Detection of Elements in Organic Compounds

Real-Life Hook: Chemical Fingerprinting

Imagine you’re a forensic scientist analyzing a mysterious powder found at a crime scene. Or a pharmaceutical quality control expert verifying that a drug contains only safe ingredients. How do you know what elements are present? Elemental analysis is your answer!

From drug safety testing (detecting harmful chlorine in medicines) to environmental monitoring (identifying toxic compounds) to archaeological chemistry (analyzing ancient dyes), detecting elements in organic compounds is fundamental to:

• Quality control: Ensuring product purity and composition
• Drug discovery: Confirming molecular formulas of new compounds
• Forensics: Identifying unknown substances
• Environmental science: Detecting pollutants
• Academic research: Characterizing newly synthesized compounds

The simple tests you’ll learn have helped convict criminals, save lives by detecting contamination, and confirm the structures of Nobel Prize-winning molecules!

Introduction to Elemental Analysis

Elemental analysis determines which elements are present in an organic compound.

Why Detect Elements?

1. Confirming molecular formula: C, H, N, O, S, halogens
2. Identifying unknown compounds: Different elements = different compounds
3. Checking purity: Unexpected elements indicate impurities
4. Understanding properties: Elements determine reactivity
5. Safety: Detecting toxic elements

Common Elements in Organic Compounds

Always present:

• Carbon (C): By definition (organic = carbon-containing)
• Hydrogen (H): In almost all organic compounds

Often present:

• Oxygen (O): Alcohols, ethers, carboxylic acids, etc.
• Nitrogen (N): Amines, amides, proteins, alkaloids
• Sulfur (S): Proteins (cysteine), some drugs
• Halogens (F, Cl, Br, I): Many pharmaceuticals, pesticides

Rarely present (special compounds):

• Phosphorus (P): DNA, RNA, pesticides, nerve agents

General Strategy for Detection

Two-Stage Process

Stage 1: Lassaigne’s Test (Qualitative)

• Converts elements from organic (covalent) to inorganic (ionic) form
• Uses sodium fusion
• Detects: N, S, and halogens
• Quick and sensitive

Stage 2: Specific Tests

• Confirms presence/absence
• Uses color reactions, precipitations
• Identifies specific element

Why Lassaigne’s Test?

Problem: In organic compounds, elements exist as covalent bonds:

• Not easily ionizable
• Don’t give typical inorganic reactions
• Difficult to detect directly

Example:

• NaCl (inorganic): Na⁺ and Cl⁻ → easy to detect with AgNO₃
• Chlorobenzene (organic): C-Cl covalent bond → doesn’t react with AgNO₃ directly

Solution - Lassaigne’s Test:

• Fuse organic compound with sodium metal
• Converts covalent elements to ionic form
• Extract ions in water
• Test using inorganic qualitative analysis

Named after: French chemist Jean Louis Lassaigne (1800s)

Lassaigne’s Test - Sodium Fusion Extract

Principle

Heat organic compound with sodium metal at high temperature (~300-400°C):

• Sodium reduces and converts covalent C, N, S, X into ionic forms
• Forms water-soluble sodium salts
• Can be tested by inorganic qualitative analysis

Reactions During Fusion

For Nitrogen:

(Sodium cyanide formed)

For Sulfur:

(Sodium sulfide formed)

For Halogens (Cl, Br, I):

(Sodium halide formed)

For both N and S together:

Then they react:

(Sodium thiocyanate formed)

Detailed Procedure

Materials Required:

• Sodium metal (fresh, small piece ~50 mg)
• Organic compound (~100 mg)
• Fusion tube (thick-walled ignition tube)
• Bunsen burner
• Beaker with distilled water (50 mL)
• Mortar and pestle
• Filter paper and funnel

Safety First:

• ⚠️ Sodium is highly reactive with water → violent reaction
• ⚠️ Work in fume hood
• ⚠️ Wear safety goggles and gloves
• ⚠️ Keep water and sodium separate until fusion is complete
• ⚠️ Never add water to hot fusion tube containing excess sodium

Step-by-Step Procedure:

Step 1: Preparing Fusion Tube

1. Take clean, dry ignition tube
2. Cut small piece of sodium metal (size of small pea)
3. Dry sodium piece on filter paper (remove kerosene)
4. Place sodium in fusion tube

Step 2: Adding Organic Compound

1. Add small amount (50-100 mg) of organic compound
2. Mix gently by tapping tube
3. If compound is liquid, add dropwise

Step 3: Fusion (Heating)

1. Heat tube gently first (sodium melts at 98°C)
2. Sodium melts and spreads in tube
3. Heat strongly until tube becomes red hot
4. Organic compound chars (becomes black)
5. Continue heating for 2-3 minutes
6. Remove from flame

Interactive Demo: Visualize Elemental Detection

See how different elements are detected through color reactions and tests.

Step 4: Quenching (CRITICAL SAFETY STEP)

1. Let tube cool for 10-15 seconds (NOT completely cool!)
2. Prepare beaker with 30-50 mL distilled water
3. While tube is still hot (but not red hot), plunge it into water
4. Tube breaks with a mild explosion
5. Sodium reacts: 2Na + 2H₂O → 2NaOH + H₂↑

Why plunge while hot?

