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Question 56

Consider the following reactions:

'A' is:

Reaction with Ag2O / Heat (Tollens'-like test for Alkynes)

When compound A is treated with silver oxide (Ag2O) and heat, it forms a precipitate (ppt).

  • Deduction: This is a characteristic test for terminal alkynes (alkynes with a -C triple bond CH group). The acidic terminal hydrogen is replaced by a silver ion to form an insoluble silver acetylide precipitate.
  • Therefore, A must be a terminal alkyne.

2. Hydration of Alkyne A to form B

When compound A (a terminal alkyne) undergoes hydration in the presence of Hg2+ / H+:

  • Markovnikov addition of water occurs across the triple bond to form an enol, which quickly tautomerizes into a stable methyl ketone.
  • Reaction path: R-C triple bond CH --> [ R-C(OH)=CH2 ] --> R-CO-CH3 (Compound B)

3. Reduction of B to C

Compound B (the methyl ketone) is reduced using sodium borohydride (NaBH4):

  • NaBH4 selectively reduces ketones to secondary (2 degree) alcohols.
  • Reaction path: R-CO-CH3 --> R-CH(OH)-CH3 (Compound C)

4. Reaction of C with Lucas Reagent (ZnCl2 / conc. HCl)

Compound C is treated with Lucas reagent, giving turbidity within 5 minutes.

  • Lucas Test Rules:
    • 3 degree alcohols give turbidity immediately.
    • 2 degree alcohols give turbidity within 5 minutes.
    • 1 degree alcohols do not give turbidity at room temperature.
  • Since C gives turbidity within 5 minutes, it confirms that C is a secondary alcohol, which perfectly matches the deduction from Step 3.

Summary of Compounds

  • A: A terminal alkyne (R-C triple bond CH)
  • B: A methyl ketone (R-CO-CH3)
  • C: A secondary alcohol (R-CH(OH)-CH3)

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