Alkanes and Alkenes
Alkanes are saturated. All the C's are bonded to 4 other atoms and so it is relativily stable.
Alkenes are Un-Saturated and contain a C=C bond. These bonds are likely to be broken and react.
Use Aqueous Bromine.
Simpily add in a few drops of Aqueous Bromine
Alkanes and Alkenes
Alkane- No change, solution will stay Brown as the alkane is Saturated
Alkene- Change in the solutions colour from Brown to Colourless. The Alkene is Un-Saturated an so it can react with the Bromine.
Aldehydes and Ketones
Both contain a C=O group.
Aldehydes have it on the end, and their names end in -al.
Ketones contain the C=O group in the middle, with their names ending in -one.
Test 1: Add a few drops of Tollens Reagent and gently warm.
Test 2: Add a few drops of Fehlings Solution, and gently warm.
Test 3: Add in Accidified Potasium Dichromate.
Aldehydes and Ketones Test 1
The Aldehydes will give a silver mirror on the Test tube.
The following reaction will take place;
Ag+(aq)+ e- → Ag(s)
Ketones will give no reaction, hence no Silver Mirror.
Aldehydes and Ketones Test 2
Aldehydes will turn the solution from blue to a brick red precipitate. The following reaction will occour;
2Cu2+(aq) + 2e- + 2OH-(aq) → Cu2O(s) + H2O(l)
Ketones will not react with the Fehlings Solution.
Aldehydes and Ketones Test 3
Aldehydes will turn the solution from Orange to Green as the Chromium is reduced from Cr2O72 to Cr3+ ions.
Contains the -COOH group on the end.
They are a weak acid with a pH of about 3.
Test 1- Use Universal Indicator.
Test 2- React with a Carbonate and bubble any gas though lime water.
Test 1- Since the pH is about 3, the universal indicator will turn to orange or red. However, most acids will do this.
Test 2- Carboxylic acids react with carbonates to form a salt, water and carbon dioxide. The carbon dioxide will cause the lime water to turn cloudy.
They contain the -OH group, and can be either primary, secondary or tertiary.
Test part 1- Heat with Accidified Potassium Dichromate.
Test part 2- Use the tests to identify between Aldehydes and Ketones.
Part 1- Teriary alcohols are not easily oxidised and so there will be no colour change.
Primary and Secondary alcohols will be oxidised, causing a change from orange to green.
Part 2- If a colour change occoured, then you will be left with an Aldehyde or a Ketone.
Primary alcohols go to an Aldehyde.
Secondary alcohols from Ketones.
Therfore, use the standard test to identify what is left.
These are atoms found in group 7 and consist of F, Cl, Br, and I at A level.
Part 1- Warm with NaOH to relase the Halide ion.
Part 2- Acidify the mixture with nitric acid. (HCl would give off Chlorine Gas!)
Part 3- Add silver nitrate.
Part 4- Add in ammonia to fully identify the Halide ion.
Haloalkanes Parts 1 and 2
Part 1- This step is only there to remove the Halide ion from the Haloalkane.
Part 2- Without this step, the next reactions would not occour. Make sure that you use something like nitric acid and not HCl as HCl will give off Chlorine gas which is hazardous.
Haloalkanes Part 3
When the silver nitrate is added a precipitate may occour;
No Precipitate --- Flouride ions are present
White Precipitate --- Chloride ions are present
Cream Precipitate --- Bromide ions are present
Yellow Precipitate --- Iodide ions are present
Haloalkanes Part 4
The previous part identified the Halide ions that were present. This just confirms any results.
If a precipitate was discovered first add dilute ammonia. The white, chloride precipitate should disolve. Other precipitates will remain.
Add Concentrated ammonia to any remaining precipitates. The bromide precipitate should now disolve.
The iodide ion precipitate should not disolve at all.