Import Question JSON

Current Question (ID: 21309)

Question:

$\text{The increasing order of the reactivity of the following compounds}$ $\text{toward electrophilic aromatic substitution reactions (EASR) is:}$ $\begin{array}{ccc} \text{Cl} & \text{CH}_3 & \text{COCH}_3 \\ (\text{I}) & (\text{II}) & (\text{III}) \end{array}$

Options:

1. $\text{III} < \text{II} < \text{I}$
2. $\text{III} < \text{I} < \text{II}$
3. $\text{I} < \text{II} < \text{III}$
4. $\text{I} < \text{III} < \text{II}$

Solution:

$\text{Hint: Electron donating group increases the reactivity of the benzene}$ $\text{ring towards EASR.}$ $\text{Step 1:}$ $\text{As the electron-donating group is attached to the benzene ring the}$ $\text{electron density in the benzene ring increases and the reactivity of the}$ $\text{benzene ring toward EASR increases. But in the case of the}$ $\text{electron-withdrawing group reactivity of the benzene ring toward EASR}$ $\text{decreases.}$ $\text{A CH}_3 \text{ group is an electron-donating group hence it activates the}$ $\text{benzene ring toward EASR. The COCH}_3 \text{ group is a strong electron-}$ $\text{withdrawing group it withdraws electrons by -R effect and deactivates the}$ $\text{benzene ring towards EASR.}$ $\text{The chlorobenzene has intermediate reactivity because it shows a}$ $\text{strong -I effect and weak +R effect. Hence, Cl is moderately}$ $\text{deactivating.}$ $\text{Hence the reactivity order is as follows:}$ $\text{(II) > (I) > (III)}$

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
}