Import Question JSON

Current Question (ID: 10928)

Question:

$\text{The hydrocarbons having the lowest dipole moment among the following is:}$

Options:

1. $\text{H}_3\text{C}\text{CH}=\text{CHCH}_3$ (2-butene)
2. $\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3$ (2-butyne) (Correct)
3. $\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2$ (1-butene)
4. $\text{CH}_2=\text{CH}-\text{C}\equiv\text{CH}$ (1-buten-3-yne)

Solution:

$\text{Hint: Linear symmetrical structures have zero dipole moment.}$ $\text{Explanation:}$ $\text{(i) The dipole moment of an alkene depends on whether it's symmetrical or unsymmetrical:}$ $\text{(1) Symmetrical: In a symmetrical trans-alkene, the dipole moments of individual groups are equal and opposite, resulting in a zero dipole moment.}$ $\text{(2) Unsymmetrical: In an unsymmetrical alkene, the individual groups are not equal, resulting in a net dipole moment.}$ $\text{(ii) Terminal alkynes have small dipole moments that are slightly larger than the dipole moments of terminal alkenes. Internal alkynes have no dipole moment.}$ $\text{2-Butyne is a linear structure. Due to its linear structure and symmetrical nature, the dipole moment is zero.}$ $\text{All other are unsymmetrical compounds and will have a dipole moment. So, option 2 is the correct answer.}$

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
}