The correct order of increasing acidic strength is – Class 12 Chemistry

NCERT Exemplar · Q3

Q. The correct order of increasing acidic strength is _____________.

(i) Phenol < Ethanol < Chloroacetic acid < Acetic acid

(ii) Ethanol < Phenol < Chloroacetic acid < Acetic acid

(iii) Ethanol < Phenol < Acetic acid < Chloroacetic acid

(iv) Chloroacetic acid < Acetic acid < Phenol < Ethanol

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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.

Acidic strength depends on the stability of the conjugate base after losing H⁺. Here, the order is: Ethanol (weakest) < Phenol < Acetic acid < Chloroacetic acid (strongest). The correct option is (iii).

Why this order? The concept of acidic strength

Acidity is all about who wants to give away a proton (H⁺) the most. The stronger the acid, the more willingly it donates H⁺, and the more stable its conjugate base (the anion left behind). So to compare acidic strengths, we compare the stability of the anions: more stable anion → stronger acid.

Three key factors come into play here:

Inductive effect – electron-withdrawing groups (like –Cl) pull electron density away from the negative charge, stabilising the anion. Electron-donating groups (like –CH₂CH₃) do the opposite, destabilising it.
Resonance – if the negative charge can be delocalised over multiple atoms (especially oxygen atoms in a carboxylate group, or into an aromatic ring), the anion is much more stable.
Hybridisation – the more s-character in the orbital holding the lone pair, the closer the electrons are held to the nucleus, making the anion more stable.

Let’s apply these to the four compounds.

Step-by-step reasoning

1. Ethanol ( $\ce{C2H5OH}$ ) – the weakest acid

Ethanol is an alcohol. When it loses H⁺, the conjugate base is the ethoxide ion ( $\ce{C2H5O-}$ ). The negative charge is localised entirely on one oxygen atom. There is no resonance to spread it, and the ethyl group is weakly electron-donating (it pushes electrons toward the oxygen, making the negative charge less stable). So the ethoxide ion is quite unstable, meaning ethanol is a very weak acid — weaker than water itself (pKa ≈ 16). Among the four, ethanol is definitely the weakest.

2. Phenol ( $\ce{C6H5OH}$ ) – stronger than ethanol, weaker than carboxylic acids

Phenol looks like an alcohol, but the –OH group is attached to a benzene ring. When phenol loses H⁺, the phenoxide ion ( $\ce{C6H5O-}$ ) forms. Here, the negative charge on oxygen can be delocalised into the aromatic ring via resonance — the lone pair on oxygen interacts with the π-system, spreading the charge over the ortho and para positions of the ring. This resonance stabilisation makes phenoxide much more stable than ethoxide. So phenol (pKa ≈ 10) is a stronger acid than ethanol.

Watch out

A common mistake is to think that because phenol is an alcohol, it should be as weak as ethanol. But the aromatic ring changes everything — resonance stabilisation of the conjugate base is the key. Never ignore the effect of the attached group.

3. Acetic acid ( $\ce{CH3COOH}$ ) – a typical carboxylic acid

Acetic acid is a carboxylic acid. Its conjugate base, the acetate ion ( $\ce{CH3COO-}$ ), has the negative charge equally shared between two oxygen atoms through resonance (the carboxylate group is a classic example of resonance stabilisation). This delocalisation makes acetate very stable. In fact, the two oxygen atoms are equivalent, so the charge is spread over a larger volume. Acetic acid (pKa ≈ 4.76) is therefore a much stronger acid than phenol.

4. Chloroacetic acid ( $\ce{ClCH2COOH}$ ) – the strongest of the four

Chloroacetic acid is just acetic acid with one hydrogen on the α-carbon replaced by a chlorine atom. Chlorine is highly electronegative and exerts a strong electron-withdrawing inductive effect (through the sigma bonds). This pulls electron density away from the carboxylate group, further stabilising the already-resonance-stabilised conjugate base. The negative charge becomes even more dispersed, making chloroacetic acid (pKa ≈ 2.87) stronger than acetic acid.

Tip

The inductive effect of chlorine weakens with distance — here it's on the α-carbon, right next to the carboxyl group, so its effect is maximal. If the chlorine were further away (e.g., on the β-carbon), the effect would be smaller.

