(i) Complete the following reaction and suggest a suitable mechanism for the reaction : $CH_3CH_2OH \xrightarrow{H^+,\ 443\ K}$ (ii) Why ortho-Nitrophenol is steam volatile while para-Nitrophenol is less volatile ?

(i) Complete the following reaction and suggest a suitable mechanism for the reaction : CH₃CH₂OH xrightarrowH^

Question 135 of 135

CH_3CH_2OH \xrightarrow{H^+,\ 443\ K}

(ii) Why ortho-Nitrophenol is steam volatile while para-Nitrophenol is less volatile ?

Bihar BsebCBSE Class XII Board 2018Subjective· 3mImportance★★★★★

100% · 135/135 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

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

Concept understanding — 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. …

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

(i) Ethanol dehydrates to ethene by an E1 mechanism. (ii) Intramolecular H-bonding makes o-nitrophenol steam volatile; intermolecular H-bonding makes p-nitrophenol less volatile.

Concept. Acid-catalysed dehydration of alcohols and H-bonding in phenols (CBSE Class-12 alcohols-phenols-and-ethers).

(i) Reaction + mechanism.

$CH_3CH_2OH \xrightarrow[443\ \text{K}]{\text{conc. } H_2SO_4} CH_2=CH_2 + H_2O$

Mechanism (E1):

Protonation of the $-OH$ : $CH_3CH_2OH + H^+ \rightarrow CH_3CH_2\!-\!\overset{+}{O}H_2$ .
Loss of water gives the ethyl carbocation: $CH_3\overset{+}{C}H_2 + H_2O$ .
Loss of a $\beta$ -proton from the carbocation forms the double bond: $CH_2=CH_2 + H^+$ (catalyst regenerated). …

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

←Previous questionQ8 — (a) How do you convert the following: (i) Phenol to Anisole (ii) Ethan…→Next chapterCh 12Aldehydes, Ketones and Carboxylic Acids

Q.(i) Complete the following reaction and suggest a suitable mechanism for the reaction : CH3CH2OH→H+, 443 KCH_3CH_2OH \xrightarrow{H^+,\ 443\ K}CH3​CH2​OHH+, 443 K​

🎓Unlock everything free for 14 days

Q.(i) Complete the following reaction and suggest a suitable mechanism for the reaction :
$CH_3CH_2OH \xrightarrow{H^+,\ 443\ K}$

Unlock everything free for 14 days