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.

Showing the 12 most recent of 14 on this concept.

AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$n-C_4H_9OH$
›Reveal solutionSolution
This tests the classic boiling-point trend among an alcohol, a 1° amine, a 2° amine, and a 3° amine of comparable molecular weight, driven by hydrogen-bonding capacity. The match is A-III, B-IV, C-II, D-I.

Concept and Intuition

For molecules of similar size, boiling point tracks how strongly molecules can hydrogen-bond to each other:

Alcohols (O–H) hydrogen-bond most strongly (O is more electronegative than N, and the O–H bond is highly polarized), so they have the highest boiling points among comparably-sized compounds.

Primary amines have two N–H bonds per molecule available for intermolecular hydrogen bonding — next highest.

Secondary amines have only one N–H bond — weaker hydrogen bonding, lower boiling point than primary amines.

Tertiary amines have no N–H bond at all (nitrogen's lone pair can still accept a hydrogen bond from something else, but the molecule itself cannot donate one), so they rely mainly on weaker dipole–dipole and dispersion forces — lowest boiling point of the four.

Step-by-Step Solution

A. $n$ - $C_4H_9OH$ (n-butanol): a primary alcohol — strongest H-bonding → highest boiling point among the four, 390.3 K → list item III. A-III.

C. $n$ - $C_4H_9NH_2$ (n-butylamine): a primary amine, two N–H bonds → next highest, 350.8 K → list item II. C-II.

B. $(C_2H_5)_2NH$ (diethylamine): a secondary amine, one N–H bond → lower still, 329.3 K → list item IV. B-IV.

D. $C_2H_5N(CH_3)_2$ (N,N-dimethylethylamine): a tertiary amine, no N–H bond → lowest boiling point, 310.5 K → list item I. D-I.

Ranking confirms: $390.3 > 350.8 > 329.3 > 310.5$ , i.e. alcohol > 1° amine > 2° amine > 3° amine, exactly as the hydrogen-bonding argument predicts.

Common Mistakes

Ranking amines purely by molecular weight/size rather than by how many N–H bonds are available for hydrogen bonding.

Assuming a secondary amine boils higher than a primary amine of similar formula weight — it's actually lower, because a 2° amine has only one N–H donor versus two for a 1° amine.

✓Final answer
$The correct option is (C) — A-III, B-IV, C-II, D-I.$

ANSWER: C
AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$\underline{A}$
›Reveal solutionSolution
Boiling points of group 15 hydrides dip after ammonia (loss of H-bonding) then rise with increasing molar mass; highest is BiH3, lowest is PH3.

Concept and Intuition

$NH_3$ has strong intermolecular hydrogen bonding (N is small and highly electronegative), giving it an unusually high boiling point for its size. Once H-bonding is lost going to $PH_3$ , boiling point drops sharply because only weak van der Waals (London dispersion) forces operate. As you continue down the group ( $AsH_3\to SbH_3\to BiH_3$ ), molecular size and mass increase steadily, so van der Waals forces strengthen again and boiling point rises — eventually exceeding even $NH_3$ .

Step-by-Step Solution

Approximate boiling points: $NH_3\approx -33°C$ , $PH_3\approx -87.7°C$ , $AsH_3\approx -55°C$ , $SbH_3\approx -17°C$ , $BiH_3\approx +17°C$ .

Lowest of these is $PH_3$ (the H-bonding of $NH_3$ is gone, and molecular mass is still small).

Highest of these is $BiH_3$ (largest, heaviest molecule with strongest dispersion forces).

So A (highest) = $BiH_3$ , B (lowest) = $PH_3$ .

Common Mistakes

Assuming $NH_3$ must be the highest-bp hydride of the whole series just because of H-bonding — $BiH_3$ actually surpasses it due to its much greater size.

Picking $NH_3$ as the lowest by mistakenly thinking H-bonding lowers boiling point.

