Methods Of Purification Of Organic Compounds

Given,

of Hg dry gas pressure







From the combine gas equation,







of of at STP

of of

mass of

Step 1: Identify Molecular Geometry and Dipole Moments
- and are symmetric → No dipole moment.
- (bent) and (pyramidal) → Have dipole moment. Step 2: Correct Pair

Step 1: Understanding Peptisation Peptisation is the process of converting a precipitate into a colloidal solution by adding a suitable electrolyte. Step 2: Explanation - In peptisation, the electrolyte adsorbs onto the surface of the precipitate, causing it to break into smaller particles that remain suspended in the solution.
- It is widely used in the preparation of colloidal sols.
Thus, the correct answer is Peptisation.

Step 1: Identify the Reagents for Partial Oxidation
The conversion of toluene to benzaldehyde involves partial oxidation. The effective reagents for this reaction are:
- Chromyl chloride ( ) in CS , H (Etard reaction)
- Chlorine ( ) in the presence of light and water Step 2: Explanation
- → Effective (Partial oxidation of toluene to benzaldehyde)
- → Ineffective (Leads to benzoic acid)
- → Ineffective (Complete oxidation to benzoic acid)
- → Effective (Etard reaction forms benzaldehyde)
Thus, the correct answer is .

Step 1: Naming Compound A (Amine)
- The longest chain has 4 carbons → Butene.
- The double bond starts at C-1 → But-1-ene.
- The amine (-NH ) is at C-3 → But-1-en-3-amine. Step 2: Naming Compound B (Aromatic Amine)
- The benzene ring has a chloro (-Cl) group at C-4.
- The dimethylamine (-N(CH ) ) is at C-1.
- Correct IUPAC name → 4-Chloro-N, N-dimethylbenzenamine.
Thus, the correct answer is .

Step 1: Identifying the Complex
- The green complex formed is Cl .
- This means that for each molecule of the complex, 3 chloride ions are available for precipitation. Step 2: Moles of Complex and Cl Ions
- Given: 100 mL of 1M solution Moles of complex = 0.1 moles.
- Each mole of the complex releases 3 moles of Cl . Step 3: Reaction with AgNO
- AgNO reacts with Cl to form AgCl precipitate:
- Since each Ag reacts with one Cl , the number of moles of AgCl precipitated is 0.3 moles.
Thus, the correct answer is 0.3 moles.

Step 1: Identifying X and Y
- Given reactions: - The compounds formed are:
- is Phosphorous acid ( )
- is Hypophosphorous acid ( ) Step 2: Identifying P–OH bonds in each acid - The structure of Phosphorous acid ( ) contains two P–OH bonds.
- The structure of Hypophosphorous acid ( ) contains one P–OH bond. Step 3: Correcting the order
- The given reactions indicate that and (Phosphoric acid).
- Phosphoric acid ( ) has four P–OH bonds.
- Thus, the number of P–OH bonds in are .

Step 1: Analyze Each Statement Statement 1: Basic Nature of Hydroxides - The basicity of hydroxides increases down the group due to decreasing ionization energy and increasing lattice solubility.
- The correct order: - This statement is correct. Statement 2: Thermal Stability of Carbonates
- Incorrect because the thermal stability of carbonates increases down the group.
- BeCO₃ is the least thermally stable, while BaCO₃ is the most stable.
- Correct order: - This statement is incorrect, making it the correct answer to the question. Statement 3: Solubility of Sulphates
- Solubility decreases down the group as hydration energy decreases.
- Correct order: - This statement is correct. Statement 4: Nitrate Decomposition - All alkaline earth metal nitrates decompose on heating to form metal oxides, NO₂, and O₂.
- Example reaction: - This statement is correct. Final Answer: Statement 2 is incorrect.

Step 1: Identify Deactivating Groups
- Deactivating (-I or -M effects):
- , , - Activating (+M or +I effects):
- , , Step 2: Count Deactivating Groups There are 3 deactivating groups.

Step 1: Periodic Trends of Atomic Radius - Across a period: Atomic radius decreases. - Down a group: Atomic radius increases. Step 2: Apply the Trends - and are in the same period, but has a smaller radius. - and are in the same period, but has a slightly larger radius. - Down the group, radius increases. Step 3: Correct Order

Step 1: Apply Fajan’s Rule Covalent character increases with: - Higher cation charge density. - Larger anion size. Step 2: Arrange Based on Covalent Character

Step 1: Compute Initial Moles of H₂SO₄ Step 2: Convert Removed Molecules to Moles Step 3: Compute Remaining Moles Thus, the correct answer is .

To prepare 1-iodopropane, we can follow these methods:

• Method (A): The reaction of ethene ( ) with hydrogen iodide ( ) forms 1-iodopropane. This is an addition reaction, but it is not a typical method for preparing 1-iodopropane directly from alcohol or alkyl halides.
• Method (B): The reaction of propanol ( ) with sodium iodide ( ) in the presence of 95% phosphoric acid ( ) is a known method for preparing 1-iodopropane. The alcohol is first converted to the alkene, and then iodination occurs.
• Method (C): The reaction of ethyl chloride ( ) with sodium iodide in dry acetone is a classic example of the nucleophilic substitution (SN2) reaction, which is used to prepare 1-iodopropane. This is the correct and commonly used method.
• Method (D): The reaction of propane with iodine ( ) in the presence of ultraviolet light ( ) leads to the formation of an iodopropane, but this process generally results in a mixture of products, including 2-iodopropane, not selectively 1-iodopropane.

Thus, the correct methods are (B) and (C).

Step 1: Understanding the Oxidizing Nature of KMnO
- In an acidic medium, KMnO is a strong oxidizing agent.
- In a neutral medium, its oxidizing power decreases but still oxidizes some species. Step 2: Oxidation Reactions in Neutral Medium
- KMnO oxidizes I to IO in neutral medium:
- Other reactions:
- Oxidation of oxalate ion occurs in acidic medium.
- Precipitation of sulfur from occurs in acidic medium.
- Fe to Fe oxidation occurs in acidic medium.
Thus, the correct answer is .