• Ensures complete reaction of residual sodium
• Cold tube might trap unreacted sodium inside
• Unreacted sodium + water = dangerous reaction later

Step 5: Boiling and Filtering

1. Boil the solution for 2-3 minutes (ensures complete extraction)
2. Cool to room temperature
3. Filter through filter paper
4. Lassaigne’s filtrate (or sodium fusion extract) is ready

Step 6: Testing

• Use this filtrate for detecting N, S, and halogens
• Divide filtrate into portions for different tests

Memory Trick: “Safe Fusion - SMART”

Sodium piece (small, dry) Mix with organic compound Apply heat (red hot) Rapidly plunge in water (while hot!) Test the filtrate

Detection of Nitrogen (Prussian Blue Test)

Principle

If nitrogen is present → NaCN formed during fusion

NaCN is converted to Prussian blue (ferric ferrocyanide) - an intense blue precipitate

Reactions

During fusion:

Testing (two steps):

Step 1: Formation of sodium ferrocyanide

(Add FeSO₄ to ensure excess CN⁻ combines with Fe²⁺)

Step 2: Formation of Prussian blue

Overall balanced reaction:

Detailed Procedure

Reagents:

• Lassaigne’s filtrate
• Freshly prepared FeSO₄ solution
• Dilute H₂SO₄
• FeCl₃ solution

Steps:

1. Take 2-3 mL of Lassaigne’s filtrate in test tube
2. Add few drops of freshly prepared FeSO₄ solution
3. Add 1-2 drops of dilute H₂SO₄
4. Boil for 1-2 minutes (to ensure complete reaction)
5. Cool to room temperature
6. Add 2-3 drops of FeCl₃ solution
7. Observe:
   • Positive test: Intense Prussian blue or blue-green precipitate
   • Negative test: No blue color (remains brown/yellow from FeCl₃)

Why FeSO₄ and then FeCl₃?

Question: Why not add FeCl₃ directly to Lassaigne’s filtrate containing CN⁻?

Answer:

• FeCl₃ (Fe³⁺) would react with CN⁻, but won’t form Prussian blue directly
• Need [Fe(CN)₆]⁴⁻ complex first
• Fe²⁺ (from FeSO₄) forms ferrocyanide: [Fe(CN)₆]⁴⁻
• Then Fe³⁺ combines with ferrocyanide → Prussian blue

Two-stage process ensures sensitive test!

Interferences and Precautions

1. Sulfur Interference: If both N and S are present:

• NaSCN (sodium thiocyanate) forms instead of NaCN
• SCN⁻ + Fe³⁺ → [Fe(SCN)]²⁺ (blood red color)
• Masks Prussian blue!

Solution: Remove sulfur before testing for nitrogen

• Add lead acetate to Lassaigne’s filtrate
• Pb²⁺ + S²⁻ → PbS↓ (black precipitate)
• Filter to remove PbS
• Use filtrate for nitrogen test

2. Halogen Interference: Halogens don’t interfere with nitrogen test (different chemistry)

3. FeSO₄ must be freshly prepared:

• Old FeSO₄ solution oxidizes to Fe³⁺
• Fe³⁺ doesn’t form ferrocyanide
• Use freshly prepared (colorless to pale green) solution

Common Mistakes

Mistake 1: Adding FeCl₃ before FeSO₄ Problem: Wrong complex forms, no Prussian blue Solution: Always add FeSO₄ first, boil, cool, then add FeCl₃

Mistake 2: Using old FeSO₄ solution Problem: Oxidized to Fe³⁺, doesn’t work Solution: Prepare fresh FeSO₄ just before use

Mistake 3: Not boiling after adding FeSO₄ Problem: Incomplete reaction Solution: Boil for 1-2 minutes, then cool before adding FeCl₃

Mistake 4: Testing when sulfur is present Problem: Blood red SCN⁻ complex masks blue color Solution: Remove sulfur with lead acetate first

Detection of Sulfur

Principle

If sulfur is present → Na₂S formed during fusion

Na₂S gives two positive tests:

1. Sodium nitroprusside test: Purple/violet color
2. Lead acetate test: Black precipitate

Test 1: Sodium Nitroprusside Test

Reaction:

(Purple/violet colored complex)

Procedure:

1. Take 2 mL Lassaigne’s filtrate
2. Add 2-3 drops sodium nitroprusside solution
3. Observe:
   • Positive: Purple or violet color appears
   • Negative: No color change

Sensitivity: Very sensitive test, can detect traces of sulfur

Test 2: Lead Acetate Test

Reaction:

(Black precipitate)

Procedure:

1. Take 2 mL Lassaigne’s filtrate
2. Add 2-3 drops lead acetate solution
3. Observe:
   • Positive: Black precipitate (PbS)
   • Negative: No precipitate

Note: PbS is black, highly insoluble

Test 3: Acidification Test (Alternate)

Procedure:

1. Take Lassaigne’s filtrate
2. Acidify with dilute acetic acid
3. Add lead acetate solution
4. Positive: Black precipitate of PbS

Why acidify?: Prevents precipitation of other lead salts

When Both N and S are Present

Problem: If both nitrogen and sulfur are present:

