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Current Question (ID: 21345)

Question:
$\text{Seven compounds are given below. How many of the following will give Friedel Craft's reaction?}$
Options:
  • 1. $3$
  • 2. $7$
  • 3. $6$
  • 4. $1$
Solution:
$\text{Hint: Strong deactivating groups (like -NO}_2 \text{ and -NH}_2\text{) do not allow Friedel-Crafts alkylation or acylation.}$ $\text{Step 1: Friedel-Crafts Reaction Restrictions:}$ $\text{(1) Strong deactivators (like -NO}_2 \text{ and -NH}_2\text{) do not allow Friedel-Crafts alkylation or acylation due to their electron-withdrawing effects.}$ $\text{(2) Halogens (-Cl) are deactivating but still allow the Friedel-Crafts reaction because they are ortho/para directing.}$ $\text{(3) Alkyl group (-CH}_3\text{, -C}_2\text{H}_5\text{) are activating and favor Friedel-Crafts reaction.}$ $\text{Step 2: Friedel Crafts' reaction is given by}$ $\text{(i) Aniline (-NH}_2\text{ group) } \rightarrow \text{ Does NOT undergo Friedel-Crafts due to strong Lewis acid-base interaction with AlCl}_3\text{, which deactivates the ring.}$ $\text{(ii) Nitrobenzene (-NO}_2\text{ group) } \rightarrow \text{ does NOT undergo Friedel-Crafts due to strong deactivation.}$ $\text{(iii) m-Dinitrobenzene (-NO}_2\text{ groups) } \rightarrow \text{ Does NOT undergo Friedel-Crafts for the same reason as above.}$ $\text{(iv) p-Xylene (-CH}_3\text{ groups) } \rightarrow \text{ Undergoes Friedel-Crafts (activating group).}$ $\text{(v) Chlorobenzene (-Cl group) } \rightarrow \text{ Undergoes Friedel-Crafts (deactivating but still reactive).}$ $\text{(vi) p-Nitrochlorobenzene (-NO}_2\text{ and -Cl groups) } \rightarrow \text{ Does NOT undergo Friedel-Crafts due to strong deactivation by -NO}_2\text{.}$ $\text{(vii) Cumene } \rightarrow \text{ Undergoes Friedel-Crafts (activating group).}$ $\text{Out of the given compounds, only three will give Friedel-Crafts reaction. Hence, option 1 is the correct answer.}$

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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
}