Import Question JSON

Current Question (ID: 21277)

Question:

$\text{How many of the following compounds are optically active?}$ $\text{(I) CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{Cl}$ $\text{(II) CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{OH}$ $\text{(III)}$ \begin{array}{c} \text{CH}_3 - \text{CH} - \text{CH}_2 - \text{CH}_2 - \text{Cl} \\ \quad \vert \\ \text{CH}_3 \end{array} $ $\text{(IV)}$ \begin{array}{c} \text{CH}_3 \\ \vert \\ \text{H} - \text{C} - \text{OH} \\ \vert \\ \text{H} - \text{C} - \text{OH} \\ \vert \\ \text{CH}_3 \end{array} $ $\text{(V)}$ \begin{array}{c} \text{CH}_3 - \text{C} - \text{C} - \text{C} - \text{CH}_3 \\ \quad \vert \quad \vert \quad \vert \\ \text{OH} \quad \text{OH} \quad \text{OH} \end{array} $

Options:

1. $\text{Zero}$
2. $\text{(I), (II), (III)}$
3. $\text{(IV), (III), (V)}$
4. $\text{(II), (III)}$

Solution:

$\text{Hint: Meso compounds are optically inactive.}$ $\text{Step 1: A compound to be optically active must possess one or more chiral centers (an atom that has four unique atoms or groups attached to it).}$ $\text{Achiral compounds are optically inactive. Among the given compounds, (I), (II), and (III) are optically inactive.}$ $\text{Step 2: Meso form is optically inactive because the molecules in meso form have a plane of symmetry due to which the optical rotations of upper and lower parts are equal and in the opposite direction which is balanced internally and the compound becomes optically inactive. This property is called internal compensation.}$ $\text{(IV) and (V) are meso compounds and optically inactive. All are optically inactive. Hence, option 1 is the correct answer.}$

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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
}