Jacaranda 9.2 – Laboratory analysis: Tests for C=C, -OH and -COOH functional groups
Learning Intention
To learn about the different laboratory tests used to detect key functional groups in organic compounds:
Carbon–carbon double bonds (C=C)
Hydroxyl groups (-OH)
Carboxyl groups (-COOH)
Emphasis on how these tests inform the chemical and physical properties of substances, including polarity and intermolecular forces.
The Study Design
Focus: qualitative tests for the presence of C=C, -OH, and -COOH functional groups.
Applications and principles include:
Verifying components and purity of consumer products.
Melting point determination and distillation (simple and fractional).
Measurement of the degree of unsaturation using the iodine number.
Redox titrations for quantitative analysis (including calculations of excess and limiting reactants; back titrations excluded).
Introduction
Targeted functional groups:
Carbon–carbon double bonds, C=C
Hydroxyl groups, R–OH
Carboxyl groups, R–COOH
Relevance:
These groups determine polarity and intermolecular forces.
Detection helps predict chemical and physical properties of the substance.
Detection of carbon-to-carbon double bonds
The bromine water test
Purpose: Qualitative determination of degree of unsaturation (presence of C=C).
Mechanism: Bromine (Br2) reacts across C=C bonds, decolorizing the Br2 solution.
Colour change: Br2 solution is orange-brown; it fades and can become colourless if Br2 is consumed by reaction with an unsaturated compound.
Interpretation: Decolorization indicates the presence of C=C bonds (unsaturation).
Quantitative hint: The test can be used to estimate the iodine number via stoichiometry (next section).
The potassium permanganate test (KMnO4)
Reagent: Acidified KMnO4 (aq) tests for the presence of double or triple bonds.
Reaction with unsaturated compounds:
Addition of acidified KMnO4 reduces purple ext{MnO}_4^- to colorless solution.
Alternative (alkaline KMnO4): Can be reduced to a dark green solution with a brown precipitate of ext{MnO}_2 .
Interpretation: Decolorization/green-to-brown changes indicate unsaturation (C=C or C≡C).
The Iodine number test
Purpose: Quantitative determination of unsaturation by measuring how much iodine reacts with 100 g of substance.
Concept: Iodine ext{I}_2 adds across C=C bonds via an addition reaction. Stoichiometry: 1 mole of ext{I}_2 reacts with 1 mole of C=C.
Higher iodine number → greater degree of unsaturation (more C=C bonds).
Applications: Determine whether fats/oats/oils are saturated, monounsaturated, or polyunsaturated.
Calculate the Iodine number
Key idea: 100 g of a compound reacts completely with ext{I}_2 .
How to determine the iodine number:
Measure mass of iodine that reacted, ext{m(I}_2) .
Iodine number, I.V. = (mass of ext{I}_2 that reacted in grams) × 100 / (mass of sample in grams).
General reaction:
ext{I}_2 + ext{C=C} \rightarrow ext{diiodinated product}Example workflow (qualitative data to a numeric value):
Let ext{m(sample)} be the sample mass used (often 100 g for the standard method).
Determine ext{m(I}_2) that has reacted from the mass gain or titration/detection method.
Compute ext{I.V.} = \frac{ ext{m(I}_2)}{ ext{m}_{\text{sample}}} \times 100
Example interpretation steps (as shown in the transcript):
Iodine adds across all C=C sites in the sample; the more C=C bonds, the more ext{I}_2 is consumed, hence a higher I.V.
If a sample of fat is used, determine the mass change due to iodine addition and apply the I.V. formula to gauge saturation vs unsaturation.
Practice: Iodine number example structure (from transcript guidance)
Step 1: Assume 100 g of a fat reacts completely with ext{I}_2 . Determine the mass of ext{I}_2 that reacted, ext{m(I}_2) .
Step 2: Calculate ext{n(I}_2) = \text{m(I}_2) / \text{M(I}_2) , with \text{M(I}_2) = 253.8 \text{ g/mol} .
