Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain.

NCERT Exemplar · Q26

Q. Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain.

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Concept understanding — Steam Distillation Volatility

Steam Distillation Volatility – From Intuition to Precision

Imagine you have a pot of water boiling on the stove. Now imagine you drop a few drops of a fragrant oil — say, clove oil — into the water. The oil doesn't dissolve; it floats as a separate layer. Yet, as the water boils, you smell the clove oil strongly in the steam. How did that oil, which boils at a much higher temperature than water, get carried into the vapour?

That is the core puzzle that steam distillation volatility explains.

The Intuition: Two Liquids That Don't Mix

When two immiscible liquids (like water and oil) are heated together, they do not behave like a single liquid. Each liquid exerts its own vapour pressure independently, as if the other weren't there. The total vapour pressure above the mixture is simply the sum of the two individual vapour pressures.

Note

This is completely different from a solution of two miscible liquids (like ethanol and water), where the vapour pressure of each is lowered by the presence of the other (Raoult's law). For immiscible liquids, each acts alone.

Now, boiling occurs when the total vapour pressure equals the surrounding atmospheric pressure. Because the two vapour pressures add up, the mixture reaches atmospheric pressure at a temperature lower than the boiling point of either pure liquid.

That is the key: the mixture boils at a temperature below 100°C (if water is one component) — often well below the boiling point of the organic compound. The organic compound, which would normally require a much higher temperature to boil, now gets carried over in the steam at this lower temperature.

The Precise Statement

P_{\text{total}} = P_{\text{water}} + P_{\text{organic}} = P_{\text{atm}}

When $P_{\text{total}}$ equals atmospheric pressure, the mixture boils. The temperature at which this happens is always less than the boiling point of pure water (100°C at 1 atm) and far less than the boiling point of the pure organic compound.

The vapour that distills over contains both water and the organic compound. The ratio of the masses of the two components in the distillate is given by:

\frac{m_{\text{organic}}}{m_{\text{water}}} = \frac{P_{\text{organic}} \times M_{\text{organic}}}{P_{\text{water}} \times M_{\text{water}}}

where $P$ is the vapour pressure of each component at the distillation temperature, and $M$ is the molar mass.

Why This Matters for Exams

Steam distillation volatility is not a property of the compound alone — it is a property of the mixture with water. A compound is "steam volatile" if it is immiscible with water and has a measurable vapour pressure at 100°C (or below).
The compound does not need to have a low boiling point. Many high-boiling natural oils (like eugenol from clove, boiling point ~254°C) are steam volatile because they have enough vapour pressure at ~99°C to be carried over.
The key condition: the compound must be immiscible with water. If it dissolves even slightly, the simple additive vapour pressure model breaks down. …

Why this formula?

Steam Distillation Volatility: Why the Formula Holds

Steam distillation is a technique used to separate immiscible liquids — typically an organic compound (like an essential oil) and water. The key idea is that the mixture boils when the sum of the vapor pressures equals the external pressure, even though each component's individual boiling point is higher.

The Core Formula

For a mixture of two immiscible liquids (A and water), the total vapor pressure at a given temperature is:

P_{\text{total}} = P_A^\circ + P_{\text{water}}^\circ

where $P_A^\circ$ and $P_{\text{water}}^\circ$ are the vapor pressures of the pure components at that temperature.

The mixture boils when:

P_{\text{total}} = P_{\text{atm}}

Why This Works — The Reasoning

1. Immiscibility → No Mutual Solubility

Since the two liquids do not mix, each exists as a pure phase (not a solution). There is no Raoult's law deviation — each liquid exerts its own pure vapor pressure independently.

In a solution, the vapor pressure of a component is lowered by the presence of the other (Raoult's law).
In an immiscible mixture, each liquid behaves as if the other is not there — they are separate layers.

2. Vapor Pressure Adds Independently

Because the liquids are immiscible, the vapor above the mixture contains molecules from both pure phases. The total pressure is simply the sum:

P_{\text{total}} = P_A^\circ + P_{\text{water}}^\circ

This is Dalton's law of partial pressures applied to two independent pure vapors.

3. Boiling Occurs When Total Pressure Equals Atmospheric Pressure

Boiling happens when the vapor pressure of the liquid equals the external pressure. Here, the "liquid" is the two-phase system. So:

P_A^\circ + P_{\text{water}}^\circ = P_{\text{atm}}

This temperature is lower than the boiling point of either pure component — because each contributes only part of the required pressure.

The Composition of the Distillate

The mole fraction of each component in the vapor (and hence in the distillate) is given by:

y_A = \frac{P_A^\circ}{P_{\text{total}}}, \quad y_{\text{water}} = \frac{P_{\text{water}}^\circ}{P_{\text{total}}}

Since the vapor is in equilibrium with the pure liquids, the mass ratio in the distillate is:

\frac{m_A}{m_{\text{water}}} = \frac{P_A^\circ \cdot M_A}{P_{\text{water}}^\circ \cdot M_{\text{water}}}

where $M_A$ and $M_{\text{water}}$ are molar masses. …

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Steam distillation exploits the volatility difference between isomers. o-Nitrophenol is more volatile than p-nitrophenol because intramolecular hydrogen bonding in the ortho isomer reduces its effective intermolecular forces, lowering its boiling point and increasing its vapour pressure.

The Concept: What Makes a Molecule Volatile?

Volatility is a measure of how readily a substance vaporises. At a given temperature, a more volatile compound has a higher vapour pressure and a lower boiling point. The key factor is the strength of intermolecular forces in the liquid phase — weaker forces mean molecules escape more easily.

For organic compounds, hydrogen bonding is the strongest intermolecular force. When molecules can form hydrogen bonds with each other (intermolecular H-bonding), they stick together tightly, requiring more energy (higher temperature) to boil. But if a molecule can form a hydrogen bond within itself (intramolecular H-bonding), it ties up its own polar groups, reducing its ability to bond with neighbours. This makes the liquid less cohesive and more volatile.

Step-by-Step Reasoning

1. Identify the structural difference

Both o-nitrophenol and p-nitrophenol have the formula $\ce{C6H4(NO2)(OH)}$ . The difference is the position of the nitro group ( $\ce{-NO2}$ ) relative to the hydroxyl group ( $\ce{-OH}$ ):

ortho (o-): $\ce{-NO2}$ and $\ce{-OH}$ are adjacent (positions 1 and 2).
para (p-): $\ce{-NO2}$ and $\ce{-OH}$ are opposite (positions 1 and 4).

This positional change dramatically alters the hydrogen bonding pattern.

2. Hydrogen bonding in p-nitrophenol

In p-nitrophenol, the $\ce{-OH}$ group and the $\ce{-NO2}$ group are far apart. The $\ce{-OH}$ hydrogen can only form hydrogen bonds with the oxygen of another molecule's $\ce{-OH}$ or $\ce{-NO2}$ group. This is intermolecular hydrogen bonding — it links many molecules together in a network.

Watch out

A common mistake is to think that the $\ce{-NO2}$ group in p-nitrophenol forms an intramolecular bond with $\ce{-OH}$ . It cannot — the groups are too far apart (para position). The bond must be intermolecular.

This strong intermolecular association means p-nitrophenol has a high boiling point (around $279\,^\circ\text{C}$ ) and low vapour pressure at ordinary temperatures. It is not very volatile.

