Import Question JSON

Current Question (ID: 21322)

Question:

$\text{Consider the reaction,}$ $\text{(i) Fe, Br}_2$ $\text{(ii) Cl}_2/\Delta$ $\text{(iii) alc. KOH}$ $\rightarrow \text{major product}$ $\text{Major product is:}$

Options:

1. $\begin{array}{c} \text{Br} \\ \text{C}_6\text{H}_4\text{CH}_2\text{CH}_3 \end{array}$
2. $\begin{array}{c} \text{C} \equiv \text{CH} \\ \text{C}_6\text{H}_4\text{Br} \end{array}$
3. $\begin{array}{c} \text{Cl} \\ \text{C}_6\text{H}_4\text{CH}_2\text{Br} \end{array}$
4. $\begin{array}{c} \text{C} \equiv \text{C} - \text{Cl} \\ \text{C}_6\text{H}_4 \end{array}$

Solution:

$\text{Hint: Second reaction is free radical substitution reaction occur in benzyl position}$ $\text{Step 1:}$ $\text{In the first step, electrophilic aromatic substitution reaction takes place.}$ $\text{The ethyl group is an electron donating group hence, reaction occur at ortho and para position.}$ $\text{The para position product is major.}$ $\text{In the second step, free radical substitution reaction occur in the benzyl position.}$ $\text{The reaction follows free radical substitution reaction.}$ $\text{In the last reaction, elimination reaction occur and alkene is obtained as a product.}$ $\text{The reaction follows E2 mechanism.}$ $\text{Step 2:}$ $\text{The reactions are as follows:}$ $\begin{array}{c} \text{(i) Fe, Br}_2 \\ \text{(ii) Cl}_2/\Delta \\ \text{(iii) alc. KOH} \end{array}$ $\rightarrow \text{C}_6\text{H}_4\text{CH}_2\text{CH}_3$ $\text{The correct option is first option.}$

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
}