Identifying An Unknown Organic Compound

Ricardo Paixao

Analysis:

Compound A

Through the use of Wet tests I am able to determine the likely identity of the organic unknown, I shall start with the reactions of Compound A.

When Blue litmus paper is used to test compound A there is no observable change in colour. Since Blue Litmus is the test for a proton donor1 it tells me that compound A is not an acid. If compound A were an acid then the blue litmus paper would turn red.

The test with acidified potassium dichromate turned out negative. This means that compound A can possibly be a tertiary alcohol. This is because if the compound were a primary alcohol an aldehyde would be formed. The aldehyde would cause the dichromate ions to change to chromium ions (a green crystalline ionic compound). If compound A were a secondary alcohol then a reaction would also take place whereby chromium ions would be formed.

Phosphorous pentachloride is a chemical, used to test a compound for the presence of an OH (hydroxyl) group2 since compound A does not react it leads to the conclusion that there is no OH group present in compound A.

The test for the presence of a carbonyl group uses 2,4 dinitrophenylhydrazine3 (2,4 DNP). This chemical will form an orange precipitate if there is a carbonyl group present in the compound hydrazone4 will be formed. The equation for this reaction is5:

When sodium carbonate is added to compound A not reaction takes place. This indicates that compound A is not a carboxylic acid. A positive test for carboxylic acid would result in the production of carbon dioxide gas, which could be tested with limewater. Since there is no reaction between compound A and sodium carbonate, it leads to the conclusion that compound A is not a carboxylic acid.

Compound A does not react with neutral iron (III) chloride. This is a test for phenols. If compound A was a phenol then a colour change to purple would be noted since this is not the case then the conclusion that compound A is not a phenol can be reached.

When compound A is added to Tollen’s reagent there is no reaction. Although compound A is a carbonyl compound, in this case it is not an aldehyde. This is because aldehydes would react with the silver ions in the Tollen’s reagent to form silver metal to be precipitated. A reaction of such a case is given below6:

CH3CHO + 2Ag+ + 3OH- → CH3CO2- + 2Ag + H2O

Since this reaction doesn’t take place it can be concluded that compound A is not an aldehyde.

Compound A does not react with sodium metal, this confirms what the phosphorous pentachloride had indicated, that compound A has no OH group. This eliminates the chances of compound A being an alcohol.

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CH3CHO + 2Ag+ + 3OH- → CH3CO2- + 2Ag + H2O

Since this reaction doesn’t take place it can be concluded that compound A is not an aldehyde.

There is no reaction between bromine water and compound A. If compound A were an alkene then electrophilic addition would occur and a bromoalkane would be formed7. The bromine water would go colourless. It is important to carry out this reaction in the dark to ensure if compound A is an alkene or not. If compound A were an alkane then in the presence of UV light a reaction would take place. Since there is no reaction with compound A and bromine water, I can conclude that the compound is not an alkene.

When compound A is heated with ethanoic acid and sulphuric acid no reaction takes place. This means that compound A is not an alcohol because if it were then an ester would be formed and the test tube would contain a fruity smelling product.

Fehling’s solution did not react with compound A. This supports the Tollen’s reagent test, in that no reaction with Fehling’s solution indicates that compound A is not an aldehyde otherwise a brick-red precipitate would form.

All these wet tests lead me to believe that compound A is actually a ketone. I will now make use of the spectra of compound A to try and decide upon a structure for this ketone.

Using the mass spectrum I can tell that compound A has the total molecular mass of 58, this is shown by the peak furthest to the right. The molecular mass can be worked out by adding the 43 (from the tallest peak) and the 15 (from the peak on the far left).

The NMR spectrum of compound A shows one peak at chemical shift (δ) = 2.2. this NMR peak also shows that there are six protons present at that peak. Both these pieces of information help me conclude that there are 6 hydrogen atoms in compound A. These six hydrogen atoms are in the same environment. Looking at the chemical shift table indicates that the proton type at δ = 2.2 is R-C(=O)CH3. This piece of information together with the mass spectrum leads me to believe that compound A is the ketone, propanone with formula CH3C(=O)CH3. The C=O wouldn’t show up on NMR because it does not consist of a proton.

Using the Infra-red (IR) spectrum further supports my conclusion of the identity of compound A. The IR spectrum of compound A shows an absorption at 1705 cm-1 this indicates the presence of a C=O in compound A.

Since (12*3)+(1*6)+16 = 58 my conclusion for the structure of compound A is supported by the mass spectrum.