✓Final answer
$BiH_3, PH_3$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$Q. The correct order of boiling points of the compounds given below is A) Methoxy ethane B) Propan-1-ol C) Propanal D) Propanone (A) C > B > A > D (B) B > D > C > A (C) B > C > D > A (D) C > A > B > D$
›Reveal solutionSolution
Tests ranking boiling points by intermolecular forces: H-bonding alcohol > dipolar ketone > dipolar aldehyde > weakly-polar ether.

Concept and Intuition

For molecules of similar molar mass, boiling point is set by the strength of intermolecular forces. An –OH group enables strong hydrogen bonding (raising b.p. sharply above similarly-sized non-alcohols). A C=O group gives a fairly strong permanent dipole (ketones/aldehydes), but weaker than H-bonding. An ether has a weaker net dipole (bond dipoles partly oppose) and no H-bond donor, so it boils at the lowest temperature of the four functional classes here.

Step-by-Step Solution

B) Propan-1-ol, $\mathrm{CH_3CH_2CH_2OH}$ : extensive intermolecular H-bonding via $-\mathrm{OH}$ gives it the highest boiling point of the four.

D) Propanone (acetone), $\mathrm{CH_3COCH_3}$ : a symmetric ketone with a strong dipole from C=O but no H-bond donor — boils next highest.

C) Propanal, $\mathrm{CH_3CH_2CHO}$ : also has a polar C=O, but the aldehyde's dipole/packing gives it a slightly lower boiling point than the ketone of the same carbon count.

A) Methoxyethane, $\mathrm{CH_3-O-C_2H_5}$ : an ether — only weak dipole-dipole/van der Waals forces, no H-bonding — has by far the lowest boiling point.

Order (highest to lowest): B > D > C > A.

Common Mistakes

Assuming aldehydes always boil higher than ketones of the same size — in this size range the ketone (acetone) actually boils a little higher than the aldehyde (propanal).

Forgetting that ethers, despite having an oxygen atom, cannot hydrogen-bond with each other and so boil much lower than alcohols of similar mass.

✓Final answer
$The correct option is (B) — B > D > C > A.$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$Q. Observe the following substances. Ethanol, acetic acid, ethylamine, trimethylamine, salicylic acid, ethanal. In the above list, the number of substances with H-bonding is (A) 4 (B) 3 (C) 5 (D) 2$
›Reveal solutionSolution
Tests recognizing which functional groups (O–H, N–H) enable hydrogen bonding; 4 of the 6 substances qualify.

Concept and Intuition

Hydrogen bonding needs a hydrogen atom covalently bonded to a small, highly electronegative atom — O, N, or F — so that the H carries a strong partial positive charge able to interact with a lone pair on a neighbouring electronegative atom. A carbonyl oxygen (as in an aldehyde) or a nitrogen with no attached H (as in a fully substituted tertiary amine) cannot act as an H-bond donor themselves.

Step-by-Step Solution

Ethanol ( $\mathrm{C_2H_5OH}$ ): has an O–H group → capable of H-bonding.

Acetic acid ( $\mathrm{CH_3COOH}$ ): has a carboxylic O–H group → capable of H-bonding.

Ethylamine ( $\mathrm{C_2H_5NH_2}$ ): a primary amine with N–H bonds → capable of H-bonding.

Trimethylamine ( $\mathrm{N(CH_3)_3}$ ): a tertiary amine — nitrogen has no attached H, so it cannot donate a hydrogen bond → excluded.

Salicylic acid: has both a carboxylic O–H and a phenolic O–H → capable of H-bonding.

Ethanal ( $\mathrm{CH_3CHO}$ ): an aldehyde with no O–H or N–H bond → excluded.

Total qualifying substances: ethanol, acetic acid, ethylamine, salicylic acid = 4.

Common Mistakes

Assuming any amine can hydrogen-bond, without checking whether it actually has an N–H bond (tertiary amines don't).

Assuming a carbonyl compound like an aldehyde can hydrogen-bond just because it contains oxygen.

