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Current Question (ID: 21310)

Question:

$\text{Consider the following reactions:}$ $\text{Ag}_2\text{O} \xrightarrow{\Delta} \text{ppt}$ $\text{Hg}^{2+}/\text{H}^+ \xrightarrow{\text{NaBH}_4} \text{ZnCl}_2$ $\text{H}_2\text{O} \rightarrow \text{Turbidity within 5 minutes}$ $\text{conc. HCl}$

Options:

1. $\text{CH}_3 - \text{C} \equiv \text{CH}$
2. $\text{CH}_2 = \text{CH}_2$
3. $\text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_3$
4. $\text{CH} \equiv \text{CH}$

Solution:

$\text{Hint: A is a terminal alkyne}$ $\text{From the option it is clear that A is an alkyne. Now A can be terminal alkyne or a non-terminal alkyne. In the reaction it is given that A reacts with Ag}_2\text{O and gives white ppt. Only terminal reacts with Ag}_2\text{O and give white ppt.}$ $\text{A generate C and C reacts with lucas reagent and generate white turbidity at 5 min. It indicate that C is a secondary alcohol and ketone generate secondary alcohol when reacts with NaBH}_4.$ $\text{Hence, B is a ketone. Propyne reacts with Hg}^{2+} \text{ ion and generate acetone as a product.}$ $\text{Ethyne cannot be the answer because ethyne will generate aldehyde (acetaldehyde) with Hg}^{2+} \text{ ion.}$ $\text{The overall reaction are as follows:}$ $\text{Ag}_2\text{O} \rightarrow \text{CH}_3 - \text{C} \equiv \text{C} - \text{Ag} \downarrow \text{ppt}$ $\text{Hg}^{2+} \rightarrow \text{O} \text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_3$ $\text{NaBH}_4 \rightarrow \text{OH} \text{CH}_3 - \text{C} - \text{C} - \text{CH}_3$ $\text{ZnCl}_2/\text{HCl} \rightarrow \text{Turbidity in 5 minutes}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}