(Sodium thiocyanate forms)

Detection: Thiocyanate can be detected by blood red coloration with FeCl₃:

(Blood red complex)

Procedure:

1. Take Lassaigne’s filtrate
2. Add few drops of FeCl₃ solution
3. Positive: Blood red color (indicates both N and S present)

To test N and S individually when both present:

For Sulfur:

• Sodium nitroprusside test (no interference from nitrogen)
• Or lead acetate test

For Nitrogen:

• Remove sulfur first with lead acetate
• Filter
• Do Prussian blue test on filtrate

Memory Trick: Sulfur Tests “PLS”

Purple with nitroprusside Lead acetate → black SCN gives red (if N also present)

Detection of Halogens (Cl, Br, I)

Overview

Three halogens can be detected:

• Chlorine (Cl): Most common in organic compounds
• Bromine (Br): Common in pharmaceuticals
• Iodine (I): Less common
• Fluorine (F): Difficult to detect by this method (different chemistry)

Principle

During fusion:

(where X = Cl, Br, or I)

NaX tested with silver nitrate (AgNO₃) → colored precipitate

Interference from CN⁻ and S²⁻

Problem:

• If nitrogen present → CN⁻ in Lassaigne’s filtrate
• CN⁻ + Ag⁺ → AgCN (white ppt) - FALSE positive!
• If sulfur present → S²⁻ in filtrate
• S²⁻ + Ag⁺ → Ag₂S (black ppt) - FALSE positive!

Solution: Remove CN⁻ and S²⁻ before testing for halogens

Detailed Procedure

Step 1: Removing CN⁻ and S²⁻

Method: Boil Lassaigne’s filtrate with dilute HNO₃

Reactions:

(HCN escapes as gas - highly toxic! Do in fume hood!)

(H₂S escapes as gas - rotten egg smell)

Procedure:

1. Take Lassaigne’s filtrate in test tube
2. Add dilute HNO₃ to acidify
3. Boil gently for 2-3 minutes (in fume hood!)
4. HCN and H₂S gases escape
5. Cool to room temperature

⚠️ Safety: HCN is extremely poisonous! Always work in fume hood!

Step 2: Testing with Silver Nitrate

Procedure:

1. Cool the acidified, boiled filtrate
2. Add 2-3 drops of AgNO₃ solution
3. Observe the precipitate:

Reactions:

(white)

(pale yellow)

(yellow)

Distinguishing Between Halogens

If white precipitate (could be Cl):

1. Add dilute (2M) NH₃ solution
2. If dissolves → Chlorine confirmed
3. If doesn’t dissolve → Not chlorine (false positive from other sources)

Reaction:

(Soluble complex)

If pale yellow precipitate (could be Br):

1. Try adding dilute NH₃ → Won’t dissolve
2. Add concentrated NH₃ → Partial dissolution
3. Confirms bromine

If yellow precipitate (could be I):

1. Add NH₃ (even concentrated) → No dissolution
2. Confirms iodine

Alternative Test for Halogens: Beilstein Test

Principle: Organic halogen compounds give green flame when heated on copper wire

Procedure:

1. Clean copper wire by heating in flame until no color
2. Dip wire in organic compound
3. Heat in Bunsen flame
4. Observe:
   • Positive: Blue-green or emerald green flame
   • Negative: No green color

Reason: Volatile copper halides (CuCl₂, CuBr₂, CuI₂) form and give green color

Advantage: Quick, no need for sodium fusion

Disadvantage:

• Less specific (some non-halogen compounds also give green color)
• Cannot distinguish which halogen
• Semi-quantitative only

Comparison: Beilstein vs. Lassaigne

Best Practice:

• Use Beilstein for quick screening
• Confirm with Lassaigne’s test and AgNO₃

Memory Trick for Halogen Precipitates: “Clever Babies are Impressive”

Chlorine → White precipitate → Clean/Clear Bromine → Pale yellow → Butter/Banana (pale) Iodine → Deep yellow → Intense/Imperial yellow

Solubility in NH₃: “Chlorine Loves Ammonia, Iodine Ignores”

• Cl: Loves (soluble in dilute NH₃)
• Br: Lukewarm (needs concentrated NH₃)
• I: Ignores (insoluble even in conc. NH₃)

Detection of Carbon and Hydrogen

These are NOT done by Lassaigne’s Test!

C and H are detected by oxidation, not sodium fusion.

Detection of Carbon

Principle: Carbon oxidizes to CO₂ when organic compound is heated with copper(II) oxide

Reaction:

CO₂ is detected by passing through lime water (turns milky)

Detailed Procedure:

Apparatus: Dry test tube with delivery tube connected to lime water

Steps:

1. Mix organic compound with excess CuO (copper(II) oxide, black powder)
2. Place mixture in dry test tube
3. Attach delivery tube leading to lime water (Ca(OH)₂ solution)
4. Heat strongly
5. Observe: Lime water turns milky

Reaction with lime water:

(White precipitate - milky appearance)

Confirmation: Continue passing CO₂

(Soluble - milkiness disappears)

Positive test: Lime water turns milky, then clears on excess CO₂

Detection of Hydrogen

Principle: Hydrogen oxidizes to H₂O when organic compound is heated with CuO

Reaction:

Water detected by anhydrous copper sulfate (turns blue) or cobalt chloride paper (turns pink)

Detailed Procedure:

Method 1: Anhydrous CuSO₄ Method

Steps:

1. Mix organic compound with CuO in test tube
2. Heat strongly
3. Water vapor forms
4. Hold white anhydrous CuSO₄ at mouth of tube
5. Observe: White CuSO₄ turns blue

Reaction:

Method 2: Cobalt Chloride Paper Method

Steps:

1. Same heating with CuO
2. Hold blue cobalt chloride paper at mouth
3. Observe: Blue paper turns pink

Reaction:

Combined Test for C and H

Usually both are tested simultaneously:

[Test tube with compound + CuO]
         ↓ (heating)
    Water vapor + CO₂
         ↓
    [Anhydrous CuSO₄] → turns blue (H detected)
         ↓
    [Lime water] → turns milky (C detected)

Memory Trick: “Copper Helps Detect”

• Copper oxide oxidizes
• H₂O → blue copper sulfate
• Dioxide (CO₂) → milky lime water

Detection of Phosphorus

Principle: Organic compound is oxidized to phosphate (PO₄³⁻), which is detected by yellow precipitate with ammonium molybdate

Oxidation Reaction:

Detection:

(Canary yellow precipitate - ammonium phosphomolybdate)

Procedure:

1. Heat organic compound with conc. HNO₃ (oxidation)
2. Cool and dilute
3. Add excess ammonium molybdate solution
4. Warm gently
5. Positive: Canary yellow precipitate

Note: Not commonly asked in JEE, but good to know

Quantitative Estimation

Beyond Detection: How Much?

Qualitative tests tell IF an element is present. Quantitative analysis tells HOW MUCH is present.

1. Carbon and Hydrogen Estimation (Liebig’s Method)

Principle:

• Organic compound completely oxidized by heating with CuO
• C → CO₂ (absorbed in KOH solution, weighed)
• H → H₂O (absorbed in anhydrous CaCl₂ or Mg(ClO₄)₂, weighed)

Apparatus: Combustion tube with absorption bulbs

Calculations:

For Carbon:

• Mass of CO₂ absorbed = m₁ g
• Molar mass of CO₂ = 44 g/mol
• Moles of CO₂ = m₁/44
• Moles of C = m₁/44 (1 mol CO₂ has 1 mol C)
• Mass of C = (m₁/44) × 12 = m₁ × (12/44) = m₁ × 0.2727

For Hydrogen:

• Mass of H₂O absorbed = m₂ g
• Molar mass of H₂O = 18 g/mol
• Moles of H₂O = m₂/18
• Moles of H = 2 × (m₂/18) (1 mol H₂O has 2 mol H)
• Mass of H = 2 × (m₂/18) × 1 = m₂ × (2/18) = m₂ × 0.1111

2. Nitrogen Estimation (Dumas Method)

Principle:

• Organic compound heated with CuO
• N → N₂ gas (collected and volume measured)
• Use gas equation to find moles of N

Reaction:

CO₂ is absorbed in KOH, pure N₂ collected over water

Calculation:

3. Nitrogen Estimation (Kjeldahl’s Method)

Principle:

• Organic compound heated with conc. H₂SO₄
• N → (NH₄)₂SO₄
• NH₃ liberated by adding NaOH
• NH₃ absorbed in known volume of standard acid
• Back-titrated to find amount of NH₃

Reaction:

Limitation: Doesn’t work for:

• Nitro compounds (-NO₂)
• Azo compounds (-N=N-)
• Nitrogen in ring (pyridine, etc.)

Works only for amino nitrogen (amines, amides, proteins)

4. Sulfur Estimation (Carius Method)

Principle:

• Organic compound heated with fuming HNO₃ in sealed tube
• S → SO₄²⁻
• Precipitated as BaSO₄
• Weighed

Reaction:

Calculation:

(1:1 ratio)

5. Halogen Estimation (Carius Method)

Principle: Same as sulfur - oxidation in sealed tube

Reactions:

Calculation:

For chlorine (AgCl, M = 143.5):

For bromine (AgBr, M = 188):

For iodine (AgI, M = 235):

Practice Problems

Level 1 - JEE Main (Basics)

Problem 1: Why is the organic compound fused with sodium metal in Lassaigne’s test?

Solution:

Reason for Sodium Fusion:

In organic compounds, elements like N, S, and halogens exist in covalent form:

• C-N bonds (amines, nitriles)
• C-S bonds (thiols, sulfides)
• C-X bonds (alkyl halides)

Problem: Covalent forms don’t give typical inorganic qualitative tests

• Example: Chlorobenzene (C₆H₅Cl) doesn’t react with AgNO₃ (no ionic Cl⁻)

Solution: Sodium fusion converts covalent to ionic form

How sodium works:

1. Strong reducing agent: Sodium is highly electropositive
2. High temperature: Breaks C-N, C-S, C-X bonds
3. Forms ionic compounds: NaCN, Na₂S, NaX
4. Water-soluble: Can be extracted and tested

Reactions:

• Nitrogen: C,N + Na → NaCN (water-soluble, ionic)
• Sulfur: C,S + Na → Na₂S (water-soluble, ionic)
• Halogen: C,X + Na → NaX (water-soluble, ionic)

Answer: Sodium fusion converts covalent organic forms of N, S, and halogens into water-soluble ionic compounds (NaCN, Na₂S, NaX) that can be detected by standard inorganic qualitative tests.