✓Final answer
$The correct option is (A) — 4.$

ANSWER: A
AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$Q. What is the correct boiling point order of the following haloalkanes? i) 2-chloro 2-methylpropane ii) 1-Cholobutane iii) 2-Chlorobutane (A) i > ii > iii (B) ii > iii > i (C) i < ii < iii (D) i > iii > ii$
›Reveal solutionSolution
Among isomeric C₄H₉Cl haloalkanes, the straight-chain isomer boils highest and the most branched (tertiary) isomer boils lowest, giving the order ii > iii > i.

Concept and Intuition

For a set of structural isomers with the same molecular formula, boiling point is governed mainly by the strength of intermolecular van der Waals (London dispersion) forces, which depend on the surface area available for molecules to contact each other. A straight (unbranched) chain packs closely and has more surface contact, giving stronger dispersion forces and a higher boiling point. Branching makes the molecule more compact/spherical, reducing surface area and intermolecular contact, and hence lowering the boiling point.

Step-by-Step Solution

Identify the three isomers, all of formula $C_4H_9Cl$ : (i) 2-chloro-2-methylpropane (tert-butyl chloride) — most branched, chlorine on a tertiary carbon; (ii) 1-chlorobutane — straight (unbranched) chain, chlorine on a primary carbon; (iii) 2-chlorobutane — chlorine on a secondary carbon, slightly branched.

Rank by branching (least to most): (ii) unbranched < (iii) one branch point < (i) most branched (quaternary-like carbon skeleton around the C–Cl carbon).

Since boiling point decreases with increasing branching for isomers, the order (highest bp to lowest) is: (ii) > (iii) > (i).

This matches known experimental values: 1-chlorobutane (~78.5°C) > 2-chlorobutane (~68°C) > tert-butyl chloride (~51°C).

Common Mistakes

Assuming a more "substituted" (tertiary) halide would have a higher boiling point by analogy with stability of carbocations — boiling point trends for branching go the opposite way from carbocation stability trends.

✓Final answer
$The correct option is (B) — ii > iii > i.$

ANSWER: B
AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$Q. Para-nitro phenol has higher boiling point than ortho-nitrophenol. This is due to (A) The presence of intermolecular hydrogen bonding between Para-nitro phenol molecules (B) The presence of intramolecular hydrogen bonding in Para-nitro phenol molecules (C) The presence of intermolecular hydrogen bonding between ortho-nitro phenol molecules (D) The absence of intramolecular hydrogen bonding between ortho-nitro phenol molecules$
›Reveal solutionSolution
Para-nitrophenol boils higher than ortho because ortho forms intramolecular H-bonding (chelation) while para is forced into intermolecular H-bonding, which needs more energy to break.

Concept and Intuition

Boiling point depends on the strength of the forces holding molecules together in the liquid. Ortho-nitrophenol's $\mathrm{-OH}$ and $\mathrm{-NO_2}$ are adjacent, so they hydrogen-bond to each other within the same molecule (a six-membered ring "chelate"). This uses up the $\mathrm{-OH}$ 's hydrogen-bonding capacity internally, so ortho-nitrophenol molecules interact with each other only weakly (via van der Waals forces) — it boils low and is even steam-volatile. In para-nitrophenol the groups are on opposite ends of the ring and cannot reach each other, so the $\mathrm{-OH}$ of one molecule instead hydrogen-bonds to the $\mathrm{-NO_2}$ / $\mathrm{-OH}$ of a neighbouring molecule — building an extended, harder-to-break intermolecular network, hence a higher boiling point.

Step-by-Step Solution

Identify the substitution pattern: ortho places $\mathrm{-OH}$ and $\mathrm{-NO_2}$ next to each other; para places them across the ring.

Ortho: intramolecular H-bond forms a stable ring — no need for the molecule to H-bond with neighbours.

Para: no intramolecular H-bond is geometrically possible, so $\mathrm{-OH}$ groups H-bond between different molecules.

Intermolecular H-bonds must all be broken simultaneously to vaporise the liquid → higher boiling point for para.

Hence para-nitrophenol has the higher boiling point due to intermolecular H-bonding among its own molecules.

Common Mistakes

Assuming intramolecular H-bonding always raises boiling point — it's the opposite; it lowers it by removing the drive to associate with other molecules.

