Question

Two compounds $\left(A\right)$ and $\left(B\right)$ have $s{p}^2$ hybridized carbon atoms $\left(A\right)$ can decolourise Baeyer's reagent but $\left(B\right)$ cannot $\left(A\right)$ and $\left(B\right)$ could be:

• A. ethylene and acetylene
• B. propylene and acetylene
• C. benzene and acetylene
• D. ethylene and benzene

Answer 1

Answer: (D)

ethylene and benzene

This decolourising of Baeyer's reagent is a test for unsaturated compounds. Apart from benzene all other compounds are unsaturated but due to delocalisation of pi electrons benzene doesn't give test of unsaturation and doesn't undergo addition reaction. So B is benzene.