Problem 2: Why must Lassaigne’s filtrate be acidified with dilute HNO₃ before testing for halogens?

Solution:

Purpose of Acidification:

Problem 1 - Remove CN⁻:

• If nitrogen is present → NaCN in filtrate
• CN⁻ + Ag⁺ → AgCN↓ (white precipitate)
• False positive for chlorine!

Removal:

HCN gas escapes (toxic! use fume hood)

Problem 2 - Remove S²⁻:

• If sulfur is present → Na₂S in filtrate
• S²⁻ + Ag⁺ → Ag₂S↓ (black precipitate)
• False positive for iodine/halogen!

Removal:

H₂S gas escapes (rotten egg smell)

Why HNO₃ specifically?:

1. Strong acid: Effectively decomposes NaCN and Na₂S
2. Oxidizing: Ensures complete conversion
3. Doesn’t interfere: NO₃⁻ doesn’t precipitate with Ag⁺
4. Can’t use HCl or H₂SO₄:
   • HCl would add Cl⁻ (interferes with halogen test!)
   • H₂SO₄ → SO₄²⁻ + Ag⁺ → Ag₂SO₄ (slightly soluble, interference)

Answer: Acidification with dilute HNO₃ removes CN⁻ (as HCN gas) and S²⁻ (as H₂S gas) which would otherwise give false positive tests with AgNO₃. HNO₃ is used because it doesn’t introduce interfering anions.

Problem 3: A compound gives a blue precipitate in Lassaigne’s test for nitrogen. What is the composition and color of this precipitate?

Solution:

The blue precipitate is Prussian blue (Ferric ferrocyanide)

Composition: Fe₄[Fe(CN)₆]₃

Systematic name: Ferric ferrocyanide or Iron(III) hexacyanoferrate(II)

Structure: Complex coordination compound

• Contains both Fe³⁺ and [Fe(CN)₆]⁴⁻
• Fe³⁺ ions coordinated to CN groups of ferrocyanide

Color: Intense blue (sometimes blue-green)

Formation:

1. NaCN (from nitrogen in organic compound) + FeSO₄ → Na₄[Fe(CN)₆]
2. Na₄[Fe(CN)₆] + FeCl₃ → Fe₄[Fe(CN)₆]₃↓ (Prussian blue)

Properties:

• Insoluble in water
• Intense blue color (very sensitive test)
• Same compound used as blue pigment in paints and inks

Answer: Composition: Fe₄[Fe(CN)₆]₃ (Ferric ferrocyanide); Color: Intense Prussian blue or blue-green

Level 2 - JEE Main (Application)

Problem 4: An organic compound gives the following results:

• Lassaigne’s test: Blood red color with FeCl₃
• Sodium nitroprusside test: Purple color
• AgNO₃ test: No precipitate

What elements are present in the compound?

Solution:

Analysis of Tests:

Test 1: Blood red with FeCl₃

• Indicates: SCN⁻ (thiocyanate) present
• SCN⁻ + Fe³⁺ → [Fe(SCN)]²⁺ (blood red complex)
• SCN⁻ forms when both N and S are present: $$\ce{NaCN + Na2S -> NaSCN + Na}$$
• Conclusion: Both nitrogen and sulfur present

Test 2: Purple color with sodium nitroprusside

• Confirms: Sulfur present
• S²⁻ + [Fe(CN)₅NO]²⁻ → purple complex
• Confirmation of sulfur

Test 3: No precipitate with AgNO₃

• Indicates: No halogens (Cl, Br, I)
• If halogen present → AgX precipitate would form
• Conclusion: No halogens

Summary:

• ✓ Nitrogen (from blood red SCN⁻ test)
• ✓ Sulfur (from nitroprusside + SCN⁻)
• ✗ Halogens (no AgX precipitate)
• ✓ Carbon and Hydrogen (assumed in organic compound)

Answer: The compound contains Nitrogen (N) and Sulfur (S) (in addition to C and H). No halogens are present.

Example compounds: Thiourea (NH₂-CS-NH₂), Thiocyanates, Amino acids with sulfur (cysteine, methionine)

Problem 5: Why is freshly prepared FeSO₄ solution used in the test for nitrogen?

Solution:

Requirement: FeSO₄ must be fresh (colorless to pale green)

Problem with old FeSO₄:

1. Oxidation: Fe²⁺ oxidizes to Fe³⁺ in presence of air

   $$\ce{4FeSO4 + O2 + 2H2O -> 4Fe(OH)SO4}$$ $$\ce{4Fe^{2+} + O2 + 4H^+ -> 4Fe^{3+} + 2H2O}$$

2. Color change:

   • Fresh FeSO₄: Colorless to pale green (Fe²⁺)
   • Old FeSO₄: Brown/rust colored (Fe³⁺)

Why Fe²⁺ is needed:

• Nitrogen test requires formation of ferrocyanide: [Fe(CN)₆]⁴⁻
• Only Fe²⁺ forms ferrocyanide: $$\ce{Fe^{2+} + 6CN^- -> [Fe(CN)6]^{4-}}$$ ✓
• Fe³⁺ forms ferricyanide (different complex): $$\ce{Fe^{3+} + 6CN^- -> [Fe(CN)6]^{3-}}$$ (Wrong complex!)

