While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be

Exercises · 7.8

Q. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

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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 separates compounds based on volatility in steam. Ortho-nitrophenol is steam volatile due to intramolecular hydrogen bonding, while para-nitrophenol is not because of intermolecular hydrogen bonding.

The Concept: Why Steam Distillation Works

Steam distillation is a technique used to separate or purify organic compounds that are immiscible with water. The key principle is that a mixture of immiscible liquids boils at a temperature lower than the boiling point of either pure component. For a compound to be carried over by steam, it must have appreciable vapour pressure at $100^\circ\text{C}$ (the boiling point of water). This is what we mean by "steam volatile."

But the real question is: why are some isomers steam volatile while others are not? The answer lies in how their molecules interact with each other — specifically, the type of hydrogen bonding present.

Step-by-Step Reasoning

1. Identify the isomers and their structures

Ortho-nitrophenol ( $o$ -nitrophenol) has the nitro group ( $-\text{NO}_2$ ) and the hydroxyl group ( $-\text{OH}$ ) adjacent to each other on the benzene ring. Para-nitrophenol ( $p$ -nitrophenol) has these groups opposite each other.

2. Understand the hydrogen bonding patterns

In ortho-nitrophenol, the $-\text{OH}$ and $-\text{NO}_2$ groups are close enough to form an intramolecular hydrogen bond — a hydrogen bond within the same molecule. The hydrogen of the $-\text{OH}$ group bonds with an oxygen of the $-\text{NO}_2$ group, creating a stable six-membered ring.

In para-nitrophenol, the groups are far apart. Intramolecular hydrogen bonding is impossible. Instead, the $-\text{OH}$ group of one molecule forms intermolecular hydrogen bonds with the $-\text{NO}_2$ group of another molecule. This links many molecules together in a network.

Watch out

A common mistake is to think that hydrogen bonding always increases boiling point. While that's true for intermolecular hydrogen bonding, intramolecular hydrogen bonding actually decreases boiling point because it prevents molecules from associating with each other.

3. Connect hydrogen bonding to volatility

Intermolecular hydrogen bonding (as in $p$ -nitrophenol) creates strong attractive forces between molecules. These forces must be overcome to vaporise the compound, which requires a high boiling point. At $100^\circ\text{C}$ , $p$ -nitrophenol has negligible vapour pressure — it does not volatilise with steam. …

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Here are the common mistakes students make on this Steam Distillation question, along with the correct reasoning and how to avoid each.

The Correct Answer First

Isomer: ortho-nitrophenol ( $o$ -nitrophenol).

Reason: $o$ -nitrophenol forms an intramolecular hydrogen bond (within the same molecule). This "locks" the $-$ OH group, preventing it from forming strong intermolecular hydrogen bonds with water. As a result, it is less soluble in water and has a higher vapor pressure, making it steam volatile.

$p$ -nitrophenol, on the other hand, forms intermolecular hydrogen bonds with other $p$ -nitrophenol molecules and with water. This makes it more water-soluble and less volatile.

Common Mistake #1: Confusing the Isomer (Saying para is volatile)

The Mistake: Students often assume that because $p$ -nitrophenol has a lower melting point or is more "symmetric," it should be more volatile.

Why it’s wrong: Volatility in steam distillation depends on vapor pressure and resistance to dissolving in water. $p$ -nitrophenol’s intermolecular H-bonding makes it dissolve in steam (like a solid dissolving), reducing its effective vapor pressure.

How to Avoid:

Remember the key phrase: "Intramolecular H-bonding = less water interaction = more volatile."
Draw the structures. $o$ -nitrophenol has the $-$ OH and $-$ NO $_2$ groups close together (ortho position), allowing a chelate ring (internal bond). $p$ -nitrophenol has them far apart, forcing external bonds.

Common Mistake #2: Confusing the Reason (Intermolecular vs Intramolecular)

The Mistake: Students write "ortho has intermolecular H-bonding" or "para has intramolecular H-bonding."

Why it’s wrong: This is the exact opposite of the truth.

How to Avoid:

Use a memory trick: "Intra = inside (same molecule)" — ortho groups are close, so they bond inside. "Inter = between (different molecules)" — para groups are far apart, so they bond with neighbors.
Draw the H-bond: In $o$ -nitrophenol, draw a curved arrow from the H of $-$ OH to the O of $-$ NO $_2$ (within the same ring). In $p$ -nitrophenol, draw a dotted line from the H of one molecule to the O of another.

Common Mistake #3: Ignoring the Role of Water (Solubility)

The Mistake: Students only talk about "boiling point" or "melting point" without mentioning interaction with steam.

Why it’s wrong: Steam distillation is about co-distillation with water. If a compound dissolves easily in water (like $p$ -nitrophenol), it stays in the liquid phase and doesn't vaporize as readily.

How to Avoid:

Always connect H-bonding to water solubility: Intermolecular H-bonding with water $\rightarrow$ high solubility $\rightarrow$ low volatility in steam. …

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Q.While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

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: Why Steam Distillation Works

Step-by-Step Reasoning

Method: Steam Distillation Volatility Analysis

Step 1: Identify the isomers

Step 2: Analyze hydrogen bonding

Step 3: Connect to volatility in steam

The Correct Answer First

Common Mistake #1: Confusing the Isomer (Saying para is volatile)

Common Mistake #2: Confusing the Reason (Intermolecular vs Intramolecular)

Common Mistake #3: Ignoring the Role of Water (Solubility)

Q.While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

The Concept: Why Steam Distillation Works

Step-by-Step Reasoning

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