3. Hydrogen bonding in o-nitrophenol

In o-nitrophenol, the $\ce{-OH}$ and $\ce{-NO2}$ groups are right next to each other. The hydrogen of $\ce{-OH}$ can bend around and form a hydrogen bond with the oxygen of the $\ce{-NO2}$ group on the same molecule. This is intramolecular hydrogen bonding, forming a stable six-membered ring:

\ce{O-H \cdots O=N-O}

This internal bond "uses up" the $\ce{-OH}$ hydrogen. The molecule now has a much weaker ability to form hydrogen bonds with its neighbours. The intermolecular forces are reduced to weaker dipole-dipole and London forces. …

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

TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$_3$
›Reveal solutionSolution
The nitration of benzene is an electrophilic aromatic substitution reaction where the active electrophilic species is the nitronium ion ( $\text{NO}_2^+$ ), generated from the reaction of nitric acid and sulfuric acid. The correct option is (C).

The nitration of benzene is a classic example of an electrophilic aromatic substitution reaction. In such reactions, the electron-rich benzene ring is attacked by an electron-deficient species, known as an electrophile. The role of the mixture of concentrated nitric acid ( $\text{HNO}_3$ ) and concentrated sulfuric acid ( $\text{H}_2\text{SO}_4$ ) is to generate this strong electrophile. Nitric acid itself is not a sufficiently strong electrophile to react with benzene under these conditions. Sulfuric acid, being a stronger acid, protonates nitric acid, leading to the formation of the highly reactive electrophile.

Here is the step-by-step generation of the active species:

Protonation of Nitric Acid: Concentrated sulfuric acid ( $\text{H}_2\text{SO}_4$ ) is a stronger acid than concentrated nitric acid ( $\text{HNO}_3$ ). Therefore, in their mixture, sulfuric acid acts as a proton donor, and nitric acid acts as a base, accepting a proton. The oxygen atom of the hydroxyl group in $\text{HNO}_3$ gets protonated.

$\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{NO}_3^+ + \text{HSO}_4^-$
The species $\text{H}_2\text{NO}_3^+$ is protonated nitric acid.

2. Formation of Nitronium Ion: The protonated nitric acid ( $\text{H}_2\text{NO}_3^+$ ) is unstable. It readily loses a molecule of water ( $\text{H}_2\text{O}$ ) to form the nitronium ion ( $\text{NO}_2^+$ ). This step is highly favorable because water is a stable leaving group, and the resulting nitronium ion is resonance-stabilized.
$\text{H}_2\text{NO}_3^+ \rightleftharpoons \text{NO}_2^+ + \text{H}_2\text{O}$
The $\text{NO}_2^+$ ion is the nitronium ion.

3. Overall Reaction for Electrophile Generation: Combining the above steps, the overall reaction for the generation of the electrophile is:

$\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-$ …
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$Q. Given below are two statements Statement - I: A mixture of aniline and chloroform can be separated by differential extraction Statement - II: Aniline can be separated from a mixture of aniline and water by steam distillation (A) Both statements I and II are correct (B) Statement I is correct but statement II is not correct (C) Statement I is not correct but statement II is correct (D) Both statements I and II are not correct$
›Reveal solutionSolution
The key idea is that differential extraction works only when the two substances have very different solubilities in the chosen solvent, while steam distillation works for compounds that are immiscible with water and volatile. Aniline and chloroform are miscible, so Statement I is false; aniline is immiscible with water and steam-volatile, so Statement II is true. Hence the correct option is (C).

Concept and Intuition

This question tests two common separation techniques in organic chemistry.

Differential extraction relies on partitioning a solute between two immiscible solvents. It works when the two substances you want to separate have very different solubilities in the extracting solvent. If they are both soluble in the same solvent, you cannot separate them this way.

Steam distillation is used to separate compounds that are immiscible with water but have appreciable vapor pressure at 100 °C. The mixture distills at a temperature below 100 °C, carrying the organic compound with the steam. Aniline is a classic example: it is only slightly soluble in water and boils at 184 °C, but it co-distills with steam.

Now let’s examine each statement.

Statement I: “A mixture of aniline and chloroform can be separated by differential extraction.”

Differential extraction requires two immiscible solvents (e.g., water and ether). You add the mixture to one solvent, then shake with the other; the components distribute based on their partition coefficients.