Putting it all together

From weakest to strongest acid:

Compound	Conjugate base	Stabilising factors	Relative strength
Ethanol	Ethoxide ion	None (localised charge, +I group)	Weakest
Phenol	Phenoxide ion	Resonance into ring	Intermediate
Acetic acid	Acetate ion	Resonance over two O atoms	Strong
Chloroacetic acid	Chloroacetate ion	Resonance + strong –I effect of Cl	Strongest

So the order is: Ethanol < Phenol < Acetic acid < Chloroacetic acid.

This matches option (iii).

✓Final answer

The correct order of increasing acidic strength is Ethanol < Phenol < Acetic acid < Chloroacetic acid, which corresponds to option (iii) .

Method: Inductive Effect & Resonance Stabilisation Analysis

This method compares acidic strength by analysing how well the conjugate base (after losing H⁺) is stabilised.

Step 1 — Identify the acidic proton in each compound

Ethanol ( $\text{CH}_3\text{CH}_2\text{OH}$ ): O–H bond
Phenol ( $\text{C}_6\text{H}_5\text{OH}$ ): O–H bond
Acetic acid ( $\text{CH}_3\text{COOH}$ ): O–H of carboxyl group
Chloroacetic acid ( $\text{ClCH}_2\text{COOH}$ ): O–H of carboxyl group

Step 2 — Compare conjugate base stability

Compound	Conjugate base	Stabilising factor	Relative strength
Ethanol	$\text{CH}_3\text{CH}_2\text{O}^-$	Only negative charge on oxygen	Weakest
Phenol	$\text{C}_6\text{H}_5\text{O}^-$	Negative charge delocalised into benzene ring (resonance)	Stronger than ethanol
Acetic acid	$\text{CH}_3\text{COO}^-$	Negative charge delocalised over two oxygen atoms (resonance)	Stronger than phenol
Chloroacetic acid	$\text{ClCH}_2\text{COO}^-$	Same resonance as acetic acid + electron-withdrawing Cl (inductive effect)	Strongest

Step 3 — Arrange in increasing order

Weakest acid → Strongest acid:

Ethanol < Phenol < Acetic acid < Chloroacetic acid

Step 4 — Match with options

This corresponds to option (C).

Final Answer: $\boxed{\text{C}}$

Here are the most common mistakes students make when ranking acidic strength (especially for the given question) and how to avoid each.

Mistake 1: Confusing Boiling Point Trends with Acidic Strength

The Error: Students see "Phenol" and "Ethanol" and immediately think about hydrogen bonding. They remember that alcohols have higher boiling points than phenols (due to packing), and incorrectly apply this logic to acidity.
Why it’s wrong: Boiling point depends on intermolecular forces (H-bonding, van der Waals). Acidity depends on the stability of the conjugate base after losing $H^+$ . These are completely different concepts.
How to Avoid: When you see "acidic strength," immediately switch your mental framework to conjugate base stability. Ask: Which anion is most stable? (More stable anion = stronger acid).

Mistake 2: Forgetting the Inductive Effect of Chlorine

The Error: Students rank Acetic acid ( $CH_3COOH$ ) as stronger than Chloroacetic acid ( $ClCH_2COOH$ ) because they think "acid" is always stronger than "chloro" compounds.
Why it’s wrong: Chlorine is highly electronegative. It pulls electron density away from the O–H bond via the inductive effect ( $-I$ effect). This stabilizes the conjugate base (the carboxylate anion) by spreading the negative charge.
How to Avoid: Remember: Electron-withdrawing groups (EWG) like $-Cl$ , $-NO_2$ , $-F$ increase acidity. Electron-donating groups (EDG) like $-CH_3$ , $-OCH_3$ decrease acidity. So $ClCH_2COOH$ is stronger than $CH_3COOH$ .

Mistake 3: Misjudging Phenol vs. Ethanol

The Error: Students think Ethanol is more acidic than Phenol because Ethanol is a "stronger" alcohol or because they confuse it with basicity.
Why it’s wrong: Phenol is more acidic than Ethanol. The phenoxide ion (conjugate base of phenol) is stabilized by resonance — the negative charge is delocalized into the benzene ring. The ethoxide ion (conjugate base of ethanol) has no such resonance; the negative charge is localized on oxygen.
How to Avoid: Draw the resonance structures of phenoxide. Count how many atoms share the negative charge. More resonance = more stable = stronger acid.

