Which of the following compounds will have the highest melting point and why? ![Three isomeric dichloro-dimethylbenzenes (I), (II), (III)](/figures/haloalkanes-and-haloarenes/diagram-6-q4682-dichloro-dimethylbenzenes.svg) All three are isomeric dichloro-dimethylbenzenes (both $\mathrm{CH_3}$ groups para to each other, at ring positions 1 and 4): (I) 2,3-dichloro-1,4-dimethylbenzene (the two Cl atoms on adjacent carbons) (II) 2,5-dichloro-1,4-dimethylbenzene (the two Cl atoms para to each other — fully symmetrical) (III) 2,6-dichloro-1,4-dimethylbenzene (the two Cl atoms flanking the same $\mathrm{CH_3}$)

Which of the following compounds will have the highest melting point and why? All three are isomeric dichloro-

NCERT Exemplar · Q58

Q. Which of the following compounds will have the highest melting point and why?

Three isomeric dichloro-dimethylbenzenes (I), (II), (III) — Figure

\mathrm{CH_3}

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Concept understanding — Boiling Point Trends

Boiling Point Trends (Organic Compounds)

A substance's boiling point is set by how much energy is needed to overcome the attractive forces HOLDING its molecules together in the liquid — the stronger those intermolecular forces, the higher the boiling point.

The Forces, Weakest to Strongest

Van der Waals (London dispersion) forces — present in every molecule, and they grow stronger as the molecule gets bigger (more electrons, larger surface area of contact between neighbouring molecules) and more polarisable.
Dipole–dipole forces — present in polar molecules, add an extra attraction on top of dispersion forces.
Hydrogen bonding — present when H is bonded directly to N, O, or F; much stronger than ordinary dipole–dipole attraction, and it raises the boiling point sharply compared to a similarly-sized molecule without it.

Trend 1: Down a Series of Halogens (Same Alkyl Group)

For a fixed R group, boiling point rises as the halogen gets heavier: $R{-}I > R{-}Br > R{-}Cl > R{-}F$ . This looks surprising at first, since electronegativity (and so bond polarity/dipole moment) actually DECREASES down the group — but boiling point here is dominated by the growing size and polarisability of the halogen atom (stronger dispersion forces), which outweighs the shrinking dipole contribution.

The measured values for the methyl, ethyl and propyl halides show this rise clearly:

Measured boiling points of methyl, ethyl and propyl chlorides, bromides and iodides (NCERT Fig 6.1), rising steadily from R-Cl to R-Br to R-I for every alkyl group

Trend 2: Chain Length and Branching

Longer chains (more carbons) have more surface area for van der Waals contact between neighbouring molecules, so boiling point rises with chain length within a homologous series.
Branching LOWERS boiling point compared to a straight-chain isomer of the same molecular formula — a more compact, spherical shape has less surface-to-surface contact with neighbouring molecules, weakening the dispersion forces. (E.g. neopentane boils well below n-pentane.)

Trend 3: Hydrogen Bonding Beats Molecular Mass …

Why this formula?

Boiling Point Trends: Why They Happen

Boiling point is the temperature at which a liquid's vapor pressure equals the external atmospheric pressure. To understand why boiling points follow certain trends, we must first understand what determines vapor pressure.

The Core Idea: Intermolecular Forces

A liquid boils when its molecules have enough kinetic energy to overcome the intermolecular forces (IMFs) holding them together in the liquid phase. Stronger IMFs → harder to escape → lower vapor pressure at a given temperature → higher boiling point.

There is no single "formula" for boiling point, but the relationship is captured by the Clausius–Clapeyron equation, which links vapor pressure ( $P$ ) to temperature ( $T$ ) and the enthalpy of vaporization ( $\Delta H_{\text{vap}}$ ):

\ln P = -\frac{\Delta H_{\text{vap}}}{R} \cdot \frac{1}{T} + C

Where:

$P$ = vapor pressure
$\Delta H_{\text{vap}}$ = enthalpy of vaporization (energy needed to vaporize 1 mole)
$R$ = gas constant
$T$ = absolute temperature (Kelvin)
$C$ = constant (depends on substance)

Why this formula makes sense

$\Delta H_{\text{vap}}$ is large when IMFs are strong — more energy is needed to separate molecules.
At boiling point, $P = P_{\text{atm}}$ (usually 1 atm). So a substance with larger $\Delta H_{\text{vap}}$ needs a higher $T$ to reach that pressure.

Thus, boiling point ∝ strength of intermolecular forces.

The Four Key Trends (with Reasoning)

1. Trend across a period (e.g., Period 2: CH₄ → NH₃ → H₂O → HF)

Molecule	IMFs present	Boiling point (°C)
CH₄	London dispersion only	-161
NH₃	Dispersion + H-bonding	-33
H₂O	Dispersion + H-bonding (2 per molecule)	100
HF	Dispersion + H-bonding	19

Why?

CH₄ is nonpolar — only weak London dispersion forces.
NH₃, H₂O, HF have hydrogen bonding (strongest IMF).
H₂O forms two H-bonds per molecule (donor + acceptor), while NH₃ forms one and HF forms one — hence H₂O has the highest boiling point.