Prussian blue formation:

• Requires ferrocyanide + Fe³⁺: $$\ce{[Fe(CN)6]^{4-} + Fe^{3+} -> Fe4[Fe(CN)6]3 v}$$ (Prussian blue)
• Ferricyanide doesn’t form Prussian blue!

Result if old FeSO₄ used:

• No ferrocyanide formation
• No Prussian blue precipitate
• False negative for nitrogen!

How to ensure fresh:

• Prepare FeSO₄ solution just before use
• Add few drops of dilute H₂SO₄ to prevent oxidation
• Store under layer of oil (prevents air contact)
• Check color: should be colorless to pale green, NOT brown

Answer: Fresh FeSO₄ contains Fe²⁺ which is required to form ferrocyanide [Fe(CN)₆]⁴⁻. Old FeSO₄ contains Fe³⁺ (due to oxidation) which forms ferricyanide instead, preventing Prussian blue formation and giving false negative result for nitrogen.

Problem 6: A compound gave a white precipitate with AgNO₃ after proper treatment of Lassaigne’s filtrate. The precipitate dissolved in dilute NH₃. However, the Beilstein test was negative. Explain.

Solution:

Analysis:

Test 1: White precipitate with AgNO₃

• Could be: AgCl (chlorine) OR AgCN (nitrogen)
• Not conclusive alone

Test 2: Dissolves in dilute NH₃

• AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻ (soluble) ✓
• AgCN also dissolves in NH₃ (forms complex)
• Still not conclusive!

Test 3: Beilstein test negative

• No green flame when heated with copper wire
• Beilstein test is positive for halogens
• Negative result means NO halogen (including Cl)

Contradiction Analysis:

If white ppt dissolves in NH₃ BUT Beilstein is negative:

• Cannot be chlorine (Beilstein would be positive)
• Must be AgCN (from nitrogen)

But the procedure says “proper treatment”:

• Proper treatment = boiling with HNO₃ to remove CN⁻
• CN⁻ should have been removed as HCN gas
• So AgCN shouldn’t form!

Explanation:

Either:

1. Improper treatment: Didn’t boil sufficiently with HNO₃
   • CN⁻ not completely removed
   • Residual CN⁻ → AgCN

Or: 2. False positive from other source:

• Some other compound giving white ppt
• Not actually halogen or CN⁻

Or (Most likely): 3. The compound contains nitrogen, NOT chlorine:

• Despite “proper treatment,” some CN⁻ remained
• Beilstein test correctly shows NO halogen
• The precipitate is AgCN, not AgCl

Conclusion: The compound contains nitrogen, not chlorine. The white precipitate is AgCN (not AgCl). The Beilstein test correctly identified absence of halogens. The “proper treatment” was incomplete - should have boiled longer with HNO₃ to completely remove CN⁻.

Answer: The compound contains nitrogen, not chlorine. The white precipitate is AgCN (from residual CN⁻), not AgCl. Incomplete acidification failed to remove all CN⁻. The Beilstein test correctly shows no halogen present. This demonstrates why both tests and proper procedure are essential for accurate identification.

Level 3 - JEE Advanced (Conceptual & Numerical)

Problem 7: 0.30 g of an organic compound gave 0.66 g of CO₂ and 0.36 g of H₂O on complete combustion. Calculate the percentage of C and H. If the compound contains only C, H, and O, find the percentage of oxygen.

Solution:

Given:

• Mass of organic compound = 0.30 g
• Mass of CO₂ produced = 0.66 g
• Mass of H₂O produced = 0.36 g

Step 1: Calculate percentage of Carbon

Molar mass of CO₂ = 44 g/mol Molar mass of C = 12 g/mol

(1 mol CO₂ contains 1 mol C)

Step 2: Calculate percentage of Hydrogen

Molar mass of H₂O = 18 g/mol Molar mass of H = 1 g/mol

(1 mol H₂O contains 2 mol H)

Step 3: Calculate percentage of Oxygen

If compound contains only C, H, and O:

Verification: Total = 60 + 13.33 + 26.67 = 100% ✓

Answer:

• Carbon: 60%
• Hydrogen: 13.33%
• Oxygen: 26.67%

Finding Empirical Formula (bonus):

Dividing by smallest (1.67):

Empirical formula: C₃H₈O (possibly propanol or methoxyethane)

Problem 8: In Kjeldahl’s method, 0.5 g of an organic compound was digested and ammonia evolved was absorbed in 50 mL of 0.5 M H₂SO₄. The excess acid required 30 mL of 0.5 M NaOH for neutralization. Calculate the percentage of nitrogen.

