Import Question JSON

Current Question (ID: 21323)

Question:

$\text{Consider the following reaction:}$ $\text{Bu} - \equiv \text{CH} \xrightarrow{1.\ \text{nBuLi}} \xrightarrow{2.\ \text{n-C}_5\text{H}_{11}\text{Cl}} \xrightarrow{3.\ \text{H}_2\text{-Pd, BaSO}_4} \text{Major Product}$ $\text{The major product in the above reaction is:}$

Options:

1. $\text{n-Bu} - \text{C}_5\text{H}_{11}$
2. $\text{n-Bu} - \text{C}_5\text{H}_{11}$
3. $\text{n-Bu} - \text{C}_5\text{H}_{11}$
4. $\text{n-Bu} - \text{C}_5\text{H}_{11}$

Solution:

$\text{Hint: Third reaction Lindlar's catalyst is used.}$ $\text{Step 1:}$ $\text{In the first reaction, an acid-base reaction takes place and carbanion is obtained as a product. Terminal alkyne hydrogen is acidic in nature.}$ $\text{In the next step, a nucleophilic substitution reaction takes place and an alkyne is obtained as a product.}$ $\text{In the last step, the formed alkyne reacts with Lindlar's catalyst, and cis alkene is obtained as a product.}$ $\text{Step 2:}$ $\text{The reactions are as follows:}$ $\text{Bu} - \equiv \text{CH} \xrightarrow{1.\ \text{nBuLi}} \text{Bu} - \equiv \text{C}^- \xrightarrow{2.\ \text{n-C}_5\text{H}_{11}\text{Cl}} \text{Bu} - \equiv \text{C} - \text{C}_5\text{H}_{11} \xrightarrow{3.\ \text{H}_2\text{-Pd, BaSO}_4} \text{Bu} - \text{C} = \text{C} - \text{C}_5\text{H}_{11}$ $\text{Hence, an option first is the correct answer}$

Import JSON File

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
}