Question

A compound responds to Baeyer’s test but does not give white precipitate when treated with ammoniacal silver nitrate solution, the compound should be:
[A] ${{C}_{2}}{{H}_{6}}$
[B] ${{C}_{3}}{{H}_{4}}$
[C] ${{C}_{2}}{{H}_{2}}$
[D] ${{C}_{2}}{{H}_{4}}$

Answer 1

HINT: Baeyer’s test is given by alkenes and alkynes. Tollen’s test is given by alkynes and aldehydes. To solve this question you have to find out whether the given compound is an alkene or an alkyne. It will help you determine if it will undergo the given tests or not.


COMPLETE STEP BY STEP SOLUTION: In the question, two tests are given to us. The first is the Baeyer’s test and the second is the Tollen’s test i.e. treatment with ammoniacal silver nitrate. So, let’s discuss which compounds give these tests and then try to find the answer.
We use Baeyer’s test for the determination of presence of alkenes and alkynes bonds in a compound. This test doesn’t work on alkanes and aromatic compounds.
Tollen’s test is also known as the silver mirror test. We use it to determine the presence of aldehyde and aromatic aldehydes. It is also used for the detection of some alpha-hydroxy ketones which undergo tautomerization to form an aldehyde.
Now, in the question it is given to us that the compound responds to Baeyer’s test but does not give white precipitate when treated with ammoniacal silver nitrate solution.
Now, both alkenes and alkynes give the Baeyer’s test but Tollen’s test is given by alkynes and aldehydes. Therefore, the correct answer here will be an alkene.
Let us check the options to find out the correct answer.
 In the first option we have ${{C}_{2}}{{H}_{6}}$ which we can write as $C{{H}_{3}}-C{{H}_{3}}$ and this is an alkane. Therefore, this cannot be the correct answer.
Then we have ${{C}_{3}}{{H}_{4}}$. This is an alkyne named propyne therefore this is not the correct answer.
In the next option we have ${{C}_{2}}{{H}_{2}}$. This is also an alkyne so this is not the correct answer either.
And lastly we have ${{C}_{2}}{{H}_{4}}$, we can write is as $C{{H}_{2}}=C{{H}_{2}}$. We can see that this is an alkene. Therefore, this is the correct answer.

Therefore, the correct answer is option [D] ${{C}_{2}}{{H}_{4}}$.


NOTE: In Baeyer’s test, we use dilute potassium permanganate as a reagent. Potassium permanganate oxidises the double or the triple bonds by replacing them with a hydroxyl group. Potassium permanganate is a violet coloured solution and upon formation of the –OH bonds, the solution loses its colour thus proving the presence of unsaturated hydrocarbons.
In the Tollen’s test, we use reagents consisting of silver nitrate, ammonia and sodium hydroxide. The base is used to maintain the pH of the solution.