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

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:

1. 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.

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.

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:

$\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-$ …

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.

1. 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.

2. Statement II: “Aniline can be separated from a mixture of aniline and water by steam distillation.” …

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): $\text{PH}_3\ (-88\,^{\circ}\text{C}) < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\ (\approx +17\,^{\circ}\text{C})$. …

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

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

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.

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

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. …

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:

1. 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.

2. 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.

3. 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.

4. 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: …