Step 3: Because each C=C reacts with 1 mole of ext{I}_2 , set ext{n(C=C)} = ext{n(I}_2) .
Step 4: Determine ext{n(fat)} = \text{m(sample)} / \text{M(fat)} using the given molar mass ext{M(fat)} .
Step 5: Compute the total number of C=C bonds per fat molecule: Number of C=C per fat molecule = ext{n(C=C)} / ext{n(fat)} .
Step 6: Interpret results: higher value indicates greater unsaturation per fat molecule.
Practice Problem 1 (Iodine number exercise)
Task: A 100 g sample of a fat reacts with iodine, and the final mass after reaction is 200 g. ext{M(fat)} = 254.0 \text{ g/mol} .
Questions:
a) Calculate the iodine number of the fat.
b) Calculate the amount of ext{I}_2 (in mol) that reacted with the fat.
c) Calculate the amount (in mol) of C=C present in the fat sample.
d) Calculate the amount (in mol) of fat in the 100 g sample.
e) Determine the number of C=C present (degree of unsaturation) in each fat molecule.
Detection of hydroxyl groups
The sodium metal test (hydrogen pop test)
Purpose: Qualitative detection of hydroxyl (-OH) groups (often used for alcohols).
Reaction: Alcohols react with active metals (e.g., Na) to form an ionic salt and hydrogen gas:
2 \text{CH}3\text{OH (l)} + 2 \text{Na (s)} \rightarrow 2 \text{CH}3\text{ONa (l)} + \text{H}_2 \text{(g)}Notes:
Alcohols are typically liquids (not aqueous) in these tests.
Procedure: add a piece of Na metal to a sample in a test tube; observe hydrogen gas bubbles and a pop when a lit taper is introduced near the mouth.
Safety: Na is highly reactive; perform under proper supervision and with appropriate safety protocols.
The Lucas test
Purpose: Qualitative differentiation of primary, secondary, and tertiary alcohols.
Reagents: Concentrated HCl with ext{ZnCl}_2 catalyst (Lucas reagent).
Observation: Presence of an alcohol generally yields a cloudy liquid; the rate and persistence help distinguish substitution class:
Primary alcohols: slower cloudy formation
Secondary alcohols: quicker cloudiness
Tertiary alcohols: immediate/rapid cloudiness
The oxidation test
Purpose: Determine the class of alcohol after initial detection of -OH.
Reagents: Acidified dichromate ( ext{Cr}2\text{O}7^{2-} ) or permanganate ( ext{MnO}4^- ) in acidic medium ( ext{H}2\text{SO}_4 ).
Color changes (typical with ext{Cr}2\text{O}7^{2-} ):
Primary or secondary alcohols under reflux: ext{Cr}2\text{O}7^{2-} (orange) → Green ( ext{Cr}^{3+} )
Tertiary alcohols: ext{Cr}2\text{O}7^{2-} remains orange (no reaction)
Color changes (typical with ext{MnO}_4^- ):
Primary or secondary alcohols: ext{MnO}_4^- (pink) → colorless ( ext{Mn}^{2+} )
Tertiary alcohols: ext{MnO}_4^- pink persists (no reaction)
Note: Reflux is used to apply heat over an extended period for the reaction.
Interpretation: Green color indicates oxidation of primary/secondary alcohols; persistence of orange or pink indicates tertiary alcohols or no reaction, respectively.
Esterification context for alcohol detection
Alcohols can be identified via esterification (with a carboxylic acid) using concentrated sulfuric acid as a catalyst.
Procedure: add an alcohol and a carboxylic acid to a tube with concentrated ext{H}2\text{SO}4 and heat.
Observation: Fruity smell indicates ester formation, confirming the presence of an alcohol reacting with a carboxylic acid.
Esterification role in alcohol testing (summary)
Esterification tests can help confirm alcohol presence by producing esters with carboxylic acids (fruit-like odors).