Key insight: Hydrogen bonding dominates over molecular mass in small molecules.

2. Trend down a group (e.g., Halogens: F₂ → Cl₂ → Br₂ → I₂)

Molecule	Molar mass (g/mol)	Boiling point (°C)
F₂	38	-188
Cl₂	71	-34
Br₂	160	59
I₂	254	184

Why?

All are nonpolar — only London dispersion forces.
Dispersion force strength increases with number of electrons (larger molar mass → more polarizable electron cloud → stronger temporary dipoles).
So boiling point increases down the group.

Key insight: For nonpolar molecules, molar mass (electron count) is the primary factor.

3. Branching in alkanes (e.g., C₅H₁₂ isomers)

Isomer	Boiling point (°C)
n-pentane (straight chain)	36
2-methylbutane (branched)	28
2,2-dimethylpropane (highly branched)	10

Why?

All have same molecular formula — same molar mass. …

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The melting point of an organic solid depends on how efficiently molecules pack in the crystal lattice. Symmetrical molecules pack more tightly, raising the melting point. Here, compound (II) is the most symmetrical and will have the highest melting point.

The 3 isomeric dichloro-dimethylbenzenes and their ring symmetry

Why symmetry controls melting point

Melting is the process where a crystal lattice breaks down into a disordered liquid. The stronger the intermolecular forces and the better the molecules fit together in the solid, the more energy (heat) is needed to melt them. For non-polar or weakly polar molecules like these dichloro-dimethylbenzenes, the dominant forces are London dispersion forces. These forces depend on surface contact — the more closely and extensively molecules can nestle together, the stronger the overall attraction.

Symmetry is the key. A highly symmetrical molecule packs into a crystal with fewer voids and more uniform contacts. Think of stacking identical bricks versus stacking oddly shaped stones: the bricks form a denser, more stable pile. The same principle applies here.

All three compounds are isomers with the same molecular formula, so their electron counts and polarizabilities are similar. The difference in melting point comes almost entirely from packing efficiency.

Step-by-step analysis

1. Identify the molecular structures

All three have two methyl groups fixed at positions 1 and 4 (para to each other). The only variation is where the two chlorine atoms sit on the remaining four ring carbons (positions 2, 3, 5, 6).

(I) 2,3-dichloro-1,4-dimethylbenzene: Cl atoms on adjacent carbons (2 and 3). This puts both Cl atoms on the same side of the ring — a crowded arrangement (though the molecule still keeps one mirror plane, between the two Cl atoms).
(II) 2,5-dichloro-1,4-dimethylbenzene: Cl atoms at positions 2 and 5, which are para to each other. The molecule has a centre of symmetry: a 180° rotation about the centre gives the same arrangement. This is the most symmetrical isomer.
(III) 2,6-dichloro-1,4-dimethylbenzene: Cl atoms at positions 2 and 6, which are meta to each other and flank one of the methyl groups. The molecule has a plane of symmetry (through the 1-4 axis) but no centre of symmetry.

2. Rank the symmetry

Tip

A molecule with a centre of symmetry (inversion centre) often packs into a more ordered, higher-melting crystal than one with only a plane of symmetry. The centre of symmetry forces the molecule to be "balanced" — no dipole moment, and a shape that fits neatly into a lattice.

(II) has a centre of symmetry (inversion centre) — clearly the most symmetrical of the three.
(III) has one mirror plane (through the C1–C4 axis, mapping the two flanking Cl atoms onto each other) but no centre of symmetry. …

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🧠 The Core Concept

Melting point in isomeric aromatic compounds depends on symmetry and packing efficiency in the crystal lattice.

Higher symmetry → better packing → stronger intermolecular forces (mostly van der Waals) → higher melting point.
Lower symmetry → poorer packing → lower melting point.

For these three isomers, all have the same molecular formula and the same number of substituents — only the positions of the two Cl atoms differ.

✗ Common Mistake #1: Assuming "more polar" means higher melting point

Why it's wrong:

Polarity (dipole moment) is not the main factor here. All three have C–Cl bonds, but the net dipole varies. Students often think a larger dipole means stronger intermolecular forces → higher melting point.

But melting point depends more on how well molecules pack in the solid state, not just dipole strength.

Example in this problem:

Compound (II) is fully symmetrical (Cl atoms para to each other, CH₃ groups para to each other). → Net dipole is zero (or very small). → Yet it has the highest melting point because it packs perfectly.

How to avoid:

✓ Focus on symmetry and crystal packing, not just polarity.

✓ Remember: symmetrical molecules often have higher melting points even if nonpolar.

✗ Common Mistake #2: Thinking "more steric hindrance" always lowers melting point

Why it's wrong:

Steric hindrance can affect packing, but symmetry can override it.

Compound (I) has two Cl atoms on adjacent carbons — a crowded arrangement with only one mirror plane and no centre of symmetry. → Packs less efficiently than (II) → melts below (II).
Compound (III) has Cl atoms flanking one CH₃ — also only one mirror plane, no centre. → Also melts below (II). Since (I) and (III) have comparable symmetry (one plane each), symmetry alone cannot rank them against each other — don't assert a specific (I)-vs-(III) order.

