StreakPeaked· Practice

ExamsJEE AdvancedChemistry

How many of the following compounds will give a white precipitate of silver salt when treated with ammoniacal AgNO3 solution (Tollens-type test for terminal alkynes and active halides)? (a) CH3-CH2-C≡CH (1-butyne) (b) CH3-C≡C-CH3 (2-butyne) (c) CH2=CH-CH2-Cl (allyl chloride) (d) CH2=C(Cl)-CH3 (2-chloropropene, vinyl type) (e) 3-chlorocyclohexene (allylic chloride) (f) Ph-CH2-Cl (benzyl chloride) (g) 1-iodocycloheptatriene (tropylium-type; iodocycloheptatrienyl halide)

  1. 3
  2. 4
  3. 5
  4. 6

Correct answer: 4

Solution

Ammoniacal AgNO3 reacts with: (1) Terminal alkynes: the terminal C-H is acidic (pKa ~25), forming Ag-C≡C- precipitate. (2) Halides that can undergo SN1 (active halides): allylic, benzylic, propargylic halides form carbocations readily and react with Ag+. Vinyl halides (sp2 C-X) are unreactive toward SN1. Internal alkynes have no acidic C-H. Analysis: (a) 1-butyne (terminal alkyne) -> YES (white/buff AgC≡C salt). (b) 2-butyne (internal alkyne) -> NO. (c) Allyl chloride (allylic) -> YES. (d) 2-chloropropene (vinyl halide) -> NO. (e) 3-chlorocyclohexene (allylic chloride) -> YES. (f) Benzyl chloride (benzylic) -> YES. (g) Iodocycloheptatriene: C7H7I where the cation would be the aromatic tropylium (C7H7+, 6-pi electrons) -> extremely stable -> YES. Count: a, c, e, f, g = 5 compounds. Answer: 5.

Related JEE Advanced Chemistry questions

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →