Import Question JSON

Current Question (ID: 21330)

Question:

$\text{The carbonyl compounds that cannot be prepared by addition of water on an alkyne in the presence of HgSO}_4 \text{ and H}_2\text{SO}_4, \text{ among the following, is:}$

Options:

1. $\text{CH}_3\text{C} \equiv \text{C} \text{CH}_3 \xrightarrow{\text{HgSO}_4, \text{H}_2\text{SO}_4} \text{O} = \text{C} \text{CH}_3$
2. $\text{Cyclohexanone}$
3. $\text{H}_2\text{C} = \text{C} = \text{O}$
4. $\text{CH}_3\text{C} \equiv \text{C} \text{CH}_2\text{CH}_3 \xrightarrow{\text{HgSO}_4, \text{H}_2\text{SO}_4} \text{CH}_3\text{C} = \text{O} \text{CH}_2\text{CH}_3$

Solution:

$\text{Hint: HgSO}_4/\text{dil. H}_2\text{SO}_4 \text{ reaction with alkyne only given ketone accept only aldehyde that is ethanol.}$ $\text{The reaction of HgSO}_4/\text{dil. H}_2\text{SO}_4 \text{ with alkyne gives the addition of water as per markonioff's rule.}$ $1. \text{HC} \equiv \text{CH} \xrightarrow{\text{HgSO}_4, \text{H}_2\text{SO}_4} \text{CH} = \text{CH} \Rightarrow \text{CH}_3 - \text{CH} = \text{O}$ $2.$ $3. \text{CH}_3 - \text{C} \equiv \text{CH} \xrightarrow{\text{HgSO}_4, \text{H}_2\text{SO}_4} \text{CH}_3 - \text{C} = \text{CH}_2 \Rightarrow \text{CH}_3 - \text{C} = \text{O} - \text{CH}_2 - \text{CH}_3$ $\text{Hence, CH}_3 - \text{CH}_2 - \text{CHO cannot be form.}$ $4. \text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_3 \xrightarrow{\text{HgSO}_4, \text{H}_2\text{SO}_4} \text{CH}_3 - \text{C} = \text{CH} - \text{CH}_3$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}