This test is often discussed alongside the dehydration esterification reaction:
\text{Alcohol} + \text{Carboxylic acid} \rightarrow \text{Ester} + \text{H}_2\text{O}
Detection of carboxyl groups
Carboxylic acids as weak acids
Carboxylic acids ionize in aqueous solutions to release ext{H}^+ ions, lowering pH.
Detection methods include:
pH measurement using indicators or pH probes.
Hydrogen carbonate limewater tests to confirm ext{CO}_2 evolution.
The hydrogen carbonate test (carbonate test)
Procedure: add powdered ext{NaHCO}_3 (or ext{Na}_2\text{CO}_3 ) to a sample containing a carboxylic acid.
Observation: production of carbon dioxide gas ( ext{CO}_2 ) bubbles.
Confirmation: ext{CO}_2 will turn limewater ( ext{Ca(OH)}_2 ) cloudy due to the formation of calcium carbonate ( ext{CaCO}_3 ).
Other relevant reactants: metal hydrogen carbonates and metal carbonates can also react to produce ext{CO}_2 .
The esterification test (carboxyl group detection context)
Carboxylic acids can be detected by reacting with an alcohol in the presence of a concentrated sulfuric acid catalyst to produce an ester, which has a characteristic fruity smell.
This test parallels the alcohol detection test, reinforcing the presence of -COOH groups when ester formation occurs.
Observations summary for carboxyl detection
Acidic pH indicates presence of carboxyl group due to acidity of carboxylic acids.
Hydrogen carbonate reaction with ext{CO}_2 confirms carboxyl groups via gas evolution and limewater cloudiness.
Fruity smell from esterification supports the presence of a carboxyl-containing compound that reacts with an alcohol to form an ester.
Summary of tests and observations (as in the endpoint table)
Bromine water test (C=C detection):
Unsaturated: ext{Br}_2 color fades to colorless; saturation: color persists as orange-brown.
Sodium metal test (OH detection):
Hydrogen gas evolution indicates presence of an -OH group (alcohol) or carboxyl group (both can react with Na).
Oxidation tests ( ext{Cr}2\text{O}7^{2-} or ext{MnO}_4^- ) for alcohols:
Primary/secondary alcohols: ext{Cr(VI)} reduction to ext{Cr(III)} (orange ext{Cr}2\text{O}7^{2-} → green ext{Cr}^{3+} ); ext{MnO}4^- reduction (pink ext{MnO}4^- → colorless ext{Mn}^{2+} ).
Tertiary alcohols: little to no change (orange persists for ext{Cr}2\text{O}7^{2-} ; ext{MnO}_4^- pink persists).
Esterification test (presence of alcohol or carboxyl):
Fruity smell indicates ester formation, confirming alcohol and carboxyl group interaction.
Carboxyl tests:
pH measurement shows acidic solution for carboxyl-containing compounds.
Hydrogen carbonate test produces ext{CO}_2 ; limewater turns cloudy.
Esterification can confirm carboxyl presence via ester formation.
Observations across tests help identify:
C=C presence (bromine decolorization, ext{KMnO}_4 reaction)
-OH presence (sodium metal, Lucas, oxidation tests)
-COOH presence (acidic pH, bicarbonate/limewater tests, esterification outcomes)
Sample Problem 2 (interpretation exercise)
Situation: Unknown organic compound is a hydrocarbon, alcohol, or carboxylic acid; tests performed:
1) Bromine water: orange-brown colour disappeared (decolorized).
2) pH probe: pH = 4.
3) Sodium metal: hydrogen gas bubbles observed.THINK (interpretation):
a) Decolorization of bromine water indicates unsaturation (C=C) present.
b) Hydrogen gas with sodium metal can indicate hydroxyl-containing compounds (alcohols) but carboxylates may also respond similarly under some conditions.
c) pH 4 indicates acidity consistent with a carboxyl group (carboxylic acids are weak acids and lower the pH).
d) The molecule likely contains a carboxyl group and a C=C bond (unsaturation) and may also contain an -OH group (if alcohol present).
