Import Question JSON

Current Question (ID: 10906)

Question:

$\text{Which carboxylate ion among the following is the most stable?}$

Options:

1. $\text{CH}_3 - \text{C} - \text{O}^-$ with $\text{O}$ double bonded above $\text{C}$
2. $\text{Cl} - \text{CH}_2 - \text{C} - \text{O}^-$ with $\text{O}$ double bonded above $\text{C}$
3. $\text{F} - \text{CH}_2 - \text{C} - \text{O}^-$ with $\text{O}$ double bonded above $\text{C}$
4. $\text{F}_2\text{CH} - \text{C} - \text{O}^-$ with $\text{O}$ double bonded above $\text{C}$ (Correct)

Solution:

$\text{Hint: Resonance effect and inductive effect both working}$ $\text{In all the given carbocations, the negative charge is dispersed which stabilises these carbocations. Here, the negative charge is dispersed by two factors, i.e., +R effect of the carboxylate ion (conjugation) and -I effect of the halogens (fluorine).}$ $\text{These effect are shown below in the carbocations}$ $\text{As it is clearly evident from the above structures, that +R-effect is common in all the four structures, therefore, overall dispersal of negative charge depends upon the number of halogen atoms and electronegativity. Since, F has the highest electronegativity and two F-atoms are present in option (4), thus, dispersal of negative charge is maximum in option (4).}$

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
}