Arrange the following compounds in increasing order of acidity and give a suitable explanation. Phenol, o-nitr

NCERT Exemplar · Q29

Q. Arrange the following compounds in increasing order of acidity and give a suitable explanation. Phenol, o-nitrophenol, o-cresol

CBSEShort· 2mImportance★★★★★

60% · 81/135 Questions

🔒 Locked · start free trial →

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:

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. …

Start your 14-day free trial to unlock the full solution →

Acidity is controlled by the stability of the conjugate base after proton loss. Electron-withdrawing groups (like –NO₂) increase acidity by stabilizing the phenoxide ion; electron-donating groups (like –CH₃) decrease it. The correct order is: o-cresol < phenol < o-nitrophenol.

Why this order? The concept of conjugate base stability

Acidity is not about how willing a molecule is to lose a proton — it’s about how stable the resulting anion is. For phenols, the acidic proton is the –OH hydrogen. When it leaves, we get a phenoxide ion (ArO⁻). The more stable this phenoxide ion, the stronger the acid.

Phenol itself is more acidic than an alcohol because the negative charge on oxygen can be delocalized into the aromatic ring via resonance. Now, any substituent on the ring will either help or hinder this delocalization.

An electron-withdrawing group (EWG) like –NO₂ pulls electron density away from the ring. This further stabilizes the negative charge on the phenoxide oxygen, making the acid stronger.
An electron-donating group (EDG) like –CH₃ pushes electron density toward the ring. This destabilizes the negative charge, making the acid weaker.

So the order is: weakest acid (most electron-donating substituent) < phenol (no substituent) < strongest acid (most electron-withdrawing substituent).

Step-by-step reasoning

Identify the substituent effect in each compound
- Phenol: No substituent. The phenoxide ion is stabilized only by resonance with the ring. This is our reference point.
- o-Nitrophenol: The –NO₂ group is a strong electron-withdrawing group. It stabilizes the phenoxide ion by both inductive effect (pulling through sigma bonds) and resonance effect (pulling through pi bonds). Importantly, the ortho position allows the nitro group to directly participate in resonance with the phenoxide oxygen, creating additional resonance structures that spread the negative charge onto the electronegative oxygen atoms of –NO₂.
- o-Cresol: The –CH₃ group is an electron-donating group (hyperconjugation + inductive effect). It pushes electron density toward the ring, which increases the negative charge density on the phenoxide oxygen, making the ion less stable.
Compare the stability of the conjugate bases
- The phenoxide from o-nitrophenol is the most stable because the negative charge is delocalized onto the nitro group. …

Start your 14-day free trial to unlock the full solution →

Here are the most common mistakes students make when tackling this specific acidity comparison, along with the conceptual fixes to avoid them.

Mistake 1: Confusing Boiling Point Logic with Acidity Logic

The Error: Students often try to rank acidity based on molecular weight or the strength of intermolecular forces (like hydrogen bonding). They might think o-nitrophenol has the highest acidity because it has the highest boiling point (due to intramolecular H-bonding reducing intermolecular bonding).

The Fix: Acidity is about the stability of the conjugate base (the phenoxide ion), not the boiling point of the parent molecule.

Boiling Point depends on intermolecular forces (how molecules stick to each other).
Acidity depends on how well the molecule can donate a proton ( $H^+$ ) and how stable the resulting negative charge is on the oxygen.

Mistake 2: Forgetting the Role of the Nitro Group ( $-NO_2$ )

The Error: Students often treat the nitro group as just an "electron-withdrawing group" without specifying how it withdraws electrons. They might incorrectly rank o-nitrophenol as less acidic than phenol because of "steric hindrance" or "ortho effect" confusion.

The Fix: The nitro group is a strong electron-withdrawing group via both inductive effect (through sigma bonds) and resonance effect (through pi bonds).

Resonance Withdrawal: The $-NO_2$ group can delocalize the negative charge from the phenoxide ion onto its own oxygen atoms. This makes the conjugate base much more stable.
Result: o-Nitrophenol is more acidic than phenol.

Mistake 3: Misinterpreting the Effect of the Methyl Group ( $-CH_3$ )

The Error: Students think any substituent on the ring makes the compound more acidic. They might place o-cresol as more acidic than phenol.

The Fix: The methyl group ( $-CH_3$ ) is an electron-donating group (via hyperconjugation and inductive effect).

