amines, hydrogen-bonding capacity (number of N–H bonds) dominates over simple branching/molecular-weight effects.
   • Mixing up which Roman numeral corresponds to which amine class.

   ✓Final answer

   The correct option is (D) — II > I > III.

   ANSWER: D

8. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.The correct order of boiling points of following molecules is

   (i) n – Hexane

   (ii) 2-methylpentane

   (iii) 2,3 – dimethylbutane (A) i > ii > iii (B) iii > ii > i (C) iii > i > ii (D) i > iii > ii

   ›Reveal solutionSolution

   Boiling point falls as branching increases among isomeric alkanes, so n-hexane > 2-methylpentane > 2,3-dimethylbutane.

   Concept and Intuition

   All three compounds are isomers of hexane ($C_6H_{14}$), so they have identical molecular formula and hence similar total van der Waals attraction potential — but the shape of the molecule matters. A straight, extended chain (n-hexane) has more surface-to-surface contact with neighbouring molecules, maximizing van der Waals (London dispersion) forces. Branching makes the molecule more compact and spherical, reducing effective surface contact and hence the strength of intermolecular attractions, which lowers the boiling point.

   Step-by-Step Solution

   1. n-Hexane: a straight, unbranched 6-carbon chain — largest surface area for intermolecular contact — highest boiling point among the three.
   2. 2-Methylpentane: one methyl branch — somewhat more compact than n-hexane — intermediate boiling point.
   3. 2,3-Dimethylbutane: two methyl branches, the most compact/spherical of the three — smallest surface area for contact — lowest boiling point.
   4. Order (decreasing boiling point): n-hexane (i) > 2-methylpentane (ii) > 2,3-dimethylbutane (iii).

   Common Mistakes

   • Assuming boiling point depends only on molecular weight (all three isomers have the same molecular weight, so shape/branching is the deciding factor here).
   • Reversing the trend and thinking more branching increases boiling point (branching actually decreases it, unlike its effect on some other properties like octane number).

   ✓Final answer

   The correct option is (A) — i > ii > iii.

   ANSWER: A

9. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.At 298 K, if the vapour pressure of pure liquids toluene, benzene, chloroform and dichloromethane are 60, 160, 200 and 415 torr respectively. Then which liquid is having high boiling point? (A) Toluene (B) Benzene (C) Chloroform (D) Dichloromethane

   ›Reveal solutionSolution

   Boiling point and vapour pressure (at fixed T) are inversely related; toluene's lowest vapour pressure (60 torr) means it has the highest boiling point.

   Concept and Intuition

   Vapour pressure measures how readily a liquid's molecules escape into the gas phase at a given temperature — it is a direct measure of volatility. Boiling point is the temperature at which vapour pressure equals atmospheric pressure. A liquid that already has a low vapour pressure at a reference temperature needs to be heated more to reach atmospheric pressure, so lower vapour pressure at a fixed T corresponds to a higher boiling point.

   Step-by-Step Solution

   1. List the vapour pressures at 298 K: toluene 60 torr, benzene 160 torr, chloroform 200 torr, dichloromethane 415 torr.
   2. Rank from lowest to highest vapour pressure: toluene < benzene < chloroform < dichloromethane.
   3. Since boiling point ranks inversely to vapour pressure (at the same reference temperature), the ranking of boiling points (highest to lowest) is: toluene > benzene > chloroform > dichloromethane.
   4. The liquid with the highest boiling point is therefore toluene.

   Common Mistakes

   • Assuming higher vapour pressure means higher boiling point (it's the opposite — high vapour pressure means the substance evaporates easily, i.e., low boiling point).
   • Not recognizing that this comparison is valid because all values are given at the same temperature (298 K).

   ✓Final answer

   The correct option is (A) — Toluene.

   ANSWER: A

10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.Arrange the hydrides $\mathrm{NH_3}$, HF, $\mathrm{H_2O}$, HCl in the increasing order of their boiling points (A) $\mathrm{HF < NH_3 < HCl < H_2O}$ (B) $\mathrm{H_2O < HF < HCl < NH_3}$ (C) $\mathrm{NH_3 < HCl < H_2O < HF}$ (D) $\mathrm{HCl < NH_3 < HF < H_2O}$

    ›Reveal solutionSolution

    Boiling points of these hydrides are governed mainly by hydrogen bonding strength/extent, giving the increasing order $HCl<NH_3<HF<H_2O$.

    Concept and Intuition

    Among simple hydrides, boiling point is strongly influenced by hydrogen bonding, which occurs when H is bonded to a small, highly electronegative atom (N, O, F). $HCl$'s Cl is not electronegative/small enough to hydrogen bond significantly, so it relies only on weaker dipole-dipole/dispersion forces and has the lowest boiling point among these four. Among the hydrogen-bonded species, $H_2O$ forms an extensive 3-D hydrogen-bonded network (2 lone pairs and 2 H atoms per molecule, ideal for a 3-D network) giving it the highest boiling point, while $HF$ and $NH_3$ form more limited (chain-like or less networked) hydrogen bonding.

    Step-by-Step Solution

    1. $HCl$: negligible hydrogen bonding (Cl is not electronegative/small enough) — lowest boiling point among the four ($\approx-85^\circ C$).
    2. $NH_3$: hydrogen bonds via N, but only one lone pair per molecule to hydrogen bond with $\Rightarrow$ boiling point $\approx-33^\circ C$.
    3. $HF$: strong hydrogen bonding via a highly electronegative F, but limited to one H and three lone pairs (only one bond forms per molecule in the chain) $\Rightarrow$ boiling point $\approx19.5^\circ C$.
    4. $H_2O$: two H atoms and two lone pairs allow each molecule to participate in up to 4 hydrogen bonds, forming an extensive 3-D network $\Rightarrow$ highest boiling point, $100^\circ C$.
    5. Increasing order of boiling point: $HCl<NH_3<HF<H_2O$.

