Identification of an unknown Organic Compound

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Chemistry A2 C/W                Komail Rajani

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Identification of an unknown Organic Compound

The organic compound that I have to identify can have several functional groups. These are:

  • Alkene
  • Phenol
  • Carboxylic acid
  • Aldehyde
  • Ketone
  • Primary alcohol
  • Tertiary alcohol
  • Ester

The functional group in an alkene is that it has a double bond. This function allows an alkene to undergo addition reactions as it has an area of high electron density that attracts electrophiles. Therefore, in order to test for alkenes, I will use the Bromine water test as bromine can act as an electrophile. A bromine molecule may have instantaneous dipole between each other. This causes the + end of the dipoles to attack the high electron density area. If the unknown compound is an alkene, than the bromine water which is originally brown will decolourise. The mechanism for this reaction shown below;

However, phenol will also decolourise bromine water as well as form white precipitant. This is an indication that the unknown compound is a phenol rather than an alkene. Phenol is an aromatic compound similar to Benzene but one of the hydrogen atoms is substituted by and OH group.

Phenol molecules form hydrogen bonds resulting into white crystalline solid. When aqueous bromine reacts with a phenol molecule, the hydrogen atoms in the phenol are substituted by the bromine atoms. Thus the final product is 2,4,6-tribromophenol. The reason this reaction is a tri-substitution is because of the OH group in the phenol. It releases electrons onto the aromatic ring in the phenol. This activates the ring allowing it to be more reactive to electrophiles. This is one of the fundamental differences between Benzene and Phenol. This is why electrophiles such as bromine react readily with phenol than in benzene. The mechanism for this reaction is shown below;

To test for a carboxylic acid, I will react the unknown organic compound with sodium carbonate. Positive results will show carbon dioxide produced. This is because the general equation is;

Carboxylic acid + Metal Carbonate                        Salt + Water + Carbon dioxide

However, to test for CO2, I will use the limewater test. Limewater, Ca(OH)2 will react with CO2 to form white precipitate, CaCO3, and water. It is the CaCO3 that turns the limewater milky. This identifies that CO2 is present.

As aldehydes and ketones are classified as carbonyl compounds, I will perform a general test to show whether the unknown organic compound is a carbonyl or not. To test for this, I will add Brady’s reagent. Brady’s reagent is 2,4-dinitrophenylhydrazine (2,4-DNPH) mixed with either methanol or sulphuric acid.

Brady’s reagent is to test for a C=O double bond. The reason for using this test is because both aldehydes and ketones have a C=O. As the unknown compound is added to Brady’s reagent and a bright orange precipitant is observed, than a carbonyl is present.

However, as mentioned above, this will only show us whether the unknown compound is a carbonyl or not. I will have to do further tests in order to distinguish whether the carbonyl is an aldehyde or a ketone. To do this, I will perform the silver mirror test.

The silver mirror test – known as Tollen’s reagent is a solution of silver nitrate dissolved in aqueous ammonia. This alkaline solution provides Ag+ that can be reduced by the aldehyde. The following equations show that if an aldehyde is present, it will be oxidised, thus allowing the Ag+ to be reduced as they will gain an electron.

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RCHO + [O]                  RCOOH                Aldehyde is oxidised

Ag+ + e-                 Ag                        Silver is reduced

If the carbonyl is an aldehyde, a layer of metallic silver mirror will form in the test tube. Ketones can’t be oxidised under normal conditions, thus there will be no reaction. Therefore, this test will distinguish whether the carbonyl is an aldehyde or a ketone.

To test for alcohols, I will oxidise the unknown compound with excess potassium dichromate (oxidising agent) and concentrated sulphuric acid. The mixture will also be refluxed. This process is the complete oxidation of alcohols. However, only primary and secondary alcohols can ...

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