CH3CHO and C6H5CH2CHO can be distinguished chemically by:
CH3CHO and C6H5CH2CHO can be distinguished chemically by:
The correct option is B Iodoform test
The explanation for the correct option:
Iodoform test:
- The iodoform test is used to check the presence of carbonyl compounds with the structure R-CO-CH3 or alcohols in a given unknown structure.
- When iodine with sodium hydroxide is mixed with any compound which contains the carbonyl group a yellow precipitate forms which confirm the presence of the carbonyl group.
- The rest all the tests are related to only aldehydes so we can not distinguish between both the above compound only by the iodoform test.
The explanation for the incorrect option:
Option A:
Benedict test: This test is for the detection of reducing sugars(the sugars are carbohydrates that have a free aldehyde or ketone functional group in their molecular structure).
Benedict test or benedict solution is a bright blue solution prepared by mixing copper sulphate pentahydrate, sodium carbonate, and sodium citrate in water.
Option C:
Tollen's reagent test: Tollen's reagent is used to distinguish between aldehyde and ketone along with some alpha-hydroxy ketones which can tautomerize into aldehydes. The reagent consists of a solution of silver nitrate, ammonia, and some sodium hydroxide.
Option D:
Fehling solution test: Fehling solution is a deep blue alkaline solution that is used to identify the presence of aldehyde or groups that contain any aldehyde functional group(-CHO). Fehling solution is solution A, which is 70gm of cupric sulfate pentahydrate per litre of solution, and a colourless Fehling's solution B, which is about 350gm of Rochelle salt(potassium sodium tetrahydrate) and 100gm of sodium hydroxide per litre of the solution.
- (acetaldehyde) and phenylacetaldehyde) both have the same functional group that is the aldehyde group(-CHO) so they all will give the positive test for the Benedict test, Tollen's reagent test and Fehling solution test as these tests are associated with an aldehydic group only.
Therefore the correct option is Option B.
Iodoform test
The explanation for the correct option:
Iodoform test:
- The iodoform test is used to check the presence of carbonyl compounds with the structure R-CO-CH3 or alcohols in a given unknown structure.
- When iodine with sodium hydroxide is mixed with any compound which contains the carbonyl group a yellow precipitate forms which confirm the presence of the carbonyl group.
- The rest all the tests are related to only aldehydes so we can not distinguish between both the above compound only by the iodoform test.
The explanation for the incorrect option:
Option A:
Benedict test: This test is for the detection of reducing sugars(the sugars are carbohydrates that have a free aldehyde or ketone functional group in their molecular structure).
Benedict test or benedict solution is a bright blue solution prepared by mixing copper sulphate pentahydrate, sodium carbonate, and sodium citrate in water.
Option C:
Tollen's reagent test: Tollen's reagent is used to distinguish between aldehyde and ketone along with some alpha-hydroxy ketones which can tautomerize into aldehydes. The reagent consists of a solution of silver nitrate, ammonia, and some sodium hydroxide.
Option D:
Fehling solution test: Fehling solution is a deep blue alkaline solution that is used to identify the presence of aldehyde or groups that contain any aldehyde functional group(-CHO). Fehling solution is solution A, which is 70gm of cupric sulfate pentahydrate per litre of solution, and a colourless Fehling's solution B, which is about 350gm of Rochelle salt(potassium sodium tetrahydrate) and 100gm of sodium hydroxide per litre of the solution.
- (acetaldehyde) and phenylacetaldehyde) both have the same functional group that is the aldehyde group(-CHO) so they all will give the positive test for the Benedict test, Tollen's reagent test and Fehling solution test as these tests are associated with an aldehydic group only.
Therefore the correct option is Option B.