Aniline and chloroform are completely miscible with each other (both are organic liquids). If you try to extract aniline from chloroform using water, aniline is only slightly soluble in water, so very little moves into the water layer. Moreover, chloroform itself is somewhat soluble in water, and the two liquids do not form two distinct layers — they mix.

Therefore, you cannot set up a two-phase system where one component preferentially moves into a second solvent. Differential extraction fails when the two substances are mutually soluble.

Conclusion: Statement I is not correct.

Statement II: “Aniline can be separated from a mixture of aniline and water by steam distillation.” …
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TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$_3$
›Reveal solutionSolution
Highest boiling point $=\text{BiH}_3$ (largest size, strongest van der Waals forces); lowest $=\text{PH}_3$ (option C).

Boiling points of group-15 hydrides do not vary monotonically:

$\text{NH}_3$ boils at $\approx -33\,^{\circ}\text{C}$ — anomalously high because of intermolecular hydrogen bonding (small, highly electronegative N).

From $\text{PH}_3$ onward there is no significant hydrogen bonding, so boiling point rises with molecular size (increasing van der Waals / London dispersion forces): $\text{PH}_3\ (-88\,^{\circ}\text{C}) < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\ (\approx +17\,^{\circ}\text{C})$ . …
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TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
$Q. Identify the reagent which is used to distinguish primary, secondary and tertiary amines (A) p-Toluene sulphonyl chloride (B) p-Toluene benzoyl chloride (C) p-Amino sulphonic acid (D) p-nitro phenol$
›Reveal solutionSolution
Hinsberg's reagent (p-toluenesulphonyl chloride) reacts differently with 1°, 2°, and 3° amines, producing distinguishable solubility patterns in alkali. The answer is (A).

The key to distinguishing amines by class lies in their reactivity toward sulphonylation and the solubility of the resulting products. Primary, secondary, and tertiary amines differ in the number of hydrogen atoms attached to nitrogen, which controls both whether they can react with certain reagents and whether the products are acidic enough to dissolve in base.

Hinsberg's test exploits exactly this difference. When p-toluenesulphonyl chloride (tosyl chloride, TsCl) reacts with amines in the presence of aqueous alkali, each class behaves distinctly:

How each amine responds

Primary amines ( $\ce{RNH2}$ ) have two hydrogens on nitrogen. They react with tosyl chloride to form N-alkyl-p-toluenesulphonamide:

$\ce{RNH2 + TsCl ->[KOH] RNH-SO2-C6H4-CH3}$
This sulphonamide still has one N–H bond, making the hydrogen acidic (the electron-withdrawing sulphonyl group stabilizes the conjugate base). The product dissolves in aqueous KOH to form a clear solution. Acidification regenerates the insoluble sulphonamide as a precipitate.

Secondary amines ( $\ce{R2NH}$ ) have one hydrogen on nitrogen. They also react with tosyl chloride:

$\ce{R2NH + TsCl ->[KOH] R2N-SO2-C6H4-CH3}$
But now there is no N–H left, so the product is not acidic. It remains insoluble in aqueous KOH, forming an oily or solid precipitate that does not dissolve. …
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TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$K_a$
›Reveal solutionSolution
The acidity of phenols is primarily determined by the stability of their conjugate base (phenoxide ion). Electron-withdrawing groups like $-\text{NO}_2$ increase acidity, with the para-isomer being the most acidic due to strong resonance stabilization and the ortho-isomer being less acidic than para due to intramolecular hydrogen bonding. The decreasing order of p $K_a$ (which is the increasing order of acidity) is A > C > B > D.

The acidity of a phenol is determined by the stability of its conjugate base, the phenoxide ion. A more stable phenoxide ion means a stronger acid and thus a lower p $K_a$ value. Electron-withdrawing groups (EWGs) stabilize the phenoxide ion by dispersing the negative charge, thereby increasing acidity. Electron-donating groups (EDGs) destabilize it, decreasing acidity. The $-\text{NO}_2$ group is a strong electron-withdrawing group, exhibiting both a negative inductive effect (-I) and a negative resonance effect (-R).

Let's analyze each phenol:

Phenol (A)

This serves as our baseline. Its conjugate base, the phenoxide ion, is stabilized by resonance with the benzene ring, delocalizing the negative charge into the ortho and para positions.

meta-Nitrophenol (C)

The $-\text{NO}_2$ group is at the meta position.

Inductive Effect (-I): The nitro group exerts a strong electron-withdrawing inductive effect, which helps stabilize the phenoxide ion by pulling electron density through the sigma bonds. This increases acidity.

Resonance Effect (-R): At the meta position, the $-\text{NO}_2$ group cannot directly participate in resonance with the phenoxide oxygen. The negative charge on the oxygen cannot be delocalized onto the nitro group via resonance from the meta position. Therefore, the resonance effect is negligible here.

Overall, the acidity of meta-nitrophenol is enhanced primarily by the -I effect of the nitro group.

ortho-Nitrophenol (B)

The $-\text{NO}_2$ group is at the ortho position.

Inductive Effect (-I): The nitro group exerts a strong electron-withdrawing inductive effect, stabilizing the phenoxide ion.

Resonance Effect (-R): The nitro group also exerts a strong electron-withdrawing resonance effect. The negative charge on the phenoxide oxygen can be effectively delocalized onto the oxygen atoms of the nitro group through resonance, significantly stabilizing the conjugate base.

Intramolecular Hydrogen Bonding: A crucial factor here is the formation of intramolecular hydrogen bonding between the phenolic hydrogen and an oxygen atom of the nitro group in the neutral ortho-nitrophenol molecule. This hydrogen bonding stabilizes the neutral molecule, making it more difficult to remove the proton. This effect reduces the acidity compared to what would be expected from just the -I and -R effects.

para-Nitrophenol (D)

The $-\text{NO}_2$ group is at the para position.

Inductive Effect (-I): The nitro group exerts a strong electron-withdrawing inductive effect, stabilizing the phenoxide ion.

Resonance Effect (-R): The nitro group exerts a very strong electron-withdrawing resonance effect. The negative charge on the phenoxide oxygen can be extensively delocalized onto the oxygen atoms of the nitro group through resonance. This is the most effective stabilization due to resonance among the nitrophenols.

No Intramolecular Hydrogen Bonding: Due to the distance between the hydroxyl and nitro groups, intramolecular hydrogen bonding is not possible in para-nitrophenol.

Overall, para-nitrophenol is the most acidic because of the combined strong -I and -R effects, with no counteracting intramolecular hydrogen bonding.

Comparing Acidity

Based on the analysis: …
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Q.Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain.

Concept understanding — Steam Distillation Volatility

Steam Distillation Volatility – From Intuition to Precision

The Intuition: Two Liquids That Don't Mix

The Precise Statement

Why This Matters for Exams

Why this formula?

Steam Distillation Volatility: Why the Formula Holds

The Core Formula

Why This Works — The Reasoning

1. Immiscibility → No Mutual Solubility

2. Vapor Pressure Adds Independently

3. Boiling Occurs When Total Pressure Equals Atmospheric Pressure

The Composition of the Distillate

The Concept: What Makes a Molecule Volatile?

Step-by-Step Reasoning

Method: Intermolecular Force Analysis (Hydrogen Bonding & Volatility)

Steps

Mistake 1: Guessing based on boiling point alone

Mistake 2: Confusing volatility with solubility

Mistake 3: Forgetting the role of intermolecular forces

Mistake 4: Not linking to steam distillation principle

How each amine responds

Comparing Acidity

Q.Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain.

The Concept: What Makes a Molecule Volatile?

Step-by-Step Reasoning

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