Question:
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
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Ortho-substituted aromatic compounds are often steam volatile due to weaker intermolecular interactions.
Updated On: Feb 4, 2026
- Simple distillation
- Sublimation
- Fractional distillation
- Steam distillation
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The Correct Option is D
Solution and Explanation
Concept:
Electrophilic substitution reactions on benzene often give mixtures of positional isomers.
The choice of separation technique depends on physical properties such as:
Volatility
Solubility in water
Difference in boiling points
Step 1: Identify compound \(A\) Benzene reacts with bromine in presence of \(\mathrm{FeBr_3}\) via electrophilic aromatic substitution: \[ \text{Benzene} \xrightarrow{\mathrm{Br_2/FeBr_3}} \text{Bromobenzene } (A) \]
Step 2: Nitration of bromobenzene Bromobenzene undergoes nitration with concentrated \(\mathrm{HNO_3/H_2SO_4}\). Since bromine is an ortho–para directing group, nitration gives:
Ortho-nitrobromobenzene (B)
Para-nitrobromobenzene (C) Thus, the products \(B\) and \(C\) are positional isomers.
Step 3: Compare physical properties of B and C
Ortho-nitrobromobenzene is {steam volatile}
Para-nitrobromobenzene is {less volatile} due to better molecular packing and higher melting point This difference in volatility makes steam distillation an effective separation method.
Step 4: Eliminate other options
Simple distillation: Boiling points are too close for efficient separation.
Fractional distillation: Not ideal for separating solids/liquids with similar boiling points and differing steam volatility.
Sublimation: Neither compound readily sublimes. Final Answer: \[ \boxed{\text{Steam distillation}} \]
Volatility
Solubility in water
Difference in boiling points
Step 1: Identify compound \(A\) Benzene reacts with bromine in presence of \(\mathrm{FeBr_3}\) via electrophilic aromatic substitution: \[ \text{Benzene} \xrightarrow{\mathrm{Br_2/FeBr_3}} \text{Bromobenzene } (A) \]
Step 2: Nitration of bromobenzene Bromobenzene undergoes nitration with concentrated \(\mathrm{HNO_3/H_2SO_4}\). Since bromine is an ortho–para directing group, nitration gives:
Ortho-nitrobromobenzene (B)
Para-nitrobromobenzene (C) Thus, the products \(B\) and \(C\) are positional isomers.
Step 3: Compare physical properties of B and C
Ortho-nitrobromobenzene is {steam volatile}
Para-nitrobromobenzene is {less volatile} due to better molecular packing and higher melting point This difference in volatility makes steam distillation an effective separation method.
Step 4: Eliminate other options
Simple distillation: Boiling points are too close for efficient separation.
Fractional distillation: Not ideal for separating solids/liquids with similar boiling points and differing steam volatility.
Sublimation: Neither compound readily sublimes. Final Answer: \[ \boxed{\text{Steam distillation}} \]
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