Arrange the following compounds in increasing order of their boiling points. (a) mathrm(CH₃)₂CH-CH₂Br (b) math

NCERT Exemplar · Q8

Q. Arrange the following compounds in increasing order of their boiling points.

(a)

\mathrm{(CH_3)_2CH-CH_2Br}

(b)

\mathrm{CH_3CH_2CH_2CH_2Br}

(c)

\mathrm{(CH_3)_3C-Br}

(i)

(b) <

(a) <

(c)

(ii)

(a) <

(b) <

(c)

(iii)

(a) <

(b)

(iv)

(b) < (a)

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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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Boiling points of alkyl halides depend on molecular weight and branching. More branching lowers boiling point due to reduced surface area. The correct order is (c) < (a) < (b), which corresponds to option (iii).

Boiling point trends in organic compounds are governed by two main factors: molecular mass and intermolecular forces. For isomeric alkyl halides like these three C₄H₉Br compounds, the molecular weight is identical (137 g/mol), so the deciding factor is the strength of van der Waals forces — specifically, how well the molecules can pack together.

The key insight: more branching means a more compact, spherical molecule. A compact molecule has less surface area available for intermolecular contact, so the van der Waals forces are weaker, and the boiling point is lower. A straight-chain molecule, by contrast, is long and flexible, allowing many points of contact between neighbouring molecules.

Let’s examine each compound:

Compound (c): 2-bromo-2-methylpropane — This is the most branched. The central carbon is bonded to three methyl groups and one bromine. The molecule is nearly spherical. Very little surface area for neighbouring molecules to “grip” each other. This will have the lowest boiling point.
Compound (a): 1-bromo-2-methylpropane — Here the longest chain is only three carbons (a propane backbone) with a methyl branch on the middle carbon and the bromine on C-1: $\mathrm{(CH_3)_2CHCH_2Br}$ . The single branch makes the molecule less elongated than the straight chain, yet it is still not as compact as (c). Its surface area is intermediate, so its boiling point falls in the middle. …

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

COMEDK 2026Set 2026-A1 markMCQ
$D<A<B<C$
›Reveal solutionSolution
Boiling point rises with chain length and falls with branching; the increasing order is n-butane $<$ 2,2-dimethylpropane $<$ 2-methylbutane $<$ n-pentane, i.e. $D<A<B<C$ — option (A).

Concept

Alkanes are held together by London dispersion forces, which strengthen with (i) more electrons — longer chains — and (ii) larger surface contact — straighter chains. So for a fixed carbon count, more branching lowers the boiling point because the molecule becomes more compact.

Solution

Identify each compound:

D — n-butane: $\mathrm{C_4}$ , straight; fewest electrons $\Rightarrow$ lowest boiling point overall ( $\approx -0.5^\circ\mathrm{C}$ ).

A — 2,2-dimethylpropane (neopentane): $\mathrm{C_5}$ , most branched $\Rightarrow$ lowest among the $\mathrm{C_5}$ 's ( $\approx 9.5^\circ\mathrm{C}$ ).

B — 2-methylbutane (isopentane): $\mathrm{C_5}$ , one branch ( $\approx 28^\circ\mathrm{C}$ ). …
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COMEDK 2024Set 2024-M1 markMCQ
$Q. Above figure represents Vapour pressure versus Temperature graphs of 2 pure volatile liquids and a solution formed by the 2 liquids.$
(i) Which curve represents the solution?
(ii) Which curve represents the liquid with the strongest intermolecular forces of attraction? (A)
(i) B
(ii) C (B)
(i) A
(ii) C (C)
(i) C
(ii) A (D)
(i) B
(ii) A
›Reveal solutionSolution
The solution’s vapour-pressure curve lies between those of the pure liquids, so it is curve B. The liquid with the strongest intermolecular forces has the lowest vapour pressure at a given temperature, so it is curve C. Hence the answer is option (A).

Concept and intuition

Vapour pressure is a measure of how readily molecules escape from the liquid into the gas phase. For a pure liquid, stronger intermolecular forces (like hydrogen bonding or dipole-dipole interactions) mean molecules are held more tightly, so fewer escape — giving a lower vapour pressure at any given temperature. For a solution of two volatile liquids, Raoult’s law tells us that the total vapour pressure is a weighted average of the pure vapour pressures (assuming ideal behaviour). Therefore, the solution’s vapour-pressure curve must lie between the curves of the two pure components. In the graph, the three curves are ordered from highest vapour pressure (curve A) to lowest (curve C) at any fixed temperature. So:

Curve A: highest vapour pressure → weakest intermolecular forces.

Curve C: lowest vapour pressure → strongest intermolecular forces.

