Q.Arrange the following compounds in increasing order of boiling point.
Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →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:
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 (ΔHvap):
lnP=−RΔHvap⋅T1+C
Where:
- P = vapor pressure
- ΔHvap = 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
- ΔHvap is large when IMFs are strong — more energy is needed to separate molecules.
- At boiling point, P=Patm (usually 1 atm). So a substance with larger ΔHvap 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. …
Boiling point of alcohols depends on chain length (more carbons -> higher bp) and branching (more branching -> lower bp, due to reduced surface area for intermolecular forces).
- By chain length: pentan-1-ol (5C) > butan-1-ol (4C) > propan-1-ol (3C). …
Boiling point of alcohols depends on molecular size (chain length) and branching. Longer chains raise boiling point; branching lowers it. The correct increasing order is propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol, matching option (i).
Reasoning …
Method: Hydrogen Bonding + Branching Effect Analysis
This method uses two key factors that determine boiling points of alcohols:
- Chain length → more carbons = more surface area = stronger London forces = higher boiling point
- Branching → more branching = more compact shape = weaker intermolecular forces = lower boiling point
Step 1: Arrange by carbon chain length
| Compound | Carbon atoms |
|---|---|
| Propan-1-ol | 3 |
| Butan-1-ol | 4 |
| Butan-2-ol | 4 |
| Pentan-1-ol | 5 |
Longer chain → higher boiling point, so:
propan-1-ol<butan-1-ol, butan-2-ol<pentan-1-ol
Step 2: Compare isomers (same chain length)
Butan-1-ol and butan-2-ol both have 4 carbons, but:
- Butan-1-ol (primary alcohol) — straight chain, no branching
- Butan-2-ol (secondary alcohol) — branched at carbon 2
More branching → lower boiling point, so: …
Common Mistakes Students Make on Boiling Point Trends (Alcohols)
Mistake 1: Ignoring the Effect of Branching on Boiling Point
The error: Students assume that all isomers of the same molecular formula have the same boiling point. They place butan-1-ol and butan-2-ol at the same level or in the wrong order.
Why it happens: They focus only on molecular mass and forget that branching reduces surface area and weakens intermolecular forces (van der Waals forces).
How to avoid: Remember:
- Straight-chain alcohols have higher boiling points than their branched isomers.
- Butan-1-ol (straight chain) has a higher boiling point than butan-2-ol (branched).
- So the correct order is: butan-2-ol < butan-1-ol.
Mistake 2: Confusing Boiling Point with Melting Point Trends
The error: Students apply melting point logic (where branching can increase melting point due to better packing) to boiling points.
Why it happens: The two trends are opposite in some cases — branching lowers boiling point but can raise melting point.
How to avoid: For boiling points, always think about surface area and intermolecular forces in the liquid state. Branching → less surface area → weaker van der Waals → lower boiling point.
Mistake 3: Overlooking the Dominant Role of Hydrogen Bonding
The error: Students think that van der Waals forces are the only factor and ignore that hydrogen bonding between alcohol molecules is the strongest intermolecular force.
Why it happens: They treat alcohols like hydrocarbons.
How to avoid: All four compounds are alcohols — they all form hydrogen bonds. So the order is decided by the combined effect of:
- Hydrogen bonding (similar for all — one –OH group each)
- van der Waals forces (increase with chain length and decrease with branching)
Mistake 4: Forgetting That Chain Length Increases Boiling Point
The error: Students place pentan-1-ol before propan-1-ol or butan-1-ol. …
- AHSEC Higher Secondary (HS) Final Examination 2022Set ANNUAL1 markQ.Arrange the following compounds in the increasing order of their boiling points: CH3-CHO, CH3-CH2-OH, CH3-CH2-CH3
›Reveal solutionSolution
Boiling point rises with the strength of intermolecular forces: dispersion-only < dipole-dipole < hydrogen bonding.
All three compounds have comparable molar mass (propane 44, ethanal 44, ethanol 46), so the difference in boiling point comes almost entirely from the type of intermolecular attraction each can form:
- CH3-CH2-CH3 (propane): a non-polar hydrocarbon — molecules are held together only by weak London (dispersion) forces. Lowest boiling point (−42°C).
- CH3-CHO (ethanal): the C=O group is polar, so molecules attract each other by dipole-dipole interactions, which are stronger than dispersion forces alone. Intermediate boiling point (≈21°C). …
- AHSEC Higher Secondary (HS) Final Examination 2018Set ANNUAL1 markQ.Arrange the following compounds in increasing order of their boiling points: CH3CHO, CH3CH2OH, CH3-O-CH3, CH3CH2CH3.
›Reveal solutionSolution
CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH (increasing boiling point).
Boiling point rises with stronger intermolecular forces (for comparable molar masses).
- CH3CH2CH3 (propane): non-polar, only weak London dispersion forces → lowest b.p. (≈ −42 °C).
- CH3–O–CH3 (dimethyl ether): polar (dipole–dipole) but no O–H, so no hydrogen bonding → higher than propane (≈ −24 °C). …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.