Advertisements
Advertisements
Questions
Propanal and Propanone
Give a chemical test to distinguish between propanal and propanone.
Give a simple chemical test to distinguish between propanal and propanone.
Advertisements
Solution 1
Propanal and propanone can be distinguished by the following tests:
- Tollen’s test: Propanal is an aldehyde. Thus, it reduces Tollen’s reagent. But, propanone being a ketone does not reduce Tollen’s reagent.
\[\ce{\underset{Propanal}{CH3CH2CHO} + \underset{Tollen's reagent}{2[Ag(NH3)2]+} + 3OH- -> \underset{Propanoate ion}{CH3CH2COO-} + \underset{Silver mirror}{Ag\downarrow} + NH3 + 2H2O}\] - 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.
\[\ce{\underset{Propanal}{CH3CH2CHO} + 2Cu^2+ + 5OH- -> \underset{Propanoate ion}{CH3CH2COO-} + \underset{(Red-brown ppt)}{\underset{Cuprous oxide}{Cu2O\downarrow}}+ 2H2O}\] - Iodoform test: Aldehydes and ketones with at least one methyl group linked to the carbonyl carbon atom respond to the iodoform test. They are oxidized by sodium hypoiodite (NaOI) to give iodoforms. Propanone being a methyl ketone responds to this test, but propanal does not.
\[\ce{\underset{Propanone}{CH3COCH3} + 3NaOI -> \underset{Sodium acetate}{CH3COONa} + \underset{(yellow ppt)}{\underset{Iodoform}{CHI3}}+ 2NaOH}\]
Solution 2
| Reagent | Propanal | Propanone |
| Tollen’s Reagent | On heating with Tollen’s reagent, a silver mirror is formed on the inner walls of the test tube. Aldehyde groups reduce the Tollen’s reagent. | On heating in a water bath with Tollen’s reagent does not show any reaction. |
APPEARS IN
RELATED QUESTIONS
Distinguish between:
C6H5-COCH3 and C6H5-CHO
Give a simple chemical test to distinguish between the following pair of compounds:
Phenol and Benzoic acid
Alkenes decolourise bromine water in presence of CCl4 due to formation of ______.
Which of the following compounds gives a positive Tollen's test but negative Fehling's test?
Solvent used for dewaxing of petroleum products are
Acetone and acetaldehyde are differentiated by
Acetaldehyde cannot show?
A hydrocarbon (A) with molecular formula C5H10 on ozonolysis gives two products (B) and (C). Both (B) and (C) give a yellow precipitate when heated with iodine in presence of NaOH while only (B) give a silver mirror on reaction with Tollen’s reagent.
- Identify (A), (B) and (C).
- Write the reaction of B with Tollen’s reagent.
- Write the equation for iodoform test for C.
- Write down the equation for aldol condensation reaction of B and C.
Write chemical test to distinguish between the following compounds:
Phenol and Benzoic acid

Which among the above compound/s does/do not form Silver mirror when treated with Tollen's reagent?
The major products formed in the following reaction sequence A and B are:

Choose the reaction which is not possible:
In Tollen's test for aldehyde, the overall number of electrons(s) transferred to the Tollen's reagent formula \[\ce{[Ag(NH3)2]+}\] per aldehyde group to form silver mirror is ______. (Round off to the nearest integer)
An organic compound 'A' with the molecular formula C4H8O2 undergoes acid hydrolysis to form two compounds 'B' and 'C'. Oxidation of 'C' with acidified potassium permanganate also produces 'B'. Sodium salt of 'B' on heating with soda lime gives methane.
- Identify 'A', 'B' and 'C'.
- Out of 'B' and 'C', which will have higher boiling point? Give reason.
You are given four organic compounds “A”, “B” , “C” and “D”. The compounds “A”, “B” and “C” form an orange-red precipitate with 2, 4 DNP reagent. Compounds “A” and “B” reduce Tollen’s reagent while compounds “C” and “D” do not. Both “B” and “C” give a yellow precipitate when heated with iodine in the presence of NaOH. Compound “D” gives brisk effervescence with sodium bicarbonate solution. Identify “A”, “B”, “C” and “D” given the number of carbon atoms in three of these carbon compounds is three while one has two carbon atoms. Give an explanation for our answer.
Benzaldehyde is obtained from Rosenmund’s reduction of:
