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JEE Advance - Chemistry (2023 - Paper 2 Online - No. 13)

The reaction of 4-methyloct-1-ene $(\mathbf{P}, 2.52 \mathrm{~g})$ with $\mathrm{HBr}$ in the presence of $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2$ gives two isomeric bromides in a $9: 1$ ratio, with a combined yield of $50 \%$. Of these, the entire amount of the primary alkyl bromide was reacted with an appropriate amount of diethylamine followed by treatment with aq. $\mathrm{K}_2 \mathrm{CO}_3$ to give a non-ionic product $\mathbf{S}$ in $100 \%$ yield.

The mass (in mg) of $\mathbf{S}$ obtained is ________.

[Use molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$ ) : $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{Br}=80$ ]
Answer
1791

Explanation

JEE Advanced 2023 Paper 2 Online Chemistry - Practical Organic Chemistry Question 1 English Explanation

  1. First, calculate the moles of 4-methyloct-1-ene. The molar mass of 4-methyloct-1-ene is 126 g/mol. So, the moles of 4-methyloct-1-ene are :


    $$n_{4-\text{methyloct-1-ene}} = \frac{2.52 \, g}{126 \, g/mol} = 0.02 \, mol$$


  2. Given that the combined yield is 50%, the amount of isomeric bromides produced will be half of the initial amount, which is 0.01 mol.


  3. The isomeric bromides are produced in a 9 : 1 ratio, so 90% of the product, or 0.009 mol, is the primary alkyl bromide. This is the compound that reacts further with diethylamine.


  4. The reaction with diethylamine leads to the non-ionic product S with 100% yield, so we will still have 0.009 mol of product S.


  5. Based on the chemical structures provided, the molar mass of S is 199 g/mol.


  6. We can now calculate the mass of product S using the equation :


    $$ \text{Mass of S} = \text{moles of S} \times \text{molar mass of S} = 0.009 \, mol \times 199 \, g/mol = 1.791 \, g$$


  7. To convert this into milligrams, we multiply by 1000, yielding a mass of S obtained equal to 1791 mg.

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