Identification of an organic unknown.

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IDENTIFICATION OF AN ORGANIC UNKNOWN.

PLANNING:

Aim:

To outline a sequence of simple chemical tests that I could use to identify each of the following groups: •an alcohol, •an aldehyde, •a ketone, •a carboxylic acid, •an ester and •a phenol.

Introduction:

I will be given one of the above organic unknowns and it is my job to investigate which one of them I have via methods of qualitative analysis. Firstly I will need to familiarise myself with each of the groups. Below is a diagram detailing how one group can be transformed into another. Such processes will need to be duly noted when planning the method for the identification of the unknown as we may unintentionally change the chemical structure of the unknown.

I must now begin to investigate tests that identify particular groups. These I will have to string together and make into a flow diagram whereby each test is dependant on the result of the previous test.

Name (and example)

Functional Group

Test for the presence of this specific group

Alcohol

(methanol)

Heat and reflux with acidified potassium dichromate (K2Cr2O7 - orange). Unless alcohol is a tertiary alcohol, solution will form Cr3+ ions - blue - and there will be a colour change.

Aldehyde

(methanal - formerly 'formaldehyde')

Aldehydes and ketones (carbonyls) will turn 2,4-dinitrophenylhydrazine (2,4-DNPH) orange in sulphuric acid and methanol.

Adding an aldehyde to a clear solution which contains ions of [Ag(NH3)2]+ in a test tube in warm water will make a shiny silver mirror form. An alternative is to use copper (II) ions in an alkaline solution instead of the silver (I). The solution should turn from blue to red-brown.

Ketone

(propanone - formerly 'acetone')

2,4-DNPH undergoes a condensation reaction with aldehydes and ketones. Adding a ketone to the above mentioned chemicals ([Ag(NH3)2]+ and complexed Cu2+) would produce no results such as a shiny mirror surface or a colour change from blue to red-brown.

Carboxylic acid

(methanoic acid - formerly 'formic acid')

The evolution of CO2 when reacted with NaHCO3 (sodium hydrogencarbonate) can distinguish between carboxylic acids and weaker acids such as phenol. Adding carboxylic acid to solid or aqueous sodium carbonate will also liberate carbon dioxide, which can be tested for later.

Ester

(ethyl methanoate)

As well as a sweet aromatic smell, esters, after being treated with H2NOH will form a hydroxamic acid. These will form blue-red complexes on addition with a few drops of iron (III) chloride.

Phenol

- or hydroxybenzene

Iron (III) chloride when reacted with phenol will produce a violet/purple colour (in fact, any colour other than yellow indicates a type of phenol is present.)

Method - logically explained:

Firstly I will need to gather groups that have the same reacting atoms. In other words, bring together the functional groups that have similar structures. The six functional groups have either a '-OH' or an '=O' element in them (or in the case of the carboxylic acid, both elements are present.) A simple test can be done to determine which group is present. By adding the alkaline metal sodium to the organic unknown, it will tell us whether or not a '-OH' group is present as sodium will react with this, liberating hydrogen gas. No hydrogen gas will be released when sodium is reacted with the '=O' group. Instead sodium will react with the water, producing sodium hydroxide and this will react with the '=O' but without the liberation of hydrogen gas.

I now have, in essence, a hydroxyl group and a carbonyl group. Each one will have to face further tests to whittle them down. I will begin with the carbonyl group.

In the carbonyl group, I have three possible functional groups that the organic unknown can be: an aldehyde, a ketone or an ester. The odd one out is obviously ester and so there is probably a test that can identify an aldehyde and a ketone from an ester. This is achieved by adding 2,4-dinitrophenylhydrazine (2,4-DNPH) to the unknown organic and recording any colour change. If the products of the reaction have turned orange, than an aldehyde or a ketone is present. Ester will not produce any colour change with addition to 2,4-DNPH. Below is the reaction between 2,4-DNPH and an aldehyde or ketone.
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To further distinguish between the aldehyde and the ketone, we must use the fact that for one of the groups, the carbonyl group is at the end of the alkane chain (aldehyde), and for the other group the carbonyl group is in the middle of the alkane chain (ketone). The former is more reactive than the latter chiefly due to the "presence of two alkyl groups in ketones hinders the approach of attacking reagents to the carbonyl group. Another factor is that alkyl groups are electron-donating and reduce the partial positive charge on the carbonyl atom." (Ramsden, E. ...

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