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Fehling’s test is one of the most frequently used tests to determine, estimate and identify the sugars which can be both reducing or non-reducing in nature. The formulation of this test was made by the eminent German chemist Hermann von Fehling. The main utilization of this test is for the differentiation between functional groups of ketone and water-soluble carbohydrates.
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Keyterms: Ketone, Carbohydrates, Copper II Sulphate, Sodium Hydroxide, Alkali, Fehling’s A, Fehling’s B, Aldehyde
Fehling’s Solution
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Fehling’s test is carried out with a solution that is freshly prepared in the laboratories. The solution is initially present in the form of two solutions known as Fehling’s A and Fehling’s B.
Fehling’s A Solution contains copper (II) sulphate. It is a deep blue liquid in nature.
Fehling’s B solution contains potassium sodium tartrate (Rochelle salt) along with a strong alkali, most commonly sodium hydroxide. It is a clear liquid in nature. These solutions are separately prepared and subsequently stored.
Later on, both Fehling’s A solution and Fehling’s B solution are mixed so that the final Fehling’s solution can be derived. The equal volumes of these solutions need to be taken into consideration while mixing. The solution is deep blue in colour. The deep blue ingredient that is being taken is a tartrate complex of Cu2+. The tartrate tetra-anions serve as a chelating agent in the solution.

Fehling’s Test
If the outcome of the test is negative, it results in the formation of Ketone. If the outcome of the test is positive, it results in the formation of Aldehyde. Therefore, Fehling’s test is important to distinguish between aldehydes and ketones.
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Fehling’s Test Procedure
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The process for the conducting of the Fehling’s test is as follows:
- Take a dry test tube and add the sample to it.
- Keep distilled water in another test tube as a control.
- Add the Fehling’s solution in both these test tubes
- The tubes are recommended to be kept or held in water baths
- Observations are to be made and any growth or development of red precipitate needs to be noted
If there is a production or formation of a reddish-brown precipitate, the result is positive. If there is no such kind of transition, the result is negative.
Once the mixture of both the solutions is heated, Fehling’s B solution prevalent in the reagent carries out the task of chelation. If the solution of the test presents a sugar molecule or an aldehyde or chelated compound which is reddish-brown in colour along with Fehling’s reagent. If such reddish-brown precipitate is formed, it indicates that a reduced sugar or an aldehyde group is prevalent. The determination of reduced sugar is the primary aim of performing Fehling’s test. To ensure the efficiency and effectiveness of the test, it is required to prepare new reagents. As far as Fehling’s reagent test is concerned, lactose, fructose and glucose being sugars give positive results.
Reactions of Fehling’s Test
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The reaction between the copper (II) ions and an aldehyde is presented as:
RCHO + 2Cu2+ + 5OH− → RCOO. + Cu2O + 3H2O
Once there is an addition of tartrate, it is presented as:
RCHO + 2Cu(C4H4O6)22− + 5 OH− → RCOO− + Cu2O + 4 C4H4O62− + 3H2O
Applications of Fehling’s Test
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Fehling’s test can be taken into consideration for various kinds of purposes.
- The most popular application of Fehling’s test is to identify and determine whether the carbonyl group concerned is a ketone or an aldehyde. Aldehydes tend to be oxidised and display positive results. Ketones, except the alpha-hydroxy-ketones, do not display any kind of reaction.
- Fehling’s test is also generally used for monosaccharides, commonly known as simple sugars, along with other reducing sugars. In terms of monosaccharides, it displays positive results not only for aldose monosaccharides but also for ketose monosaccharides, since ketoses are transformed into aldoses through the base contained in the reagent.
- It is used to determine the presence of glucose in the urine. It is applied to identify diabetes as being prevalent in the person.
- Fehling’s test is also applied for the breaking down of starch to convert it into glucose syrup along with maltodextrins.
- To determine the Formic acid. The formic acid (HCO2H) gives a positive result for Fehling’s Test.
Things to Remember
- To carry out the Fehling’s Test, the substance is heated in the presence of Fehling’s Solution.
- Fehling’s Solution is prepared freshly in the laboratory.
- It is prepared by mixing an equal amount of Fehling’s Solution A and Fehling’s Solution B.
- If a reddish-brown precipitate is formed then the result is positive. If there is no such kind of changes, the result is negative.