Using the results from the wet test and the spectra I can conclude that compound A is the ketone propanone. Both the wet test and the spectroscopy techniques support this conclusion strongly.

Compound E

I will now use the same method as used above to determine the structure and name of compound E.

The blue litmus test indicates that compound E is not an acid. This is because the blue litmus paper does not change to a red colour.

A reaction with acidified potassium dichromate does not occur. This indicates that compound E may be a tertiary alcohol. If it were any other alcohol there would be a reaction and green chromium ion crystals would form.

When phosphorous pentachloride is added to compound E, steamy white fumes evolve which turn moist blue litmus paper red. This happens because compound E contains an OH group. When the phosphorous pentachloride is added to the unknown compound then, because the OH group is present hydrogen chloride is evolved8. The hydrogen chloride turns the blue litmus paper red because it is an acidic gas. An example of this reaction is given below:

CH3CH2OH + PCl5 → CH3CH2Cl + PCl3 + HCl

This quality of compound E leads me to believe that it may be an alcohol.

When 2,4 DNP is added to compound E, no reaction takes place. This means that there is no carbonyl group present in compound E. therefore it cannot be an aldehyde or a ketone.

Compound E does not react with sodium carbonate to form carbon dioxide. Since carbon dioxide is not formed, compound E cannot be a carboxylic acid.

Neutral iron (III) chloride is used to test a compound for phenols. The positive test for a phenol would be a colour change from colourless to purple. This colour change occurs because a purple product is formed. Since no reaction occurs between iron (III) chloride and compound E occurs the conclusion that compound E is not a phenol is reached.

Tollen’s reagent is used to test for an aldehyde. Since compound E tested negative for a carbonyl compound it should also test negative for this reaction. This is the case. No reaction takes place therefore compound E is not an aldehyde.

When sodium metal is used to test compound E, effervescence occurs and gas is evolved. This supports the findings from the earlier experiment with phosphorous pentachloride. In this reaction the sodium metal liberates the hydrogen from the present OH group. Sodium ethoxide is formed and hydrogen gas is given off. The hydrogen gas turns the litmus paper red because it contains hydrogen. An example of the formation of sodium ethoxide being formed is given below:

CH3CH2OH + Na → CH3CH2O-Na+ + ½H2

This reaction will determine the presence of the OH group.

There is no reaction between compound E and bromine water, therefore compound E cannot be an alkene because otherwise decolourisation would occur.

When compound E is warmed with ethanoic acid and sulphuric acid, an oily, sweet smelling liquid is formed when product is added to water. The sweet smell indicates the presence of an ester. When alcohols are heated with ethanoic acid and sulphuric acid esters are formed. The reaction for such a reaction is given below:

CH3CH2OH + CH3C(=O)OH → CH3C(=O)OCH2CH3 + H2O

The above equation shows an esterification reaction of alcohol. Compound E reacts with a carboxylic acid, in the presence of a sulphuric acid catalyst, to form an ester. This is a strong indication that compound E is an alcohol.

There is no reaction between compound E and Fehling’s solution therefore compound E cannot be an aldehyde.

I believe that compound E may be an alcohol. The wet tests seem to indicate this rather strongly. I will now analyse the spectra for compound E to determine its structure.

The mass spectrum of compound E shows that it has the molecular mass of 74.

The NMR of compound E shows that there are two different environments in this compound. Alcohols often contain two different environments because they contain an OH proton type and often alkyl type protons. The peak at δ = 1.2 suggests that proton type R-CH3 is present. Since there are 9 protons detected at this chemical shift it suggests that there are 3 of these types of protons present in the compound. δ = 2 suggests the presence of an R-OH proton type. Since there are 3 proton type R-CH3 and one R-OH proton type. The structure of this alcohol seems to be a tertiary one. 2-methylpropan-2-ol (CH3C(CH3)(OH)CH3) seems to be compound E and since it does not react with acidified potassium dichromate, my conclusion is supported.

The IR spectrum shows that there is a broad, strong absorption at 3400cm-1 and since this is the expected brad range for an OH group, it supports my conclusion. The absorption at 1300cm-1 shows a C-O bond and this is present in all alcohols since the carbon bonds to the oxygen atom in the hydroxyl group.

Since (12*4)+(1*10)+16 = 74 it supports the findings from the mass spectrum.

Using both the wet tests and the spectroscopy techniques I have been able to successfully conclude the structure of compound E as CH3C(CH3)(OH)CH3 (2-methylpropan-2-ol).