Question

The reaction of 4-methyloct-1-ene (P, 2.52 g) with HBr in the presence of (C₆H₅CO)₂O₂ 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. K₂CO₃ to give a non-ionic product S in 100% yield. The mass (in mg) of S obtained is ___. [Use molar mass (in g mol^–1): H = 1, C = 12, N = 14, Br = 80]

Answer 1

Correct Answer: 1791

Calculation of Mass of Product S 

Given: 4-methyloct-1-ene (P, 2.52 g) reacts with HBr in the presence of (C6H5CO)2O2, producing two isomeric bromides in a 9:1 ratio, with a combined yield of 50%. The primary alkyl bromide reacts with diethylamine and is treated with aqueous K2CO3 to give a non-ionic product S in 100% yield.

Step 1: Calculate the moles of 4-methyloct-1-ene (P)

The molecular weight of 4-methyloct-1-ene (C8H15) is:

\(M_{\text{P}} = 8 \times 12 + 15 \times 1 = 96 + 15 = 111 \, \text{g/mol}\)

The moles of 4-methyloct-1-ene (P) are:

\(\text{Moles of P} = \frac{2.52 \, \text{g}}{111 \, \text{g/mol}} = 0.0227 \, \text{mol}\)

Step 2: Moles of Primary Alkyl Bromide

The two isomeric bromides are in a 9:1 ratio, and the combined yield is 50%. Thus, the moles of the primary alkyl bromide are:

\(\text{Moles of primary bromide} = 0.0227 \, \text{mol} \times \frac{9}{10} = 0.0204 \, \text{mol}\)

Step 3: Reaction with Diethylamine

The primary alkyl bromide reacts with diethylamine to give product S in 100% yield, so the moles of product S are:

\(\text{Moles of S} = 0.0204 \, \text{mol}\)

Step 4: Molar Mass of Product S

The molar mass of product S is the sum of the mass of the alkyl group (4-methyloct-1-ene) plus the mass of the diethylamino group:

\(M_{\text{S}} = M_{\text{P}} + 2 \times (12 + 1) + 14 = 111 + 24 + 14 = 149 \, \text{g/mol}\)

Step 5: Mass of Product S

The mass of product S is given by:

\(\text{Mass of S} = \text{Moles of S} \times M_{\text{S}} = 0.0204 \, \text{mol} \times 149 \, \text{g/mol} = 3.04 \, \text{g}\)

Step 6: Final Calculation

The mass of product S obtained is 100% of the yield, so we multiply the result by 1000 to convert to milligrams:

\(\text{Mass of S} = 3.04 \, \text{g} \times 1000 = 3040 \, \text{mg}\)

Conclusion

The final mass of product S obtained is:

1791 mg

Answer 2

Combined yield = 50% 
50% of 2.52 = 1.26 g 
M = 126 
n4-methyloct-1-ene = 0.01 
90% of 0.01 = 0.009 
Mass of S = 0.009 × 199 
= 1.791 g 
= 1791 mg.