Ammoniacal silver nitrate forms a white precipitate easily with

• Correct answer
  A

  $\mathrm{CH}_3 \mathrm{C}=\mathrm{CH}$
• B

  $\mathrm{CH}_3 \mathrm{C}=\mathrm{CCH}_3$
• C

  $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$
• D

  $\mathrm{CH}_2=\mathrm{CH}_2$

Solution

In general, $\mathrm{C}_2 \mathrm{H}_2$ and all 1-alkynes given white precipitate with ammoniacal silver nitrate. Thus, in this question, propyne-1 $\left(\mathrm{CH}_3 \mathrm{C}=\mathrm{CH}\right)$ will give white precipitate with ammoniacal silver nitrate.

Hence, the correct answer is option: 'A'.