Unit 11: Qualitative Analysis (Organic)
Using simple chemical tests to identify common organic functional groups.
11.12 Tests for Alkenes, Carboxylic Acids, & Alcohols
Simple, observable chemical reactions can be used in the laboratory to identify the main functional group in an unknown organic compound.
-
Test for Alkenes (Unsaturation):
Reagent: Bromine water ($Br_2(aq)$).
Procedure: Add a few drops of orange-brown bromine water to the sample and shake.
Positive Result: The solution is rapidly decolourised (turns from orange-brown to colourless). This happens because the alkene undergoes an addition reaction with the bromine. -
Test for Carboxylic Acids:
Reagent: A carbonate or hydrogencarbonate solution or solid (e.g., $Na_2CO_3$ or $NaHCO_3$).
Procedure: Add the carbonate to the sample.
Positive Result: Effervescence (fizzing) is observed. This is due to the production of carbon dioxide gas from the acid-carbonate reaction. -
Test for Alcohols:
Reagents: A carboxylic acid (e.g., ethanoic acid) and a strong acid catalyst (conc. $H_2SO_4$).
Procedure: Gently warm the sample with the carboxylic acid and a few drops of the catalyst.
Positive Result: A characteristic sweet, fruity smell is produced due to the formation of an ester.
Solved Examples:
- An unknown liquid decolourises bromine water. What functional group is
present?
Solution: A C=C double bond (the compound is an alkene). - What would you observe if you added sodium carbonate to a solution of
ethanoic acid?
Solution: Fizzing (effervescence). - A student warms an unknown liquid with ethanoic acid and a catalyst and
notices a smell of pineapple. What type of compound was the unknown
liquid?
Solution: An alcohol. - What is the name of the reaction that produces the fruity smell in the test
for an alcohol?
Solution: Esterification. - Why does bromine water lose its colour when mixed with an
alkene?
Solution: The coloured bromine molecule ($Br_2$) is consumed as it adds across the double bond, forming a colourless dibromoalkane. - Could the test with sodium carbonate be used to identify any acid, not just
carboxylic acids?
Solution: Yes, this is a general test for acids. However, in the context of simple organic compounds, it is the characteristic test for a carboxylic acid. - What is the gas produced when a carboxylic acid reacts with sodium
carbonate?
Solution: Carbon dioxide. - Would an alkane react with bromine
water?
Solution: No, not under normal lab conditions (it requires UV light for a slow substitution reaction). - A student has three unlabelled bottles: hexane, hexene, and hexanoic acid.
How could they identify the hexanoic
acid?
Solution: By adding sodium carbonate to a sample from each bottle. The one that fizzes is hexanoic acid. - How could the student then distinguish between hexane and
hexene?
Solution: By adding bromine water to the remaining two samples. The one that decolourises the bromine water is hexene.
11.13 Distinguishing Primary/Secondary from Tertiary Alcohols
While esterification confirms a compound is an alcohol, it doesn't distinguish between the different classes (primary, secondary, tertiary). This can be done using a mild oxidising agent.
The Test:
Reagent: Acidified potassium dichromate(VI) solution ($K_2Cr_2O_7 /
H_2SO_4$).
Procedure: Add a few drops of the orange dichromate(VI) solution to the
alcohol and warm the mixture gently in a water bath.
Results:
- Primary ($1^\circ$) and Secondary ($2^\circ$) Alcohols: Are oxidised by the reagent. The dichromate(VI) ion ($Cr_2O_7^{2-}$) is reduced to the green chromium(III) ion ($Cr^{3+}$). The solution will change colour from orange to green.
- Tertiary ($3^\circ$) Alcohols: Are resistant to oxidation. There is no reaction, and the solution remains orange.
This test provides a clear visual way to distinguish tertiary alcohols from primary and secondary ones.
Solved Examples:
- What is the positive result for the oxidation of an alcohol with acidified
dichromate(VI)?
Solution: The colour changes from orange to green. - Which classes of alcohol give a positive result with this
test?
Solution: Primary and secondary alcohols. - An unknown alcohol is warmed with acidified potassium dichromate(VI) and the
solution stays orange. What can you
conclude?
Solution: The unknown alcohol is a tertiary alcohol. - What is the oxidising agent in this
test?
Solution: The dichromate(VI) ion ($Cr_2O_7^{2-}$). - What is the role of the alcohol in this
reaction?
Solution: It acts as a reducing agent. - What is the green ion responsible for the colour
change?
Solution: The chromium(III) ion ($Cr^{3+}$). - What is the change in oxidation state of chromium during this
reaction?
Solution: It is reduced from +6 in $Cr_2O_7^{2-}$ to +3 in $Cr^{3+}$. - Could you use this test to distinguish between propan-1-ol and
propan-2-ol?
Solution: No. Both are oxidised (one is primary, one is secondary), so both would turn the solution green. - Why are tertiary alcohols resistant to
oxidation?
Solution: The carbon atom attached to the -OH group does not have a hydrogen atom bonded to it, which is necessary for this type of oxidation to occur. - What other oxidising agent could be used for this test, and what would be
the colour change?
Solution: Acidified potassium manganate(VII) ($KMnO_4$). The colour change would be from purple to colourless.
11.14 Iodoform Test Revisit
The iodoform test is a more specific test that can distinguish some primary and secondary alcohols from others.
Reagents: Iodine ($I_2$) and sodium hydroxide solution ($NaOH(aq)$).
Procedure: Warm the sample with the reagents.
Positive Result: The formation of a pale yellow
precipitate of triiodomethane ($CHI_3$), which has an antiseptic smell.
Structural Requirement:
An alcohol will only give a positive iodoform test if it contains the $CH_3CH(OH)-$ group.
- The only primary alcohol that works is ethanol ($CH_3CH_2OH$).
- All secondary alcohols where the -OH group is on the second carbon of a chain (methyl-secondary alcohols) will give a positive result (e.g., propan-2-ol, butan-2-ol).
- No tertiary alcohols give a positive result.
This test is therefore very useful for identifying ethanol or distinguishing between positional isomers like propan-1-ol (negative) and propan-2-ol (positive).
Solved Examples:
- What is the visible positive result of the iodoform
test?
Solution: A pale yellow precipitate. - What is the chemical name of the yellow
precipitate?
Solution: Triiodomethane or iodoform ($CHI_3$). - What specific structural feature must an alcohol have to give a positive
iodoform test?
Solution: The $CH_3CH(OH)-$ group. - Which is the only primary alcohol to give a positive iodoform
test?
Solution: Ethanol. - A student has two isomers of butanol: butan-1-ol and butan-2-ol. Which test
could distinguish them?
Solution: The iodoform test. Butan-2-ol would give a yellow precipitate, while butan-1-ol would not. - Will pentan-3-ol give a positive iodoform
test?
Solution: No. The -OH group is on the third carbon, so it does not have the required $CH_3CH(OH)-$ structure. - Will 2-methylpropan-1-ol give a positive iodoform
test?
Solution: No, it is a primary alcohol other than ethanol. - An unknown alcohol gives a positive iodoform test. What can you conclude
about its structure?
Solution: It is either ethanol or a secondary alcohol with the -OH group on the second carbon of a chain. - What are the reagents for this test?
Solution: Iodine and sodium hydroxide solution. - The iodoform test also works for aldehydes and ketones with a $CH_3C=O$
group. Which ketone would give a positive test: propanone or
pentan-3-one?
Solution: Propanone ($CH_3COCH_3$), as it has the required methyl ketone structure.