Q.(i) Complete the following reaction and suggest a suitable mechanism for the reaction :
CH3CH2OHH+, 443 K
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🔒 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.
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
Ptotal=Pwater+Porganic=Patm
When Ptotal 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:
mwatermorganic=Pwater×MwaterPorganic×Morganic
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:
Ptotal=PA∘+Pwater∘
where PA∘ and Pwater∘ are the vapor pressures of the pure components at that temperature.
The mixture boils when:
Ptotal=Patm
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:
Ptotal=PA∘+Pwater∘
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:
PA∘+Pwater∘=Patm
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:
yA=PtotalPA∘,ywater=PtotalPwater∘
Since the vapor is in equilibrium with the pure liquids, the mass ratio in the distillate is:
mwatermA=Pwater∘⋅MwaterPA∘⋅MA
where MA and Mwater are molar masses. …
Ethanol heated with acid at 443 K undergoes acid-catalysed dehydration to ethene (an E1 pathway via a carbocation); ortho-nitrophenol is steam volatile because of intramolecular H-bonding, while para forms intermolecular H-bonds. …
(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.
CH3CH2OHconc. H2SO4443 KCH2=CH2+H2O
Mechanism (E1):
- Protonation of the −OH: CH3CH2OH+H+→CH3CH2−O+H2.
- Loss of water gives the ethyl carbocation: CH3C+H2+H2O.
- Loss of a β-proton from the carbocation forms the double bond: CH2=CH2+H+ (catalyst regenerated). …
- AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQQ.At 290 K, a vessel (I) contains equal moles of three liquids (A, B, C). The boiling points of A, B and C are 350, 373 and 308 K respectively. Vessel (I) is heated to 300 K and vapours were collected into vessel (II). Identify the correct statements. (Assume vessel (I) contains liquids and vapours and vessel (II) contains only vapours) I. Vessel – I is rich in liquid B II. Vessel – II is rich in vapour of C III. The vapour pressures of A, B, C in Vessel (I) at 290 K follows the order C>A>B (A) I, II, III (B) I, II only (C) I, III only (D) II, III only
›Reveal solutionSolution
All three statements are consistent with basic vapour-pressure/boiling-point reasoning: lower boiling point means higher vapour pressure and preferential vaporisation.
Concept and Intuition
Boiling point is the temperature at which a liquid's vapour pressure equals atmospheric pressure. A liquid with a lower boiling point therefore has a higher vapour pressure at any common (lower) temperature — it is more volatile. When a mixture of liquids is heated below all their boiling points, the vapour that escapes is richer in the most volatile component, and the liquid left behind becomes richer in the least volatile one.
Step-by-Step Solution
- Rank boiling points: C=308 K (lowest), A=350 K, B=373 K (highest).
- Since lower boiling point = higher vapour pressure at the same temperature, vapour pressure order at 290 K is C>A>B — confirms Statement III.
- C, being most volatile, evaporates preferentially into vessel II, so the collected vapour is richest in C — confirms Statement II. …
- AP EAPCET 2021Set ap-2021-09-07-FN1 markMCQQ.o-nitrophenol & p-nitrophenol are separated by ________ (A) Crystallization (B) Fractional distillation (C) Vaporization (D) Steam distillation
›Reveal solutionSolution
o- and p-nitrophenol are separated by steam distillation because intramolecular H-bonding makes the ortho isomer steam-volatile while intermolecular H-bonding keeps the para isomer non-volatile.
Concept and Intuition
In o-nitrophenol, the −OH and −NO2 groups are close enough (on adjacent carbons of the benzene ring) to form an intramolecular hydrogen bond, creating a stable six-membered chelate ring. This internally satisfies the hydrogen-bonding capacity of the molecule, so o-nitrophenol molecules don't hydrogen-bond to each other as strongly -- it behaves as a discrete, relatively low-boiling, steam-volatile molecule. In p-nitrophenol, the −OH and −NO2 groups are on opposite ends of the ring and cannot reach each other, so instead they hydrogen-bond between different molecules (intermolecular H-bonding), building up an associated, higher-boiling, non-volatile solid. This difference in volatility (not solubility or crystallizability alone) is exploited directly by steam distillation.
Step-by-Step Solution
- Identify the structural difference: ortho-substitution allows intramolecular H-bonding; para-substitution forces intermolecular H-bonding.
- Relate bonding type to volatility: intramolecular H-bonding lowers boiling point/raises vapour pressure (steam-volatile); intermolecular H-bonding (association) raises boiling point and suppresses volatility. …
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