A comprehensive question bank for Qualitative and Quantitative Organic Analysis - PART-A

M.Sc. Chemistry Practical Viva Voce Questions

A comprehensive question bank for Qualitative and Quantitative Organic Analysis


Note to Students: The following questions are designed to test your understanding of the principles behind the experiments, not just the procedure. Prepare these well for your internal and semester-end viva.

Part A: Qualitative Analysis of Organic Binary Mixtures

Course Code: 24CHE1C2P | Focus: Separation & Identification (Solid+Solid)

I. The Logic of Separation

  • Q1: On what specific property is the separation of two solids based?
    Answer: Difference in solubility (physical) or difference in acidic/basic character (chemical).
  • Q2: Why use ether or ethyl acetate instead of ethanol for extraction?
    Answer: We need a water-immiscible solvent to create two distinct layers. Ethanol mixes with water and won't separate.
  • Q3: How do you determine if a mixture is Acidic, Basic, Phenolic, or Neutral?
    Answer: By performing solubility tests in Water, NaHCO3, NaOH, and HCl.
  • Q4: What is a Eutectic Mixture?
    Answer: A mixture of substances that melts at a single temperature lower than the melting point of either individual constituent.

II. Chemical Reactions & Extraction

  • Q5: Why use NaHCO3 for acids but NaOH for phenols?
    Answer: Carboxylic acids are stronger acids and can displace CO2 from bicarbonate. Phenols are weaker and require a stronger base (NaOH) to form phenoxide ions.
  • Q6: How do you regenerate the acid from the aqueous layer?
    Answer: Acidification with concentrated HCl.
    R-COO-Na+ + HCl → R-COOH ↓ + NaCl
  • Q7: How do you separate an amine from a neutral compound?
    Answer: Dissolve the mixture in dilute HCl. The amine forms a water-soluble salt (R-NH3+Cl-), while the neutral compound remains in the organic layer.
  • Q8: How do you break a stubborn emulsion in the separating funnel?
    Answer: Add brine (saturated NaCl), wait patiently, or swirl gently instead of shaking vigorously.

III. Identification & Derivatives

  • Q9: Why perform Lassaigne’s Test (Sodium Fusion) after separation?
    Answer: To identify exactly which component contains the heteroatoms (N, S, Halogens). Testing the mixture gives ambiguous results.
  • Q10: Why is the "Mixed Melting Point" the ultimate proof of identity?
    Answer: If you mix your unknown with a pure authentic sample and the MP remains the same, it confirms identity. If the MP drops (depression), they are different compounds.

Part B: Quantitative Analysis of Organic Functional Groups

Course Code: 24CHE3C10P | Focus: Estimations & Titrations

I. Core Concepts (The Must-Knows)

  • Q1: What is a "Blank Titration" and why is it necessary?
    Answer: It is a titration performed without the sample to determine the exact amount of reagent available initially. The difference between the Blank and Sample values tells us how much reagent reacted with the compound.
  • Q2: Why do we use "Back Titration" for Esters?
    Answer: Esters are neutral and react slowly. We add excess NaOH, reflux to hydrolyze completely, and then titrate the leftover NaOH with acid.

II. Functional Group Specifics

  • Q3: In Hydroxyl estimation, what is the role of Pyridine?
    Answer: It acts as a solvent, a catalyst, and a base to neutralize the acid formed, driving the reaction forward.
  • Q4: Why use Phenolphthalein for Organic Acids vs. NaOH?
    Answer: The endpoint lies in the slightly basic range (pH 8-9) because the product is a salt of a weak acid and strong base.
  • Q5: In Aniline estimation (Bromination), where does the Bromine come from?
    Answer: It is generated in situ. Acidified Bromate reacts with Bromide to release Br2.
    BrO3- + 5Br- + 6H+ → 3Br2 + 3H2O

🧮 Sample Calculation: Estimation of an Ester

Scenario: You are determining the % purity of Ethyl Acetate (MW = 88).

Data:

  • Weight of Sample (W): 1.20 g
  • Blank Titer (Vblank): 49.5 mL
  • Sample Titer (Vsample): 22.3 mL
  • Normality of Acid/Base used (N): 0.5 N

Step 1: Calculate Volume Consumed
Vconsumed = Vblank - Vsample
Vconsumed = 49.5 - 22.3 = 27.2 mL

Step 2: Calculate Weight of Pure Ester
Formula: (Normality × Vol. Consumed × Eq. Wt) / 1000
Calculation: (0.5 × 27.2 × 88) / 1000
Result = 1.1968 g (This is the actual amount of ester present)

Step 3: Calculate % Purity
Formula: (Calculated Weight / Weight Taken) × 100
Calculation: (1.1968 / 1.20) × 100
Final Result = 99.73%


Good luck with your exams! Focus on safety, cleanliness, and understanding the chemistry behind every step.

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