Curve B: intermediate → must be the solution.

Now let’s confirm step by step.

Identify which curve is the solution For an ideal solution of two volatile liquids, the total vapour pressure at a given temperature is

$P_{\text{soln}} = \chi_A P_A^* + \chi_B P_B^*$
where $P_A^*$ and $P_B^*$ are the vapour pressures of the pure liquids and $\chi$ are mole fractions (which sum to 1). Since $P_{\text{soln}}$ is a weighted average, it always lies between $P_A^*$ and $P_B^*$ . Therefore, on the graph, the solution’s curve must be the one that sits between the other two at every temperature. The curves are labelled A (leftmost, highest vapour pressure), B (middle), and C (rightmost, lowest vapour pressure). So curve B is the solution.

Identify which pure liquid has the strongest intermolecular forces …
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KCET 2020Set A-11 markMCQ

Q. Which of the following polymer has strongest intermolecular forces of attraction ? (A) Polystyrene (B) Neoprene (C) Terylene (D) Polythene

›Reveal solution

Fibres have the strongest intermolecular forces of all polymer classes, and Terylene (a polyester) is the only fibre among the four.

Step 1 — Classify polymers by intermolecular force (the concept).

The molecular-force classification explains a polymer's entire mechanical behaviour:

Class	Intermolecular forces	Behaviour	Examples
Elastomers	Weakest — chains coiled, held by only a few crosslinks	highly elastic, stretch and snap back	rubber, neoprene, buna-S
Plastics	Intermediate	mouldable on heating	polythene, polystyrene, PVC
Fibres	Strongest — H-bonding / strong dipole–dipole; chains pack closely and crystallise	high tensile strength, thread-forming	nylon-6,6, Terylene (Dacron), silk

Step 2 — Why Terylene sits at the top.

Terylene is a polyester made from ethylene glycol and terephthalic acid. Its backbone carries repeating polar ester groups ( $-\!COO\!-$ ), whose $C{=}O$ dipoles set up strong dipole–dipole attractions (and hydrogen-bonding-type interactions) between neighbouring chains. These forces let the chains pack into a close, crystalline arrangement — giving the high tensile strength, high melting point and thread-forming ability that define a fibre.

Step 3 — Why the other three are weaker. …

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KCET 2018Set A-11 markMCQ
$Q. Phenol can be distinguished from ethanol by the reagent (A) Bromine water (B) Sodium metal (C) Iron metal (D) Chlorine water$
›Reveal solutionSolution
Phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromophenol, while ethanol does not — this makes bromine water the correct distinguishing reagent.

The key here is that both phenol and ethanol have an —OH group, so they share some reactions (like with sodium metal). But phenol is an aromatic alcohol — the —OH is attached directly to a benzene ring. That changes its chemistry dramatically. The ring activates the ortho and para positions toward electrophilic substitution, which ethanol, being aliphatic, cannot do.

Bromine water is a classic test for phenols because it exploits this difference. Let's see why each option works or fails.

Bromine water (Option A) — This is the correct choice. Phenol reacts instantly with bromine water at room temperature to form a white precipitate of 2,4,6-tribromophenol. The reaction is:

$C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH \downarrow + 3HBr$
Ethanol does not react with bromine water under these conditions — no decolourisation, no precipitate. So you get a clear visual distinction.

Sodium metal (Option B) — Both phenol and ethanol react with sodium metal to liberate hydrogen gas:

$2C_6H_5OH + 2Na \rightarrow 2C_6H_5ONa + H_2 \uparrow$ $2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \uparrow$
This test cannot distinguish between them — both give the same observation.

Iron metal (Option C) — Iron does not react with either phenol or ethanol under ordinary conditions. No useful test here. …
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Q.Arrange the following compounds in increasing order of their boiling points.

Concept understanding — Boiling Point Trends

Boiling Point Trends (Organic Compounds)

The Forces, Weakest to Strongest

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

Trend 2: Chain Length and Branching

Trend 3: Hydrogen Bonding Beats Molecular Mass …

Why this formula?

Boiling Point Trends: Why They Happen

The Core Idea: Intermolecular Forces

Why this formula makes sense

The Four Key Trends (with Reasoning)

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

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

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

Concept: Boiling Point Trends in Alkyl Halides

Method: Branching Effect on Boiling Points

Steps:

Core Concept: Boiling Points of Alkyl Halides

Correct Order

Common Mistakes & How to Avoid Them

✗ Mistake 1: Assuming higher branching increases boiling point

✗ Mistake 2: Ignoring that all are isomers

✗ Mistake 3: Confusing boiling point with melting point trends

Concept

Solution

Q.Arrange the following compounds in increasing order of their boiling points.

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