WRITE (concise conclusions):
a) The unknown is unsaturated (due to ext{Br}_2 decolorization).
b) Tests consistent with hydroxyl group: sodium metal test ( ext{H}_2 gas).
c) Acidic pH (pH = 4) is consistent with a carboxyl group.
d) Functional groups present: C=C (unsaturation) and -COOH (carboxyl), with potential -OH contribution.
Practice Problem 2 (interpretation)
Observations:
1) Bromine water: orange-brown colour persisted (no decolorization).
2) Acidified dichromate oxidation: orange colour persisted (no oxidation).
3) Reaction with an alcohol in the presence of concentrated sulfuric acid: fruity smell was not detected.
4) Reaction with a carboxylic acid: fruity smell produced.Questions and answers:
a) Saturated vs unsaturated: The orange-brown ext{Br}_2 colour persisted, so the substance is saturated.
b) Tests showing carboxyl group: test 4 (reaction with carboxylic acid) indicates carboxyl presence; acidic behavior could be inferred if pH were measured, but here the fruity smell confirms esterification context.
c) Test observation that could only be produced by an alcohol: the fruity smell from ester formation (test 3 would yield such a smell if an alcohol reacted with a carboxyl group; in this case, no fruity smell was detected, so alcohol presence is unlikely).
d) Functional groups present: carboxyl group (–COOH) is present (test 4), and the lack of unsaturation and lack of alcohol-reactivity in test 3 reduces likelihood of an alcohol; overall, the data point to a carboxylic acid without an alcohol or with limited reactive alcohol content in the setup.
Conclusion
This section highlighted the main laboratory tests for detecting C=C, -OH, and -COOH functional groups, including qualitative and quantitative approaches.
You should now be able to:
Identify the different tests used to detect C=C, -OH, and -COOH groups.
Understand how these tests are conducted in the laboratory.
Use qualitative test data to identify unknown organic compounds.
Homework: Complete Jacaranda 9.2 questions and seek help if stuck.
Connections to foundational principles and real-world relevance
These tests illustrate core organic qualitative analysis, linking molecular structure (functional groups) to observable properties (color changes, gas evolution, smells).
Understanding degree of unsaturation informs properties such as melting/boiling behavior, reactivity, and nutritional aspects for fats/oils.
Practical skills developed include acid–base concepts (pH changes), redox chemistry ( ext{Cr(VI)} and ext{MnO}_4^- tests), and reaction mechanisms (esterification).
Safety and ethics in the laboratory (practical implications)
Handle reactive metals (e.g., sodium) with care; they react vigorously with water and air; use appropriate PPE and work in a suitable fume hood.
Work with concentrated acids ( ext{H}2\text{SO}4 ) and oxidizing agents ( ext{Cr}2\text{O}7^{2-} , ext{MnO}_4^- ) under supervision and with proper waste disposal protocols.
Properly label and dispose of chemical wastes, especially organic/inorganic oxidizers and halogen-containing reagents.
Document observations accurately and cite uncertainties in qualitative assessments.
Key formulas and equations to remember
Iodine addition across C=C (qualitative):
\text{I}_2 + \text{C=C} \rightarrow \text{diiodinated product}Iodine number (I.V.) definition (per 100 g sample):
\text{I.V.} = \frac{\text{m}(\text{I}2 \text{ reacted})}{\text{m}{\text{sample}}} \times 100Molar relation for \text{I}2 in the iodine test: \text{n}(\text{I}2) = \frac{\text{m}(\text{I}2)}{\text{M}(\text{I}2)} \text{ with } \text{M}(\text{I}_2) = 253.8 \text{ g/mol}
Alcohol oxidation (general):
With \text{Cr}2\text{O}7^{2-} : orange \text{Cr}2\text{O}7^{2-} → green \text{Cr}^{3+} for oxidizable primary/secondary alcohols.
With \text{MnO}4^- : pink \text{MnO}4^- → colorless \text{Mn}^{2+} for oxidizable primary/secondary alcohols.
Esterification (condensation):
\text{Alcohol} + \text{Carboxylic acid} \rightarrow \text{Ester} + \text{H}_2\text{O}