Import Question JSON

Current Question (ID: 10877)

Question:

$\text{The compounds that show geometrical isomerism among the following are:}$

Options:

1. $\text{a, b}$
2. $\text{c, d}$
3. $\text{a, b, c}$
4. $\text{a, c}$ (Correct)

Solution:

$\text{Hint: These isomers are obtained due to restricted rotation in a molecule. Geometrical isomerism two structures are possible, that is, cis and trans. The structures of isomers are as follows:}$ $\text{For geometrical isomerism to occur, a compound must have:}$ $\text{1. A double bond (C=C) with restricted rotation}$ $\text{2. Two different groups attached to each carbon of the double bond}$ $\text{Analysis of each compound:}$ $\text{a. 2-Butene: } \text{H}_3\text{C}-\text{CH}=\text{CH}-\text{CH}_3 \text{ - Has different groups on each carbon of double bond, shows cis-trans isomerism}$ $\text{b. Propene: } \text{H}_3\text{C}-\text{CH}=\text{CH}_2 \text{ - One carbon has two H atoms, cannot show geometrical isomerism}$ $\text{c. 1-Phenylpropene: Has a phenyl group and different substituents on the double bond carbons, shows cis-trans isomerism}$ $\text{d. 2-Methylbut-2-ene: Has two methyl groups on one carbon of the double bond, cannot show geometrical isomerism}$ $\text{Only two compounds can show geometrical isomerism, that is, 2-butene and 1-phenylpropene.}$ $\text{Therefore, the correct answer is option 4: a, c}$

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
}