JEE MainChemistryHydrocarbons
Given below are two statements for the following reaction sequence: Acetylene [ (ii) CH ₃ I ] (i) NaNH ₂ P [ (ii) CH ₃ I ] (i) NaNH ₂ Q Q dil. H ₂ SO ₄, Hg ²⁺ S Q Red hot Fe tube, 873 K R Statement I: Compound S gives a yellow precipitate with I ₂/ NaOH but does not reduce Tollens' reagent. Statement II: Compound R has 12 C-C sigma bonds and all its carbon atoms are sp ^2 hybridized. In the light of the above stateme
Options
- ABoth Statement I and Statement II are true
- BStatement I is true but Statement II is false
- CStatement I is false but Statement II is true
- DBoth Statement I and Statement II are false
Correct answer
B. Statement I is true but Statement II is false
Step-by-step solution
Acetylene reacts with 1 equivalent of NaNH ₂ and CH ₃ I to form propyne (P). Propyne reacts with another equivalent of NaNH ₂ and CH ₃ I to form 2-butyne (Q). Hydration of 2-butyne (Q) in the presence of dil. H ₂ SO ₄ and Hg ²⁺ yields 2-butanone (S). 2-Butanone is a methyl ketone, so it gives a positive iodoform test (yellow precipitate with I ₂/ NaOH ). Being a ketone, it does not reduce Tollens' reagent. Thus, Statement I is true. Passing 2-butyne (Q) through a red-hot iron tube at 873 K causes cyclic polymerizat