Effect: It pushes electron density towards the ring and the oxygen of the phenoxide ion.
Result: This destabilizes the negative charge on the conjugate base. A less stable conjugate base means the parent acid is weaker.
Conclusion: o-Cresol is less acidic than phenol.

Mistake 4: Ignoring the "Ortho Effect" for Nitro Group

The Error: Students assume that para-nitrophenol and o-nitrophenol have exactly the same acidity. While para is often more acidic due to perfect resonance, ortho is still very strong.

The Fix: In o-nitrophenol, the nitro group is very close to the $-OH$ group. This creates a strong intramolecular hydrogen bond between the $-OH$ hydrogen and the nitro group's oxygen. …

Start your 14-day free trial to unlock the full solution →

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 24 on this concept.

CBSE 2026Set 56/2/11 markMCQ
$Q. Identify the correct increasing order of boiling points of the given compounds : (A) Propan-1-ol < butan-1-ol < butan-2-ol < pentan-1-ol (B) Pentan-1-ol < butan-1-ol < butan-2-ol < Propan-1-ol (C) Propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol (D) Butan-1-ol < Butan-2-ol < Propan-1-ol < Pentan-1-ol$
›Reveal solutionSolution
Boiling points of alcohols depend on chain length (more carbons → higher bp) and branching (more branching → lower bp). The correct order is Propan-1‑ol < butan‑2‑ol < butan‑1‑ol < pentan‑1‑ol, which matches option (C).

Why boiling points of alcohols behave this way

Alcohols boil at much higher temperatures than hydrocarbons of similar mass because of hydrogen bonding between the –OH groups. Two factors control the boiling point within a family of alcohols:

Chain length – A longer carbon chain means more surface area for van der Waals forces. These weak attractions add up, so a larger molecule needs more energy (higher temperature) to escape into the vapour phase. For straight‑chain alcohols, boiling point rises steadily as the number of carbons increases.

Branching – When the –OH group is attached to a secondary or tertiary carbon (as in butan‑2‑ol), the molecule becomes more compact. A compact shape reduces the surface area available for van der Waals interactions, so the boiling point drops compared to its straight‑chain isomer. The hydrogen‑bonding ability is roughly the same for all isomers (one –OH per molecule), so the difference comes from the weaker London forces in the branched form.

Watch out

A common mistake is to think that branching increases boiling point because the molecule looks “more crowded”. In reality, branching decreases the surface area and therefore weakens the intermolecular forces. Always compare chain length first, then branching.

Step‑by‑step reasoning

Identify the compounds and their carbon counts

Propan‑1‑ol: 3 carbons, straight chain.

Butan‑1‑ol: 4 carbons, straight chain.

Butan‑2‑ol: 4 carbons, branched (the –OH is on carbon 2).

Pentan‑1‑ol: 5 carbons, straight chain.

Order by chain length

Longer chain → higher boiling point. So the 5‑carbon alcohol (pentan‑1‑ol) should have the highest bp, and the 3‑carbon alcohol (propan‑1‑ol) the lowest. The two 4‑carbon alcohols will sit in between.

Compare the two C₄ isomers …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): Boiling point of alkanes decreases with increase in molecular mass. Reason (R): Intermolecular Vander Waals forces increase with increase in molecular size or surface area of the molecules.$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution

The Assertion has the trend backwards — alkane boiling points INCREASE with molecular mass — while the Reason correctly describes why (stronger van der Waals forces with larger surface area).

Assertion: 'Boiling point of alkanes decreases with increase in molecular mass' — this is false. In reality, as molecular mass (chain length) increases, boiling point increases (e.g. methane $-162°C$ → butane $-0.5°C$ → octane $126°C$ ).

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$Q. Two compounds 'A' and 'B' were being tested for their boiling points. It was observed that 'A' started boiling after 'B', when both were subjected to same conditions. If the compound 'B' is acetone, which of the following can be compound 'A'?$
(a) Propanal
(b) Propan-1-ol
(c) Methoxyethane
(d) n-Butane
›Reveal solutionSolution

"A boils after B" → A has the higher boiling point; among the options only propan-1-ol (H-bonding) boils higher than acetone, so A = propan-1-ol.

