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Current Question (ID: 10885)

Question:
$\text{The optically active compound among the following is-}$
Options:
  • 1. $\text{Isobutyric acid}$
  • 2. $\text{beta-Chloropropionic acid}$
  • 3. $\text{Propionic acid}$
  • 4. $\text{alpha-Chloropropionic acid}$ (Correct)
Solution:
$\text{Hint: If a molecule contains a chiral centre then it is optically active.}$ $\text{Step 1:}$ $\text{The structure of given compounds are as follows:}$ $\text{Isobutyric acid: } \text{H}_3\text{C}-\text{CH}(\text{CH}_3)-\text{COOH}$ $\text{beta-Chloropropionic acid: } \text{Cl}-\text{CH}_2-\text{CH}_2-\text{COOH}$ $\text{Propionic acid: } \text{H}_3\text{C}-\text{CH}_2-\text{COOH}$ $\text{alpha-Chloropropionic acid: } \text{H}_3\text{C}-\text{CH}(\text{Cl})-\text{COOH}$ $\text{(asterisk represents chiral carbon)}$ $\text{So, alpha-Chloropropionic acid is optically active because it contains an asymmetric carbon atom.}$

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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
}