Ethene and ethyne can be distinguished by employing the test
a. $\ce{Br2}$ in organic solvent
b. Baeyer's test
c. Phenyl hydrazine
d. Tollen's reagant
According to me it must be both a and b, but there's no such option.
Ethene and ethyne can be distinguished by employing the test
a. $\ce{Br2}$ in organic solvent
b. Baeyer's test
c. Phenyl hydrazine
d. Tollen's reagant
According to me it must be both a and b, but there's no such option.
I see two possible answers. Not (A) and (B) but rather (A) and (D).
Chakravarthy Kalyan is responsible for reminding us that (D) can be an answer too (refer to comments).
For option (D):
$\ce{HC ≡ CH + 2AgNO3 + 2NH4OH→ AgC ≡ CAg↓ + 2NH4NO3 + 2H2O}$
The greyish precipitate of silver acetylide will readily form when ethyne is passed through freshly prepared tollen's reagent.
You may read more from this answer by @ron: Does acetylene give a positive Tollens' test? If not, why is this so?.
For option (A):
While $\ce{Br2}$ looks like a seemingly unsuitable candidate, it can actually be used as a crude way to differentiate them.
Take the same volumes of equal strength of bromine water in two vessels and allow equal amounts of ethyne in one and ethane in the other to pass through them (this shouldn't be difficult, just passing equal volumes at STP will do the trick).
Carefully observing which vessel has the more intense color will help you identify which one is ethene, which would consume half the amount of bromine consumed by ethyne.
If you're not convinced, put the final solutions in a cuvette each and measure the absorption or transmittance in a spectrophotometer at 530 nm.
P.S. Make sure that you have a sufficient amount of bromine water, or you'll end up comparing two colorless solutions.
P.P.S. You could also use titration to find which one reaches the end point first, which would be ethene.