B is acetone (propanone), b.p. $\approx 56^\circ\text{C}$ . "A started boiling after B" means A needs a higher temperature, i.e. A has a higher boiling point than acetone. Comparing approximate boiling points of the options (all $\text{C}_3$ / $\text{C}_4$ molecules of similar mass):

Compound Approx. b.p. Reason
(a) Propanal $\approx 49^\circ\text{C}$ dipole–dipole only
(b) Propan-1-ol $\approx 97^\circ\text{C}$ strong intermolecular H-bonding
Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$Q. Select the correct order of melting points of isomeric dichlorobenzenes.$
(a) o-dichlorobenzene > m-dichlorobenzene > p-dichlorobenzene
(b) p-dichlorobenzene > m-dichlorobenzene > o-dichlorobenzene
(c) p-dichlorobenzene > o-dichlorobenzene > m-dichlorobenzene
(d) m-dichlorobenzene > o-dichlorobenzene > p-dichlorobenzene
›Reveal solutionSolution
Melting point depends on how well molecules pack in the crystal; the symmetrical para isomer packs best (highest m.p.), giving the order para > ortho > meta.

For isomeric dichlorobenzenes, melting point is governed mainly by crystal packing / molecular symmetry rather than by intermolecular force magnitude:

p-dichlorobenzene is the most symmetrical, so it packs most efficiently into the crystal lattice and has the highest melting point ( $\approx 53^\circ\text{C}$ ).

o-dichlorobenzene ( $\approx -17^\circ\text{C}$ ) packs better than the meta isomer. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$Q. Sufficient amount of 2-methylpropan-2-ol heated with 20% phosphoric acid at 358 K gives main product 'X' with the elimination of water and tert-butyl alcohol undergoes dehydration when it is passed over heated copper at 573 K gives 'Y' Pick the correct statement regarding X and Y.$
(a) The boiling points of 'X' and 'Y' are equal
(b) The boiling point of 'X' is greater than the boiling point of 'Y'
(c) The boiling point of 'X' is lesser than the boiling point of 'Y'
(d) At room temperature both 'X' and 'Y' exists as a solids
›Reveal solutionSolution
Acid dehydration and passing over hot copper both convert 2-methylpropan-2-ol to the same alkene (2-methylpropene), so X = Y and their boiling points are equal — option (a).

2-Methylpropan-2-ol is a tertiary alcohol (tert-butyl alcohol), $\text{(CH}_3)_3\text{C–OH}$ .

With 20% phosphoric acid at 358 K it undergoes acid-catalysed dehydration (elimination of water) to give the alkene: $\text{(CH}_3)_3\text{C–OH} \rightarrow \text{(CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O}$ , so X = 2-methylpropene (isobutylene). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set D1 markMCQ
$Q. At room temperature, formaldehyde is$
(a) gas
(b) liquid
(c) solid
(d) none of these
›Reveal solutionSolution

Formaldehyde, the first member of the aldehyde series, is a gas at ordinary temperature (b.p. about −19 °C).

Formaldehyde (methanal, HCHO) is the lowest aldehyde. It has a very low boiling point (about −19 °C), so at room temperature it exists as a colourless, pungent-smelling gas. Its 40% aqueous solution is calle …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: The boiling point of methanol is ________ K.$
›Reveal solutionSolution

Methanol (CH3OH), the smallest alcohol, boils at about 338 K (64.7 degrees C) at atmospheric pressure.

Methanol's boiling point of ~338 K is relatively low among common alcohols because of its small molecular size (weaker van der Waals/London forces), even though, like other alcohols, it is capable of intermolecu …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following compounds has highest melting point?$
(a) 1,2-dichlorobenzene (ortho-dichlorobenzene, structure drawn)
(b) 1,3-dichlorobenzene (meta-dichlorobenzene, structure drawn)
(c) 1,4-dichlorobenzene (para-dichlorobenzene, structure drawn)
(d) All have same melting point.
›Reveal solutionSolution

Melting point depends on how efficiently molecules pack into a crystal lattice, not just molecular weight — the highly symmetric para isomer packs far better than the less symmetric ortho and meta isomers, giving it a much higher melting point.

All three dichlorobenzenes have the same molecular formula and molecular weight, so their melting-point difference comes purely from crystal packing efficiency. p-Dichlorobenzene is linear and symmetric, so molecules stack very closely and regularly in the solid lattice, maximising van der Waals contact — this raises its melting point sharply (~53°C). The or …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$CH_3CHO$
›Reveal solutionSolution
Boiling point here tracks the strength of intermolecular forces: propane (only weak van der Waals forces) boils lowest, dimethyl ether (weak dipole-dipole, no H-bonding) next, acetaldehyde (stronger dipole-dipole from the polar C=O) next, and ethanol (hydrogen-bonded) boils highest.