Solution:

Given:

• Mass of organic compound = 0.5 g
• Volume of H₂SO₄ used = 50 mL of 0.5 M
• Volume of NaOH for back titration = 30 mL of 0.5 M

Step 1: Calculate moles of H₂SO₄ taken initially

Equivalent of acid = 2 × 0.025 = 0.05 mol H⁺ (H₂SO₄ is diprotic)

Step 2: Calculate moles of NaOH used in back titration

Equivalent = 0.015 mol OH⁻

Step 3: Calculate moles of acid that reacted with NH₃

Step 4: Calculate moles of NH₃

Reaction: NH₃ + H⁺ → NH₄⁺

(1:1 ratio)

Step 5: Calculate moles of Nitrogen

Step 6: Calculate mass of Nitrogen

Step 7: Calculate percentage

Wait, 98% is unrealistically high! Let me recalculate…

Rechecking Step 1: H₂SO₄ is diprotic, but when calculating equivalents for back titration:

• Moles of H₂SO₄ = 0.025 mol
• But H₂SO₄ provides 2H⁺ per molecule
• However, for reaction with NH₃ (monoprotic base), we count: NH₃ + H⁺ → NH₄⁺

Let me recalculate properly:

Correct Calculation:

Total equivalents of acid = Moles of H₂SO₄ × 2 = 0.025 × 2 = 0.05 equivalents

Equivalents of NaOH used = Moles of NaOH × 1 = 0.015 equivalents

Equivalents of acid consumed by NH₃ = 0.05 - 0.015 = 0.035 equivalents

Moles of NH₃ = 0.035 mol (since NH₃ + H⁺ → NH₄⁺ is 1:1)

Moles of N = 0.035 mol

Mass of N = 0.035 × 14 = 0.49 g

% of N = (0.49/0.5) × 100 = 98%

Analysis: 98% nitrogen is extremely high but not impossible - could be:

• Hydrazine (N₂H₄): N% = 87.5%
• Urea (CH₄N₂O): N% = 46.7%
• Melamine (C₃H₆N₆): N% = 66.7%

The 98% suggests calculation is correct but scenario is extreme, OR there’s an error in problem statement.

Answer: 98% nitrogen

Note: If this seems unreasonable, check problem values. Typical organic compounds have 10-30% nitrogen.

Problem 9: An organic compound containing bromine gave 0.376 g of AgBr when 0.2 g of it was treated by Carius method. Calculate the percentage of bromine in the compound. (Atomic mass: Ag = 108, Br = 80)

Solution:

Given:

• Mass of organic compound = 0.2 g
• Mass of AgBr formed = 0.376 g
• Atomic mass: Ag = 108, Br = 80
• Molar mass of AgBr = 108 + 80 = 188 g/mol

Method 1: Using Mole Concept

Step 1: Calculate moles of AgBr

Step 2: Calculate moles of Br

(1:1 ratio)

Step 3: Calculate mass of Br

Step 4: Calculate percentage

Method 2: Using Direct Formula

Answer: 80% bromine

Verification: 80% is very high for an organic compound. This suggests:

• Simple halogen compound
• Example: CBr₄ (tetrabromomethane)
  • Molar mass = 12 + 4(80) = 332 g/mol
  • % Br = (320/332) × 100 = 96.4%
• Or CHBr₃ (bromoform)
  • Molar mass = 12 + 1 + 3(80) = 253 g/mol
  • % Br = (240/253) × 100 = 94.9%

So 80% is reasonable for poly-brominated compound.

Problem 10: Explain why Kjeldahl’s method cannot be used for determining nitrogen in nitrobenzene (C₆H₅NO₂) but works for aniline (C₆H₅NH₂).

Solution:

Kjeldahl’s Method Principle:

• Organic compound heated with conc. H₂SO₄
• Nitrogen converted to (NH₄)₂SO₄
• NH₃ liberated and titrated

Critical Requirement: Nitrogen must be in amino form (-NH₂, -NH-, etc.)

Case 1: Nitrobenzene (C₆H₅NO₂)

Structure: Benzene ring with -NO₂ group

• Nitrogen in nitro group (-NO₂)
• Nitrogen is in +5 oxidation state
• Highly oxidized form

What happens in Kjeldahl’s:

Problem:

• Nitro nitrogen is NOT converted to NH₃
• Instead, complex reactions occur:
  • Ring may sulphonate
  • Nitro group may partially reduce to nitroso (-NO)
  • But NOT to ammonia!
• Nitrogen remains in oxidized form
• Cannot be converted to (NH₄)₂SO₄ by simple heating with H₂SO₄

Why?:

• Kjeldahl’s method is reduction of nitrogen to -3 oxidation state (NH₃)
• Nitro nitrogen is in +5 state
• H₂SO₄ is not strong enough reducing agent
• Would need special reducing agents (Zn dust, etc.)

