An important class of organic compounds containing the carbonyl group comprises aldehydes and ketones. Aldehyde has the RCH(=O) structure, while R2C(=O) has the structure of a ketone.
In this article, we will study the following concepts in detail.
Test for aldehydes and ketones,
How aldehyde and ketone can be distinguished by various tests.
Identify the presence of functional group aldehydes or ketones in the given organic compound.
The test of aldehyde and ketone-
Silver nitrate solution
Fehling’s solutions A
Fehling’s solutions B
Dilute ammonium hydroxide solution
Test tube holder
Aldehydes and ketones are volatile compounds with low molecular weights. Aldehyde and ketone identification is based on two types of reactions, the double bond addition reaction, and the oxidation reaction.
In aldehydes, a hydrogen atom and aliphatic or aromatic radicals are attached to the carbonyl group. Formaldehyde is an exceptional case where two hydrogen atoms are attached to the carbonyl that is present in formaldehyde. In ketones, two aliphatic or aromatic groups are affixed to the carbonyl group.
Tests to Detect the Presence of Aldehyde and Ketone
Aldehyde and ketone react with a 2,4-dinitrophenylhydrazine test to give a yellow to orange color precipitate. The aldehyde test reaction is given below-
Sodium Bisulfite Test
When aldehydes and ketone react with Sodium bisulfite, crystalline precipitate.
Aldehyde and Ketone can be Distinguished By
In some dye formulation reactions, such as the reaction between sodium bisulfite and fuchsine, the Schiff reagent is the product produced. It is used to detect an aldehyde's presence. A magenta-coloured dye with a chemical formula C20H20N3·HCl, which is decolourized by sodium bisulfate, is fuchsine or rosaniline hydrochloride. A pink colour indicates the presence of an aldehyde group.
The Fehling test consists of a solution that in laboratories is normally freshly prepared. The solution occurs in two different forms known as Fehling's A and Fehling's B. Fehling's A is a blue copper(II) sulfate solution. A clear liquid consisting of potassium sodium tartrate (Rochelle salt) and a powerful alkali, normally sodium hydroxide, is Fehling's B. Solutions A and B are prepared separately.
The deep blue ingredient is the bis(tartrate) complex of Cu2+. Cu2+ is reduced to Cu+ when the aldehyde compound is treated with Fehling's solution, and the aldehyde is reduced to acids. A red precipitate is formed during the reaction.
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The Tollens test is a chemical test used to distinguish sugar reducers from non-sugar reducers. This test is known as the silver mirror test. Aldehydes react by giving a grey-black precipitate or a silver mirror to the Tollens reagent. A freshly prepared reagent from Tollen should always be used. In the Tollens reagent, aldehydes are oxidized to the corresponding acid, and silver is reduced from the +1 oxidation state to its elemental form. Ketones usually do not respond to this test.
RCHO + 2[Ag(NH3)2]OH → R-COONH4+ 3NH3 + H2O + 2Ag↓(silver mirror)
Test with Chromic Acid
A green to blue precipitate is given by aldehydes reacting with chromic acid. With chromic acid, ketones do not react. Some of the primary and secondary alcohols are also tested.
R-CHO + 2CrO3 + 3H2SO4 → 3R-C(O)-OH + 3H2O + Cr2(SO4)3(green colour)
Sodium Nitroprusside Test:
Ketones only give this test and not aldehyde. An anion is formed when ketone reacts with an alkali, further this anion reacts with sodium nitroprusside to form a coloured complex. The red colour shows the presence of ketones.
CH3 + OH–→ CH3COCH2-+ H2O
[Fe(CN)5NO]2- + CH3COCH2- → [Fe(CN)5NO.CH3COCH2]3-
Procedure and Observation:
The given organic compound has an aldehyde/ ketone functional group present.
conduct the experiment, the reagents should be freshly prepared.
Not directly on the flame to heat the reaction mixture.
Wash the test tube with nitric acid to destroy the silver mirror after running the Tollen test, since it is explosive material.
Did you know?
Formaldehyde has an odor that is unpleasant. It is difficult to handle in a gaseous state due to its reactivity. Therefore, it is dissolved in water for many applications and marketed as a 37 percent to 40 percent aqueous solution called formalin. Proteins are denatured by formaldehyde, making them insoluble in water and resistant to bacterial decay. For this reason, in embalming solutions and in preserving biological specimens, formalin is used.
The use of the reagent 2, 4-dinitrophenylhydrazine allows imines to be formed from ketones or aldehydes (DNPH). The carbonyl of an aldehyde or ketone is attacked by the main amino group of the DNPH in an environment with an acidic nature in this addition-elimination reaction. A hydrazone is formed as a result of the condensation reaction, and it precipitates out of the solution.
Non-conjugated ketones or aldehydes are indicated by yellow precipitates, whereas conjugated systems are indicated by red-orange precipitates. This test is used to distinguish ketones and aldehydes from alcohols and esters, which do not react with DNPH and therefore do not generate a precipitate. The ketone or aldehyde derivatives are crystalline solids with well-defined melting temperatures that have been documented in the literature and can be used to identify specific compounds.
The haloform test is another method for determining whether a ketone is methyl ketone. A methyl group is attached to the carbonyl carbon in methyl ketones (RCOCH3). Hydrogen (methyl aldehyde), an alkyl group, or an aryl group can be used as the R group.
Under basic conditions, the haloform test response mechanism occurs. The ketone first passes through a keto-enol tautomerization process. The nucleophilic enolate then attacks the iodine, resulting in the formation of an iodine ion and a halogenated ketone. This process is repeated three times more until all hydrogens in the ketone methyl group have been replaced by iodine. The hydroxide then combines with the electrophilic carbon core of the carbonyl to generate a tetrahedral intermediate with a negatively charged oxygen that is single bonded to the carbonyl. The trihalomethane group departs as a stable leaving group when the C-O double bond is formatted. Finally, the negatively charged trihalomethane group deprotonates the produced carboxylic acid, yielding the haloform — trihalomethane — and a carboxylate. If the halogen is iodine, the precipitate will be a distinctive yellow tint. Only methyl ketones will undergo this reaction, and any other carbonyl-containing molecule will have a negative result.