Mistake 4: Ignoring the Order of Carboxylic Acids vs. Phenol

The Error: Students place Phenol above Acetic acid in strength.
Why it’s wrong: Carboxylic acids (like Acetic acid) are stronger acids than Phenol. The carboxylate anion ( $RCOO^-$ ) is stabilized by two equivalent resonance structures (the negative charge is shared equally between two oxygen atoms). Phenoxide has resonance, but the negative charge is on one oxygen and the ring — less effective than two oxygens.
How to Avoid: Remember the general order: Carboxylic acid > Phenol > Alcohol. This is a standard hierarchy for organic acids.

Mistake 5: Misreading the Question (Increasing vs. Decreasing)

The Error: The question asks for increasing acidic strength (weakest to strongest). Students often pick the option that lists strongest first.
Why it’s wrong: They see "Chloroacetic acid" at the end and think it's correct, but the order is reversed.
How to Avoid: Circle the word "increasing" in the question. Write a quick arrow: $\text{weakest} \rightarrow \text{strongest}$ . Then match the options.

The Correct Answer & Reasoning

Rank each species by conjugate-base stability:

Ethanol: Weakest (no resonance, no EWG).
Phenol: Stronger than ethanol (resonance in conjugate base).
Acetic acid: Stronger than phenol (two equivalent resonance structures in the carboxylate conjugate base).
Chloroacetic acid: Strongest (carboxylate resonance plus the $-I$ effect of Cl).

So the correct increasing order of acidity is:

Ethanol < Phenol < Acetic acid < Chloroacetic acid

That matches Option (C).

Final Answer: $\boxed{C}$

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

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markMCQ
$Q. Among the structural isomers of hexane, the isomer having highest boiling point is -$
(a) CH3-C(CH3)2-C2H5 (2,2-dimethylbutane)
(b) CH3-CH(CH3)-CH2-CH2-CH3 (2-methylpentane)
(c) CH3-CH(CH3)-CH(CH3)-CH3 (2,3-dimethylbutane)
(d) CH3-CH2-CH2-CH2-CH2-CH3 (n-hexane)
›Reveal solutionSolution
The least-branched (straight-chain) isomer, n-hexane, has the highest boiling point among the hexane isomers.

Boiling point of alkane isomers is governed mainly by van der Waals (London dispersion) forces, which depend on the surface area of contact between molecules.

A straight-chain molecule like n-hexane has an elongated shape, maximizing surface contact with neighbouring molecules, so intermolecular attractions (and hence boiling point) are strongest.

Branching makes a molecule more compact/spherical, reducing surface contact and weakening van der Waals forces, which lowers the boiling point.

Ordering by branching: n-hexane (0 branches) > 2-methylpentane (1 branch) > 2,3-dimethylbutane / 2,2-dimethylbutane (2 branches, most compact, lowest bp).

✓Final answer
$CH_3CH_2CH_2CH_2CH_2CH_3$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 mark
$Q. Why does 2,2-dimethylpropane have lower boiling point than its isomer n-pentane?$
›Reveal solutionSolution

More branching → smaller surface area of intermolecular contact → weaker van der Waals forces → lower boiling point, even though both compounds are isomers of C5H12.

2,2-Dimethylpropane (neopentane) and n-pentane are structural isomers, both C5H12, so they have identical molecular formula and molar mass. The difference in boiling point arises purely from molecular shape.

n-Pentane is a straight, extended chain. Such elongated molecules can lie close alongside each other over a large surface area, maximising the induced-dipole (van der Waals/London dispersion) attractions between them.

2,2-Dimethylpropane, in contrast, is highly branched and adopts a compact, nearly spherical shape. A sphere has the smallest possible surface area for a given volume, so neighbouring molecules can only touch at a few points — this drastically reduces the total van der Waals contact area and hence the strength of intermolecular attraction.

Weaker intermolecular forces mean less thermal energy is needed to separate the molecules into the gas phase, so 2,2-dimethylpropane boils at a lower temperature (9.5 °C) than n-pentane (36 °C).

✓Final answer
$Because branching makes 2,2-dimethylpropane more compact/spherical (less surface contact, weaker van der Waals forces) than the elongated straight-chain n-pentane, it has a lower boiling point despite being an isomer with the same molecular formula.$

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Q.The correct order of increasing acidic strength is _____________.

Why this order? The concept of acidic strength

Step-by-step reasoning

Putting it all together

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