Result: False negative - nitrogen present but not detected

Case 2: Aniline (C₆H₅NH₂)

Structure: Benzene ring with -NH₂ group

• Nitrogen in amino group
• Already in -3 oxidation state

What happens in Kjeldahl’s:

Success:

• Amino nitrogen easily converted to ammonium sulfate
• Ring gets sulphonated/charred (doesn’t matter)
• Nitrogen quantitatively converted to (NH₄)₂SO₄
• On adding NaOH: NH₃ released and titrated

Result: Accurate determination of nitrogen

Summary - Kjeldahl’s Method Works For:

✓ Amines: R-NH₂ (like aniline) ✓ Amides: R-CO-NH₂ (like urea) ✓ Proteins: Contain amino acids ✓ Amino acids: NH₂-CHR-COOH

Kjeldahl’s Method Does NOT Work For:

✗ Nitro compounds: R-NO₂ (like nitrobenzene) ✗ Nitroso compounds: R-NO ✗ Azo compounds: R-N=N-R ✗ Nitrogen in ring: Pyridine, quinoline (heterocyclic N) ✗ Nitriles: R-CN (some variants can, but standard Kjeldahl fails) ✗ Diazo compounds: R-N₂⁺

Modified Kjeldahl’s (with reducing agents like Zn, Se) can handle some of these, but standard method cannot.

Answer: Kjeldahl’s method works only for amino nitrogen (already in -3 oxidation state). Aniline has -NH₂ group which is easily converted to (NH₄)₂SO₄ and then to NH₃ for titration. Nitrobenzene has nitrogen in nitro group (-NO₂, oxidation state +5) which cannot be reduced to NH₃ by simple heating with H₂SO₄. Therefore, Kjeldahl’s fails for nitrobenzene but succeeds for aniline.

Practical Implication: Always check nitrogen type before choosing Kjeldahl’s method. For nitro/azo/ring nitrogen, use Dumas method instead.

Cross-Links to Related Topics

Organic Chemistry Connections

1. Crystallization: After detection of elements, purify compound by crystallization

2. Distillation: Purify liquid organic compounds before elemental analysis

3. Chromatography: Separate mixtures before element detection; TLC to check purity

4. Organic Principles: Understanding molecular formulas and functional groups based on elements present

5. Hydrocarbons: Detection of C and H; understanding C:H ratios

6. Nitrogen Compounds: Detecting N in amines, amides, nitro compounds

7. Halogens Compounds: Detection of halogens in alkyl/aryl halides

Inorganic Chemistry Connections

1. Qualitative Analysis: Similar precipitation and color tests; Ag⁺, Pb²⁺ reagents

2. Coordination Compounds: Understanding complex formation (Prussian blue, nitroprusside complexes)

3. d-Block Elements: Iron complexes (ferrocyanide, ferricyanide), copper reactions

4. Redox Reactions: Understanding oxidation states of nitrogen in different compounds

Practical Chemistry Connections

1. Qualitative Analysis: Salt analysis techniques applied to organic compounds

2. Volumetric Analysis: Back titration in Kjeldahl’s method

3. Organic Tests: Identifying functional groups after knowing elements present

4. Organic Preparations: Confirming product composition after synthesis

Key Takeaways

1. Lassaigne’s Test: Converts covalent elements (N, S, X) to ionic form (NaCN, Na₂S, NaX) using sodium fusion

2. Nitrogen Detection: Prussian blue test (intense blue precipitate of Fe₄[Fe(CN)₆]₃)

3. Sulfur Detection: Sodium nitroprusside (purple color) or lead acetate (black PbS precipitate)

4. Halogen Detection: AgNO₃ after removing CN⁻ and S²⁻ with HNO₃

   • AgCl: White, soluble in dilute NH₃
   • AgBr: Pale yellow, soluble in conc. NH₃
   • AgI: Yellow, insoluble in NH₃

5. Carbon Detection: Heating with CuO → CO₂ → turns lime water milky

6. Hydrogen Detection: Heating with CuO → H₂O → anhydrous CuSO₄ turns blue

7. Interference Removal:

   • Remove S²⁻ before N test (lead acetate)
   • Remove CN⁻ and S²⁻ before halogen test (boil with HNO₃)

8. Safety: Sodium fusion dangerous - hot tube plunged in water; HCN gas toxic - work in fume hood

9. Quantitative Analysis: Combustion method for C,H; Kjeldahl’s for amino-N; Carius for S and halogens

10. Kjeldahl’s Limitation: Only works for amino nitrogen (-NH₂, -NH-), NOT for nitro/azo/ring nitrogen

Quick Revision Points

✓ Lassaigne’s: Na fusion converts organic to inorganic ionic forms ✓ Nitrogen: FeSO₄ + FeCl₃ → Prussian blue (Fe₄[Fe(CN)₆]₃) ✓ Sulfur: Nitroprusside → purple; Lead acetate → black PbS ✓ N+S together: FeCl₃ → blood red (SCN⁻ complex) ✓ Halogens: Boil with HNO₃, then AgNO₃

• Cl: White AgCl (soluble in dilute NH₃)
• Br: Pale yellow AgBr (soluble in conc. NH₃)
• I: Yellow AgI (insoluble in NH₃) ✓ Carbon: CuO → CO₂ → milky lime water ✓ Hydrogen: CuO → H₂O → blue CuSO₄ ✓ Beilstein: Green flame = halogen present (quick test) ✓ Quantitative: Combustion (C,H), Kjeldahl (amino-N), Carius (S, halogens) ✓ Safety: Fresh FeSO₄, HCN toxic, hot tube → water

Master element detection, and you unlock the identity of any organic compound!

Interactive Demo: Visualize Titration Process

Watch how titration curves change during quantitative elemental analysis experiments.