How to avoid:

✓ Compare the overall molecular shape — not just local crowding.

✓ Draw the molecules and check for mirror planes or rotational symmetry.

✗ Common Mistake #3: Ignoring the "para-dimethyl" backbone

Why it's wrong:

All three have the two CH₃ groups fixed at positions 1 and 4 (para to each other).

Students sometimes forget this and treat the compounds as if the methyl groups can move — but they are fixed.

This means the only variable is the Cl positions.

How to avoid:

✓ Always note the fixed substituents first.

✓ Then focus on the variable part — here, the Cl atoms.

✓ Correct Answer & Reasoning

Highest melting point: Compound (II) — 2,5-dichloro-1,4-dimethylbenzene.

Why:

It has a center of symmetry and a mirror plane — highest symmetry.
Molecules pack very efficiently in the crystal lattice. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$n-C_4H_9OH$
›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

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.

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.

B. $(C_2H_5)_2NH$ (diethylamine): a secondary amine, one N–H bond → lower still, 329.3 K → list item IV. B-IV. …
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AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$\underline{A}$
›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

Approximate boiling points: $NH_3\approx -33°C$ , $PH_3\approx -87.7°C$ , $AsH_3\approx -55°C$ , $SbH_3\approx -17°C$ , $BiH_3\approx +17°C$ .

Lowest of these is $PH_3$ (the H-bonding of $NH_3$ is gone, and molecular mass is still small). …
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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

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.

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.

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. …
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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

Ethanol ( $\mathrm{C_2H_5OH}$ ): has an O–H group → capable of H-bonding.

Acetic acid ( $\mathrm{CH_3COOH}$ ): has a carboxylic O–H group → capable of H-bonding.

Ethylamine ( $\mathrm{C_2H_5NH_2}$ ): a primary amine with N–H bonds → capable of H-bonding.

Trimethylamine ( $\mathrm{N(CH_3)_3}$ ): a tertiary amine — nitrogen has no attached H, so it cannot donate a hydrogen bond → excluded.

Salicylic acid: has both a carboxylic O–H and a phenolic O–H → capable of H-bonding. …
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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

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.

Rank by branching (least to most): (ii) unbranched < (iii) one branch point < (i) most branched (quaternary-like carbon skeleton around the C–Cl carbon). …
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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

Identify the substitution pattern: ortho places $\mathrm{-OH}$ and $\mathrm{-NO_2}$ next to each other; para places them across the ring.

Ortho: intramolecular H-bond forms a stable ring — no need for the molecule to H-bond with neighbours. …
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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

Classify each compound: Butanamine (II) = $\mathrm{CH_3CH_2CH_2CH_2NH_2}$ , a primary amine (2 N–H bonds).

N-Ethylethanamine (I) = diethylamine, $\mathrm{(C_2H_5)_2NH}$ , a secondary amine (1 N–H bond).

N,N-Dimethylethanamine (III) = $\mathrm{CH_3CH_2N(CH_3)_2}$ , a tertiary amine (0 N–H bonds). …
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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

n-Hexane: a straight, unbranched 6-carbon chain — largest surface area for intermolecular contact — highest boiling point among the three.

2-Methylpentane: one methyl branch — somewhat more compact than n-hexane — intermediate boiling point.

2,3-Dimethylbutane: two methyl branches, the most compact/spherical of the three — smallest surface area for contact — lowest boiling point. …
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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

List the vapour pressures at 298 K: toluene 60 torr, benzene 160 torr, chloroform 200 torr, dichloromethane 415 torr.

Rank from lowest to highest vapour pressure: toluene < benzene < chloroform < dichloromethane. …
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AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$\mathrm{NH_3}$
›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

$HCl$ : negligible hydrogen bonding (Cl is not electronegative/small enough) — lowest boiling point among the four ( $\approx-85^\circ C$ ).

$NH_3$ : hydrogen bonds via N, but only one lone pair per molecule to hydrogen bond with $\Rightarrow$ boiling point $\approx-33^\circ C$ .

$HF$ : 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 $\approx19.5^\circ C$ . …
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AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$\mathrm{CH_3CH(OH)CH_2CH_3}$
›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

Butan-2-ol: contains -OH, capable of strong intermolecular hydrogen bonding $\Rightarrow$ highest boiling point among these four.

Butan-2-one: a ketone, polar C=O but no H-bond donor $\Rightarrow$ moderate boiling point (dipole-dipole), lower than the alcohol.

Ethyl propyl ether: polar C-O-C but no H-bond donor either $\Rightarrow$ boiling point similar to or slightly below the ketone. …
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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

Butan-1-ol (a): $-\mathrm{OH}$ group, strongest hydrogen bonding of the four compounds → highest boiling point.

Butan-1-amine (b), a primary amine: two N–H bonds, extensive intermolecular hydrogen bonding → next highest. …
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Q.Which of the following compounds will have the highest melting point and why?

Why symmetry controls melting point

Step-by-step analysis

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