WAEC often brings out the analysis of a salt in qualitative analysis, but this year they decided to change it to the qualitative analysis of an organic compound labelled Cn (propan-2-ol). See the image of the specimen for proof.
So it is important you have a knowledge on how to analyze an organic compound, in this case, propan-2-ol and this I have done with my students in the laboratory.
The writer, "Osasbio" will explain the concept of the qualitative analysis of propan-2-ol before giving you possible questions and answers you may likely see on that day.
Cn as provided as WAEC is propan-2-ol. Propan-2-ol is a secondary alcohol with two alkyl groups. It has an hydroxyl group, two methyl group, and an hydrogen atom bonded to a central carbon atom. Below is an image of the structure of the compound;
(Propan-2-ol, chemical formula CH3CHOHCH3)
There are several tests that can be used to distinguished between primary, secondary and tertiary alcohols, some are; the use of Lucas reagent (a reagent consisting of Hydrochloric acid and Zinc chloride) and reaction with potassium dichromate solution (K2Cr2O7).
For the qualitative analysis of propan-2-ol, we are going to carryout two test to confirm that the organic compound is propan-2-ol, these tests are:
1. Reaction with acidified potassium dichromate solution.
2. Iodoform test.
Reaction with acidified potassium dichromate solution.
Potassium heptaoxodichromate (vi) (K2Cr2O7) is an oxidizing agent and it oxidizes primary alcohols first to aldehydes (alkanals) and then to carboxylic acids (alkanoic acids). The color of the potassium dichromate solution is orange and after the reaction, it will change from orange to green due to the reduction of Cr6+ to Cr3+.
Potassium dichromate solution also oxidizes secondary alcohols, but they form ketones (alkanone) instead of aldehydes. Tertiary alcohols can not be oxidized by potassium dichromate solution, so it gives a negative test.
In summary, primary and secondary alcohols give a positive test for potassium dichromate solution.
Propan-2-ol reacts with acidified K2Cr2O7 to form propanone (acetone) and water. In the reaction, an oxygen atom from K2Cr2O7 oxidizes propan-2-ol to propanone.
CH3CHOHCH3 + [O] from K2Cr2O7 ----------> CH3COCH3 + H2O
This is a test used to check for the presence of carbonyl compounds (RCOCH3) or secondary alcohols (RCHOHCH3) in a given unknown substance.
NB: Primary alcohols do not give a positive test for Iodoform, but the exception is that of Ethanol which forms Iodoethane and gives a positive test.
Methyl ketones, aldehydes, ethanol and secondary alcohols with a methyl group at the carbon atom bearing the hydroxyl group all give a positive test for Iodoform.
The appearance of a pale yellow precipitate with an antiseptic smell indicates positive test.
First, propan-2-ol reacts with dilute sodium hydroxide and iodine to form propanone (acetone, CH3COCH3), sodium iodide and water.
CH3CHOHCH3 + 2NaOH +I2 ------> CH3COCH3 + 2NaI + 2H2O.
Upon addition of excess iodine, the acetone is converted to triiodoacetone, which is observed due to the production of a pale yellow precipitate.
CH3COCH3 + 3I2 --------> CH3COCI3 + 3HI
1. Since it is an alcohol, they might want you to check if it is soluble in water. If you at asked to do this, then you to mix "Cn", in this case which is propan-2-ol with about 10cm³ of distilled water and shake thoroughly.
2. For the second test, you may be asked to react "Cn" with acidified potassium dichromate solution.
3. For the third test, you may be asked to react "Cn" with 2cm³ of distilled water, 2cm³ of 10% NaOH, and excess 10% Iodine solution.
This is how your table will look like:
NB: This questions and answers are tentative and are subject to changes, it is just a practical guide on what you are to expect which was done by me.
Please ensure you share this to everyone and drop your questions in the areas you are not clear on and I will answer. Don't be sceptical I am a Chemistry tutor.
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