    Common Mistakes

    • Assuming $HF$, having the strongest single hydrogen bond, must have the highest boiling point overall — but the extent of hydrogen-bond networking (as in water) matters more than single-bond strength.
    • Forgetting $HCl$ essentially lacks hydrogen bonding altogether and placing it above $NH_3$.

    ✓Final answer

    The correct option is (D) — $HCl < NH_3 < HF < H_2O$.

    ANSWER: D

11. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.Which among the following will have the highest boiling point ? (A) Butan-2-ol ($\mathrm{CH_3CH(OH)CH_2CH_3}$) (B) Butan-2-one ($\mathrm{CH_3COCH_2CH_3}$) (C) n-Butane ($\mathrm{CH_3CH_2CH_2CH_3}$) (D) Ethyl propyl ether ($\mathrm{CH_3CH_2-O-CH_2CH_2CH_3}$)

    ›Reveal solutionSolution

    Among an alcohol, a ketone, an alkane, and an ether of comparable size, the alcohol has the highest boiling point because only it can hydrogen-bond between its own molecules.

    Concept and Intuition

    Boiling point depends on the strength of intermolecular forces that must be overcome to vaporise the liquid. Alcohols (-OH group) can form hydrogen bonds with each other, a strong, directional intermolecular force. Ketones and ethers only have permanent dipole-dipole interactions (no O-H or N-H to hydrogen-bond with each other), which are weaker than hydrogen bonding. Alkanes have only weak, non-polar van der Waals (London dispersion) forces, the weakest of all.

    Step-by-Step Solution

    1. Butan-2-ol: contains -OH, capable of strong intermolecular hydrogen bonding $\Rightarrow$ highest boiling point among these four.
    2. Butan-2-one: a ketone, polar C=O but no H-bond donor $\Rightarrow$ moderate boiling point (dipole-dipole), lower than the alcohol.
    3. Ethyl propyl ether: polar C-O-C but no H-bond donor either $\Rightarrow$ boiling point similar to or slightly below the ketone.
    4. n-Butane: non-polar, only weak dispersion forces $\Rightarrow$ lowest boiling point (in fact a gas near room temperature).
    5. Hence Butan-2-ol has the highest boiling point.

    Common Mistakes

    • Assuming molecular weight alone determines boiling point, ignoring the type of intermolecular force present.
    • Forgetting that ethers, despite having an oxygen, cannot hydrogen-bond with themselves (no O-H bond) the way alcohols can.

    ✓Final answer

    The correct option is (A) — Butan-2-ol.

    ANSWER: A

12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.Arrange the following in decreasing order of their boiling points

    (a) $CH_3CH_2CH_2CH_2OH$ (butan-1-ol)

    (b) $CH_3CH_2CH_2CH_2NH_2$ (butan-1-amine, a primary amine)

    (c) a tertiary amine, $(CH_2CH_3)$ chain with an N bearing two other alkyl branches (drawn as a small N with two branches, i.e. a trialkylamine)

    (d) a secondary amine with an N-H, drawn as two ethyl-type chains joined through an N-H (a secondary amine) (A) a > b > d > c (B) a > c > d > b (C) b > c > d > a (D) c > a > b > d

    ›Reveal solutionSolution

    Boiling point here tracks hydrogen-bonding ability: the alcohol (strongest H-bonding) is highest, then primary amine (two N–H), then secondary amine (one N–H), then tertiary amine (no N–H, weakest): a > b > d > c.

    Concept and Intuition

    For molecules of comparable molecular weight, boiling point is governed largely by the strength and extent of intermolecular hydrogen bonding. Oxygen is more electronegative than nitrogen, so O–H···O hydrogen bonds are stronger than N–H···N hydrogen bonds — alcohols therefore boil higher than amines of similar size. Among amines themselves, hydrogen bonding requires an N–H bond to donate; a primary amine has two N–H bonds (most extensive hydrogen-bonded network), a secondary amine has only one N–H bond (less association), and a tertiary amine has none (cannot hydrogen-bond to itself at all, only weaker dipole-dipole/van der Waals forces), giving it the lowest boiling point of the three.

    Step-by-Step Solution

    1. Butan-1-ol (a): $-\mathrm{OH}$ group, strongest hydrogen bonding of the four compounds → highest boiling point.
    2. Butan-1-amine (b), a primary amine: two N–H bonds, extensive intermolecular hydrogen bonding → next highest.
    3. The secondary amine (d): only one N–H bond, weaker/less extensive hydrogen bonding than a primary amine → next.
    4. The tertiary amine (c): no N–H bond at all, cannot hydrogen bond with itself, relies only on weaker dipole-dipole and dispersion forces → lowest boiling point.
    5. Overall order: a > b > d > c.

    Common Mistakes

    • Ranking by molecular weight alone instead of hydrogen-bonding capacity.
    • Assuming all amines hydrogen-bond equally regardless of how many N–H bonds are present.
    • Placing the tertiary amine above the secondary amine by mistake.

    ✓Final answer

    The correct option is (A) — a > b > d > c.

    ANSWER: A