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Current Question (ID: 21342)

Question:
$\text{The total number of secondary carbon atoms present in the given compound is:}$ $\text{CH}_3\text{C(CH}_3\text{)}\text{CH}_2\text{CH(CH}_3\text{)}\text{CH}_3$
Options:
  • 1. 1 (Correct)
  • 2. 2
  • 3. 3
  • 4. 4
Solution:
$\text{Hint: A carbon atom that is bonded to two other carbon atoms.}$ $\text{Explanation:}$ $(\text{i}) \text{Primary carbons (1°), are carbons attached to one other carbon and three hydrogens. Also known as a methyl (CH}_3\text{)}$ $(\text{ii}) \text{Secondary carbons (2°) are attached to two other carbons and two hydrogens. Also known as methylene (CH}_2\text{) carbons.}$ $(\text{iii}) \text{Tertiary carbons (3°) are attached to three other carbons and one hydrogen. Also known as methine (R}_3\text{CH) carbons.}$ $(\text{iv}) \text{Finally, quaternary carbons (4°) are attached to four other carbons.}$ $\text{In the given compounds, the carbon atoms are labeled as:}$ $\text{Only one carbon atom is a secondary carbon atom. Hence, option 1 is the correct answer.}$

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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
}