Mixing up which isomer has which type of bonding.

✓Final answer
$The correct option is (A) — The presence of intermolecular hydrogen bonding between para-nitrophenol molecules.$

ANSWER: A
AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$Q. Arrange the following in increasing order of their boiling points N-Ethylethanamine - I Butanamine - II N,N-dimethylethanamine - III (A) III > II > I (B) III > I > II (C) II > III > I (D) II > I > III$
›Reveal solutionSolution
Boiling point of amines of the same formula falls as 1° > 2° > 3°, because more N–H bonds mean stronger intermolecular hydrogen bonding.

Concept and Intuition

All three compounds share the molecular formula $C_4H_{11}N$ , so molecular weight/dispersion forces are essentially comparable; the boiling-point differences are governed by hydrogen bonding capacity. A primary amine ( $-NH_2$ ) has two N–H bonds and can form the most extensive intermolecular hydrogen-bond network, giving it the highest boiling point among the three classes for a given carbon count. A secondary amine ( $-NH-$ ) has only one N–H bond, so it hydrogen-bonds less extensively (lower bp than the primary isomer). A tertiary amine has no N–H bond at all, so it cannot hydrogen-bond with itself, relying only on weaker dipole–dipole and dispersion forces, giving it the lowest boiling point.

Step-by-Step Solution

Classify each compound: Butanamine (II) = $\mathrm{CH_3CH_2CH_2CH_2NH_2}$ , a primary amine (2 N–H bonds).

N-Ethylethanamine (I) = diethylamine, $\mathrm{(C_2H_5)_2NH}$ , a secondary amine (1 N–H bond).

N,N-Dimethylethanamine (III) = $\mathrm{CH_3CH_2N(CH_3)_2}$ , a tertiary amine (0 N–H bonds).

Ranking by H-bonding strength (and hence boiling point): primary > secondary > tertiary, i.e. II > I > III.

Checking each option against this true relative ranking (II's bp highest, then I, then III lowest), only option (D) has both pairwise relations (II > I and I > III) correctly stated.

Common Mistakes

Assuming more substitution (like more alkyl branching) always raises boiling point — for amines, hydrogen-bonding capacity (number of N–H bonds) dominates over simple branching/molecular-weight effects.

Mixing up which Roman numeral corresponds to which amine class.

✓Final answer
$The correct option is (D) — II > I > III.$

ANSWER: D
AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$Q. The correct order of boiling points of following molecules is$
(i) n – Hexane
(ii) 2-methylpentane
(iii) 2,3 – dimethylbutane (A) i > ii > iii (B) iii > ii > i (C) iii > i > ii (D) i > iii > ii
›Reveal solutionSolution
Boiling point falls as branching increases among isomeric alkanes, so n-hexane > 2-methylpentane > 2,3-dimethylbutane.

Concept and Intuition

All three compounds are isomers of hexane ( $C_6H_{14}$ ), so they have identical molecular formula and hence similar total van der Waals attraction potential — but the shape of the molecule matters. A straight, extended chain (n-hexane) has more surface-to-surface contact with neighbouring molecules, maximizing van der Waals (London dispersion) forces. Branching makes the molecule more compact and spherical, reducing effective surface contact and hence the strength of intermolecular attractions, which lowers the boiling point.

Step-by-Step Solution

n-Hexane: a straight, unbranched 6-carbon chain — largest surface area for intermolecular contact — highest boiling point among the three.

2-Methylpentane: one methyl branch — somewhat more compact than n-hexane — intermediate boiling point.

2,3-Dimethylbutane: two methyl branches, the most compact/spherical of the three — smallest surface area for contact — lowest boiling point.

Order (decreasing boiling point): n-hexane (i) > 2-methylpentane (ii) > 2,3-dimethylbutane (iii).

Common Mistakes

Assuming boiling point depends only on molecular weight (all three isomers have the same molecular weight, so shape/branching is the deciding factor here).