All four compounds have comparable molar mass (propane 44, dimethyl ether 46, acetaldehyde 44, ethanol 46 g mol $^{-1}$ ), so the boiling-point order is decided almost entirely by the type of intermolecular attraction available, not by size:

$CH_3CH_2CH_3$ (propane): a non-polar hydrocarbon; molecules are held together only by weak instantaneous dipole–induced dipole (London/van der Waals) forces. Lowest boiling point (real value ≈ −42 °C).

$CH_3OCH_3$ (dimethyl ether): the C–O–C linkage gives the molecule a small permanent dipole, so molecules attract each other by dipole–dipole forces, stronger than propane's dispersion forces alone but the ether oxygen has no O–H bond, so no hydrogen bonding is possible. Boils higher than propane (real value ≈ −24 °C). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The correct order of boiling points of alcohols having the same number of Carbon atoms is ...................... .$
(a) 2° > 1° > 3°
(b) 1° > 2° > 3°
(c) 3° > 1° > 2°
(d) 3° > 2° > 1°
›Reveal solutionSolution

Among isomeric alcohols, boiling point falls as branching increases, because branching reduces the effective surface area available for intermolecular hydrogen bonding and van der Waals interactions.

All isomeric alcohols with the same molecular formula can hydrogen-bond through their –OH group, but a straight-chain (primary) alcohol packs more efficiently and has a larger surface area for van der Waals contact between molecules than a branched (te …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following has the highest melting point?$
(a) o-xylene
(b) m-xylene
(c) p-xylene
(d) Toluene
›Reveal solutionSolution

p-Xylene has the highest melting point because its high molecular symmetry allows the most efficient crystal packing.

Among the three xylene isomers (o-, m-, p-) and toluene, melting point depends heavily on how symmetrically the molecules can pack into a solid lattice (not just on molecular weight or boiling point). p-Xylene, with its methyl groups symmetrically placed at opposite (1,4) positions on the ring, packs most efficiently into a crystal lattice, giving it a distinct …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$Q. Why has propanol higher boiling point than propane?$
›Reveal solutionSolution

Boiling point depends on the strength of intermolecular forces that must be overcome; propanol's -OH group enables hydrogen bonding between molecules, a much stronger force than the weak van der Waals (London dispersion) forces that are all propane has.

Propane (CH3-CH2-CH3) is a non-polar hydrocarbon with no functional group capable of hydrogen bonding. Its molecules are held together only by weak van der Waals (induced-dipole) forces, so relatively little energy is needed to separate them - it boils at a very low temperature (-42 degree C) and is a gas at room temperature.

…

Start your 14-day free trial to see the full solution →

Compound	Approx. b.p.	Reason
(a) Propanal	$\approx 49^\circ\text{C}$	dipole–dipole only
(b) Propan-1-ol	$\approx 97^\circ\text{C}$	strong intermolecular H-bonding

←Previous questionQ28 — When phenol is treated with bromine water, white precipitate is obtain…→Next questionQ30 — Alcohols react with active metals e.g. Na, K etc. to give correspondin…

Q.Arrange the following compounds in increasing order of acidity and give a suitable explanation.
Phenol, o-nitrophenol, o-cresol

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)

Why this order? The concept of conjugate base stability

Step-by-step reasoning

Method: Resonance & Inductive Effect Analysis for Acidity of Phenols

Step 1: Write the general reaction

Step 2: Analyse each compound

1. Phenol (reference)

2. o-Nitrophenol

3. o-Cresol

Step 3: Arrange in increasing order of acidity

Mistake 1: Confusing Boiling Point Logic with Acidity Logic

Mistake 2: Forgetting the Role of the Nitro Group ( $-NO_2$ )

Mistake 3: Misinterpreting the Effect of the Methyl Group ( $-CH_3$ )

Mistake 4: Ignoring the "Ortho Effect" for Nitro Group

Why boiling points of alcohols behave this way

Step‑by‑step reasoning

Q.Arrange the following compounds in increasing order of acidity and give a suitable explanation. Phenol, o-nitrophenol, o-cresol

Why this order? The concept of conjugate base stability

Step-by-step reasoning

🎓Unlock everything free for 14 days

Q.Arrange the following compounds in increasing order of acidity and give a suitable explanation.
Phenol, o-nitrophenol, o-cresol

Unlock everything free for 14 days