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Current Question (ID: 10862)

Question:
$\text{The type of structural isomerism shown by given compounds is-}$ $\text{CH}_3-\text{S}-\text{CH}_2-\text{CH}_2-\text{CH}_3$ $\text{and}$ $\text{CH}_3-\text{S}-\text{CH}(\text{CH}_3)_2$
Options:
  • 1. $\text{Tautomerism}$
  • 2. $\text{Positional isomerism}$ (Correct)
  • 3. $\text{Functional isomerism}$
  • 4. $\text{Ring Chain isomerism}$
Solution:
$\text{Hint: positional isomers have the same carbon skeleton and the same functional groups but present at different positions.}$ $\text{Explanation:}$ $\text{Step 1: The two isomers which differ in the position of the functional group on the carbon skeleton are called position isomers and this phenomenon as position isomerism.}$ $\text{Thus, (A) and (B) may be regarded as position isomers and further they cannot be regarded as metamers since metamers are those isomers that have a different number of carbon atoms on either side of the functional group. But here, the number of carbon atoms on either side of sulphur atom (functional group) is the same i.e., 1 and 3.}$ $\text{Step 2: See the position of the same functional group below present at different position}$ $\text{Compound (A): CH}_3-\text{S}-\text{CH}_2-\text{CH}_2-\text{CH}_3$ $\text{Methyl n-propylthio ether}$ $\text{Compound (B): CH}_3-\text{S}-\text{CH}(\text{CH}_3)_2$ $\text{Isopropylthio ether}$ $\text{Both compounds have the same molecular formula and the sulfur atom is connected to different carbon arrangements, showing positional isomerism.}$

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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
}