Reversing the trend and thinking more branching increases boiling point (branching actually decreases it, unlike its effect on some other properties like octane number).

✓Final answer
$The correct option is (A) — i > ii > iii.$

ANSWER: A
AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$Q. At 298 K, if the vapour pressure of pure liquids toluene, benzene, chloroform and dichloromethane are 60, 160, 200 and 415 torr respectively. Then which liquid is having high boiling point? (A) Toluene (B) Benzene (C) Chloroform (D) Dichloromethane$
›Reveal solutionSolution
Boiling point and vapour pressure (at fixed T) are inversely related; toluene's lowest vapour pressure (60 torr) means it has the highest boiling point.

Concept and Intuition

Vapour pressure measures how readily a liquid's molecules escape into the gas phase at a given temperature — it is a direct measure of volatility. Boiling point is the temperature at which vapour pressure equals atmospheric pressure. A liquid that already has a low vapour pressure at a reference temperature needs to be heated more to reach atmospheric pressure, so lower vapour pressure at a fixed T corresponds to a higher boiling point.

Step-by-Step Solution

List the vapour pressures at 298 K: toluene 60 torr, benzene 160 torr, chloroform 200 torr, dichloromethane 415 torr.

Rank from lowest to highest vapour pressure: toluene < benzene < chloroform < dichloromethane.

Since boiling point ranks inversely to vapour pressure (at the same reference temperature), the ranking of boiling points (highest to lowest) is: toluene > benzene > chloroform > dichloromethane.

The liquid with the highest boiling point is therefore toluene.

Common Mistakes

Assuming higher vapour pressure means higher boiling point (it's the opposite — high vapour pressure means the substance evaporates easily, i.e., low boiling point).

Not recognizing that this comparison is valid because all values are given at the same temperature (298 K).

✓Final answer
$The correct option is (A) — Toluene.$

ANSWER: A
AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$\mathrm{NH_3}$
›Reveal solutionSolution
Boiling points of these hydrides are governed mainly by hydrogen bonding strength/extent, giving the increasing order $HCl<NH_3<HF<H_2O$ .

Concept and Intuition

Among simple hydrides, boiling point is strongly influenced by hydrogen bonding, which occurs when H is bonded to a small, highly electronegative atom (N, O, F). $HCl$ 's Cl is not electronegative/small enough to hydrogen bond significantly, so it relies only on weaker dipole-dipole/dispersion forces and has the lowest boiling point among these four. Among the hydrogen-bonded species, $H_2O$ forms an extensive 3-D hydrogen-bonded network (2 lone pairs and 2 H atoms per molecule, ideal for a 3-D network) giving it the highest boiling point, while $HF$ and $NH_3$ form more limited (chain-like or less networked) hydrogen bonding.

Step-by-Step Solution

$HCl$ : negligible hydrogen bonding (Cl is not electronegative/small enough) — lowest boiling point among the four ( $\approx-85^\circ C$ ).

$NH_3$ : hydrogen bonds via N, but only one lone pair per molecule to hydrogen bond with $\Rightarrow$ boiling point $\approx-33^\circ C$ .

$HF$ : strong hydrogen bonding via a highly electronegative F, but limited to one H and three lone pairs (only one bond forms per molecule in the chain) $\Rightarrow$ boiling point $\approx19.5^\circ C$ .

$H_2O$ : two H atoms and two lone pairs allow each molecule to participate in up to 4 hydrogen bonds, forming an extensive 3-D network $\Rightarrow$ highest boiling point, $100^\circ C$ .

Increasing order of boiling point: $HCl<NH_3<HF<H_2O$ .

Common Mistakes

Assuming $HF$ , having the strongest single hydrogen bond, must have the highest boiling point overall — but the extent of hydrogen-bond networking (as in water) matters more than single-bond strength.

Forgetting $HCl$ essentially lacks hydrogen bonding altogether and placing it above $NH_3$ .

✓Final answer
$HCl < NH_3 < HF < H_2O$

ANSWER: D
AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$\mathrm{CH_3CH(OH)CH_2CH_3}$
›Reveal solutionSolution
Among an alcohol, a ketone, an alkane, and an ether of comparable size, the alcohol has the highest boiling point because only it can hydrogen-bond between its own molecules.