- Fehling’s test is most commonly used to determine whether the carbonyl group is a ketone or an aldehyde.
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Sample Questions
Ques: (a) Write the products formed when CH3CHO reacts with the following reagents:
(i) HCN (ii) H2N–OH (iii) CH3CHO in the presence of dilute NaOH
(b) Give simple chemical tests to distinguish between the following pairs of compounds :
(i) Benzoic acid and Phenol (ii) Propanal and Propanone. (All India 2014, 5 Marks)
Ans:
(b) (i) Benzoic acid and Phenol: On adding NaHCO3 to both solutions carbon dioxide gas is evolved with benzoic acid while phenol does not form CO2.
(ii) Propanal and Propanone: Propanal gives a positive test with Fehling’s solution in which a red ppt. of cuprous oxide is obtained while propanone does not respond to the test.
Ques: (a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanone.
(ii) Pentan-2-one and Pentan-3-one.
(b) Arrange the following compounds in increasing order of their acid strength: Benzoic acid, 4- Nitrobenzoic acid, 3,4 -Dinitrobenzoic acid, 4- Methoxybenzoic acid. (All India 2015, 3 Marks)
Ans: (a) (i) Acetaldehyde (Ethanal) and Acetone (Propanone)
Acetaldehyde gives the silver mirror with Tollen’s reagent while Acetone does not give this test.
Acetaldehyde forms red ppt with Fehling’s solution. Acetone does not form a precipitate with Fehling’s Solution.
(ii) Pentan-2-one and Pentan-3-one
Pentan-2-one forms yellow ppt with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.
CH3COCH2CH2CH3 + 3I2+ 4NaOH → CH3CH2CH2COONa + CHI3 + 3H2O + 3NaI
Pentan-2-one gives white ppt with sodium bisulphite while pentan-3-one does not.
(b) 4- Methoxybenzoic add < Benzoic acid < 4- Nitrobenzoic acid < 3,4-Dinitrobenzoic acid
Ques: Give a simple chemical test to distinguish between Acetophenone and Benzophenone. (2 Marks)
Ans: The test which can distinguish between Acetonphenone and Benzophenone is Fehling’s Test.
Aldehydes respond to Fehling’s test, but ketones do not.
Propanal being an aldehyde reduces Fehling’s solution to a red-brown precipitate of Cu2O, but propanone being a ketone does not react to Fehling’s solution.
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Ques: Fehling solution does not oxidize benzaldehyde but Tollen's reagent oxidizes benzaldehyde. Give reasons. (2 Marks)
Ans: Aldehydes such as benzaldehyde does not contain alpha hydrogens and cannot form an enolate. Therefore, they do not give a positive test with Fehling's solution which is also a weaker oxidizing agent than Tollen's reagent. That is why Fehling’s solution does not oxidize benzaldehyde and gives negative tests.
Ques: What is Fehling's solution? (1 Mark)
Ans: Fehling’s solution is used to carry out Fehling’s solution. Fehling’s solution is a mixture of an alkaline solution of copper (II) sulphate (CuSO4) and Sodium Potassium Tartrate (Rochelle Salt). The solution is freshly prepared in the laboratories. It is initially present in the form of two solutions known as Fehling’s A and Fehling’s B solution.
Ques: Which of the following aldehydes give a red precipitate with Fehling’s solution? (2 Marks)
(a) Benzaldehyde
(b) Salicyladehyde
(c) Acetaldehyde
(d) None of the above
Ans: Correct option is (C) Acetaldehyde
Explanation: As we know that only aliphatic aldehydes give red ppt of Cu2O with Fehling Solution. So, only acetaldehyde gives a red ppt with Fehling’s solution.
Ques: Few simple chemical tests are given below to differentiate between the pairs of compounds. Which of the following tests is not correct for differentiation? (2 Marks)
(a) Propanal and Propanone - Silver mirror test
(b) Acetophenone and benzophenone - Idoform test
(c) Ethanal and propanal - Fehling’s test
(d) Benzoic acid and ethyl benzoate- Sodium bicarbonate test
Ans: The correct option is (c) Ethanal and propanal - Fehling's test
Explanation: Both ethanal and propanal can not be differentiated by Fehling’s Solution test. They both will give the silver mirror test and form a red precipitate with Fehling's solution.
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