The reaction of 4 -methyloct- 1 -ene ( P , 2 . 52 g ) with HBr in the presence of C 6 H 5 CO 2 O 2 gives two isomeric bromides in a 9 : 1 ratio, with a combined yield of 50 % . Of these, the entire am

Question

The reaction of 4 -methyloct- 1 -ene ( P , 2 . 52 g ) with HBr in the presence of C 6 H 5 CO 2 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. K 2 CO 3 to give a non-ionic product S in 100 % Yield. The mass (in mg ) of S obtained is [Use molar mass (in gmol - 1 ): H = 1 , C = 12 , N = 14 , Br = 80 ]

TestHub · Accepted Answer

The 1-alkene undergo addition reaction and major product formed according to anti-Markonikov's rule.
Moles of $P$ $= \frac{Mass of P}{Molar mass of P} = \frac{2.52}{252} = 0.01 mol$
Given that isomeric bromides formed have the mole fractions $\frac{9}{10}$ and $\frac{1}{10}$
Moles of $A$ (major product) $= 0.02 \times \frac{9}{10} \times \frac{50}{100} = 0.009 mol$
Now the alkyl bromide formed undergo nucleophilic substitution reaction with diethyl amine and potassium carbonate.
In the above reaction, the product $S$ formed has the molecular weight $= 199 g {mol}^{- 1}$
Mass of $S = 0.009 \times 199$
$= 1.791 g$
$= 1791 mg$

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