Concept and Intuition

Boiling point depends on the strength of intermolecular forces that must be overcome to vaporise the liquid. Alcohols (-OH group) can form hydrogen bonds with each other, a strong, directional intermolecular force. Ketones and ethers only have permanent dipole-dipole interactions (no O-H or N-H to hydrogen-bond with each other), which are weaker than hydrogen bonding. Alkanes have only weak, non-polar van der Waals (London dispersion) forces, the weakest of all.

Step-by-Step Solution

Butan-2-ol: contains -OH, capable of strong intermolecular hydrogen bonding $\Rightarrow$ highest boiling point among these four.

Butan-2-one: a ketone, polar C=O but no H-bond donor $\Rightarrow$ moderate boiling point (dipole-dipole), lower than the alcohol.

Ethyl propyl ether: polar C-O-C but no H-bond donor either $\Rightarrow$ boiling point similar to or slightly below the ketone.

n-Butane: non-polar, only weak dispersion forces $\Rightarrow$ lowest boiling point (in fact a gas near room temperature).

Hence Butan-2-ol has the highest boiling point.

Common Mistakes

Assuming molecular weight alone determines boiling point, ignoring the type of intermolecular force present.

Forgetting that ethers, despite having an oxygen, cannot hydrogen-bond with themselves (no O-H bond) the way alcohols can.

✓Final answer
$The correct option is (A) — Butan-2-ol.$

ANSWER: A
AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$Q. Arrange the following in decreasing order of their boiling points$
(a) $CH_3CH_2CH_2CH_2OH$ (butan-1-ol)
(b) $CH_3CH_2CH_2CH_2NH_2$ (butan-1-amine, a primary amine)
(c) a tertiary amine, $(CH_2CH_3)$ chain with an N bearing two other alkyl branches (drawn as a small N with two branches, i.e. a trialkylamine)
(d) a secondary amine with an N-H, drawn as two ethyl-type chains joined through an N-H (a secondary amine) (A) a > b > d > c (B) a > c > d > b (C) b > c > d > a (D) c > a > b > d
›Reveal solutionSolution
Boiling point here tracks hydrogen-bonding ability: the alcohol (strongest H-bonding) is highest, then primary amine (two N–H), then secondary amine (one N–H), then tertiary amine (no N–H, weakest): a > b > d > c.

Concept and Intuition

For molecules of comparable molecular weight, boiling point is governed largely by the strength and extent of intermolecular hydrogen bonding. Oxygen is more electronegative than nitrogen, so O–H···O hydrogen bonds are stronger than N–H···N hydrogen bonds — alcohols therefore boil higher than amines of similar size. Among amines themselves, hydrogen bonding requires an N–H bond to donate; a primary amine has two N–H bonds (most extensive hydrogen-bonded network), a secondary amine has only one N–H bond (less association), and a tertiary amine has none (cannot hydrogen-bond to itself at all, only weaker dipole-dipole/van der Waals forces), giving it the lowest boiling point of the three.

Step-by-Step Solution

Butan-1-ol (a): $-\mathrm{OH}$ group, strongest hydrogen bonding of the four compounds → highest boiling point.

Butan-1-amine (b), a primary amine: two N–H bonds, extensive intermolecular hydrogen bonding → next highest.

The secondary amine (d): only one N–H bond, weaker/less extensive hydrogen bonding than a primary amine → next.

The tertiary amine (c): no N–H bond at all, cannot hydrogen bond with itself, relies only on weaker dipole-dipole and dispersion forces → lowest boiling point.

Overall order: a > b > d > c.

Common Mistakes

Ranking by molecular weight alone instead of hydrogen-bonding capacity.

Assuming all amines hydrogen-bond equally regardless of how many N–H bonds are present.

Placing the tertiary amine above the secondary amine by mistake.

✓Final answer
$The correct option is (A) — a > b > d > c.$

ANSWER: A

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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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