Arrange the following in decreasing order of their boiling points. (A) n-butane (B) 2-methylbutane (C) n-penta

NCERT Exemplar · Q1

Q. Arrange the following in decreasing order of their boiling points. (A) n-butane (B) 2-methylbutane (C) n-pentane (D) 2,2-dimethylpropane (A) A > B > C > D (B) B > C > D > A (C) D > C > B > A (D) C > B > D > 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

An alcohol boils noticeably higher than a haloalkane or ether of similar molecular mass, because the O–H bond can hydrogen-bond to neighbouring alcohol molecules, while a haloalkane or ether (no H directly on the electronegative atom in a donor position) cannot do the same. Comparing purely by molecular mass without checking for H-bonding capability is a common source of wrong predictions.

Watch out

Don't rank boiling points by dipole moment alone. A haloalkane's dipole moment trend and its boiling-point trend can point in different directions (see the C–X dipole note above) — boiling point is about the total intermolecular attraction (dispersion + dipole + any H-bonding), not any one factor in isolation.

Tip

When comparing boiling points, check in this order: (1) is hydrogen bonding possible for one but not the other? — usually decisive if so; (2) if neither/both can H-bond, compare size/branching (more surface area, more contact, higher boiling point); (3) only then consider polarity as a tie-breaker.

Boiling point trends, especially the role of hydrogen bonding, are discussed across the NCERT/CBSE Class 11 and 12 Organic Chemistry chapters, including Alcohols, Phenols and Ethers, and ‘boiling point comparison of isomers’ is a commonly searched important-question topic for board exams, JEE Main and NEET. Applying the hydrogen-bonding-first, then-size-and-branching approach is a strategy tested repeatedly in competitive chemistry MCQs.

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.
Branching reduces surface area → weaker London dispersion forces (less contact between molecules).
More spherical molecules pack less efficiently → lower boiling point.

Key insight: Shape matters — surface area determines dispersion force strength for same-mass molecules.

4. Polarity vs. nonpolarity (e.g., C₂H₅OH vs. C₂H₆)

Molecule	IMFs	Boiling point (°C)
Ethanol (C₂H₅OH)	H-bonding + dispersion	78
Ethane (C₂H₆)	Dispersion only	-89

Why?

Ethanol has an –OH group → hydrogen bonding.
Ethane is nonpolar — only weak dispersion.
Despite similar molar mass (46 vs. 30), ethanol boils 167°C higher.

Key insight: Polarity and hydrogen bonding dominate over mass when present.

Summary: The "Formula" is Conceptual

There is no single equation that gives boiling point directly. Instead, the Clausius–Clapeyron equation is the theoretical backbone:

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

And the boiling point is the $T$ at which $P = P_{\text{atm}}$ .

To predict trends, ask:

What IMFs are present? (Dispersion, dipole-dipole, H-bonding)
How strong are they? (More electrons → stronger dispersion; H-bonding is strongest)
How does molecular shape affect surface area?

Stronger IMFs → higher $\Delta H_{\text{vap}}$ → higher boiling point.

Boiling points of alkanes depend on molecular size and branching — larger molecules have higher boiling points, and for the same number of carbons, more branching lowers the boiling point. The decreasing order is: n-pentane > 2-methylbutane > 2,2-dimethylpropane > n-butane, which corresponds to option (D).

The boiling point of an alkane is determined by the strength of intermolecular forces — specifically, London dispersion forces. These forces increase with molecular size (more electrons, larger surface area) and decrease with branching (more compact shape reduces surface contact between molecules). So the key idea is: more carbons → higher boiling point; same carbons → less branching → higher boiling point.

Let’s identify each compound and its carbon count:

n-butane: 4 carbons, straight chain
2-methylbutane: 5 carbons, branched (one methyl group on carbon 2)
n-pentane: 5 carbons, straight chain
2,2-dimethylpropane: 5 carbons, highly branched (two methyl groups on carbon 2)

So we have one C4 compound and three C5 isomers. The C5 compounds will all boil higher than the C4 one, because they have more electrons and stronger dispersion forces. Among the C5 isomers, the straight-chain n-pentane has the largest surface area, so it boils highest. 2-methylbutane is moderately branched, so it boils lower than n-pentane but higher than the highly branched 2,2-dimethylpropane (which is nearly spherical and has the least surface contact).

Now let’s work through the ordering step by step.

Separate by carbon number. n-butane (C4) has the smallest molecule, so it will have the lowest boiling point of the four. The other three are all C5, so they will all be higher than n-butane. This already tells us that n-butane comes last in decreasing order.
Compare the C5 isomers. For molecules with the same molecular formula, boiling point decreases as branching increases. n-pentane is unbranched — maximum surface area, strongest dispersion forces. 2-methylbutane has one branch, reducing surface area. 2,2-dimethylpropane has two branches on the same carbon, making it very compact — minimum surface area, weakest dispersion forces among the three.
Establish the order among C5. So: n-pentane (highest) > 2-methylbutane > 2,2-dimethylpropane.
Insert n-butane at the bottom. Since n-butane is C4, it boils lower than all C5 isomers. So the full decreasing order is: n-pentane > 2-methylbutane > 2,2-dimethylpropane > n-butane.

Watch out

A common mistake is to think that more branching always means a higher boiling point (confusing with melting point trends, where branching can sometimes raise the melting point due to better packing in solids). For boiling points, branching always lowers the value because it reduces intermolecular contact in the liquid phase.

Tip

You can remember this as: "Straight chains stack well, branched chains stack poorly." For boiling points, think of how well the molecules can "touch" each other — more touching means more dispersion force, means higher boiling point.

Now match this order to the options given:

(A) A > B > C > D → n-butane > 2-methylbutane > n-pentane > 2,2-dimethylpropane — wrong, because n-butane is lowest, not highest.
(B) B > C > D > A → 2-methylbutane > n-pentane > 2,2-dimethylpropane > n-butane — wrong, because n-pentane should be above 2-methylbutane.
(C) D > C > B > A → 2,2-dimethylpropane > n-pentane > 2-methylbutane > n-butane — wrong, because 2,2-dimethylpropane is the lowest among C5, not the highest.
(D) C > B > D > A → n-pentane > 2-methylbutane > 2,2-dimethylpropane > n-butane — correct.

✓Final answer

The correct option is (D) .

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Q.Arrange the following in decreasing order of their boiling points. (A) n-butane (B) 2-methylbutane (C) n-pentane (D) 2,2-dimethylpropane (A) A > B > C > D (B) B > C > D > A (C) D > C > B > A (D) C > B > D > A

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Q.Arrange the following in decreasing order of their boiling points.
(A) n-butane
(B) 2-methylbutane
(C) n-pentane
(D) 2,2-dimethylpropane
(A) A > B > C > D
(B) B > C > D > A
(C) D > C > B > A
(D) C > B > D > A

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